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## Found 586 Documents (Results 1–100)

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### Discounted optimal stopping problems in first-passage time models with random thresholds. (English)Zbl 07589155

MSC:  60G40 60G44 60J65 91B25 60J60 35R35
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### The application of model predictive control on stock portfolio optimization with prediction based on geometric Brownian motion-Kalman filter. (English)Zbl 07574126

MSC:  93C95 91G10
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### Reflected BSDEs in non-convex domains. (English)Zbl 07565240

MSC:  60D05 60G65 60J60
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### Discounted optimal stopping problems in continuous hidden Markov models. (English)Zbl 07554288

MSC:  60G40 91G20 60J22
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### Optimal double stopping problems for maxima and minima of geometric Brownian motions. (English)Zbl 1490.60097

MSC:  60G40 91G20 60J70
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### Variational principles for fluid dynamics on rough paths. (English)Zbl 07537688

Reviewer: Cheng He (Beijing)
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MSC:  62-XX
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MSC:  60D05
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### Stochastic analysis and invariant subspace method for handling option pricing with numerical simulation. (English)Zbl 07527953

MSC:  60J20 65M06
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### Differential optimal investment timing decision model for nonlinear dynamic urban infrastructure planning system. (English)Zbl 1486.91036

MSC:  91B18 60J70
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### Generalization of affine feedback stock trading results to include stop-loss orders. (English)Zbl 1480.91279

MSC:  91G15 60J70 93B52
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MSC:  90Bxx
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### On the analysis of Black-Scholes equation for European call option involving a fractional order with generalized two dimensional differential transform method. (English)Zbl 07530053

MSC:  26A33 34K37
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### Pricing geometric Asian power options in the sub-fractional Brownian motion environment. (English)Zbl 07514608

MSC:  91G20 91G60 65C05
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### Anomalous diffusion in umbrella comb. (English)Zbl 07511369

MSC:  82-XX 35-XX
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### Impact of plant utilization on irreversible investment under uncertainty with application to refinery investment. (English)Zbl 1482.91220

MSC:  91G50 60J70
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### Limit theorems for prices of options written on semi-Markov processes. (English)Zbl 1480.91294

MSC:  91G20 60J70 60G44
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### Research on pricing of forward start options in jump diffusion model with stochastic interest rate. (Chinese. English summary)Zbl 1488.91141

MSC:  91G20 91G30 60J65
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### The reliability of geometric Brownian motion forecasts of S&P500 index values. (English)Zbl 1476.62229

MSC:  62P05 60J70 91G20
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### Derivatives in the mean of random processes and diffusion models in economics. (Russian. English summary)Zbl 1475.60158

MSC:  60J70 91G05 91G80
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### First-to-default and second-to-default options in models with various information flows. (English)Zbl 07384603

MSC:  91G40 60J70
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### A Talenti-type comparison theorem for $$\mathrm{RCD}(K,N)$$ spaces and applications. (English)Zbl 1480.53060

MSC:  53C23 53C21 35R01
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### Computer turbulence as a tunnelling effect. (English. Ukrainian original)Zbl 1491.65079

J. Math. Sci., New York 256, No. 5, 703-712 (2021); translation from Neliniĭni Kolyvannya 23, No. 1, 124-133 (2020).
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### Liouville quantum gravity surfaces with boundary as matings of trees. (English. French summary)Zbl 1472.60017

MSC:  60D05 60J65 60J67
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### The Tutte embedding of the mated-CRT map converges to Liouville quantum gravity. (English)Zbl 1484.60007

MSC:  60D05 60J65 60J67
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### Closed form optimal exercise boundary of the American put option. (English)Zbl 1467.91188

MSC:  91G20 60G40
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### Breaking of continuous scale invariance to discrete scale invariance: a universal quantum phase transition. (English)Zbl 1473.81062

Freiberg, Uta (ed.) et al., Fractal geometry and stochastics VI. Selected papers of the 6th conference, Bad Herrenalb, Germany, September 30 – October 6, 2018. Cham: Birkhäuser. Prog. Probab. 76, 209-238 (2021).
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### Generalized disconnection exponents. (English)Zbl 1478.60226

MSC:  60J67 60J65 60D05
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### Transforming public pensions: a mixed scheme with a credit granted by the state. (English)Zbl 1460.91210

MSC:  91G05 93E20 60J70
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### Homotopy analysis method and its applications in the valuation of European call options with time-fractional Black-Scholes equation. (English)Zbl 07511246

MSC:  26A33 34K37
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### Optimal dividend of compound Poisson process under a stochastic interest rate. (English)Zbl 1476.91131

MSC:  91G05 49L25 93E20
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### Volume of metric balls in Liouville quantum gravity. (English)Zbl 1470.60026

MSC:  60D05 60K35
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### Modeling linear nonstationary stochastic systems by spectral method. (Russian. English summary)Zbl 1459.93169

MSC:  93E03 60J70 93C05
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### Asian option pricing under sub-fractional Brownian motion with jump. (Chinese. English summary)Zbl 1463.91176

MSC:  91G20 60G22

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### Revisiting integral functionals of geometric Brownian motion. (English)Zbl 1460.60037

MSC:  60G51 60J65 91G80
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### A mating-of-trees approach for graph distances in random planar maps. (English)Zbl 1464.60011

MSC:  60D05 60G50
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### Credit default swaps in two-dimensional models with various informations flows. (English)Zbl 1444.91217

MSC:  91G40 91G20
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### The Tutte embedding of the Poisson-Voronoi tessellation of the Brownian disk converges to $$\sqrt{8/3}$$-Liouville quantum gravity. (English)Zbl 1441.60015

MSC:  60D05 60K35
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### Extended Black and Scholes model under bankruptcy risk. (English)Zbl 1427.91275

MSC:  91G20 62P05
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MSC:  82-XX
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### Double AR model without intercept: an alternative to modeling nonstationarity and heteroscedasticity. (English)Zbl 1490.62257

MSC:  62M10 62F12 62P20
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### Pricing of fixed-strike arithmetic Asian powered options. (English)Zbl 1472.91046

MSC:  91G20 60H30
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MSC:  65C30
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### The optimal thresholds of pairs trading with a stop-loss condition. (Chinese. English summary)Zbl 1449.91142

MSC:  91G15 60G40

### Tightness and line ensembles for Brownian polymers under geometric area tilts. (English)Zbl 1446.82093

Gayrard, Véronique (ed.) et al., Statistical mechanics of classical and disordered systems. Proceedings of the international conference “Advances in statistical mechanics”, CIRM, Luminy, France, August 28–31, 2018. Cham: Springer. Springer Proc. Math. Stat. 293, 241-266 (2019).
MSC:  82D60 82C24 60J65
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### Solvency of an insurance company in a dual risk model with investment: analysis and numerical study of singular boundary value problems. (English. Russian original)Zbl 1443.91251

Comput. Math. Math. Phys. 59, No. 11, 1904-1927 (2019); translation from Zh. Vychisl. Mat. Mat. Fiz. 59, No. 11, 1973-1997 (2019).
MSC:  91G05 91G60 65N99
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### Emergent dynamics of the first-order stochastic Cucker-Smale model and application to finance. (English)Zbl 1431.91447

MSC:  91G99 91G80 60J70
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### Solvency need resulting from reserving risk in a ORSA context. (English)Zbl 1452.91278

MSC:  91G05 62P05
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### Brownian cuboid and its geometric characteristics. (English)Zbl 1428.60026

MSC:  60D05 60J65 60G15
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### Convergence of Brownian motions on metric measure spaces under Riemannian curvature-dimension conditions. (English)Zbl 1444.60023

MSC:  60F17 53C23
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### Is hard Brexit detrimental to EU integration? Theory and evidence. (English)Zbl 1425.91285

MSC:  91B60 91B82 60J70
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### Decision making of mine investment based on fuzzy interval-valued GM $$(1, 1)$$ model and geometric Brown motion. (Chinese. English summary)Zbl 1438.90164

MSC:  90B50 03E72 60J65
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### Drawdown and drawup for fractional Brownian motion with trend. (English)Zbl 1478.60121

MSC:  60G22 60G15 60G70
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### Defaultable claims in switching models with partial information. (English)Zbl 1411.91599

MSC:  91G40 91G20 60J70
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### Optimal exchange rates management using stochastic impulse control for geometric Lévy processes. (English)Zbl 1410.91349

MSC:  91B64 93E20 60G51
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### Introduction to stochastic differential equations with applications to modelling in biology and finance. (English)Zbl 1425.60001

Hoboken, NJ: John Wiley & Sons (ISBN 978-1-119-16606-1/hbk; 978-1-119-16609-2/ebook). xv, 283 p. (2019).
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### Maximum likelihood estimation of first-passage structural credit risk models correcting for the survivorship bias. (English)Zbl 1411.91593

MSC:  91G40 62P05
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### On moments of integral exponential functionals of additive processes. (English)Zbl 1407.60115

MSC:  60J75 60J60 60E10
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### Sample path properties of an explosive double autoregressive model. (English)Zbl 1490.62258

MSC:  62M10 62F12 62P20
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MSC:  91Bxx
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### Exact solutions for Fokker-Planck equation of geometric Brownian motion with Lie point symmetries. (English)Zbl 1424.58007

MSC:  58D19 58J70 76M60
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### An exactly solvable multiple stochastic optimal stopping problem. (English)Zbl 1446.93078

MSC:  93E20 92D40
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### A new method for option pricing via time-fractional PDE. (English)Zbl 1459.91202

MSC:  91G20 35R11 35R60
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### Coupling in the Heisenberg group and its applications to gradient estimates. (English)Zbl 1441.60013

MSC:  60D05 60H30
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### Generating sample paths and their convergence of the geometric fractional Brownian motion. (English)Zbl 1394.60035

MSC:  60G22 60G17 60H07
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### The evaluation of geometric Asian power options under time changed mixed fractional Brownian motion. (English)Zbl 1395.91467

MSC:  91G20 91G80 60G22
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# 5.1 Continuous probability functions

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This module introduces the continuous probability function and explores the relationship between the probability of X and the area under the curve of f(X).

We begin by defining a continuous probability density function. We use the function notation $f(x)$ . Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function $f(x)$ so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one,the maximum area is also one.

For continuous probability distributions, PROBABILITY = AREA.

Consider the function $f(x)=\frac{1}{20}$ for $((0, x), \mathrm{20})$ . $x$ = a real number. The graph of $f(x)=\frac{1}{20}$ is a horizontal line. However, since $((0, x), \mathrm{20})$ , $f(x)$ is restricted to the portion between $(x, 0)$ and $(x, 20)$ , inclusive .

$f(x)=\frac{1}{20}$ for $((0, x), 20)$ .

The graph of $f(x)=\frac{1}{20}$ isa horizontal line segment when $((0, x), \mathrm{20})$ .

The area between $f(x)=\frac{1}{20}$ where $((0, x), \mathrm{20})$ and the x-axis is the area of a rectangle with base = $20$ and height = $\frac{1}{20}$ .

$((\text{AREA}, 20\cdot \frac{1}{20}), 1)$

This particular function, where we have restricted $x$ so that the area between the function and the x-axis is 1, is an example of a continuousprobability density function. It is used as a tool to calculate probabilities.

Suppose we want to find the area between $f(x)=\frac{1}{20}$ and the x-axis where $((0, x), 2)$ .

$((\text{AREA}, \left(2-0\right)\cdot \frac{1}{20}), 0.1)$

$((\left(2-0\right), 2), \text{base of a rectangle})$

$\frac{1}{20}$ = the height.

The area corresponds to a probability. The probability that $x$ is between 0 and 2 is 0.1, which can be written mathematically as $\text{P(0 .

Suppose we want to find the area between $f(x)=\frac{1}{20}$ and the x-axis where $((4, x), 15)$ .

$((\text{AREA}, \left(15-4\right)\cdot \frac{1}{20}), 0.55)$

$\left(15-4\right)=11=\text{the base of a rectangle}$

$\frac{1}{20}$ = the height.

The area corresponds to the probability $(((P\left(4, x), 15\right)), 0.55)$ .

Suppose we want to find $P\left(x=\mathrm{15}\right)$ . On an x-y graph, $x=\mathrm{15}$ is a vertical line. A vertical line has no width (or 0 width). Therefore, $P\left(x=\mathrm{15}\right)=\text{(base)}\text{(height)}=\left(0\right)\left(\frac{1}{20}\right)=0$ .

$(P\left(X, x\right))$ (can be written as $(P\left(X, x\right))$ for continuous distributions) is called the cumulative distribution function or $\mathrm{CDF}$ . Notice the "less than or equal to" symbol. We can use the $\mathrm{CDF}$ to calculate $(P\left(X, x\right))$ . The $\mathrm{CDF}$ gives "area to the left" and $(P\left(X, x\right))$ gives "area to the right." We calculate $(P\left(X, x\right))$ for continuous distributions as follows: $(((P\left(X, x\right)), 1-P\left(X), x\right))$ .

Label the graph with $\mathrm{f\left(x\right)}$ and $x$ . Scale the x and y axes with the maximum $x$ and $y$ values. $(f(x), \frac{1}{20})$ , $((0, x), \mathrm{20})$ .

$(((((P\left(2.3, x), 12.7\right)), \left(\text{base}\right)\left(\text{height}\right)), \left(12.7-2.3\right)\left(\frac{1}{20}\right)), 0.52)$

What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.

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# Convergence of sigma sin(1/n^2) . . . does it converge?

• Jul 23rd 2012, 01:06 AM
Physicsdochi
Convergence of sigma sin(1/n^2) . . . does it converge?
I want to know if sigma[sin(1/n^2)] converges...

Help!
• Jul 23rd 2012, 03:04 AM
girdav
Re: Convergence of sigma sin(1/n^2) . . . does it converge?
Yes, since $\left|\sin\left(\frac 1{n^2}\right)\leq \frac 1{n^2}$ and the series $\sum_{n\geq 1}\frac 1{n^2}$ is convergent.
• Jul 24th 2012, 08:04 PM
Physicsdochi
Re: Convergence of sigma sin(1/n^2) . . . does it converge?
I think got it! thanks for answering ^^


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# How to Compute Pearson Correlation Coefficient in PySpark?

To Compute the Pearson Correlation Coefficient in PySpark, we use the corr() function.

### Syntax –

corr(column1, column2)

### Read a Dataset –

Let’s read a dataset to work with. We will use the clothing store sales data.

from pyspark.sql import SparkSession
spark = SparkSession.builder.getOrCreate()
df = spark.read.format('csv') \
df.show(5)

### Compute Pearson Correlation Coefficient in PySpark –

Let’s compute the Pearson correlation coefficient of Net Sales and Age columns.

from pyspark.sql.functions import corr
df.select(corr("Net Sales", "Age")).show()

You can also compute it like this –

df.stat.corr("Net Sales", "Age")
#output
-0.010635891709415892

Rating: 1 out of 5.


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# Homework Help: Natural log problem

1. Jul 25, 2007

### lLovePhysics

Problem: If $$2^{x}=3$$, what does 3^x equal?

I solved this by taking the $$log_{2}$$ of both sides. However, the book's solution involves taking the natural log of both sides. Can someone show me why that would work? I don't get what the x power changes to a x term. The natural log is base 10 right?

Last edited: Jul 25, 2007
2. Jul 25, 2007

### Kurdt

Staff Emeritus
What exactly are you asking? If 23, what..... means nothing.

Do you mean:

$2^3 = x$ then what does $3^x$ equal?

3. Jul 25, 2007

### lLovePhysics

My bad... I've edited the problem

4. Jul 25, 2007

### lLovePhysics

Also, someone please answer this question. How come: $$\sqrt{(-2)^{2}}=2$$??

This is very mind boogling because (-2)^2=4 and the square root of 4 is 2... but, why is it wrong to simplify just by canceling out the power of 2 with the square root? I've always done it that way... =/

Last edited: Jul 25, 2007
5. Jul 25, 2007

### Kurdt

Staff Emeritus
Ok no probs.

Because exponentials are defined in terms of the exponential function (hence the name) and the natural logarithm is the inverse to the exponential function, it is easy to write a logarithm to the base of any number in terms of natural logs. For example, we define $\log_a (x)=y$ to be the inverse of $x= a^y$. But from our knowledge of exponentials we can say:

$$x=a^y=e^{y\ln (a)} \rightarrow \ln(x)=y\ln(a) \rightarrow y=\frac{\ln (x)}{\ln (a)}$$

6. Jul 25, 2007

### Kummer

The mistake that many people make (especially those in physics, because math majors look out for such mistakes) is that $$\sqrt{x^2}=x$$. That is wrong. For example, if $$x=(-2)^2$$ then $$\sqrt{(-2)^2}=2$$.

The correct version states that,
$$\sqrt{x^2}=|x|$$.

7. Jul 25, 2007

### lLovePhysics

So $$ln$$ is basically equal to $$log_{e}$$ right? What is $$e^{log_{e}}$$??

8. Jul 25, 2007

### lLovePhysics

Is there any reason why you can't just take cancel the sqrt. and power of 2? It is different for cube roots and cube powers right?

So the 4th root of (-4)^4 is equal to 4 right?

9. Jul 25, 2007

### Kurdt

Staff Emeritus
Yes, ln is a shorthand for natural log or loge. The natural logarithm is the inverse function of the exponential function and thus:

$$e^{\ln a} = a$$

Last edited: Jul 25, 2007
10. Jul 25, 2007

### Kurdt

Staff Emeritus
Sorry, I didn't see in your original post that you asked if natural log was log to the base ten. Natural log is log to the base e.

11. Jul 25, 2007

### Mr. Mathmatics

What Kummer said was correct about the absolute value of two. You can't just cancel it out, you may cancel it out though if it is (square root of x)^2 as long as the radical symbol is inside the parentheses then you may cancel out.

12. Jul 26, 2007

### HallsofIvy

Getting back to the first question:
If 2x= 3, then log(2x)= x log(2)= log(3) so
$$x= \frac{log(3)}{log(2)}[/itex] Now, 3x= 3log(3)/log(2). That isn't going to be any simple number (it's about 5.7). 13. Jul 26, 2007 ### VietDao29 Yup, exactly. Well, since y = x3 is a 1-to-1 function, i.e, for every [tex]x_0 \neq x_1$$, we'll obtain: $$y_0 = x_0 ^ 3 \neq y_1 = x_1 ^ 3$$, so given one y, we can find only one x such that y = x3.

But y = 2 is a different thing. 2 numbers a, and -a, are different (for a <> 0), but when squared, they'll become: (a2) = a2, and (-a)2 = (-1)2 a2 = a2, so, we have: $$(a) ^ 2 = (-a) ^ 2 = a ^ 2$$. So, for any y (not 0), there'll be 2 different x's, one positive, one negative, such that, when squared, they'll give y.

Say, y = 4, then 2 x's are 2, and -2.

Square root function will only return the non-negative value.

Say, we have: y = 4, so sqrt(y) = 2; k = 16 ~~> sqrt(k) = 4.

So, in general, we have: $$\sqrt{a ^ 2} = |a|$$

To take the negative value, we can assign the minus sign in front of it, like this:

$$t = 36 \Rightarrow - \sqrt{t} = -6$$

-------------------------

And it's the same for even-th root, like: $$\sqrt[4]{x ^ 4} = |x| , \ \sqrt[6]{x ^ 6} = |x|, ...$$, blah, blah, blah, so on. :)

Hope that's clear. :)

Last edited: Jul 26, 2007


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# Inverse Trigonometry.

• November 15th 2009, 08:11 PM
Sunyata
Inverse Trigonometry.
A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

What exactly am I meant to do for these types of questions?
• November 15th 2009, 08:20 PM
redsoxfan325
Quote:

Originally Posted by Sunyata
A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

What exactly am I meant to do for these types of questions?

To find the original function you need to integrate

$\int\frac{1-x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx-\int\frac{x}{1+x^2}\,dx$

The first integral is $\arctan(x)$ (you should find this on any table of integrals) and the second integral can be solved using a simple u-sub ( $u=1+x^2$), so you end up with $f(x)=\arctan(x)-\frac{1}{2}\ln(1+x^2)$. Then you set $f(x)=0$ to get the answer.

It seems a little strange that the question would ignore the presence of the integration constant though.


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# Importance of integral equations

Differential equations are at the heart of applied mathematics - they are used to great success in fields from physics to economics. Certainly, they are very useful in modelling a wide range of phenomena.

Integral equations, on the other hand, do not receive such attention. While I have seen some integral equations crop up in physics (Boltzmann equation or the tautochrone problem) or biology (population dynamics), their importance pales in comparison to differential equations.

Why is it that differential equations are so much more popular than integral ones? Or am I just ignorant of the matter and there actually are many examples of integral equations in applied mathematics?

It also seems that when an integral equation appears, one immediately wants to reduce it to a differential equation. So examples, where this is not possible or not done for different reasons would be welcome.

• A big subfield in PDEs is a that of non-local equations: in time and especially in space. These equations are integro-differential. web.ma.utexas.edu/mediawiki/index.php/… Nov 1, 2021 at 21:11
• In cases where both differential and integral forms of the equations exist, there are math.SE posts arguing that the integral form is what you really care about or is more fundamental because it corresponds to a conservation law. The integral form can sometimes be more useful for solving or for numerical computations. Nov 2, 2021 at 15:08
• Why the downvote? Nov 3, 2021 at 17:25
• @TimothyChow The point about fundamental conservation laws being expressed as integral equations is a really good one. Nov 5, 2021 at 18:35
• I accidentally came across the rendering equation in 3D graphics - quite an important integral equation, it seems. Dec 15, 2021 at 11:54

## 2 Answers

One important point is that differential equations encode local behaviour of a system, while integral equations typically endcode global behaviour. Local behaviour is often easier to model and to grasp intuitively. In many cases, it can also be described by much simpler formulae.

More specifically:

• Let us consider the simple example where $$p(t)$$ describes the population of a species which reproduces without any resource limit. It is very intuitive to make the assumption that the growth of the population will be proportional to the size of the population, i.e., one has $$(*) \quad \begin{cases} \dot p(t) & = c p(t), \\ p(0) & = p_0 \end{cases}$$ where $$c$$ is a constant, and $$p_0$$ is the initial size of the population. The reason why this behaviour is easy to model is that we have an intuitive understanding of growth, which is a local (with respect to time) quantity (and modelled by a derivative).

The integral equation $$(**) \qquad p(t) = p_0 + c \int_0^{t} p(s) \, ds$$ is mathematically equivalent to $$(*)$$, but its intuitive meaning is more difficult to understand, since it involves the behaviour of the population over time intervals rather than only at single instances in time.

• The local character of differential equation is reflected by the fact that initial and boundary conditions can be taken into account separately. In the initial value problem $$(*)$$, the initial condition $$p(0) = p_0$$ is separated from the differential equation, and has a clear intuitive meaning. The equivalent integral equations $$(**)$$ on the other hand, encodes both the dynamical behaviour of $$p(t)$$ and the initial condition in the same equation, which makes it more difficult to distinguish between the two effects.

• These phenomena get even more pronounced when one consides partial differential equations. For instance, the heat equation is very easy to heuristically derive locally. The behaviour at the boundary (fixed temperature = Dirichlet boundary conditions, thermal isolation = Neumann boundary conditions) can then be taken into account separately.

Reformulating the equation as an integral equations (which, for homogeneous boundary conditions, essentially comes down to computing the resolvent of the Laplace operator with the given boundary conditions) means that ones has to include the boundary conditions in the integral equation. By corollary, such an integral formulation would also need to take the geometry of the domain into account, which can be arbitrarily complicated.

On a related note, this also explains why it is impossible to explicitly compute the integral kernel of the resolvent (= Green function) of the Laplace operator on any but the most simple domains.

• This was well said. Nov 2, 2021 at 18:35
• Nice answer, but I think you forgot the $c$ in $(**)$. Nov 3, 2021 at 19:10
• @MichaelSeifert: Thanks you! Corrected. Nov 3, 2021 at 20:31

In physics, the predominance of differential over integral equations is not that obvious. Any system with a "memory", where the response at a certain time depends on the state at earlier times, requires an integral representation.

Electromagnetism is one area where actually it is the integral equations (Gauss, Ampère, Faraday) that appeared before the differential equations (Maxwell), and still today, the integral form of the equations is typically taught first.

• Do you have any specific examples of systems with a memory? :-) Nov 3, 2021 at 11:23
• for example, the frictional force in a visco-elastic medium is given by the integral $F(t)=\int_{-\infty}^{t}\gamma(t-t')\dot{x}(t')\,dt'$, with memory kernel $\gamma(t)$; Newton's equation of motion is then an integral equation, rather than a differential equation. Nov 3, 2021 at 11:55
• Anothe example of memory: dynamics of an inertial spherical passive particle in fluid flow: arxiv.org/abs/1310.2450 Nov 6, 2021 at 3:30
• Seems like there is quite a lot of integral equations connected to fluid dynamics... Nov 7, 2021 at 12:11


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# How Do I Evaluate This Definite Integral?

$$\int_0^{\pi/2}\color{red}{3}\;\cos^2\theta\;d\theta$$

I tried doing it using $u$-substitution and then getting the anti-derivative. All in all, I keep getting $-1$, which is apparently wrong. I'd appreciate any help.

-
That integral evaluates to a number. It is not a function, unless you are thinking of it as a constant function of some unnamed variable. So either there is nothing to differentiate, or you are differentiating a constant function, and in the latter case the answer is $0$. As for the integral itself, the value is $3\pi/4$. The identity $\cos(2\theta)=2\cos^2(\theta)-1$ is helpful for evaluating it. –  Jonas Meyer Feb 27 '12 at 4:44
Note that the value of the integral cannot possibly be $-1$: the function $f(\theta)=3\cos^2\theta$ is never negative, so the net area under the curve cannot be negative. –  Brian M. Scott Feb 27 '12 at 5:11

Hint: Use the double-angle formula $\cos 2\theta=2\cos^2\theta-1$. So $\cos^2\theta =\frac{\cos 2\theta +1}{2}$.

I believe you mean to ask how does one evaluate, or compute the definite integral.

-
Thanks! That did the trick. –  user754950 Feb 27 '12 at 4:54

There is a cute trick for this:

$$I = 3\int_{0}^{\pi/2} \cos^2 x \text{ dx} = 3\int_{0}^{\pi/2} \sin^2 \text{ dx}$$

by using $\int_{a}^{b} f(x) \text{ dx} = \int_{a}^{b} f(a+b-x) \text{ dx}$

$$2I = 3 \int_{0}^{\pi/2} (\cos^2 x + \sin^2 x) \text{ dx} = 3 \int_{0}^{\pi/2} 1 \text{ dx} = \frac{3\pi}{2}$$

$$\frac{3\pi}{4}$$

But, there are more general methods which can be used. See the answers here: Evaluating $\int P(\sin x, \cos x) \text{d}x$

-

$$\int_0^\frac{\pi}{2} 3 \cos^2\theta d\theta= \int_0^\frac{\pi}{2} \frac{3}{2}(1+\cos2\theta)d\theta$$ $$=\int_0^\frac{\pi}{2} \frac{3}{2}+\frac{3}{2}\cos 2\theta d\theta$$ $$=\frac{3}{2}(\frac{\pi}{2}-0)+\frac{3}{4}(\sin2\frac{\pi}{2}-\sin 0)$$ $$=\frac{3}{4}$$

-

Just for fun, you can use integration by parts to find the antiderivative of $\cos^2 x$: \eqalign{ \color{maroon}{\int\cos^2 x\,dx} =\int\underbrace{\cos x}_u\,\underbrace{\cos x\,dx}_{dv}&= \underbrace{\cos x\vphantom{)}}_u\,\underbrace{( \sin x )}_{ v}- \int\underbrace{\sin x\vphantom{)}}_v\,\underbrace{(-\sin x\,dx)}_{du}\cr &=\sin x\cos x+\int\sin^2 x\,dx\cr &=\sin x\cos x+\int(1-\cos^2 x)\,dx\cr &=\sin x\cos x+ x-\color{maroon}{\int\cos^2 x \,dx};\cr } Whence $$\int\cos^2 x\,dx={\sin x\cos x+x\over 2}+C.$$

-


text:
## And while I’m at it

Yaroslav Bulatov sent me the following nice question. Given vectors (a1,…,an) and (b1,…,bn) in Rn, is there an efficient algorithm to decide whether sgn(a1x1+…+anxn) equals sgn(b1x1+…+bnxn) for all x in {0,1}n? I could think about it myself, but wouldn’t it be more fun to call upon the collective expertise of my readers? After all, this is a Serious Blog now. I await the observation that’s eluded me for the past five minutes.

### 9 Responses to “And while I’m at it”

1. Anonymous Says:

Doesn’t this mean that the vectors are
linearly dependent, related by a positive
coefficient?

Assume that a’,b’ have this property.
Normalize these to unit vectors a,b which
will still clearly have the property.

Now
sgn(a dot (a-b))=sgn(b dot (a-b)).
Notice that if sgn(x)=sgn(y) then also
sgn(x+y)=sgn(x)=sgn(y). Applying this to
the above with x=a dot a – a dot b
and y=b dot a – b dot b we have
sgn(a dot a – a dot b)=0 and so
a dot b=1 and a=b. Thus the two
original vectors are linearly independent,
related by a positive coefficient.

–Troy

2. Kurt Says:

Scott, how are those vectors being specified? Are they actually in Q^n and not R^n? I ask that because, if you think like Andrej Bauer, sgn(x) for x ∈ R is not even computable.

3. Kurt Says:

This post has been removed by the author.

4. Anonymous Says:

Troy-

Why does sgn(a dot (a-b)) have to equal
sgn(b dot (a-b))? The equality was only
specified for vectors in {0,1}^n
which a-b might not be.

5. Anonymous Says:

oh, I didn’t see the {0,1}^n. oops, ok, more interesting now.

–Troy

6. David Says:

Yes there is. Check every pair (sgn(Ai), sgn(Bi) ) and if any are different, you can construct X s.t. it is 1 for Xi and 0 for all others.

I’ll prove this is exhaustive later.

7. Anonymous Says:

David: a=(3,-2), b=(1,-2), x=(1,1)?

There’s a reduction from subset sum. Let S=(s1, …, sn) be an instance of subset sum (integer entries, we’re looking for a subset that sums to 0). Define a and b so that:

ai = si + 1/3n
bi = si – 1/3n

for every i. Then the absolute difference between a sum from a and a sum from b is always less than one, and the signs of the sums differ if and only if the corresponding sum of entries from S sums to 0. So the problem is NP-complete.

(Any errors in this?)

8. Yaroslav Says:

Thanks Anonymous, I think your argument works

9. Scott Says:

Thanks, anonymous. I figured there was probably a reduction from Subset Sum, but I suck at reductions.


text:
# RD Sharma Solutions Class 7 Simple Interest Exercise 13.1

## RD Sharma Solutions Class 7 Chapter 13 Exercise 13.1

### RD Sharma Class 7 Solutions Chapter 13 Ex 13.1 PDF Free Download

#### Exercise 13.1

Q 1. Find the simple interest  , when :

SOLUTION :

(i) . Principal = Rs . 2000 , Rate of interest = 5% per annum , and Time = 5 years

(i) . Principal (P) = Rs 2000

Rate of interest (R) = 5% p.a.

Time (T) = 5 years

Simple interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{2000\times 5\times 5}{100} \right )$=Rs 500

(ii) . Principal = Rs . 500 , Rate of interest = 12.5% per annum , and Time = 4 years

(ii) . Principal (P) = Rs 500

Rate of interest (R) = 12.5% p.a.

Time (T) = 4 years

Simple interest = $\left ( \frac{P\times R\times T}{100} \right )$

=$\left ( \frac{500\times 12.5\times 4}{100} \right )$=Rs 250

(iii) . Principal = Rs . 4500 , Rate of interest = 4% per annum , and Time = 6 months

(iii) . Principal (P) = Rs 4500

Rate of interest (R) = 4% p.a.

Time (T) = 6 months

T = $\frac{6}{12}$ = $\frac{1}{2}$ year (1 year = 12 months)

Simple interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $(\frac{4500\times 4\times\frac{1}{2}}{100})$

=Rs 90

(iv) . Principal =Rs . 12000  , Rate of interest = 18% per annum , and Time = 4 months

(iv) . Principal (P) = Rs 12000

Rate of interest (R) = 18% p.a.

Time (T) = 4 months =$\frac{4}{12}$= $\frac{1}{3}$year

(1 year = 12 months)

Simple interest =  $\left ( \frac{P\times R\times T}{100} \right )$

= $(\frac{12000\times 18\times\frac{4}{12}}{100})$ =Rs 720

(v) . Principal = Rs . 1000 , Rate of interest = 10% per annum , and Time = 73 days

(v) . Principal (P) = Rs 1000

Rate of interest (R) = 10% p.a.

Time (T) = 73 days =$\frac{73}{365}$year (1 year = 365 days)

Simple interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $( \frac{1000\times 10\times\frac{1}{5}}{100})$ =Rs 20

Q 2 .Find the interest on Rs . 500 for a period of 4 years at the rate of 8% per annum . Also , find the amount to be paid at the end of the period .

SOLUTION :

Principal amount (P) = Rs 500

Time period (T) = 4 years

Rate of interest (R) = 8% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{500\times 8\times 4}{100} \right )$ = Rs 160

Total amount paid = Principal amount + Interest = Rs 500 + 160 = Rs 660

Q 3 . A sum of Rs . 400 is lent at the rate of 5% per annum . Find the interest at the end of 2 years .

SOLUTION :

Principal amount (P) = Rs 400

Time period (T) = 2 years

Rate of interest (R) = 5% p.a.

Interest paid after 2 years = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{400\times 5\times 2}{100} \right )$ = Rs 40

Q 4 . A sum of Rs . 400 is lent for 3 years at the rate of 6% per annum . Find the interest .

SOLUTION :

Principal amount (P) = Rs 400

Time period (T) = 3 years

Rate of interest (R) = 6% p.a.

Interest after 3 years = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{400\times 6\times 3}{100} \right )$= Rs 72

Q 5 .A person deposits Rs . 25000 in a firm who pays an interest at the rate of 20 % per annum . Calculate the income he gets from it annually .

SOLUTION :

Principal amount (P) = Rs 25000

Time period (T) = 1 year

Rate of interest (R) = 20% p.a.

Annual interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{25000\times 20\times 1}{100} \right )$ = Rs 5000

Q 6 . A man borrowed Rs . 8000 from a bank at 8% per annum . Find the amount he has to pay after $4\frac{1}{2}$ years .

SOLUTION :

Principal amount (P) = Rs 8000

Time period (T) = $4\frac{1}{2}$ years = $\frac{9}{2}$ years

Rate of interest (R) = 8% p.a.

Interest = $( \frac{8000\times 8\times\frac{9}{2}}{100})$

= 36 x 80 = Rs 2880

Total amount paid after  $4\frac{1}{2}$  years = Principal amount + Interest = Rs 8000 + Rs 2880 = Rs 10880

Q 7 . Rakesh lent out Rs . 8000 for 5 years at 15 % per annum and borrowed Rs . 6000 for 3 years at 12% per annum . How much did he gain or lose ?

SOLUTION :

Principal amount lent out by Rakesh (P) = Rs 8000

Time period (T) = 5 years

Rate of interest (R) = 15% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{8000\times 15\times 5}{100} \right )$ = Rs 6000

Principal amount borrowed by Rakesh (P) = Rs 6000

Time period (T) = 3 years

Rate of interest (R) = 12% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{6000\times 12\times 3}{100} \right )$ = Rs 2160

Amount gained by Rakesh = Rs 6000 – Rs 2160 = Rs 3840

Q 8 . Anita deposits Rs . 1000 in a savings bank account . The bank pays interest at the rate of 5 % per annum . What amount can Anita get after 1 year ?

SOLUTION :

Principal amount (P) = Rs 1000

Time period (T) = 1 year

Rate of interest (R) = 5% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{1000\times 5\times 1}{100} \right )$ = Rs 50

Total amount paid after 1 year = Principal amount + Interest = Rs 1000 + Rs 50 = Rs 1050

Q 9 . Nalini borrowed Rs . 550 from her friend at 8% per annum . She returned the amount after six months . How much did she pay ?

SOLUTION :

Principal amount (P) = Rs 550

Time period (T) = 6 months = $\frac{1}{2}$ = $\frac{1}{2}$ year                   (1 year = 12 months)

Rate of interest (R) = 8% p.a.

Interest =  $\left ( \frac{P\times R\times T}{100} \right )$

$(\frac{550\times 8\times\frac{1}{2}}{100})$ = Rs 22

Total amount paid after 6 months = Principal amount + Interest = Rs 550 + Rs 22 = Rs 572

Q 10 . Rohit borrowed Rs . 60000 from a bank at 9% per annum for 2 years . He lent this sum of money to Rohan at 10% per annum for 2 years . How much did Rohit earn from this transaction ?

SOLUTION :

Principal amount lent out by Rohit (P) = Rs. 60000

Time period (T) = 2 years

Rate of interest (R) = 10% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{60000\times 10\times 2}{100} \right )$

= Rs. 12000

Principal amount borrowed by Rohit from the bank (P) = Rs. 60000

Time period (T) = 2 years

Rate of interest (R) = 9% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{60000\times 9\times 2}{100} \right )$

= Rs. 10800

Amount gained by Rohit = Rs. 12000 – 10800 = Rs. 1200

Q 11 . Romesh borrowed Rs . 2000 at 2% per annum and Rs . 1000 at 5% per annum . He cleared his debt after 2 years by giving Rs . 2800 and a watch . What is the cost of watch ?

SOLUTION :

Principal amount borrowed by Romesh (P) = Rs. 2000

Time period (T) = 2 years

Rate of interest (R) = 2% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{2000\times 2\times 2}{100} \right )$ = Rs.80

Principal amount borrowed by Romesh (P) = Rs. 1000

Time period (T) = 2 years

Rate of interest (R) = 5% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{1000\times 5\times 2}{100} \right )$

= Rs.100

Total amount that he will have to return = Rs. 2000 + 1000 + 80 + 100 = Rs. 3180

Amount repaid = Rs. 2800

Value of the watch = Rs. 3180 – 2800 = Rs. 380

Q 12 . Mr.Garg lent Rs . 15000 to his friend . He charged 15% per annum on Rs. 12500 and 18% on the rest . How much interest does he earn in 3 years ?

SOLUTION :

Principal amount (P) = Rs 12500

Time period (T) = 3 years

Rate of interest (R) = 15% p.a.

Interest = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{12500\times 15\times 3}{100} \right )$ = Rs 5625

Rest of the amount lent = Rs 15000 – Rs 12500 = Rs 2500

Rate of interest = 18 % p.a.

Time period = 3 years

Interest =$\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{2500\times 18\times 3}{100} \right )$ = Rs 1350

Total interest earned = Rs 5625 + Rs 1350 = Rs 6975

Q 13 . Shikha deposited Rs . 2000 in a bank which pays 6% simple interest . She withdrew Rs . 700 at the end of first year .What will be her balance after 3 years ?

SOLUTION :

Principal amount deposited (P) = Rs 2000

Time period (T) = 1 year

Rate of interest (R) = 6% p.a.

Interest after 1 year = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{2000\times 6\times 1}{100} \right )$ = Rs 120

So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120

After 1 year , amount withdrawn = Rs 700

Principal amount left (P1) = Rs 2120 – Rs 700 = Rs 1420

Time period (T) = 2 years

Rate of interest (R) = 6% p.a.

Interest after 2 years = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{1420\times 6\times 2}{100} \right )$ = Rs 170.40

Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40

Q 14 . Reema took a loan of Rs . 8000 from a money lender , who charged interest at the rate of 18% per annum . After 2 years , Reema paid him Rs . 10400 and wrist watch to clear the debt . What is the price of the watch ?

SOLUTION :

Principal amount (P) = Rs 8,000

Rate of interest (R) = 18%

Time period (7) = 2 years

Interest after 2 years =  $\left ( \frac{P\times R\times T}{100} \right )$

$\left ( \frac{8000\times 18\times 2}{100} \right )$ = Rs 2,880

Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880 = Rs 10,880

Amount paid = Rs 10,400

Value of the watch = Rs 10,880 – Rs 10,400 = Rs 480

Q 15 . Mr . Sharma deposited Rs. 20000 as a fixed deposit in a bank at 10% per annum . If 30% is deducted as income tax on the interest earned , find his annual income .

SOLUTION :

Amount deposit (P) = Rs 20,000

Rate of interest (R) = 10% p.a.

Time period (T) = 1 year

Interest after 1 year = $\left ( \frac{P\times R\times T}{100} \right )$

= $\left ( \frac{20000\times 10\times 1}{100} \right )$ = Rs 2,000

Amount deducted as income tax = 30% of Rs 2,000 = 30 x $\frac{2000}{100}$ = Rs 600

Annual interest after tax deduction = Rs 2,000 – Rs 600 = Rs 1,400

#### Practise This Question

Let f: R R be function defined by f(x) = x3 + 4.  Then f is


text:
# A high intensity $^{6}$He beam for the β-beam neutrino oscillation facility

Abstract : This work presents the production and extraction of the short-lived radionuclide 6He in yet unmatched yields from the ISOLDE facility at CERN. It is the first report of 6He production using spallation neutrons via the 9Be(n, α)6He reaction. These neutrons are produced from the 1.4GeV proton beam of the Proton Synchrotron Booster (PSB) striking a tungsten converter, and are impinging on a porous BeO material. The central position of 6He in future experiments is due to its role as a necessary radioactive nucleus to realize the β-beam at CERN, a next-generation facility to study neutrino oscillation parameters, and hence neutrino masses. In the β-beam scenario, an intense beam of radioactive 6He nuclei will be produced, accelerated to multi-GeV energies and stored in a dedicated storage ring. The resulting virtually mono-directional anti-neutrino beam from the decay of the stored 6He nuclei will be directed towards a remote underground neutrino detector. A similar beam of, e.g., 18Ne will provide neutrinos, an ideal concept to test CP violation in the neutrino sector. The results of the present experiment demonstrate for the first time that the necessary conditions for the realization of the proposed β-beam scheme with anti-neutrinos can be fulfilled.
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-00697265
Contributor : Michel Lion <>
Submitted on : Tuesday, May 15, 2012 - 9:16:15 AM
Last modification on : Friday, February 12, 2021 - 10:58:07 AM

### Citation

T. Stora, E. Noah, R. Hodak, T.Y. Hirsh, M. Hass, et al.. A high intensity $^{6}$He beam for the β-beam neutrino oscillation facility. EPL - Europhysics Letters, European Physical Society/EDP Sciences/Società Italiana di Fisica/IOP Publishing, 2012, 98, pp.32001. ⟨10.1209/0295-5075/98/32001⟩. ⟨in2p3-00697265⟩

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text:
## yrivers36 3 years ago Solve each equation symbolically. With explanation please. sqrt (x)+1=sqrt (x+2)

1. Kreshnik

hint... $\LARGE \sqrt x+1=\sqrt{x+2}$ square both sides..

2. yrivers36

Im not sure but I think I should foil left side of problem

3. yrivers36

$(\sqrt{x}+1)(\sqrt{x}+1)-x+2$

4. yrivers36

thats an = in front of x

5. Kreshnik

$\LARGE (\sqrt x+1)^2=(\sqrt{x+2})^2$ $\LARGE (\sqrt x)^2+2\cdot \sqrt{x}\cdot 1+1^2=x+2$ $\LARGE x+2\cdot \sqrt{x}=x+2-1$ $\LARGE 2\cdot \sqrt{x}=x-x +2-1$ $\LARGE 2\cdot \sqrt{x}=1$ $\LARGE \sqrt{x}=\frac12$ square both sides and there you go :)

6. Kreshnik

can you do it ?

7. yrivers36

solution manuel says x = 1/4

8. Kreshnik

that's correct. Well done ! ;)

9. yrivers36

Thank you once again :)

10. Kreshnik


text:
# Questioning an assumption in calculus of variations

1. Jul 13, 2015

### hideelo

When deriving stationary points of a function defined by a 1-D integral (think lagranian mechanics, Fermat's priniciple, geodesics, etc) and arriving at the Euler Lagrange equation, there seems to me to be an unjustified assumption in the derivation. The derivations I have seen start with something along the following lines: assume some function x(t) is the function we are looking for, let x'(t) = x(t) + η(t) be a nearby path... The derivation will then go on to show the conditions for the original function x(t), namely that the function satisfy the Euler Lagrange equation.

It seems a little odd that we assume, without proof, that this function exists and then sort out its properties. How do we know such a function exists? Does it always exist? Are there conditions on this? Isn't it a little shady to be discussing properties of something if we havent proved yet that it exists?

On the other hand, once we complete the derivation, it seems clear to me that a function which satisfies the Euler Lagrange equation will be a stationary function. I think.

I'm still left feeling uncomfortable however about this. Is there some outside proof which shows that this function must exist?

I shoud give the caveat that I have only seen this derivation in physics books, I dont own any math books on calculus of variations

2. Jul 13, 2015

### Dr. Courtney

Functions that represent reasonable things in physics are real and have reasonable mathematical properties (continuous, derivatives exist, etc.)

3. Jul 13, 2015

### hideelo

I understand that, but we are asking for something more here, the existence of some extreme values. In calculus on R we can say that on compact subsets of R, the extreme values exist. I don't know what the analogy would be here when I am not looking at R, but some subset of all continuous functions.

I think I am looking for some topology on the space of functions and hope to see some compact set or something. Maybe there an easier way, I don't know.

4. Jul 13, 2015

### Terandol

The minimal function certainly does not always exist mathematically. I haven't done this type of analysis for a long time but couldn't you just take say $C^1([0,1])$ as your space of functions with an action functional given by $S(f)=\int_{0}^1 f(x) dx$? Surely this can't have a local min/max because you could always just remove a tiny portion of the original function and glue in a Gaussian of the appropriate size in a continuous way to make the integral a tiny bit bigger/smaller than any proposed min/max function.

What it is saying is simply that if the minimizer does exist, then it must satisfy these equations. So we can replace the problem of finding a minimizer with the problem of solving a differential equation which is usually more tractable. Of course not every differential equation has a solution so the nonexistence of a minimizer will manifest itself in the nonexistence of a solution to the differential equation. In the example I gave above, the Euler lagrange equations simply become 1=0 so no solution to the Euler Lagrange equation exist as expected.

If you want to know the conditions about when the existence of a minimizer is guaranteed, generally you will need to use some functional analysis (although in the one dimensional case things may be much simpler, I don't really know) and it is much more complicated than a simple compactness argument. For example, I remember a theorem from a PDE course that stated if you take a reflexive Banach space B (which is some space of functions for calculus of variations applications) with a subset $A \subseteq B$ which is weakly closed in $B$ and if $S:A\to \mathbb{R}$ is a coercive, weakly lower semicontinuous functional then it is bounded below and achieves it's minimum in $A$. I'm not sure why I remember this theorem since I don't even remember the precise definitions of the conditions anymore but in any case, the existence of the minimum is a hard problem to solve and and the answer depends quite a bit on the assumptions on your space of functions and on the properties of the functional you are trying to minimize.

Last edited: Jul 13, 2015
5. Jul 20, 2015

### Xiuh

What Terandol said. The thing is that the Euler-Lagrange equations are a necessary condition. If you look closely, what the theorem actually says, is that if the minimizer exists, then it has to satisfy the E-L equations.


text:
Question-and-Answer Resource for the Building Energy Modeling Community
Get s tarted with the Help page

# Revision history [back]

### EMS for ERV bypass (free-cooling) control

Hello,

I was trying to model a controlled mechanical ventilation with heat recovery using ZoneHVAC:EnergyRecoveryVentilator in OpenStudio. For the heat recovery using an Air-To-Air heat exchanger, I want to active the free-cooling mode (bypass) when exhaust air temperature or zone air temperature is higher a set-point temperature. However, with ERV component in OpenStudio Application I cannot do that. Regarding temperature-based control for ERV unit, what I can do is to active the free-cooling when

low limit < Toutdoor < high limit && Toutdoor < Texshaut

Does anyone know if there is anyway to create an EMS for ERV bypass controller based on zone air temperature or exhaust air temperature please?

Thanks a lot

Long

 2 retagged __AmirRoth__ 4335 ●5 ●16 http://bleedinggreenna...

### EMS for ERV bypass (free-cooling) control

Hello,

I was trying to model a controlled mechanical ventilation with heat recovery using ZoneHVAC:EnergyRecoveryVentilator in OpenStudio. For the heat recovery using an Air-To-Air heat exchanger, I want to active the free-cooling mode (bypass) when exhaust air temperature or zone air temperature is higher a set-point temperature. However, with ERV component in OpenStudio Application I cannot do that. Regarding temperature-based control for ERV unit, what I can do is to active the free-cooling when

low limit < Toutdoor < high limit && Toutdoor < Texshaut

Does anyone know if there is anyway to create an EMS for ERV bypass controller based on zone air temperature or exhaust air temperature please?

Thanks a lot

Long


text:
# Quadratic Equation Solver v2.0 in Javascript using Functional Programming

I have posted this in continuation to my earlier post Quadratic equation solver in JavaScript.

Tried to improve on the shortcomings last time and also changed the logic. It is still not covering all the input cases. Planning to improve it over time with feedbacks.

Recently started learning about Functional Programming and would also want this program to be reviewed for correct use of Functional Programming techniques.

Example:

const solutions = solveEquation('2 * x^2 - 10 * x + 12');
console.log(solutions); // [2, 3]

Solution:

/**
*  Solves a Quadratic/Linear Equation Equation
*  @param {string} equation
*  Input variants of equation:
*  ax^2 + bx + c,
*  ax^2 + bx,
*  ax^2 + c,
*  bx + c,
*  c,
*  For a = ax^2, +ax^2, -ax^2, a*x^2, axx, a*x*x, ax*x, x^2, +x^2, -x^2, 1/2 x^2 but not (1 1/2 x^2 as trimming spaces)
*  For b = bx, +bx, -bx, b*x, x, +x, -x
*  For c = c, -c, +c, 0
*  Where a,b,c are Integers and Floats
*  Spaces are removed
*  TBD for cases:
*  bx + ax^2 + c,
*  (1 * x ^ 2 + 1 * x ^ 2) === ( 2 * x ^ 2)
*  Throw valid Errors
*  @returns An array of roots, empty array if no roots
*/
// # Functional Programming
// # Higher Order Functions
// # Pure Functions
// # No Side Effects
try {
// In Functional Programming is function chaining like this correct.
return getRoots(getCoefficients(removeStar(removeSpaces(equation))));
}
catch (e) {
console.log(e);
}
}
/**
* Removes ' ' from a string
* @param {string} equation
* @returns string without spaces ' '
*/
function removeSpaces(equation) {
const _equation = equation;
return _equation.replace(/\s/g, '');
}
/**
* Removes '*' from a string
* @param {string} equation
* @returns string without stars '*'
*/
function removeStar(equation) {
const _equation = equation;
return _equation.replace(/\*/g, '');
}
/**
* Parse equation to find Coefficients
* @param {any} equation
* @returns Array of (Integers/Floats) Quadraric Equation Coefficients.
*/
function getCoefficients(equation) {
// Used eval instead of parseFloat to support Fractions
const _equation = equation;
const regxForA =
(_equation.search(/[0-9]x\^2/g) !== -1)
? /x\^2/g
: (_equation.search(/x\^2/g) !== -1)
? /x\^2/g
: (_equation.search(/[0-9]xx/g) !== -1)
? /xx/g
: (_equation.search(/xx/g) !== -1)
? /xx/g
: "ERROR"
const a = eval(getMatchingValue(_equation, regxForA));
const equationForB =
(regxForA === 'ERROR')
? _equation
: _equation.split(getRegexSource(regxForA))[1];
const regxForB =
(equationForB.search(/[0-9]x/g) !== -1)
? /x/g
: (equationForB.search(/x/g) !== -1)
? /x/g
: "ERROR";
const b = (regxForB === "ERROR") ? 0 : eval(getMatchingValue(equationForB, regxForB));
const c = (b === 0)
? eval(equationForB)
: !isNaN(eval(equationForB.split(getRegexSource(regxForB))[1]))
? eval(equationForB.split(getRegexSource(regxForB))[1])
: 0;
return [a, b, c];
}
/**
* Get Formatted Regex Source
* @param {any} regex
* @returns Source of a regex after removing any \
*/
function getRegexSource(regex) {
const _regex = regex;
return (_regex.source).toString().replace(/\\/g, '');
}
/**
* Iterates over equation to find coffecient
* @param {any} equation
* @param {any} regx
* @returns (Integers/Floats) coffecient
*/
function getMatchingValue(equation, regx) {
if (regx === 'ERROR') return 0;
const _equation = equation;
const charOccurance = _equation.search(regx);
let counter = 0;
let coefficient = '';
while (counter < charOccurance) {
coefficient = coefficient + equation.charAt(counter);
counter++;
}
return (coefficient === '-')
? '-1'
: (coefficient === '+' || coefficient === '')
? '1'
: coefficient;
}
/**
* Return Roots of the equation
* @param {any} a Coefficient
* @param {any} b Coefficient
* @param {any} c Coefficient
* @returns Array containing roots
*/
function getRoots(coeffList) {
const [a, b, c] = coeffList;
if (a === 0) {
if (b === 0) {
if (c === 0) {
return [];
} else {
return [];
}
} else {
// Linear Equation
return [-c / b];
}
}
let d = b * b - 4 * a * c;
if (d < 0) {
return [];
}
const b1 = eval(- + b);
if (d === 0) {
const r = -0.5 * b1 / a;
return [r, r];
}
d = Math.sqrt(d);
return [(b1 - d) / (2 * a), (b1 + d) / (2 * a)];
}

console.log(solveQuadraticEquation('4/2x^2 - 20/2x + 24/2'));
console.log(solveQuadraticEquation('2x^2 - 10x + 12'));
console.log(solveQuadraticEquation('0 * x^2 - 10 * x + 12'));

# What an improvement!

... and now it wont fall over at the slightest misplaced character.

## There is one bad point.

function getRegexSource(regex) {
const _regex = regex;

function getCoefficients(equation) {
// Used eval instead of parseFloat to support Fractions
const _equation = equation;

Dont copy the arguments to another variable for no reason. You copy variables to

1. create a less noisy alias, if the variable starts to pad out code lines

function getCoefficients(equation) {
const equ = equation;  // as an alias

2. If you need to preserve the variable

 function foo(someVal) {
const rollBack = someVal;

... code in a try
someVal += "some changes";

catch(e){
return rollBack;
}
return someVal

Doing it for no reason just adds noise which is never good. Also creating copies can be dangerous, particularly the underscored variety.

_underscore is not a very visible marker. We read by shape and the underscore can be missed. See your getMatchingValue function where you have used the wrong variable equation rather than the _equation . Though it does not mater in this case, it could have been a bug if the variable with the underscore was being changed, and a hard to spot one at that.

## Some style points.

• Don't use block-less statement blocks.

// this is not good style
if(foo === bar) bar ++;

// and this is even worse
if(foo === bar)
bar ++;

Always add the block delimiters {} the time saved typing is not worth the time that can be lost via this short cut (its insidious evil). As any long term C like language coder will attest to, when you modify the code you can miss the { } and create a very very hard to spot bug.

// example in some code noise to illistrate the problem
//  you may add another line as so statement block.
bar = someInput();
foo = getInfo();
if(foo === bar)
bar ++;
foo --;
show(foo);
show(bar);
// sittine inside a lot of code and with the indent the missing {}
// can easily be overlooked.

To prevent this ever happening always delimit blocks

 if(foo === bar) { bar++ }

// or
if(foo === bar) {
bar++;
}

• Encapsulate the core functionality.

You can wrap the whole thing in a object/class/export exposing only what you need to solve an equation. This will make the functionality portable and reusable.

As it is, it exposes a lot of named functions to the global scope that may clash with other code if you are creating larger projects.

• Use the "use strict" directive

For extra safety add the "use strict" directive to the top of every JS file as an extra protection against common bad coding habits (not that I spotted any). Note that if you use modules the use strict mode is automatic.

## Summing up.

This is a vast improvement over the previous version. There are some logic inefficiencies in how you handle the ternary tests to find the RegExp for each coefficient. ( RegExp have built in logic operations, you seldom need to do it in code)

If you find yourself nesting more than 2 ternary expressions there is likely a better simpler and easier to read solution.

Getting the roots is but a minor part of the code. You will get a lot more miles out of it if you think of the whole thing as a polynomial equation parser that returns the polynomial equation object. To store any polynomial all you need is the coefficients. Then it would be

const equ = polynomial.parse("x^2+3x-4"); // returns an equation object
console.log(equ.roots());

Roots is just one of the functional properties of the equation, some others like toString(), toFunction(), y(x), derivative(), antiDerivative(), minima() maxima(), and all the other useful stuff that a set of Coefficients can represent.

So all up, great work.. :)

• What you write at the last has inspired me to write another programme. :) – Shubhendu Vaid Apr 20 '18 at 14:53


text:
Table 1.
Values of different structural and dynamical parameters for the graph illustrated in Figure 5 to which edges have been removed. See note below.
edge$ΔL-%$$ΔG-pp%$$ΔG-pq%$$Δtinf%$
{1,2} − 25.4 10.0 19.2 30.5
{1,9} − 23.2 15.7 26.5 25.7
{8,9} − 21.5 17.7 29.6 16.9
{3,5} − 0.04 14.9 15.6 3.8
{2,5} + 0.89 15.9 15.0 3.8
{2,4} + 1.42 18.4 18.7 3.8
edge$ΔL-%$$ΔG-pp%$$ΔG-pq%$$Δtinf%$
{1,2} − 25.4 10.0 19.2 30.5
{1,9} − 23.2 15.7 26.5 25.7
{8,9} − 21.5 17.7 29.6 16.9
{3,5} − 0.04 14.9 15.6 3.8
{2,5} + 0.89 15.9 15.0 3.8
{2,4} + 1.42 18.4 18.7 3.8

Note. The edges correspond to the labeling of the nodes in the mentioned figure. $ΔL-%$ is the percentage of change respect to the original graph in the average communicability shortest path length. $ΔG-pp%$ and $ΔG-pq%$ are the percentage of change with respect to the original graph for the values of Gpp and Gpq averaged for the nodes and edges in the shortest communicability paths. $Δtinf%$ is the time needed by a disease factor to infect all the nodes of the corresponding graph in an SI simulation by using the approximate solution described below with β = 0.005 and initial condition xi(0) = 1/9 for all i.

Close Modal


text:
# Linear Algebra: Linear Combinations

by lockedup
Tags: algebra, combinations, linear
 P: 70 1. The problem statement, all variables and given/known data Let V = {f: $$\mathbb {R}\rightarrow\mathbb {R}$$} be the vector space of functions. Are f1 = ex, f2 = e-x (both $$\in$$ V) linearly independent? 2. Relevant equations 0 = aex + be-x Does a = b = 0? 3. The attempt at a solution My first try, I put a = e-x and b = -ex. He handed it back and told me to try again. I think the problem was that my a and b were not constants. But how to prove that there are no constants that will make the equation 0? I wrote some stuff down about the fact that, if a=0, then b = 0 (and the converse). Is that sufficient or am I way off?
 Sci Advisor HW Helper P: 1,276 One way to do it would be a proof by contradiction. Suppose there are constants, not both 0, a and b, such that aexp(x) + bexp(-x) = 0 for all x. Then aexp(x) = -bexp(-x), so -a/b * exp(2x) = 1 for all x. I won't complete it for you, but look at x=0, and see what restriction it places on a/b. Then look at a different point and you will reach a contradiction. (Also note, I've implicitly assumed b is nonzero, so you should handle that case as well)
 HW Helper P: 3,309 how about a.ex+b.e-x = 0 multiply by e-x a.e2x+b = 0 this is not true in general for all x in the reals unless a=b=0
Mentor
P: 21,067

## Linear Algebra: Linear Combinations

 Quote by lockedup 1. The problem statement, all variables and given/known data Let V = {f: $$\mathbb {R}\rightarrow\mathbb {R}$$} be the vector space of functions. Are f1 = ex, f2 = e-x (both $$\in$$ V) linearly independent? 2. Relevant equations 0 = aex + be-x Does a = b = 0?
Yes, a = b = 0 is one solution, and is always a solution regardless of whether these functions are linearly dependent or linearly independent. The real question is whether this solution, the trivial solution, is the only solution. If so, the functions are linearly independent. If not, they are linearly dependent.

 Related Discussions Calculus & Beyond Homework 2 Calculus & Beyond Homework 3 Calculus & Beyond Homework 0 Calculus & Beyond Homework 8 Calculus & Beyond Homework 6


text:
# An actual first-order formulation of ZFC?

1. May 27, 2012

### mpitluk

Can someone point me to a first-order axiomatization of ZFC?

As I've mostly seen ZFC expressed in higher-order logics.

2. May 27, 2012

### AKG

ZFC is a first order theory, where have you been seeing second order formulations? I'm sure every one of the top Google results for "ZFC axioms" will give you a first order formulation. In particular, Wikipedia.

3. May 27, 2012

### Hurkyl

Staff Emeritus
The higher-order axioms are replaced with axiom schema. e.g. the axiom schema of subsets is collection of statements
{x in A | P(x)} is a set,
one for every unary predicate P in the language of first-order set theory.

First-order ZFC requires infinitely many axioms to specify. First-order NBG, however, is an 'equivalent' set theory in an important sense, but only requires finitely many axioms.


text:
# Proving two sequences identical

I found something quite interesting while browsing around the OEIS yesterday. I have no idea how to prove this (I don't even know if it's true in general, but Mathematica tells me that it holds up to at least 50 000.) Can we prove that this sequence, that is $$\text{A095112}(n) = \sum_{\substack{p \text{ prime, } k>0 \\ p^k \mid n}} \frac{n}{p^k} \quad \overset{\underset{\mathrm{?}}{}}{=} \quad \sum_{d \mid n} \varphi(d) \Omega \left( \frac{n}{d} \right),$$ where the double sum on the left is "sum of $n/k$ over all prime powers $k>1$ which divide $n$". On the right $\varphi(n)$ denotes Euler's totient function and $\Omega(n)$ is total number of prime factors of $n$. Looking at an example we see that $$\sum_{\substack{p \text{ prime, } k>0 \\ p^k \mid 12}} \frac{12}{p^k} = \frac{12}{2} + \frac{12}{2^2} + \frac{12}{3} = 6 + 3 + 4 = 13,$$ and \begin{align*} \sum_{d \mid 12} \varphi(d) \Omega \left( \frac{12}{d} \right) &= \varphi(1)(3)+\varphi(2)(2)+\varphi(3)(2)+\varphi(4)(1)+\varphi(6)(1)+\varphi(12)(0) \\\ &= 3+2+4+2+2=13. \end{align*} How do we attack a problem like this? Is this obvious or hard? Any light shed on this would be nice!

-
Your lefthand expression doesn’t say what you want it to say: $3^2\le 12$, but $3^2\nmid 12$, so you don’t want a term $\frac{12}{9}$. You want $\sum\limits_{\substack{p\text{ prime},k>0\\p^k\mid n}}\frac{n}{p^k}$. – Brian M. Scott Nov 7 '11 at 9:43
@BrianM.Scott: I was a bit unsure about that, thanks. – Carolus Nov 7 '11 at 10:11

The identity is true. Here is a proof:

Let $*$ denote Dirichlet convolution, that is $(f*g)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right)$. Then let $\chi$ be the indicator function for the prime powers, that is $\chi(n)=1$ if $n=p^k$ and $\chi(n)=0$ otherwise. Notice that in particular $$(\chi*1)=\sum_{d|n}\chi(d) =\Omega(n).$$

Recall that $(\phi*1)(n)=\sum_{d|n}\phi(d)=n$. Hence we can write $$\sum_{p^k |n} \frac{n}{p^k}=\sum_{d|n} \chi(d)\frac{n}{d}=(\text{Id}*\chi)(n)=(\phi*1*\chi)(n)$$

$$=(\phi*\Omega)(n)=\sum_{d|n}\phi(d)\Omega\left(\frac{n}{d}\right)$$ as desired.

A similar proof with the indicator function for the primes rather then prime powers yields

$$\sum_{p|n} \frac{n}{p} =\sum_{d|n}\phi(d)\omega\left(\frac{n}{d}\right).$$

-
Wow, it was easier than I thought! Very nice explanation. By the way, did you see my question regarding this answer of yours? – Carolus Nov 7 '11 at 12:10
This is an old answer, but the second line should read: $(f*g)(n)=\sum_{d|n}f(\textbf{d})g\left(\frac{n}{d}\right)$ – 6005 May 15 '13 at 14:23

The key here is the formula $$\sum_{d | n} \varphi(d) = n.$$ Note that $\Omega(n/d)$ counts the number of non-trivial prime powers $p^k$ dividing $n/d$. Let $p^k|n$ where $k > 0$. How many times is $p^k$ counted in the right-hand side? It is counted $\varphi(d)$ times whenever $p^k|n/d$, which is to say, whenever $d|n/p^k$, and in total $$\sum_{d | n/p^k} \varphi(d) = \frac{n}{p^k} \text{ times}.$$ The identity immediately follows.

To answer your second question, how do we attack a problem like this, I suggest trying to understand what the two sides are counting. My solution (which is identical to Eric's) starts with an understanding of the right-hand side, and then uses a well-known formula to relate it to the left-hand side. So the second suggestion is: find out some basic information about the functions involved.

-


text:
# Find all natural values n, that $\sqrt{P_{2}(n)}$ is also a natural number

I have a polynomial of the second degree $a\cdot n^2 + b \cdot n + c$ and I need to find out natural numbers $n$, such that $\sqrt{a\cdot n^2 + b \cdot n + c}$ is also a natural number.

After thinking about this problem, my idea was to rewrite it into $a\cdot n^2 + b \cdot n + c = x^2$ and rewrite it to look like a Pell's equation:

$$(2an + b)^2 - 4ax^2= b^2 - 4 ac$$ which kind of resembles $x^2 - n y ^2 = 1$, but not really close enough to for me to solve it.

Then I tried to find some similarities in solutions for particular cases. For example when I took the equation $\sqrt{3\cdot n^2 - 2 \cdot n - 1}$ and wrote the program, the couple of first values were: 1, 5, 65, 901, 12545 (with no visible pattern for me).

So how should I solve this problem?

• Do you really want to have a general solution that works for all values of $a,b,c$? If you are eventually only interested in some special cases, it would be easier to tackle those. – Joonas Ilmavirta Jul 15 '15 at 7:51
• It's difficult to take your question seriously with that as your profile picture. – user230734 Jul 15 '15 at 7:53
• @JoonasIlmavirta if the general solution is too hard, you can show me how to come up with a solution for $3n^2 - 2n - 1$ – Salvador Dali Jul 15 '15 at 8:09
• If $c=k^2$ it is easy to rewrite in another Pell equation. In another case decided the same view through the Pell equation. – individ Jul 15 '15 at 8:41
• – individ Jul 15 '15 at 8:54

If you wish to have an infinite number of integer solutions to,

$$an^2+bn+c = d^2\tag1$$

but not necessarily all of them, then one way is, yes, to solve a Pell equation. First, as pointed out in the other answer, you need a initial solution. Second, if you limit it to only integers, then $a$ must not be a square.

Given an initial $n,d$ to $(1)$, then an infinite more can be found as,

$$ax^2+bx+c = (-d+py)^2$$

where,

$$x = n+qy$$

$$y = 2dp+(b+2an)q$$

and $p,q$ solve the Pell equation $p^2-aq^2 = 1$.

Example:

$$3n^2-2n-1 = d^2$$

with initial $n,d = 5,8$, you get,

$$3x^2-2x-1 = (-8+py)^2$$

where,

$$x = 5+qy$$

$$y = 16p+28q$$

and $p,q$ is any solution to $p^2-3q^2=1$.

Note: This is an easy method to get an infinite number of solutions, but not all of them.

• Any idea how can I find initial solution? – Salvador Dali Jul 15 '15 at 18:00
• @SalvadorDali: You can use mod arguments like in this answer. Or, since it is the smallest , then a simple program can quickly establish if it has a solution within a bound, say $|n|<1000$. If there is, the method above then guarantees there are an infinite more. – Tito Piezas III Jul 16 '15 at 2:49

With slight change of notation, you are looking for, given $a,b,c\in \mathbf{Z}$, points with integer co-ordinates on the curve $ax^2-y^2+bx+c=0$.

This is a conic. If you know one solution exists, then we can find infinitely many through rational parametrization. This is well-known (for example can be found in Silverman-Tate). Call this one solution point $Q_0$

Now fix a line $L$ whose equations has integer/rational coefficients. For a variable point $P\in L$ with RATONAL co-ordinates, write the equation of the line connecting $P$ with $Q_0$; as the intersection of a conic with line has 2 points you get the other point $P'$. This will be a rational point.(Convince yourself). So you get solutions you want.


text:
## Cannot assign raw object 2

My objective is to build a function using modules that runs the bisection method. I have a working bisection method but when I try to implement it as a module I keep getting this error and I can’t figure out why.

F[z_]:=3 + z + z^2 * Cos[z]   bisectionMethod[f_, a_, b_, n_] := Module[   {},   For[i = 0, i < n, i++,    Print[StringForm["Loop #: ", i + 1]];    c = (a + b)/2;    y = f[c];    If[y < 0, a = c];    If[y > 0, b = c];    Print[StringForm["a = ", N[a]]];    Print[StringForm["b  = ", N[b]]];    c = (a + b)/2;    Print[StringForm["c = ", N[c]]];    Print[N[f[c]]];    ]      ] bisectionMethod[F, -4 , 2 , 10]

After being run

  Loop #: 1  Set::setraw: Cannot assign to raw object 2.  a = -4.  b  = 2.  c = -1.  2.5403  Loop #: 2  Set::setraw: Cannot assign to raw object 2.  a = -4.  b  = 2.  c = -1.  2.5403

The issue appears to be specifically variable b but I have no idea how b is any different from the others and how to tackle this.

I did a little bit of experimentation and it seems like the line If[y > 0, b = c]; is problematic

## Randomly assign n elements to n agents such that each agent only knows its own element

### Problem

I’m working on an app that involves shuffling and distributing playing cards to players. As a challenge, I tried to solve this problem in a way that doesn’t require a trusted intermediary.

In other terms, the task is to find a distributed algorithm that

• uniquely assigns $$n$$ agents numbers $$1..n$$
• allows each agent to know nothing about the assignment but its own
• when revealing the assignment, allows other players to verify the assignment

We also assume that knowing other’s assignment is an advantage for each agent, and revealing its own prematurely a disadvantage. Agents are also assumed to be able to talk with each other in a way hidden from all other agents.

## Partial solution

The solution I came up with works only under the assumption that adversaries do not collaborate.

The idea is to create a set of $$n$$ nonces, and assign each agent exactly one nonce. The set is then passed from agent to agent in an agreed upon order, hidden from all others, until each agent received the set exactly once. Each time an agent receives the set, it swaps its nonce with a new one, memorizes the new nonce, and confirms receival of the set to the others. This entire procedure is done twice, at which point, all agents have received the set at least once after all other agents swapped their nonces, making it impossible to recognize and hence map the nonces to the other agents.

When the last agent receives the set the second time, it shares it with everyone, and all agents confirm to the others that their nonce is contained in the set. The agents then assign a number to each nonce in the set based on an agreed upon random seed, giving us the required unique assignment.

To allow ownership verification, instead of the nonces, agents put the hash value of their nonce on the set, revealing the actual nonce only when verification is required.

The problem with this solution is that if adversaries are allowed to collaborate, each time an adversary receives the set, they can compare their versions, identify changes and potentially derive which nonce belongs to other agents, allowing them to know what number got assigned to them.

All ideas are appreciated!

## How to assign a specific service to a specific provider based on location

I’m looking for a way to assign a specific service to a specific provider based on location. I’m currently developing a website with various service providers providing the same service. Do we have a way to achieve this using woocommerce? Thanks

## I am looking for a way to assign numerical values to colors (then interpolate a color scale) and then assign numerical values to comparative colors

Hello may name is Chris

I am completely new to programming (but I love it!! I am learning Swift).

I am currently looking for a solution to the mentioned problem. I’m not sure if it is a ML task or if it is better to do it in the traditional way – maybe by comparing and interpolating RGB values. How I imagine that might be easier to understand in the video.

VIDEO (is there a better way to provide a video?)

The color values below were taken before the adjustment and assigned to the assigned values 0, 2, 5, 10, 25, 50 and 100. The round section is taken from a photo (by light and shadow you can see that not every color value brings a perfect match. But that shouldn’t be a problem (I hope so) by comparing the pixels and evaluating the color value with the most matching colour at the end). My spontaneous manual evaluation gives a value that should be around 19-20.

I have already thought back and forth whether it is possible to do the whole thing using color values (e.g. R G and B) but that is not so easy since there should also be many different results depending on the color format I am working with.

On the other hand, I don’t know how to solve it using ML. It would be easy to recognize the exact color values 0, 2, 5, 10, 25, 50 and 100 – but how can I have intermediate values determined? What do you think? Which approach is the better one? If you have a tip for me, that I know in which direction I have to think further would be great! And if you have an idea how I can work on my problem I would be so grateful!!! … otherwise I think I will need the coming days to find some solution.

## construct a sql , assign it to a variable and execute inside shell script

I am trying to execute a mysql query inside my shell script, which is not executing . Here is what i am trying to do

query="\"select count(*) from information_schema.tables where TABLE_SCHEMA not in ('mysql','performance_schema','information_schema') and engine!='INNODB';\""

echo $query "select count(*) from information_schema.tables where TABLE_SCHEMA not in ('mysql','performance_schema','information_schema') and engine!='INNODB';"  Inside my shell script , i am executing below which is not working. It is not giving any error but simply not considering -e option and listing down all the valid options for mysql client. mysql -u$  username -p$password -e$  query

I even validated below in my shell prompt

echo "mysql -uroot -p -e$query" mysql -uroot -p -e"select count(*) from information_schema.tables where TABLE_SCHEMA not in ('mysql','performance_schema','information_schema') and engine!='INNODB';"  which is perfectly fine and executes without any issues. I am not sure what to do now. please help.Thanks ## Is there some mechanisms in PHP to assign “less trust” to scripts in a given dir? (not a duplicate) [closed] Kindly stop redirecting my questions to that unrelated one which doesn’t answer my question whatsoever. I’ve already read every answer there and it doesn’t help at all. If it did, why would I ask this much more specific question? This has been a continuous worry and problem for me for ages: For practical and logical reasons, I am forced to trust some third-party PHP libraries. These are installed, updated and managed with Composer, and live in C:\PHP-untrusted-external, entirely separated from my own PHP scripts, which live in C:\PHP-my-own. The scripts in C:\PHP-my-own include and make use of the libraries in C:\PHP-untrusted-external. Since there is no way that anyone, especially not I, could ever vet all the third-party code, and all updates, I’m looking for some way to “secure” or “sandbox” these in some way, even if it’s just partial. Basically, I’m worried that one day, an update will make an edit such as: unlink('C:\');  Or: phone_home_to_hacker_server($  contents_of_my_harddrive);

If that happened, the scripts would happily run and do those actions. Nothing prevents them from doing so.

Is there really no way to specify in the php.ini configuration file, something like:

security.sandbox_dir = "C:\PHP-untrusted-external"

Or:

security.refuse_network_connections_for_dir = "C:\PHP-untrusted-external" security.refuse_disk_io_for_dir = "C:\PHP-untrusted-external"

… or something like that?

I don’t understand Docker. I have tried it countless times, and it makes no sense whatsoever to me. I don’t want Docker. I don’t want to deal with containers. Correction: I can’t deal with it. I’ve tried to, but didn’t understand it. Several times.

I just want PHP to support this in itself, and it seems more than reasonable to me. Doesn’t it seem reasonable to you?

The saying that “at some point, you have to trust other people” is way too generic/vague to apply here. It’s bypassing the problem. I don’t trust people at all, and for good reason. It seems idiotic that we are (apparently) just supposed to sit around and wait for the disaster to happen. At least if I could prevent the third-party scripts form doing anything with the file system and network, that would go some way toward mitigating this issue. It still won’t make the scripts unable to lie about the numbers/data they return to me, but at least they can’t directly “phone home” or delete random files.

## Assign resources that each have a certain amount of work they can provide to tasks that require a certain amount of work

I’m attempting to do a hobby automation project and have come to a roadblock. I have a certain problem I need to solve, but can’t think of the solution nor what the name for the problem would be.

Say we have n tasks where each task requires $$x_i$$ amount of work to be done and m resources where each resource can provide $$y_j$$ work. The total amount of work required will equal the the total amount of work the resources can provide, i.e. $$\sum_{i=1}^{n} y_i = \sum_{j=1}^{m}x_j$$. For all j from 1 to m, $$y_j \in \left \{1, 2, 6, 12, 24\right \}$$ and each $$x_i = a*1 +b *2 + c*6+d*12+e*24$$. I was looking at task assignment problems, but those seemed to be a bit overkill since they were concerned with optimization where I’m just simply trying to slot the correct blocks in the right place so that I don’t have tasks that are given too few or too many resources.

My current guess is that you can iterate over each task, and give it the largest resource available that doesn’t go over the amount of work that is left for the task to be completed. It’s almost analagous to filling a jar with different sized rocks; the best way is to start with the largest rocks and then go down in size from there, so that the smaller rocks fill in the space between the larger ones. Am I not taking something into consideration that complicates this problem further? I’m sorry if this is an obvious one, but I’m a hobbyist programmer and couldn’t think of the name of the problem or of a good set of keywords to google. The closest I could find is task assignment so far.

## Psycopg2 using value SET before assign to other column

I have a table with columns as below: (id, col1, col2, col3, col4). I need to insert new row to this table, with given data is a dict: data = {'id': 1, 'col1': 1, 'col2':1, 'col3':1, 'col4': 1} for example.

I want to use the query below to insert data into table if id not existed, or update the table if id existed:

query =  '''     INSERT INTO tablename (id, col1, col2, col3, col4)     VALUES (%(id)s, %(col1)s, %(col2)s, %(col3)s, %(col4)s)     ON CONFLICT (id) DO UPDATE SET     col2 = tablename.col2 + 1;     col3 = col2/100 ''' result = cur.execute(query, data)

In the code above, I want to assign col3’s value to (value of col2 calculated before)/100. How can I do that?

## Encoding System that Assign Same Number of Bits for Each Character

I am trying to get a binary string that has been converted from text of a text file, I am able to get that but the problem is, I need each character to be represented by same number of bits, but that is not what I get (please see the below python code and corresponding output). For example, character i is represented by 1101001, which is 7 bits long, but character ! is represented by 100001, which is 6 bits long.

Is there any encoding/decoding system where each character takes same amount of bits?

content = open('a.txt', 'r').read() test_str = content # using join() + ord() + format()  ... Converting String to binary   Binary = ' '.join(format(ord(i), 'b') for i in test_str)  #Decimal=int(Binary, 2)    # printing original string   print("The original string is : " + str(test_str))  # printing result   print("The string after Binary conversion : \n" + str(Binary))

Output:

The original string is : Hi! Is there a solution? The string after Binary conversion :  1001000 1101001 100001 100000 1001001 1110011 100000 1110100 1101000 1100101 1110010 1100101 100000 1100001 100000 1110011 1101111 1101100 1110101 1110100 1101001 1101111 1101110 111111

## How to assign a big integer struct to a long n

I have the following code, it does not work properly: void assignInt(bint res ,long n) { int i; / counter / int t; / int to work with */

if (n >= 0)     res->sign = PLUS; else     res->sign = MINUS;  for (i=0; i<MAX_SIZE; i++)     res->digits[i] = (char) 0;  res->size = -1;  t = labs(n);  while (t > 0) {     res->size++;     res->digits[res->size] = (t % 10);     t = t / 10; }  if (n == 0)     res->size = 0;

} // initialzes a bint to value n


text:
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Verbal Aids for Science (Mottos)

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Posted (edited)

We're all  familiar with how certain concepts in science are difficult to grasp, or perhaps to remember.

Mottos are these pieces of wisdom that try to capture an idea and make it easier to grasp yourself, or to get across to others through an intuitive verbal formulation.

Some of them can be quite unfortunate. They shall not be mentioned here, if possible.

So my suggestion is: Let's all share those brief phrasings that have helped us remember an idea, understand it better, keep it closer to our hearts and brains. They can be our own creation or found elsewhere and treasured ever since.

One example could be (these are my own):

"Velocity is the parameter which tells you what time and space directions you're looking at when you're moving" (special relativity)

Or:

"A tensor is a product of projections of physical quantities; different observers see different projections; but if they knew the rules of rotation, they would all relate their data and say: 'Oh, we're all seeing the same thing!'" (general relativity, tensor calculus)

I invite you all to share brief formulas like these that you've found useful to capture an idea, whether you've devised them yourselves or you've found them somewhere else and retained them as useful conceptual tools. Keep in mind that the more rigorous you try to be, the less memorable the motto.

If you don't like mine, feel free to tell me. All science areas welcome.

Edited by joigus
minor modification

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A worth topic sir.  +1

Input  =  Output  +  Accumulation

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It doesn't seem to change the need for rigor, while leaving the less rigorous an invite too join in; a dear wish of mine is to play a musical instrument, but I've yet to receive an aphorism that by-passes the need to practice...

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13 minutes ago, dimreepr said:

It doesn't seem to change the need for rigor, while leaving the less rigorous an invite too join in; a dear wish of mine is to play a musical instrument, but I've yet to receive an aphorism that by-passes the need to practice...

That reminds me of Matrix. One of those drives you could download all your knowledge from. Even practical knowledge. Tensor calculus in 5 seconds. Helicopter piloting in 3. That would be nice.

1 hour ago, studiot said:

A worth topic sir.  +1

Input  =  Output  +  Accumulation

Thanks. I knew you would appreciate it. Let's see if people can provide some good ideas. Let's see if it arouses some interest. I hope so.

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First two laws of Thermodynamics:

1. You can't get something for nothing

2. Even at best you can't break even

(Hope I remembered that correctly! 😄)

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Posted (edited)
10 minutes ago, zapatos said:

First two laws of Thermodynamics:

1. You can't get something for nothing

2. Even at best you can't break even

(Hope I remembered that correctly! 😄)

LOL. That's a good one. Or two. +1

1 hour ago, studiot said:

Input  =  Output  +  Accumulation

Sorry, is it meant to be Input = Output + Accumulation or Output = Input + Accumulation?

Edited by joigus

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1 hour ago, zapatos said:

First two laws of Thermodynamics:

1. You can't get something for nothing

2. Even at best you can't break even

(Hope I remembered that correctly! 😄)

As I recall it goes like this

The First Law states that you can't win you can only break even.

The Second Law states that you can't break even.

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2 hours ago, studiot said:

Input  =  Output  +  Accumulation

In what context? I am trying to get my head round what that means!

My contribution: "Are photons particles or waves? Neither: they are photons"

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The Laws of thermodynamics had a 'gambling' analogy.

You can't win.
You can't break even.
The third was...
And you can't get out of the game.

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Well, we've got thermodynamics, SR, GR, algorithms, and QM, off to a good start. Pending some clarification on algorithms. This looks promising.

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waves/particle/fields are not what these thing are, they are just how we describe them.

The map, not the country.

(With thanks to @Strange )

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12 minutes ago, Dord said:

waves/particle/fields are not what these thing are, they are just how we describe them.

The map, not the country.

Corollary to this...

Theories are not reality; They just describe (some aspects of ) reality.
Models, not the 'real thing'.

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These are police expressions but seem equally relevant to all inquisitive professions:

Absence of evidence is not evidence of absence.

ABC - the detectives' mantra

Accept nothing (on face value)

Believe nothing (without independent corroboration)

Challenge Everything (with reasonable lines of enquiry)

ABC - radio comms & report writing

Accurate

Brief

Clear

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Posted (edited)
16 hours ago, Strange said:

In what context? I am trying to get my head round what that means!

17 hours ago, joigus said:

Sorry, is it meant to be Input = Output + Accumulation or Output = Input + Accumulation?

Input = Output + Accumulation is correct.

I learned this one because my first year flatmate mumbled it every day for a year and came top in Chemical Engineering.
He said it was the only thing he learned for the exams and covered eeverything, which is not far from the truth.
It is a general flux formula for momentum, mass energy etc.

Some more from the time when formulae were not hidden by apps.

The surveyors mantra

$\frac{{{{\left( {{\rm{something}}} \right)}^{\rm{2}}}}}{{{\rm{twicesomethingelse}}}}$

I put this one in because I know joigus likes taylor - maclaurin series.
It arises in the truncation of such a series for many purposes eg

correction to a measured line on sloping ground

${\rm{correction = }}\frac{{{{\left( {{\rm{heightdifference}}} \right)}^{\rm{2}}}}}{{{\rm{2slopedistance}}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{H}}^{\rm{2}}}}}{{{\rm{2L}}}}$

Deviation of a circular arc from the straight (tangent)

${\rm{Offset = }}\frac{{{{\left( {{\rm{straightdistance}}} \right)}^{\rm{2}}}}}{{{\rm{2circleradius}}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{L}}^{\rm{2}}}}}{{{\rm{2R}}}}$

Finally there is 'onion differentiation'  - differentiation from the oustside in.

When differentiating a complicated trigonometric expression

"First the power, then the trig, then the angle" is the rule

Edited by studiot

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Color codes for resistors:  Bad Boys rape.......  Totally out of place today and I would not repeat it now-- but it got me through my EE courses 54 years ago.

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On 8/19/2020 at 10:32 AM, studiot said:

I learned this one because my first year flatmate mumbled it every day for a year and came top in Chemical Engineering.
He said it was the only thing he learned for the exams and covered eeverything, which is not far from the truth.
It is a general flux formula for momentum, mass energy etc.

I still have some difficulty with this, but maybe it's not 'up my alley' as they say. Every trade has its lingo.

On 8/19/2020 at 10:32 AM, studiot said:

I put this one in because I know joigus likes taylor - maclaurin series.

I do. Brilliant ideas never die. But Fourier rocks too.

14 hours ago, OldChemE said:

Color codes for resistors:  Bad Boys rape.......  Totally out of place today and I would not repeat it now-- but it got me through my EE courses 54 years ago.

There seems to be a branch of mnemonics here ranging from cross-cultural to off colour.

Some of them I find more difficult to remember than the sequence of codes itself.

On 8/19/2020 at 9:25 AM, Dord said:

[...]

Absence of evidence is not evidence of absence.

ABC - the detectives' mantra

[...]

ABC - radio comms & report writing

[...]

Thank you. I find especially useful the report-writing one.

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1 hour ago, joigus said:

I still have some difficulty with this, but maybe it's not 'up my alley' as they say. Every trade has its lingo.

No its very widespread.

It is just a statement of the equation for a control volume

The base is the input.

If you subtract the output, what remians must be left inside the control volume ie is the accumulation.

Think income minus expenditure = addition to bank balance.

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1 hour ago, studiot said:

No its very widespread.

It is just a statement of the equation for a control volume

The base is the input.

If you subtract the output, what remians must be left inside the control volume ie is the accumulation.

Think income minus expenditure = addition to bank balance.

Yes, I can see that input - output = accumulation would make a bit more sense.

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17 hours ago, OldChemE said:

Color codes for resistors:  Bad Boys rape.......  Totally out of place today and I would not repeat it now-- but it got me through my EE courses 54 years ago.

Bless Violet's big heart!

Questions to ask, and answers to question ?

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On 8/18/2020 at 3:42 PM, dimreepr said:

It doesn't seem to change the need for rigor, while leaving the less rigorous an invite too join in; a dear wish of mine is to play a musical instrument, but I've yet to receive an aphorism that by-passes the need to practice...

I don't think any aphorism will bypass the need to practice. Not even with tensor calculus.

But how about this for pep talk?:

If Django Reinhardt could do it, why can't you?

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11 hours ago, joigus said:

Django Reinhardt

Are you in Belgium?

------------------------------------

If you want to remember the value of Log 2, look at the face of your colleague next to you: the ear (3) the eye (0) the nose (1) the other eye (0) the ear (3) and you get log 2 = 0,30103 (approx.)

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Posted (edited)
2 hours ago, michel123456 said:

Are you in Belgium?

No. I learnt about Django Reinhardt from within my light cone, by gathering signals.

Very nice mnemonics. +1

The left ear is inverted.

2 hours ago, michel123456 said:

Are you in Belgium?

I mentioned Django Reinhardt because he had two fingers burnt, and managed to play beautifully with only eight fingers.

Edited by joigus
Addition

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13 hours ago, joigus said:

If Django Reinhardt could do it, why can't you?

Because I'm lazy and content with that, which makes my speciality 'being efficient, in doing enough to live and have time left over to think and dream'. 😊

And for the life of me, I can't make an acronym from that.

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Posted (edited)
On 8/18/2020 at 1:43 PM, joigus said:

So my suggestion is: Let's all share those brief phrasings that have helped us remember an idea,

Nice topic! My personal motto regarding science in relation to engineering and personal interest

In mainstream science I depend
Its correctness and methods I defend
while hoping for flaws until the end

As an engineer I trust and depend on well established theories to be correct and to have predictive power within their limits. I have limited possibilities and interest to confirm their correctness, I’m depending on them to be correct. I have, in my profession, not yet observed an issue that could not be explained by correct application of mainstream physics.

I try to defend the established mainstream when required but also the methods that finds issues and drives updates; theories should be modified when there is support by observations or new evidence.

That said, I always hope for flaws to emerge. I do not expect or have hopes to be the one that discover such flaws. I don not think it is likely for something new to replace for instance General Relativity but that does not prevent me from having a (how ever small) hope that something new and even better will be established. Yet another confirming observation of GR is exciting. But genuine evidence* of something macroscopic that contradicts and requires modification or replacement of GR would be even more exciting.

*) Not counting known limitations such as Black Hole centers, Quantum effects etc.

Edited by Ghideon

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If we are getting poetic:

"A little learning is a dang'rous thing;
Drink deep, or taste not the Pierian spring:
There shallow draughts intoxicate the brain,
And drinking largely sobers us again.

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