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# Synopsis: Supercurrents in a junction—it’s all about the timing

#### Evidence for Two Time Scales in Long SNS Junctions

F. Chiodi, M. Aprili, and B. Reulet

Published October 23, 2009

Superconductor–normal-metal–superconductor (SNS) junctions are used in a variety of sensor applications, but also provide a way to study the fundamentals of superconductivity. Two current values determine the response of an SNS junction when excited by ac signals: critical current, the maximum supercurrent that can flow through the junction, and retrapping current, below which the junction switches from a resistive state back to a superconducting state. What actually causes the time response of these currents is a longstanding puzzle and the picture is more complicated in junctions than in all-superconductor samples.

In a paper published in Physical Review Letters, Francesca Chiodi, Marco Aprili, and Bertrand Reulet, at the Université Paris-Sud, Orsay, France, have performed conductivity experiments on $\text{Nb}$$\text{Al}$-based junctions. By varying the excitation frequency and measuring the current through the junctions, Chiodi et al. have identified the characteristic time scales of the critical and retrapping currents, and associate the timescales of these currents with elastic and inelastic electron-phonon scattering, respectively, giving further insight into the complicated dynamics of SNS junctions. The understanding of the dynamics of SNS junctions may offer new routes to design fast devices, such as low impedance detectors. – Daniel Ucko


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# With known ciphertext and plaintext, find other party's Three Pass Protocol Key?

If I'm the second party participating in a three pass protocol (for simplicity, let's say Shamir's Three Pass Protocol) such that:

1. Bob sends me a ciphertext
2. I encrypt it and send it back
3. He removes his encryption and sends it back to me
4. I decrypt it fully

Using the information in steps 1 (ciphertext) and 4 (plaintext), how difficult is it for me to determine the key that Bob is using?

I know that this is a somewhat esoteric use-case, but I'm curious.

## 1 Answer

how difficult is it for me to determine the key that Bob is using?

This is essentially the discrete log problem; Bob's key is a value $b$, and his encryption mechanism is computing the value $P^b \bmod p$. Given $P$, $P^b \bmod p$, recover $b$ is the definition of discrete log.

Actually, you get two pairs of values encrypted with Bob's key; the plaintext (encrypted in step 1, and recovered in step 4), and the exchanged ciphertext (which you see encrypted in step 2 and decrypted in step 3); however having multiple pairs doesn't actually make the discrete log problem any easier.

So, if the group is large enough to make the three pass algorithm secure (for example, large enough that someone hearing the values in steps 2 and 3 could not recover Bob's key), it's also large enough to make recovering the symmetric keys by you infeasible as well.

format: plain


text:
## EmmaH Group Title Solve by factoring: 3x^2 + 2x - 1 = 0 {-1, 2/3} {1, -2/3} {-1, 1/3} {1, -1/3} one year ago one year ago

1. EmmaH Group Title

please explain how i find this

2. JamesJ Group Title

well, dividing first by 3, we have $x^2 + \frac{2}{3}x - \frac{1}{3} = 0$ Now, you want factors of -1/3 that add up to 2/3

3. Hero Group Title

@EmmaH, the steps for solving this are the same as what I showed you in the previous problem.

4. EmmaH Group Title

divide what by three?

5. Hero Group Title

I'll even give you a hint, but you have to solve the rest yourself: Given: $$3x^2 + 2x - 1$$ Split the middle term: $$3x^2 + 3x - 1x - 1$$ Now factor by grouping

6. EmmaH Group Title

(3x^2+3x) + (-1x-1)?

7. Hero Group Title

Okay, now what's the next step?

8. Hero Group Title

Do you know how to factor $$3x^2 + 3x$$ ?

9. EmmaH Group Title

3x(1x+1)

10. Hero Group Title

Yes, or simply 3x(x + 1) Writing 1 before the x is not necessary. Now factor (-1x - 1). Let me know what you get.

11. EmmaH Group Title

?

12. JamesJ Group Title

Well, you have a factor of (x+1) above where you wrote 3x(x+1) Can you get a factor of (x+1) out of (-1x - 1) ?

13. EmmaH Group Title

what do you mean?


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# Tahir Hassan's Blog

My Technical Notes

## Wednesday, 12 June 2013

### Ajax Toolkit's ModalPopupExtender hiding page content

On an ASP.NET page I was working on, I put a Ajax Toolkit ModalPopupExtender instance. As part of a Button Click event I called the Show() method, after which the content of the page disappeared and thereafter the dialog appeared in its place.

The one clue I had to go by was the FireBug error which occurred each time:

TypeError: (intermediate value)(...) is not a function

It turns out that this error has something to do with the a known error to do with ValidationSummary and ModalPopupExtender. See this answer on StackOverflow by Aaron Hoffman.

The solution is to define and use a subclass of the ValidationSummary class which inserts a semi-colon in the JavaScript.

public class AjaxValidationSummary : ValidationSummary
{
protected override void OnPreRender(EventArgs e)
{
base.OnPreRender(e);
ScriptManager.RegisterStartupScript(this.Page, this.Page.GetType(), this.ClientID, ";", true);
}
}


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# Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$

1. Let $k$ be a positive integer.
2. Define $u_0 = 0\,,\ u_1 = 1\$ and $\ u_n = k\,u_{n-1}\ -\ u_{n-2}\,,\ n \geq 2$.
3. Show that for each integer $n$,
the number $u_{1}^{3} + u_{2}^{3} + \cdots + u_{n}^{3}\$ is a multiple of $\ u_{1} + u_{2} + \cdots + u_{n}$.

Computing a few terms I found \begin{align*}u_0 &= 0\\u_1 &= 1\\u_2 &= k\\u_3 &= k^2-1\\u_4 &= k(k^2-1)-k = k^3-2k\\u_5 &= k(k^3-2k)-(k^2-1) = k^4-3k^2+1\\u_6 &= k(k^4-3k^2+1)-(k^3-2k) = k^5-4k^3+3k.\end{align*}

I am not sure how we can use this to solve the question, but I think it may help. Cubing these expressions seems very computational so there must be an easier way.

• $u_3 = k^2 - 1$, not $k^2 - k$. – Robert Israel Jul 15 '16 at 17:55
• @RobertIsrael Thanks, fixed. – user19405892 Jul 15 '16 at 18:04

It appears that $$y_n = \dfrac{u_1^3 + \ldots u_n^3}{u_1 + \ldots + u_n}$$ satisfies the recurrence relation $$y_n = (k^2+k-1)(y_{n-1} - k y_{n-2} + k y_{n-3} - y_{n-4}) + y_{n-5} \ \text{for}\ n \ge 6$$ Given that $y_1, \ldots, y_5$ are integers, this would imply that all $y_n$ are integers.

EDIT: Writing $\cos(\theta) = k/2$, we have $$u_n = \frac{\sin(n\theta)}{\sin(\theta)}$$ which can be verified by induction. Don't worry about $\theta$ being real only for $|k|\le 2$.

Using this we can obtain closed-form formulas for $u_1 + \ldots + u_n$ and $u_n^3 + \ldots + u_n$, and $y_n$ (it's rather tedious if working by hand, but elementary)

$$y_n = \frac{-\cos((2n+1)\theta) + 2 \cos(\theta) - \cos((n+1)\theta) - \cos(n\theta) + 1}{\cos(\theta) - \cos(3\theta) - \cos(2\theta)+1}$$

and it can be verified directly that this satisfies the recurrence above.

• How did you find out $y_n$ satisfied the recurrence? – user19405892 Jul 15 '16 at 18:12
• Using the gfun package in Maple. – Robert Israel Jul 15 '16 at 18:12
• I am looking for an elementary way of solving this question. – user19405892 Jul 15 '16 at 18:13
• @lhf: Sorry, that was $u_n$ not $a_n$. – Robert Israel Jul 15 '16 at 20:59

Given the equation $$u_n=ku_{n-1}-u_{n-2}\tag{1}$$ where $u_0=0$ and $u_1=1$, we get the solution $$u_n=\frac{\alpha^n-\alpha^{-n}}{\alpha-\alpha^{-1}}\tag{2}$$ where $$\alpha=\frac{k+\sqrt{k^2-4}}2\tag{3}$$ except when $k=2$ where the solution is $$u_n=n\tag{4}$$ and the result for $k=2$ follows from the fact that the sum of the cubes of the first $n$ consecutive integers is the square of the sum of the first $n$ consecutive integers. A proof without words is given in this answer.

For the solution $(2)$, we get $\alpha^2+\alpha+1=(k+1)\alpha$ and \begin{align} &\sum_{j=0}^{n-1}\left(\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\right)^3\\ &=\sum_{j=0}^{n-1}\frac{\alpha^{3j}-3\alpha^j+3\alpha^{-j}-\alpha^{-3j}}{\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}}\\ &=\frac1{\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}}\left(\frac{\alpha^{3n}-1}{\alpha^3-1}-3\frac{\alpha^n-1}{\alpha-1}+3\frac{\alpha^{-n}-1}{\alpha^{-1}-1}-\frac{\alpha^{-3n}-1}{\alpha^{-3}-1}\right)\\ &=\frac1{\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}}\left(\frac{\left(\alpha^{3n}-1\right)\left(1-\alpha^{3-3n}\right)}{\alpha^3-1}-\frac{3\left(\alpha^n-1\right)\left(1-\alpha^{1-n}\right)}{\alpha-1}\right)\tag{5} \end{align} and \begin{align} &\sum_{j=0}^{n-1}\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\\ &=\frac1{\alpha-\alpha^{-1}}\left(\frac{\alpha^{n}-1}{\alpha-1}-\frac{\alpha^{-n}-1}{\alpha^{-1}-1}\right)\\ &=\frac1{\alpha-\alpha^{-1}}\left(\frac{\left(\alpha^{n}-1\right)\left(1-\alpha^{1-n}\right)}{\alpha-1}\right)\tag{6} \end{align} Therefore, we can compute the ratios \begin{align} r_{n-1} &=\left.\sum_{j=0}^{n-1}u_j^3\middle/\sum_{j=0}^{n-1}u_j\right.\\ &=\left.\sum_{j=0}^{n-1}\left(\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\right)^3\middle/\sum_{j=0}^{n-1}\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\right.\\ &=\frac1{\alpha^2-2+\alpha^{-2}}\left(\frac{\left(\alpha^{2n}+\alpha^n+1\right)\left(\alpha^{2-2n}+\alpha^{1-n}+1\right)}{(k+1)\alpha}-3\right)\\ &=\frac1{k^2-4}\left(\frac{\left(\alpha^{2n}+\alpha^n+1\right)\left(\alpha^{2n-2}+\alpha^{n-1}+1\right)}{(k+1)\alpha^{2n-1}}-3\right)\\ &=\frac1{k^2-4}\left(\frac{\alpha^{2n-1}+\alpha^{n}+\alpha^{n-1}+\alpha^{1}+1+\alpha^{-1}+\alpha^{1-n}+\alpha^{-n}+\alpha^{1-2n}}{k+1}-3\right)\\ &=\frac1{k^2-4}\left(\frac{\alpha^{2n-1}+\alpha^{n}+\alpha^{n-1}+\alpha^{1-n}+\alpha^{-n}+\alpha^{1-2n}}{k+1}-2\right)\tag{7} \end{align} Due to the equation \begin{align} &(x-1)\left(x-\alpha\right)\left(x-\alpha^{-1}\right)\left(x-\alpha^2\right)\left(x-\alpha^{-2}\right)\\ &=(x-1)\left(x^2-kx+1\right)\left(x^2-\left(k^2-2\right)x+1\right)\\ &=x^5-mx^4+kmx^3-kmx^2+mx-1\tag{8} \end{align} where $m=k^2+k-1$, the ratios $r_n$ in $(7)$ satisfy the relation $$r_n=mr_{n-1}-kmr_{n-2}+kmr_{n-3}-mr_{n-4}+r_{n-5}\tag{9}$$ Computing the first few values of $r_{n-1}$ yields \begin{align} r_{-2} &=\frac1{k^2-4}\left(\frac{\alpha^{-3}+\alpha^{-1}+\alpha^{-2}+\alpha^2+\alpha^1+\alpha^3}{k+1}-2\right)\\ &=1\\ r_{-1} &=\frac1{k^2-4}\left(\frac{\alpha^{-1}+\alpha^0+\alpha^{-1}+\alpha^1+\alpha^0+\alpha^1}{k+1}-2\right)\\ &=0\\ r_{0} &=\frac1{k^2-4}\left(\frac{\alpha^1+\alpha^1+\alpha^0+\alpha^0+\alpha^{-1}+\alpha^{-1}}{k+1}-2\right)\\ &=0\\ r_{1} &=\frac1{k^2-4}\left(\frac{\alpha^3+\alpha^2+\alpha^1+\alpha^{-1}+\alpha^{-2}+\alpha^{-3}}{k+1}-2\right)\\ &=1\\ r_{2} &=\frac1{k^2-4}\left(\frac{\alpha^5+\alpha^3+\alpha^2+\alpha^{-2}+\alpha^{-3}+\alpha^{-5}}{k+1}-2\right)\\ &=k^2-k+1 \end{align}\tag{10} The recurrence $(9)$ and the computations $(10)$ ensure that for all $n$, $r_n\in\mathbb{Z}$.

fiddling with small $k.$ Take $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$

CONCLUSION: for $k \geq 2,$ $$\color{blue}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},}$$ while $\color{blue}{x \equiv 0,1 \pmod {k+1}}.$

When $k=2,$ $$y = x^2.$$ This comes up pretty often, the sum of the consecutive cubes (starting with $1$) is the square of the sum of the consecutive numbers.

When $k=3,$ $$y = \frac{x^2 (x + 3)}{4}.$$ For this one, you need to know that $x \equiv 0,1 \pmod 4.$

This already suggests that $k=4$ gives $y = a x^4 + b x^3 + c x^2 + d x,$ with rational coefficients, and some restrictions on $x$ that make $y$ an integer. If true, the coefficients can be found by taking four $x$ points, then making and inverting a certain four by four rational matrix. NOPE, not that much effort required. Stays cubic, a three by three matrix would have been enough...

Not quite what I expected: for $k=4,$ $$y = \frac{x^2 (2 x + 3)}{5},$$ while $x \equiv 0,1 \pmod 5.$

For $k=5,$ $$y = \frac{x^2 (3 x + 3)}{6},$$ while $x \equiv 0,1 \pmod 6.$

For $k=6,$ $$y = \frac{x^2 (4 x + 3)}{7},$$ while $x \equiv 0,1 \pmod 7.$

Apparently, for $k \geq 2,$ $$\color{red}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},}$$ while $x \equiv 0,1 \pmod {k+1}.$ Recall $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$

• It's not too hard to show that $u_n = \frac{\left(\frac{1}{a}\right)^n-a^n}{\frac{1}{a}-a}$ where $a$ satisfy $a = \frac{1}{k-a}$ and from this the formula you put up is easily checked to be true. – Winther Jul 15 '16 at 23:41


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# Detailed balance of an energy function

I'm having difficulties understanding a problem about the acceptance rules for a given energy landscape.

The Problem

Suppose a system in which the energy is a function of x only: $$e^{-\beta U(x)} = \theta(x)\theta(1-x)$$ where $\theta$ is the Heaviside step function. The question asks to write down the distribution of x in the canonical ensemble for two possible algorithms.

i.Generate a random change x between [-$\delta$,$\delta$]. Accept or reject the new x according to its energy.

ii. Generate a random number $\phi$ between [1,1+$\delta$]. With a probability of 0.5 invert the value of $\phi$ thus obtained. The new value of x is obtained by multiplying x with $\phi$.

a) Derive the correct acceptance/rejection rules for both schemes.

b) What happens when the acceptance rule of method (i) is used in the algorithm of method (ii)?

My approach so far

Starting with detailed balance $$N_i\cdot\alpha_{ij}\cdot acc_{ij} =N_j\cdot\alpha_{ji}\cdot acc_{ji}$$ we can write

$$\frac{acc_{ij}}{acc_{ji}} = \frac{N_j}{N_i}$$ Supposing that $\alpha$, the probability of choosing the state in the first place, is symmetrical. Given that it is a canonical ensemble the probability per site equals

$$P = e^{ \frac{F-E}{kT} }$$ So i guess from this I have to calculate a distribution N depending on the scheme (i or ii). So the probability function can be calculated with

$$\int_a^b p(E)dE$$ I could substitute the energy, but than I wouldn't know what the integration limits would be. Also I have no idea if I'm on the right track, or I am thinking way too complicated. Anyone who could help?

• As a comment, is this potential (finite square well between x=0 and x=1) really what you're trying to study? Are there walls at x=0 or x=1? It seems like there may be a steady flow out to infinity though, I mean, maybe it can stabilize... – CR Drost Mar 2 '16 at 15:28
• its an infinite square well for U(x), right? with walls at x=0 and x=1. – Julian Mar 2 '16 at 16:15
• Ah, I see, you take $\ln(0)$. I was thinking $\theta(x) + \theta(1 - x).$ – CR Drost Mar 3 '16 at 21:51

So usually the acceptance criterion for an simulated annealing approach is "accept if $\operatorname{rand}(0,1) < \exp[(U_0 - U_1)/\tau]$," and this will always accept if $U_0 > U_1$ or occasionally accept otherwise. However if your energy goes infinite outside of this range then you get an $e^{-\infty} = 0$ result, and we instead accept with probability 1 any movement that is in the appropriate range $x\in(0,1)$. Those are apparently your acceptance rules.
Now you say that you're having trouble determining what the steady-state probability density $f(x)$ is. We know that $f(x)~dx$ is the probability of observing the particle between $x$ and $x+dx.$ Supposing indeed that we start with density $f(x),$ update with the above, and then obtain $f(x)$ again, we can work out the result. Call the random variable beforehand $Y$ and afterwards $X$.
We generate a $Z$ uniformly between $-\delta$ and $+\delta$, then calculate $X = Y + Z.$ When we want to add variables like this, we have a joint probability $j(y,z)$ that we start with (in this case the events are independent and $j(y,z) = f(y) \cdot u_\delta(z),$ of course) and we derive our new distribution $h(x)~dx$ by summing up all of the $j(y, z)~dy~dz$ which lead to an $X$ between $x$ and $x + dx.$ So $h(x)~dx$ must be equal to $j(y(x, z), z)~\frac{\partial y}{\partial x}~dx~dz,$ and in the summing case, this works out really cleanly to:$$h(x) = \int_{\Omega_z} dz~j(x-z, z).$$ Assuming the uniform distribution we find that $$h(x) = \int_{-\delta}^{\delta} dz~\frac{1}{2\delta}~f(x-z),$$ and we recover $f$ after we throw out all values which are outside $(0, 1)$:$$f(x) = \frac{h(x)~\theta(x)~\theta(1-x)}{\int_0^1 du~h(u)}.$$Solving this equation is what would be necessary to get an explicit probability distribution in closed form. You should in principle be able to get a differential equation for this thing which can possibly be solved for a suitable $f(x),$ and qualitatively it should be roughly-constant in the middle with fringes of size $\delta$ where it drops to 0.


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## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 1 - Test - Page 111: 13

#### Answer

$-6x+10$

#### Work Step by Step

First, simplify within the parentheses by distributing 3 to obtain: $=6-2[3x+3-5] \\=6-2[3x-2]$ Distribute $-2$ to obtain: $=6-6x-(-4) \\=6-6x+4$ Simplify by combining like terms: $\\=-6x+(6+4) \\=-6x+10$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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Correct calculation of recursive limit

I'm constructing a binomial (decision) tree with nodes $x_i$ according to the following recursion:

$x_{i+1} = \begin{cases} x_i e^{\mu \delta t + \sigma(x_i)\sqrt{\delta t}} & \text{for next up node} \\ x_i e^{\mu \delta t - \sigma(x_i)\sqrt{\delta t}} & \text{for next down node} \end{cases}$

with

$\sigma (x_i) = \sigma_0\left( 1 - \alpha \frac{x_i - x_0}{x_0} \right)$

Taking the limit of $x_{i+1}$ for $\delta t \rightarrow 0$, I get:

$\lim\limits_{\delta t \rightarrow 0} x_{i+1} = \begin{cases} \lim\limits_{\delta t \rightarrow 0} x_i e^{\mu \delta t + \sigma(x_i)\sqrt{\delta t}} = x_i & \text{for next up node} \\ \lim\limits_{\delta t \rightarrow 0} x_i e^{\mu \delta t - \sigma(x_i)\sqrt{\delta t}} = x_i & \text{for next down node} \end{cases}$

Therefore, $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_i$. Using this limit recursively, I yield: $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0$. Therefore:

$\lim\limits_{\delta t \rightarrow 0} \sigma (x_i) = \lim\limits_{\delta t \rightarrow 0} \sigma_0\left( 1 - \alpha \frac{x_i - x_0}{x_0} \right) = \sigma_0\left( 1 - \alpha \frac{x_0 - x_0}{x_0} \right) = \sigma_0$

Is my notation correct? Is this last expression correct?

Am I allowed to take the limit recursively to yield $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0$ ?

Let $n$ be an integer. I believe what you mean to express is the sequence of functions $\{x^{(n)}\}$ with
\begin{gather*} x^{(n)}((k+1)\delta t)=x^{(n)}(k\delta t)\exp(\mu\delta t+Y_{k}^{(n)}\sigma(x^{(n)}(k\delta t))\sqrt{\delta t})\\ \text{where }\delta t=T/n\text{ and }0\leq k<n\text{ is an integer}. \end{gather*} $Y_{k}^{(n)}$ denotes a random variable with support $\{-1,+1\}$. $x^{(n)}(0)$ is a constant independent of $n$.
When $t/\delta t$ is not an integer, $x^{(n)}(t)$ is defined by linear interpolation. The limit you care about is $$\lim_{n\rightarrow\infty}x^{(n)}(t).$$ Note that $n$ depends on $\delta t$. Another subtlety is that this is not a limit in the usual sense, since randomness is involved (look up "convergence in distribution").


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# Figuring out Signal to Noise Ratio of Spectral data?

I have some Raman spectral data of a chemical:

1. Each spectrum consist of multiple tuples of (wavenumber, intensity)
2. I have multiple instances of 1. taken with different laser power. This causes the intensity to be different across different spectra.

My goal is to evaluate which laser power provides the best reading. I would like an algorithmic way to determine this.

Assuming that I have good characterization of my sample, i.e. I know the peak wavenumbers and their relative intensity, are there any algorithms to determine the Signal to Noise ratio of each read?

• So you want to calibrate (=tune) your RAMAN spectral measurement set-up. You have some known chemicals with known spectra which you will use calibrate the set-up. After calibration, do you want to use your system to distinguish a few substances, or do you want to distinguish dozens of substances, using an RAMAN Spectral Library? – Dirk Horsten Aug 4 '15 at 21:57
• @DirkHorsten Both. I am researching the feasibility of building a general all purpose Raman spectrometer to detect presence of toxins in food – Sparrowcide Aug 5 '15 at 1:10

Bit late to the party, but here we go:

which laser power provides the best reading

As Raman spectroscopist I'd say the highest power that doesn't burn (or otherwise alter) your sample and that doesn't lead to detector saturation. The reason is that your signal to noise ratio is the better the higher your total intensity is.

are there any algorithms to determine the Signal to Noise ratio of each read

Yes and no.

The most basic algorithms involves doing repeated measurements (thus do not work for one spectrum) and calculating the signal to noise ratio as the average of the signal intensity $I$ (e.g. band height or band area) in question divided by its standard deviation: $SNR = \frac{\bar I}{sd (I)}$; the SNR is the inverse of the relative standard deviation of that band. The SNR will be different for different bands in the same spectrum.

It is important to realize that under ideal conditions the noise on a Raman spectrum depends on the signal intensity. There are (at least) two contributions to the noise: the dark current with its noise (variance) and the shot noise (photon counting noise). You can measure the variance of the dark noise for your system and measurement parameters (laser intensity should not influence it, but exposure time of the camera does). Shot noise should ideally by Poisson distributed, i.e. $var (I) = mean (I)$. Thus, assuming ideal conditions as 0th approximation you can say that the SNR of a Raman signal with intensity I is at most $SNR \leq \frac{I}{\sqrt I} = \sqrt I$. First approximation including the dark current $d$ would yield $SNR \leq \frac{I - d}{\sqrt{I + s^2 (d)}}$, where I - d is the actual Raman signal (observed intensity - dark current) and assuming dark current variance $s^2 (d)$ and shot noise are independent so that their variances add up. These calculations can be done for single spectra.

What you should not do unless you can give good reasons why this should work in your particular application is measureing the noise at the baseline of the Raman spectrum and relate this to your Raman signal intensity. This approximation works only if you have such a high background that the baseline intensity $BG$ is approximately equal to the signal intensity ($\sqrt{I + BG} \approx \sqrt{BG}$, i.e. $BG \gg I$).

There's an excellent in-depth discussion of these topics in McCreery: Raman Spectroscopy for Chemical Analysis

Of course, the signal to noise ratio for a given application is yet different as e.g.

• you may combine the information of several spectral regions in order to gain SNR
• cross senstivities (matrix spectra) can reduce the SNR and/or make some analyte bands useless for the particular application.


text:
Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 18 Mean, Median And Mode Of Ungrouped Data are provided here with simple step-by-step explanations. These solutions for Mean, Median And Mode Of Ungrouped Data are extremely popular among Class 9 students for Maths Mean, Median And Mode Of Ungrouped Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 669:

We know:

(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.
Mean of these numbers:

(ii) The first ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.
Mean of these numbers:

$\frac{1+3+5+7+9+11+13+15+17+19}{10}\phantom{\rule{0ex}{0ex}}=\frac{100}{10}\phantom{\rule{0ex}{0ex}}=10$

(iii) The first seven multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.
Mean of these numbers:

$\frac{5+10+15+20+25+30+35}{7}\phantom{\rule{0ex}{0ex}}=\frac{140}{7}\phantom{\rule{0ex}{0ex}}=20$

(iv) The factors of 20 are 1, 2, 4, 5, 10 and 20.
Mean of these numbers:

(v) The prime numbers between 50 and 80 are 53, 59, 61, 67, 71, 73 and 79.
Mean of these numbers:

$\frac{53+59+61+67+71+73+79}{7}\phantom{\rule{0ex}{0ex}}=\frac{463}{7}\phantom{\rule{0ex}{0ex}}=66.14$

#### Page No 669:

Numbers of children in 10 families = 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
Thus, we have:

$=\frac{30}{10}\phantom{\rule{0ex}{0ex}}=3$

#### Page No 670:

Numbers of books issued in the school library: 105, 216, 322, 167, 273, 405 and 346
Thus, we have:

#### Page No 670:

Daily minimum temperatures = 35.5, 30.8, 27.3, 32.1, 23.8 and 29.9
Thus, we have:

#### Page No 670:

We know that,

The first five observations are x, x + 2, x + 4, x + 6 and x + 8.

Mean of these numbers = $\frac{x+\left(x+2\right)+\left(x+4\right)+\left(x+6\right)+\left(x+8\right)}{5}$
$⇒13=\frac{5x+20}{5}\phantom{\rule{0ex}{0ex}}⇒13×5=5x+20\phantom{\rule{0ex}{0ex}}⇒65=5x+20\phantom{\rule{0ex}{0ex}}⇒5x=65-20\phantom{\rule{0ex}{0ex}}⇒5x=45\phantom{\rule{0ex}{0ex}}⇒x=\frac{45}{5}\phantom{\rule{0ex}{0ex}}⇒x=9\phantom{\rule{0ex}{0ex}}$

Hence, the value of x is 9.

Now, the last three observations are 13, 15 and 17.
Mean of these observations = $\frac{13+15+17}{3}$
= $\frac{45}{3}$
= 15

Hence, the mean of the last three observations is 15.

#### Page No 670:

The individual weights of five boys are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg.
Now,
Let the weight of the sixth boy be x kg.
We know:

Also,
Given mean = 48 kg
Thus, we have:

Therefore, the sixth boy weighs 53 kg.

#### Page No 670:

Let the marks scored by 50 students be x1, x2,...x50.
Mean = 39
We know:

Thus, we have:

Also, a score of 43 was misread as 23.

#### Page No 670:

Let the numbers be x1, x2,...x24.

We know:

Thus, we have:

After addition, the new numbers become (x1+3), (x2+3),...(x24+3).
New mean:

#### Page No 670:

Let the numbers be x1, x2,...x20.
We know:

Thus, we have:

Numbers after subtraction: (x1$-$6), (x2$-$6),...(x20$-$6)

∴ New Mean = $\frac{\left({x}_{1}-6\right)+\left({x}_{2}-6\right)+........+\left({x}_{20}-6\right)}{20}$

#### Page No 670:

Let the numbers be x1, x2,...x15
We know:

Thus, we have:

After multiplication, the numbers become 4x1, 4x2,...4x15
∴ New Mean = $\frac{4{x}_{1}+4{x}_{2}+......+4{x}_{15}}{15}$

#### Page No 670:

Let the numbers be x1, x2,...x12.
We know:

Thus, we have:

After division, the numbers become:

#### Page No 670:

Let the numbers be x1, x2,...x20.
We know:

Thus, we have:

New numbers are:

(x1 + 3), (x2 + 3),...(x10 + 3), x11,...x20

New Mean:

#### Page No 670:

Let the numbers be x1, x2,..., x6.
Mean = 23
We know:

Thus, we have:

${x}_{1}+{x}_{2}.......{x}_{6}=138$.....................(i)

If one number, say, x6, is excluded, then we have:

${x}_{1}+{x}_{2}......+{x}_{5}=100....................\left(\mathrm{ii}\right)$

Using (i) and (ii), we get:

Thus, the excluded number is 38

#### Page No 670:

We know that,

Mean of height of 30 boys = $\frac{\sum _{i=1}^{30}{x}_{i}}{30}$

It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean.

Hence, the correct mean is 151.

#### Page No 670:

Mean weight of 34 students = 46.5 kg
Sum of the weights of 34 students = kg
Increase in the mean weight when the weight of the teacher is included = 500 g = 0.5 kg
∴ New mean weight = (46.5 + 0.5) kg = 47 kg
Now,
Let the weight of the teacher be x kg.
Thus, we have:

Therefore, the weight of the teacher is 64 kg.

#### Page No 670:

Mean weight of 36 students = 41 kg
Sum of the weights of 36 students =
Decrease in the mean when one of the students left the class = 200 g = 0.2 kg
Mean weight of 35 students = (41 $-$ 0.2) kg = 40.8 kg
Now,
Let the weight of the student who left the class be x kg.

Hence, the weight of the student who left the class is 48 kg.

#### Page No 671:

Average weight of 39 students = 40 kg
Sum of the weights of 39 students =
Decrease in the average when new student is admitted in the class = 200 g = 0.2 kg
∴ New average weight = (40 $-$ 0.2) kg = 39.8 kg
Now,
Let the weight of the new student be x kg.
Thus, we have:

Therefore, the weight of the new student is 32 kg.

#### Page No 671:

Let the average weight of 10 oarsmen be x kg.
Sum of the weights of 10 oarsmen = 10x kg
∴ New average weight = (x + 1.5) kg
Now, we have:

#### Page No 671:

Mean of 8 numbers = 35
Sum of 8 numbers = $35×8=280$
Let the excluded number be x.
Now,
New mean = 35 $-$ 3 = 32
Thus, we have:

Therefore, the excluded number is 56.

#### Page No 671:

Mean of 150 items = 60
Sum of 150 items = $\left(150×60\right)=9000$
New sum = [9000 $-$ (52 + 8) + (152 + 88)] = 9180

Correct mean =

Therefore, the correct mean is 61.2.

#### Page No 671:

Mean of 31 results = 60
Sum of 31 results = $31×60=1860$
Mean of the first 16 results = 58
Sum of the first 16 results = $58×16=928$
Mean of the last 16 results = 62
Sum of the last 16 results = $62×16=992$
Value of the 16th result = (Sum of the first 16 results + Sum of the last 16 results) $-$ Sum of 31 results
= (928 + 992) $-$ 1860
= 1920 $-$ 1860
= 60

#### Page No 671:

Mean of 11 numbers = 42
Sum of 11 numbers = 42$×$11 = 462
Mean of the first 6 numbers = 37
Sum of the first 6 numbers = 37$×$6 = 222
Mean of the last 6 numbers = 46
Sum of the last 6 numbers = 46$×$6 = 276
∴ 6th number = [(Sum of the first 6 numbers + Sum of the last 6 numbers) $-$ Sum of 11 numbers]
= [(222 + 276) $-$ 462]
= [498 $-$ 462]
= 36
Hence, the 6th number is 36.

#### Page No 671:

Mean weight of 25 students = 52 kg
Sum of the weights of 25 students = (52$×$25) kg = 1300 kg
Mean weight of the first 13 students = 48 kg
Sum of the weights of the first 13 students = (48$×$13) kg = 624 kg
Mean weight of the last 13 students = 55 kg
Sum of the weights of the last 13 students = (55$×$13) kg =715 kg
Weight of the 13th student = (Sum of the weights of the first 13 students + Sum of the weights of the last 13 students) $-$ Sum of the weights of 25 students
= [(624+715)$-$1300] kg
= 39 kg
Therefore, the weight of the 13th student is 39 kg.

#### Page No 671:

Mean score of 25 observations = 80
Sum of the scores of 25 observations = 80$×$25 = 2000
Mean score of another 55 observations = 60
Sum of the scores of another 55 observations = 60$×$55 = 3300

$=\frac{2000+3300}{80}\phantom{\rule{0ex}{0ex}}=\frac{5300}{80}\phantom{\rule{0ex}{0ex}}=66.25$

Therefore, the mean score of the whole set of observations is 66.25.

#### Page No 671:

Marks scored by Arun in English = 36
Marks scored by Arun in Hindi = 44
Marks scored by Arun in mathematics = 75
Marks scored by Arun in science = x
Average marks = 50
Thus, we have:

$⇒155+x=200\phantom{\rule{0ex}{0ex}}⇒x=200-155=45$

∴ Marks scored by Arun in science = 45

#### Page No 671:

Let the distance from the starting point to the island be x km.
Speed of the ship sailing out to the island = 15 km/h
Speed of the ship sailing back to the starting point = 10 km/h
We know:

Therefore, the average speed of the ship in the whole journey was 12 km/h.

#### Page No 671:

Total students in the class = 50
Number of boys = 40
∴ Number of girls = (50 $-$ 40) = 10
Average weight of students in the class = 44 kg
Average weight of girls in the class = 40 kg
Sum of the weights of girls in the class = (40 $×$10) kg = 400 kg
Thus, we have:

#### Page No 671:

The aggregate yearly expenditure of a family = Rs (18720 × 3) + Rs (20340 × 4) + Rs (21708 × 5)
= Rs (56160 + 81360 + 108540)
= Rs 246060

The total savings during the year = Rs 35340

The average yearly income = Rs 246060 + Rs 35340
= Rs 281400

∴ The average monthly income of the family =

Hence, The average monthly income of the family is Rs 23450.

#### Page No 671:

Total salary of 75 workers = ₹ 5680 × 75
= ₹ 426000

Total salary of 25 workers = ₹ 5400 × 25
= ₹ 135000

Total salary of 30 workers = ₹ 5700 × 30
= ₹ 171000

No. of remaining workers = 75 − (25 +30)
= 20

Total salary of 20 workers = ₹ (426000 − 135000 − 171000)
= ₹ 120000

∴ The mean weekly payment of the 20 workers = $\frac{120000}{20}$
= ₹ 6000

Hence, the mean weekly payment of the remaining workers is ₹ 6000.

#### Page No 671:

Let the number of girls be x and the number of boys be y.

Mean marks of boys =

Mean marks of girls =

Mean marks of all the students =

Hence, the ratio of the number of boys to the number of girls is 2:1.

#### Page No 672:

Average monthly salary of 20 workers = Rs 45900
Sum of the monthly salaries of 20 workers =
By adding the manager's monthly salary, we get:
Average salary = Rs 49200
Now,
Let the manager's monthly salary be Rs x.
Thus, we have:

Therefore, the manager's monthly salary is Rs 115200.

#### Page No 676:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Variable (xi) 4 6 8 10 12 Frequency (fi) 4 8 14 11 3

Mean = $\frac{\left(4×4\right)+\left(6×8\right)+\left(8×14\right)+\left(10×11\right)+\left(12×3\right)}{4+8+14+11+3}$
= $\frac{16+48+112+110+36}{40}$
= $\frac{322}{40}$
= 8.05

Hence, the mean of the following distribution is 8.05 .

#### Page No 676:

We will make the following table:

 Weight (xi) No. of Workers (fi) (fi)(xi) 60 4 240 63 3 189 66 2 132 69 2 138 72 1 72 12 771

Thus, we have:

=

#### Page No 677:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Diameter (in mm) (xi) 34 37 40 43 46 Number of screws (fi) 5 10 17 12 6

Mean = $\frac{\left(34×5\right)+\left(37×10\right)+\left(40×17\right)+\left(43×12\right)+\left(46×6\right)}{5+10+17+12+6}$
= $\frac{170+370+680+516+276}{50}$
= $\frac{2012}{50}$
= 40.24

Hence, the mean diameter of the heads of the screws is 40.24 .

#### Page No 677:

We will make the following table:

 Age (xi) Frequency (fi) (fi)(xi) 15 3 45 16 8 128 17 9 153 18 11 198 19 6 114 20 3 60 40 $\sum {f}_{i}{x}_{i}=$698

Thus, we have:

#### Page No 677:

We will make the following table:

 Variable (xi) Frequency (fi) (fi)(xi) 10 7 70 30 8 240 50 10 500 70 15 1050 89 10 890 50 2750

Thus, we have:

#### Page No 677:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Daily wages (in ₹) (xi) 250 300 350 400 450 Number of workers (fi) 8 11 6 10 5

Mean = $\frac{\left(250×8\right)+\left(300×11\right)+\left(350×6\right)+\left(400×10\right)+\left(450×5\right)}{8+11+6+10+5}$
= $\frac{2000+3300+2100+4000+2250}{40}$
= $\frac{13650}{40}$
= 341.25

Hence, the mean of daily wages of 40 workers in a factory is 341.25 .

#### Page No 677:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 Variable (xi) 10 15 20 25 30 Frequency (fi) 6 8 p 10 6

Mean = $\frac{\left(10×6\right)+\left(15×8\right)+\left(20×p\right)+\left(25×10\right)+\left(30×6\right)}{6+8+p+10+6}$
$⇒20.2=\frac{60+120+20p+250+180}{30+p}\phantom{\rule{0ex}{0ex}}⇒20.2\left(30+p\right)=610+20p\phantom{\rule{0ex}{0ex}}⇒606+20.2p=610+20p\phantom{\rule{0ex}{0ex}}⇒20.2p-20p=610-606\phantom{\rule{0ex}{0ex}}⇒0.2p=4\phantom{\rule{0ex}{0ex}}⇒p=\frac{4}{0.2}\phantom{\rule{0ex}{0ex}}⇒p=\frac{40}{2}\phantom{\rule{0ex}{0ex}}⇒p=20$

Hence, the value of  p is 20.

#### Page No 677:

We will make the following table:

 (xi) (fi) (fi)(xi) 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 41 + p 303 + 9p

We know:

Given:
Mean = 8
Thus, we have:

#### Page No 677:

We will prepare the following table:

 (xi) (fi) (fi)(xi) 15 8 120 20 7 140 25 p 25p 30 14 420 35 15 525 40 6 240 50 + p 1445 + 25p

Thus, we have:

#### Page No 677:

We will make the following table:

 (xi) (fi) (fi)(xi) 8 12 96 12 16 192 15 20 300 p 24 24p 20 16 320 25 8 200 30 4 120 100 1228 + 24p

Thus, we have:

$⇒24p=432\phantom{\rule{0ex}{0ex}}⇒p=18$

#### Page No 678:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 x 10 20 30 40 50 60 Total f 4 f1 8 f2 3 4 35

Mean = $\frac{\left(10×4\right)+\left(20×{f}_{1}\right)+\left(30×8\right)+\left(40×{f}_{2}\right)+\left(50×3\right)+\left(60×4\right)}{35}$

Also, 4 + f1 + 8 + f2 + 3 + 4 = 35
⇒ 19 + f1f2 = 35
f1f2 = 35 − 19
f1f2 = 16
⇒ 26 − 2f2f 2 = 16             (from (1))
⇒ 26 − f2 = 16
⇒ 26 − 16 =  f2
⇒  f2 = 10

Putting the value of f2 in (1), we get
f1 = 26 − 2(10) = 6

Hence, the value of f1 and  f2 is 6 and 10, respectively.

#### Page No 678:

We will prepare the following table:

 (xi) (fi) (fi)(xi) 10 17 170 30 f1 30f1 50 32 1600 70 f2 70f2 90 19 1710 120 3480 + 30f1 + 70f2

Thus, we have:

Also,
G
iven:
17
+ f1 + 32 + f2 + 19 = 120
68 + f1 + f2 = 120
f1 + f2 = 52
or, f2 = 52 $-$ f1
...(ii)
By putting the value of f2
in (i), we get:
2520
= 30f1 + 70(52 $-$ f1)
2520 = 30f1 + 3640 $-$ 70f1
40f1 = 1120
f1 = 28
Substituting the value in (ii), we get:
f2 = 52 $-$ f1 = 52 $-$ 28 = 24

#### Page No 678:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 x 15 17 19 20 + p 23 f 2 3 4 5p 6

Mean = $\frac{\left(15×2\right)+\left(17×3\right)+\left(19×4\right)+\left(\left(20+p\right)×5p\right)+\left(23×6\right)}{2+3+4+5p+6}$
$⇒20=\frac{30+51+76+100p+5{p}^{2}+138}{15+5p}\phantom{\rule{0ex}{0ex}}⇒20\left(15+5p\right)=5{p}^{2}+100p+295\phantom{\rule{0ex}{0ex}}⇒300+100p=5{p}^{2}+100p+295\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=300-295\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=5\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=\frac{5}{5}\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=1\phantom{\rule{0ex}{0ex}}⇒p=±1$

Hence, the value of  p is ±1.

#### Page No 678:

We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

For the following data:

 x 10 30 50 70 90 f 17 5a + 3 32 7a – 11 19

Mean = $\frac{\left(10×17\right)+\left(30×\left(5a+3\right)\right)+\left(50×32\right)+\left(70×\left(7a-11\right)\right)+\left(90×19\right)}{17+5a+3+32+7a-11+19}$
$⇒50=\frac{170+150a+90+1600+490a-770+1710}{60+12a}\phantom{\rule{0ex}{0ex}}⇒50\left(60+12a\right)=2800+640a\phantom{\rule{0ex}{0ex}}⇒3000+600a=2800+640a\phantom{\rule{0ex}{0ex}}⇒640a-600a=3000-2800\phantom{\rule{0ex}{0ex}}⇒40a=200\phantom{\rule{0ex}{0ex}}⇒a=\frac{200}{40}\phantom{\rule{0ex}{0ex}}⇒a=5$

Hence, the value of a is 5.

Also, the frequency of 30 is 28 and the frequency of 70 is 24.

#### Page No 680:

(i) Arranging the numbers in ascending order, we get:
2, 2, 3, 5, 7, 9, 9, 10, 11
Here, n is 9, which is an odd number.
If n is an odd number, we have:

Now,

(ii) Arranging the numbers in ascending order, we get:
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, n is 9, which is an odd number.
If n is an odd number, we have:

Now,

(iii) Arranging the numbers in ascending order, we get:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, n is 11, which is an odd number.
If n is an odd number, we have:

Now,

(iv) Arranging the numbers in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, n is 13, which is an odd number.
If n is an odd number, we have:

Now,

#### Page No 680:

(i) Arranging the numbers in ascending order, we get:
9, 10, 17, 19, 21, 22, 32, 35
Here, n is 8, which is an even number.
If n is an even number, we have:

Now,

(ii) Arranging the numbers in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here, n is 10, which is an even number.
If n is an even number, we have:

Now,

(iii) Arranging the numbers in ascending order, we get:
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here, n is 12, which is an even number.
If n is an even number, we have:

Now,

#### Page No 681:

Arranging the marks of 15 students in ascending order, we get:
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here, n is 15, which is an odd number.
We know:

Thus, we have:

#### Page No 681:

Arranging the given data in ascending order:
144, 145, 147, 148, 149, 150, 152, 155, 160

Number of terms = 9 (odd)

Hence, the median height is 149.

#### Page No 681:

Arranging the weights (in kg) in ascending order, we have:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here, n is 8, which is an even number.
Thus, we have:

Hence, the median weight is 13.85 kg.

#### Page No 681:

Arranging the ages (in years) in ascending order, we have:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n is 10, which is an even number.
Thus, we have:

#### Page No 681:

10, 13, 15, 18, x+1, x+3, 30, 32, 35 and 41 are arranged in ascending order.
Median = 24
We have to find the value of x.
Here, n is 10, which is an even number.
Thus, we have:

#### Page No 681:

Arranging the given data in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93

Number of terms = 10 (even)

Hence, the value of x is 64.

#### Page No 681:

Arranging the given data in ascending order:
8, 11, 12, (2x – 8), (2x + 10), 35, 42, 50

Number of terms = 8 (even)

Hence, the value of x is 12.

#### Page No 681:

Arranging the given data in ascending order:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Number of terms = 11 (odd)

Hence, the median of the data is 58.

Now, In the above data, if 41 and 55 are replaced by 61 and 75 respectively.
Then, new data in ascending order is:
33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92

Number of terms = 11 (odd)

Hence, the new median of the data is 64.

#### Page No 683:

On arranging the items in ascending order, we get:
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Clearly, 6 occurs maximum number of times.
∴ Mode = 6

#### Page No 683:

On arranging the values in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
Clearly, 25 occurs maximum number of times.
∴ Mode = 25

#### Page No 683:

On arranging the shoe sizes in ascending order, we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Clearly, 9 occurs maximum number of times.
∴ Mode = 9

#### Page No 683:

On arranging the runs in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Clearly, 50 occurs maximum number of times.
∴ Modal score = 50

#### Page No 683:

We know that,

The given data is 3, 21, 25, 17, (x + 3), 19, (x – 4).

Mean of the given data = $\frac{3+21+25+17+\left(x+3\right)+19+\left(x-4\right)}{7}$
$⇒18\left(7\right)=84+2x\phantom{\rule{0ex}{0ex}}⇒126-84=2x\phantom{\rule{0ex}{0ex}}⇒2x=42\phantom{\rule{0ex}{0ex}}⇒x=21$

Hence, the value of x is 21.

Now, the given data is 3, 21, 25, 17, 24, 19, 17
Arranging this data in ascending order:
3, 17, 17, 19, 21, 24, 25

Here, 17 occurs maximum number of times.
∴ Mode = 17

Hence, the mode of the data is 17.

#### Page No 683:

Arranging the given data in ascending order:
52, 53, 54, 54, (2x + 1), 55, 55, 56, 57

Number of terms = 9 (odd)

Hence, the value of x is 27.

Arranging the given data in ascending order:
52, 53, 54, 54, 55, 55, 55, 56, 57

Here, 55 occurs maximum number of times.
∴ Mode = 55

Hence, the mode of the data is 55.

#### Page No 684:

Given: Mode = 25
∴ 25 occurs maximum number of times.

Arranging the given data in ascending order:
15, 20, 22, 23, 24, x + 3, 25, 26, 27, 40

x + 3 = 25
x = 25 − 3
x = 22

Hence, the value of x is 22.

Arranging the given data in ascending order:
15, 20, 22, 23, 24, 25, 25, 26, 27, 40

Number of terms = 10 (even)

Hence, the median is 24.5 .

#### Page No 684:

Arranging the given data in ascending order:
42, 43, 44, 44, (2x + 3), 45, 45, 46, 47

Number of terms = 9 (odd)

Hence, the value of x is 21.

Arranging the given data in ascending order:
42, 43, 44, 44, 45, 45, 45, 46, 47

Here, 45 occurs maximum number of times.
∴ Mode = 45

Hence, the mode of the data is 45.

#### Page No 684:

(c) 7

Mean of 5 observations = 11
We know:

#### Page No 684:

(c) 11$\frac{1}{3}$
Mean of 5 observations = 9
We know:

(b) 0

#### Page No 684:

(b) is decreased by 8

1, x2,...xn.

Now the new numbers after decreasing every number by 8 : (x1−8) , (x2−8)...,(xn−8)

Hence, mean is decreased by 8.

#### Page No 684:

(c) 53 kg

Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.

#### Page No 684:

(b) 39.4

Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = $\left(39×50\right)=1950$
Correct sum = (1950 + 43 $-$ 23) = 1970
$\therefore \mathrm{Mean}=\frac{1970}{50}=39.4\phantom{\rule{0ex}{0ex}}$

#### Page No 685:

(c) 64.91

Mean of 100 items = 64
Sum of 100 items = $64×100=6400$
Correct sum = (6400 + 36 + 90 $-$ 26 $-$ 9) = 6491

#### Page No 685:

(b) 51

Mean of 100 observations = 50
Sum of 100 observations = $100×50=5000$
It is given that one of the observations, 50, is replaced by 150.
∴ New sum = (5000 $-$ 50 + 150) = 5100
And,

#### Page No 685:

(b)

$\overline{z}=\frac{\left({x}_{1}+{x}_{2}+...+{x}_{n}\right)+\left({y}_{1}+{y}_{2}+...+{y}_{n}\right)}{2n}$

#### Page No 685:

(b) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$

#### Page No 685:

(c) 25

 x y x$×$y 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 Total 41 + p 303 + 9p

$\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Mean}=\frac{303+9p}{41+p}\phantom{\rule{0ex}{0ex}}\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mean}=8\phantom{\rule{0ex}{0ex}}\therefore \frac{303+9p}{41+p}=8\phantom{\rule{0ex}{0ex}}⇒303+9p=328+8p\phantom{\rule{0ex}{0ex}}⇒p=25$

#### Page No 685:

(b) 29

Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number.
Thus, we have:

#### Page No 685:

(c) 42 kg

Arranging the numbers in ascending order, we have:
31, 35, 36, 38, 40, 44, 45, 52, 55, 60
Here, n is 10, which is an even number.
Thus, we have:

#### Page No 686:

(c) 6

We will arrange the given data in ascending order as:
3, 4, 4, 5, 6, 7, 7, 7, 12
Here, n is 9, which is an odd number.
Thus, we have:

#### Page No 686:

(c) 54

We will arrange the data in ascending order as:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84
Here, n is 10, which is an even number.
Thus, we have:

#### Page No 686:

(b) 15

Here, 14 occurs 4 times, 15 occurs 5 times, 16 occurs 1 time, 18 occurs 1 time, 19 occurs 1 time and 20 occurs 1 time. Therefore, the mode, which is the most occurring item, is 15.


text:
# FindDistributionParameters fails with custom distribution?

Context

I would like to find the MaximumLikelihood solution of a customized PDF

dat = RandomVariate[LaplaceDistribution[2, 1], 1000];
param=FindDistributionParameters[dat, LaplaceDistribution[μ, σ],
ParameterEstimator -> {"MaximumLikelihood", Method -> "NMaximize"}]

(* {μ->2.27258,σ->0.521354} *)

Show[Plot[
PDF[LaplaceDistribution[μ, σ] /. param, x], {x, -5, 5}],
Histogram[dat, Automatic, "PDF"]]

works as expected. It finds a good estimator of $\mu$ and $\sigma$.

The problem

Now let me do the same with a customized PDF. Here I just impose that my custom PDF cannot be evaluated before it is given numerical values.

Clear[myLaplaceDistribution];
myLaplaceDistribution[μ_?NumberQ, σ_?NumberQ] :=
LaplaceDistribution[μ, σ]

Then

dat = RandomVariate[LaplaceDistribution[2, 1], 10];
FindDistributionParameters[dat, myLaplaceDistribution[μ, σ],
ParameterEstimator -> {"MaximumLikelihood", Method -> "NMaximize"}]

does not return a maximum likelihood estimate.

I am using 10.3.0 for Mac OS X x86 (64-bit) (October 9, 2015)

Question:

Any suggestions on how to make FindDistributionParameters work with unevaluated PDFs?

PS: I am aware of this https://mathematica.stackexchange.com/a/107914/1089 but here this question is a bit more general than simply a transformed distribution? And I have tried

dat = RandomVariate[LaplaceDistribution[2, 1], 10];
FindDistributionParameters[dat,
myLaplaceDistribution[μ, σ], {{μ,
Mean[dat]}, {σ, Mean[dat]}},
ParameterEstimator -> {"MaximumLikelihood", Method -> "NMaximize"}]

it does not seems to help.

Update

This related answer https://mathematica.stackexchange.com/a/61426/1089 does not seem to help.

If I define explicitly the domain for the PDF

  Clear[myLaplaceDistribution2];
myLaplaceDistribution2[μ_?NumberQ, σ_?NumberQ] :=
ProbabilityDistribution[
PDF[LaplaceDistribution[μ, σ], x], {x, -Infinity,
Infinity}, Assumptions -> (μ ∈ Reals && σ > 0)]

It still fails

dat = RandomVariate[LaplaceDistribution[2, 1], 10];
FindDistributionParameters[dat,
myLaplaceDistribution2[μ, σ], {{μ,
Mean[dat]}, {σ, Mean[dat]}},
ParameterEstimator -> {"MaximumLikelihood", Method -> "NMaximize"}]

As @J.M. points out one can use the fact that Mathematica can cope with the fact the PDF need not be normalized. As follows

Clear[myLaplaceDistribution3];
myLaplaceDistribution3[μ_, σ_] =
ProbabilityDistribution[
2 PDF[LaplaceDistribution[μ, σ],
x], {x, -∞, ∞},
Assumptions -> (μ ∈ Reals && σ > 0),
Method -> "Normalize"]

(Note the factor of 2 in front of PDF to make the PDF not normalized.)

Then

dat = RandomVariate[LaplaceDistribution[2, 1], 10];
FindDistributionParameters[dat, myLaplaceDistribution3[μ, σ],
ParameterEstimator -> {"MaximumLikelihood"}]

works.

I still think there must be situations where the PDF cannot be known before its arguments are known, and where Maximum likelihood analysis would make sense?

Note that I can always make my own:

MyFindDistributionParameters[data_, distrib_, var_] :=
NMaximize[{Total[Log@ PDF[distrib, #] & /@ data],
DistributionParameterAssumptions[distrib]}, var][[2]];

MyFindDistributionParameters[dat,LaplaceDistribution[μ, σ], {μ, σ}]

but I was hoping Mathematica would provide me with a more efficient algorithm? (this seems to be 10 times slower than the built in function).

• Have you tried using ProbabilityDistribution[] instead to express your custom distribution? – J. M.'s technical difficulties Dec 13 '16 at 7:16
• @J.M. just tried while you commented. Seem to produce the same. The reason I want to have a delayed definition is that in real life my PDF is not normalized so I need to normalize it via numerical integration. – chris Dec 13 '16 at 7:18
• Well, ProbabilityDistribution[] supports the setting Method -> "Normalize" so that the normalization is done on your behalf. – J. M.'s technical difficulties Dec 13 '16 at 7:19
• Ah! That might be a lead then. The question remains of interest? – chris Dec 13 '16 at 7:21
• Clearly I don't understand your purpose. Giving mu and sigma to FindDistributionParameters without giving them numbers is essential. What is it about having your second example work when removing ?NumberQ doesn't work for you? (I'm not trying to be sarcastic. I'm just clearly missing something.) – JimB Dec 13 '16 at 21:57

If you follow @J.M. 's advice removing ?NumberQ from the definition of the probability distribution makes everything work fine:

Clear[myLaplaceDistribution];
SeedRandom[12345];
myLaplaceDistribution[μ_, σ_] := LaplaceDistribution[μ, σ]
dat = RandomVariate[LaplaceDistribution[2, 1], 10];
FindDistributionParameters[dat, myLaplaceDistribution[μ, σ],
ParameterEstimator -> {"MaximumLikelihood", Method -> "NMaximize"}]
(* {μ -> 1.8804870321227085,σ -> 0.7153183538699862} *)

I don't know what you mean by "Here I just impose that my custom PDF cannot be evaluated before it is given numerical values." Your first example doesn't have the two parameters evaluated as numbers and it works fine:

param=FindDistributionParameters[dat, LaplaceDistribution[μ, σ],
ParameterEstimator -> {"MaximumLikelihood", Method -> "NMaximize"}]

• Thanks for this answer. I guess I have simplified too much the original problem. In my case the model PDF is the result of a convolved calculation which requires that mu and sigma are numerical. I tried to mimic this by stating that myLaplaceDistribution requires numerical values before it can be evaluated. So your answer solves the posed problem but not my problem. – chris Dec 13 '16 at 21:59
• I look forward to the new question. – JimB Dec 13 '16 at 22:00


text:
# [.net] read joystick buttons using windows api without DirectX

## Recommended Posts

Hi, I'm trying to use joyGetPosEx to read the buttons pressed. First I use joyGetNumDevs to get the number of controls, but 81269371174928 is what is returned. Shouldn't it return the number of controllers? Also, joygetposex just returns 165, which also doesnt make sense. Could someone please tell me what I'm doing wrong?

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81269371174928 is unlikely as the joyGetNumDevs() function returns a UINT, and that number would be far outside of the range of an unsigned 32 bit integer. Of course, that's assuming you're returning an unsigned integer (if it's a long/int64 then you have the function declared incorrectly). How is your p/invoke declaration for that function set up? I have mine as:
        ///         /// Function to return the number of joystick devices supported by the current driver.        ///         /// <returns>Number of joysticks supported, 0 if no driver is installed.</returns>        [DllImport("WinMM.dll"), System.Security.SuppressUnmanagedCodeSecurity]        public static extern int joyGetNumDevs();

And that works perfectly for me.

Also, you're getting error code 165 from joyGetPos? That error would be:
JOYERR_PARMS /* bad parameters */ (according to mmsystem.h)

According to MSDN, that means the ID of the joystick isn't valid.

Also make sure your joysticks showing up in the joystick control panel.

##### Share on other sites
Hi, and thanks for your response.

I tried ID's 0-15 and all of them return 165 even when the joysticks show up in the joystick control panel.

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I finally got getjoyposex to work, but it isn't returning any of the joystick information, such as the x position or the buttons pressed. So i tried my code in vb 6.0 and it works fine! Why would my code work in vb 6.0 and not in vb.net?

Public Class Form1    Public Const JOY_RETURNBUTTONS = &H80&    Public Const JOY_RETURNCENTERED = &H400&    Public Const JOY_RETURNPOV = &H40&    Public Const JOY_RETURNR = &H8&    Public Const JOY_RETURNU = &H10    Public Const JOY_RETURNV = &H20    Public Const JOY_RETURNX = &H1&    Public Const JOY_RETURNY = &H2&    Public Const JOY_RETURNZ = &H4&    Public Const JOY_RETURNALL = (JOY_RETURNX Or JOY_RETURNY Or JOY_RETURNZ Or JOY_RETURNR Or JOY_RETURNU Or JOY_RETURNV Or JOY_RETURNPOV Or JOY_RETURNBUTTONS)    Public Declare Function joyGetPosEx Lib "winmm.dll" (ByVal uJoyID As UInteger, ByRef pji As JOYINFOEX) As Integer    Public Structure JOYINFOEX        Dim dwSize As Long                      ' size of structure        Dim dwFlags As Long                     ' flags to indicate what to return        Dim dwXpos As Long                     ' x position        Dim dwYpos As Long                      ' y position        Dim dwZpos As Long                      ' z position        Dim dwRpos As Long                      ' rudder/4th axis position        Dim dwUpos As Long                      ' 5th axis position        Dim dwVpos As Long                      ' 6th axis position        Dim dwButtons As Long                 ' button states        Dim dwButtonNumber As Long              ' current button number pressed        Dim dwPOV As Long                       ' point of view state        Dim dwReserved1 As Long                 ' reserved for communication between winmm driver        Dim dwReserved2 As Long                 ' reserved for future expansion    End Structure    Dim joy As JOYINFOEX    Private Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer1.Tick        joy.dwSize = Len(joy)        joy.dwFlags = JOY_RETURNALL        joyGetPosEx(0, joy)        Label2.Text = joy.dwXpos    End SubEnd Class

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I don't know VB.NET but in C# you should declare joyGetPosEx with either "[Out]ref JOYINFOEX" or "out JOYINFOEX", otherwise it will only be marshalled from managed to unmanaged code, not the other way around.

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You don't have the [StructLayout(LayoutKind.Sequential)] attribute defined for the JOYINFOEX value type. If you don't define that attribute, the runtime will possibly rearrange your value type for a number of reasons (e.g. performance, because it's evil, etc...). Specifying sequential will tell the runtime that the fields should stay in the order in which they're declared. You can learn more about it on the MSDN.

Here's what I've got:
    ///     /// Value type containing joystick position information.    ///     [StructLayout(LayoutKind.Sequential)]    internal struct JOYINFOEX    {        /// Size of structure, in bytes.        public int Size;        /// Flags to indicate what information is valid for the device.        public JoystickInfoFlags Flags;        /// X axis.        public int X;        /// Y axis.        public int Y;        /// Z axis.        public int Z;        /// Rudder position.        public int Rudder;        /// 5th axis position.        public int Axis5;        /// 6th axis position.        public int Axis6;        /// State of buttons.        public JoystickButtons Buttons;        /// Currently pressed button.        public int ButtonNumber;        /// Angle of the POV hat, in degrees (0 - 35900, divide by 100 to get 0 - 359 degrees.        public int POV;        /// Reserved.        int Reserved1;        /// Reserved.        int Reserved2;    }

In VB.NET that attribute would be declared like this (and don't forget to import the System.Runtime.InteropServices namespace):
<StructLayout(LayoutKind.Sequential)> _Public Structure JOYINFOEX

Also I noticed you have all your fields declared as "Long" in your VB6 version. If you haven't changed those already, know that it's not going to work in VB.NET. You need to declare those fields as Integer in VB.NET. This is because in VB6, Long was a 32 bit integer, while in VB.NET Long is a 64 bit integer.

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text:
# Approximate sum of gamma distributions using normal distribution

If $X_1$, $X_2$, $X_3$, and $X_4$ are gamma distributions with $\theta=2$ and $\alpha_1=3, \alpha_2=2, \alpha_3=5, \alpha_4=3$, respectively, and $Y=X_1+X_2+X_3+X_4$, I can find P(Y$\leq$25) by integrating the gamma function Y, and I obtain 0.4810.

However, I am asked to use a normal distribution to approximate the answer.

I cannot figure out how I would use the Central Limit Theorem to do this. Are $X_1$, $X_2$, $X_3$, and $X_4$ really a random sample of size 4 from a distribution with mean 26 and variance 52? My attempt of P($\dfrac{X_{bar}-\dfrac {26}{4}}{\sqrt{52/4}/2}\leq {\dfrac{\dfrac{25}{4}-\dfrac {26}{4}}{\sqrt{52/4}/2}}$) does not work.

Also, I do not know how I would justify using a normal approximation for the gamma distribution.

I assume you are using the shape-scale parametrization $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\theta}}{\theta^\alpha \Gamma(\alpha)}, \quad x > 0$$ and that $\theta = 2$ is the common scale parameter. In such a case, $Y = X_1 + X_2 + X_3 + X_4$ would be gamma with shape $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 13$. When $\alpha \gg \theta$, we can use a normal approximation with $\mu = \alpha \theta$, $\sigma^2 = \alpha \theta^2$, to get $$\Pr[Y \le 25] \approx \Pr\left[\frac{Y - \mu}{\sigma} \le \frac{25 - 26}{2\sqrt{13}}\right] \approx \Phi(-0.138675) \approx 0.444853.$$ Is it a good approximation? No, not really.
• @heropup I guess ${\large four\ variables}$ isn't enough to claim a good approximation via Central Limit Theorem. Anyway, it's fine to know for newbies how to accomplish an exact calculation. Thanks for your advice. – Felix Marin Jan 27 '17 at 1:28


text:
# Symmetric Group on 3 Letters/Conjugacy Classes

## Conjugacy Classes of the Symmetric Group on 3 Letters

Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:

$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$

The conjugacy classes of $S_3$ are:

 $\ds$  $\ds \set e$ $\ds$  $\ds \set {\tuple {123}, \tuple {132} }$ $\ds$  $\ds \set {\tuple {12}, \tuple {13}, \tuple {23} }$

## Proof

This is the group presentation of $D_3$:

$D_3 = \gen {a, b: a^3 = b^2 = e, a b = b a^{-1} }$

First we have that Conjugacy Class of Element of Center is Singleton.

From Center of Symmetric Group is Trivial, only $e$ is in the center.

Thus $\set e$ forms one of the conjugacy classes.

We then have:

 $\ds b a b^{-1}$ $=$ $\ds b^{-1} a b$ as $b^2 = e$ $\ds$ $=$ $\ds b^{-1} b a^{-1}$ as $a b = b a^{-1}$ $\ds$ $=$ $\ds a^{-1}$ as $a b = b a^{-1}$ $\ds$ $=$ $\ds a^2$ as $a^3 = e$

and so $a^2$ is in the same conjugacy class as $a$.

From Centralizer of Group Element is Subgroup, the centralizer $\map {C_{S_3} } a$ of $a$ is a subgroup of $S_3$.

Thus $\map {C_{S_3} } a$ also contains $a^2$.

Thus the order of $\map {C_{S_3} } a$ is either $3$ or $6$, from Lagrange's Theorem.

But we have that $b a \ne a b$ and so $b \notin \map {C_{S_3} } a$.

Thus $\map {C_{S_3} } a$ contains $3$ elements.

By the Orbit-Stabilizer Theorem $a$ has $2$ conjugates.

Hence the conjugacy class of $a$ is $\set {a, a^2}$.

By a similar argument, $\map {C_{S_3} } b = \gen b$ and so contains $2$ elements.

Thus $b$ has $3$ conjugates.

As the conjugacy classes form a partition, $\card {\map {C_{S_3} } b} = 3$.

Hence the result.

$\blacksquare$


text:
# Proof by induction of Bernoulli's inequality $(1+x)^n \ge 1+nx$

I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:

Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality: $$(1+x)^n \ge 1+nx$$

Proof:

Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.

Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$.

Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$

Proof of Inductive Step: \begin{align*} (1+x)^k &\ge 1+kx \\ (1+x)(1+x)^k &\ge (1+x)(1+kx)\\ (1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\ 1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\ \Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square \end{align*}

-
Thanks for the suggestions, I'll keep your tips in mind. –  KingOliver Aug 12 '12 at 15:09
Where did you uses $x>-1$? Hint: you did use it. –  Thomas Andrews Aug 12 '12 at 15:16
When I claimed that $kx^2 >0$ –  KingOliver Aug 12 '12 at 15:18
Actually, you need it when you multiply both sides by $1+x$. Also, since $x$ can be 0, $kx^2\ge0$ –  Mike Aug 12 '12 at 16:06
Ah thank you @Mike –  KingOliver Aug 12 '12 at 17:03

\begin{align*} (1+x)^{k+1}&=(1+x)(1+x)^k\\ &\ge(1+x)(1+kx)\\ &=1+(k+1)x+kx^2\\ &\ge1+(k+1)x\;, \end{align*}
since $kx^2\ge 0$. This completes the induction step.
This looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as $$(1+x)^{k+1} \geq 1+(k+1)x+kx^2>1+(k+1)x,$$ so there is no need of the fifth line. Or even $$(1+x)^{k+1} = (1+x)(1+x)^{k} \geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x.$$


text:
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However, this is not always true, since Missing Values (MVs) can appear with the form of outliers or even wrong data (i.e. Many existing, industrial and research data sets contain Missing Values.However, this is not always true, since Missing Values (MVs) can appear with the form of outliers or even wrong data (i.e. Missing values make it difficult for analysts to perform data analysis.Let $A$ be a symmetric matrix with eigenvalue decomposition $UDU^T$. Has anything similar been done for the case where the update is of the form $A B$, where $B=uv^t vu^t$ is a rank-two symmetric matrix (note we can't just do two rank-one symmetric updates)? have shown that given such an $A$, the eigenvalue decomposition of $A \rho xx^t$ may be computed efficiently.With this observation, in this paper, we present an efficient method for updating Singular Value Decomposition of rank-1 perturbed matrix in $O(n^2 \ \text(\frac))$ time.

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text:
# choiceless grapher

Last year I said that I am learning Common-Lisp with a grapher for relationships between consequences of the Axiom of Choice in mind. A year later, the core of a first version of this program is finished and uploaded on GitHub:

I am very thankful to Prof. Paul Howard for the permission to use the data of his and Prof. Jean Rubin’s  encyclopedia of choiceless set theory: “Consequences of the Axiom of Choice”, and for sending me the latest implications data and the tex statements of the forms.

An example diagram is the featured image, depicting the relationships between the forms in the category “Alephs and their properties”. Some more examples of diagrams can be found here:
An arrow from A to B means that A implies B and a boldfaced arrow is non-reversible, i.e., there is a model of ZF where B holds and A doesn’t. Note that this diagram is made with the old implication data.

Since I uploaded this program on GitHub, I have decided that it might better (faster) to store the graph information as a matrix instead of the current graph structure. This is the structure of the data that Paul Howard sent me (book1). As soon as I finish checking my predicates against this book1, I shall upload the next version, with some drawing functionality. I will also include the complete full graph, which, judging from past experiments, I estimate to be about 3 meters wide and 2 meters tall (wallpaper anyone?).

If you can’t wait until that (my concentration is more on natural-isabelle at the moment), do send me an email with the form numbers that are of interest to you, and I will have the choiceless grapher draw a diagram with your forms, as soon as I find some time.

PS. I did notice the obvious missing arrow between Form 208 and 170. That arrow is also missing in the latest data, and it is one of my strongest arguments why we need such diagrams. :)

PS-2. All the bottom forms point to “Form 0. 0 = 0”. Right-click on the featured image and select “View Image” to see the whole diagram.

## 3 thoughts on “choiceless grapher”

1. Also the arrow from Form 25 to Form 34 should be boldfaced (many ZF models satisfy 34 but not 25), something that is missing from the database. Diagrams are so much easier to stare at and find missing information than lists of implications with codes.

Please do comment with any observations on this featured diagram if you see something!

2. There’s an arrow missing from 208 to 170.

It’s not clear to me why 104 and 182 are not equivalent (cofinality of an ordinal is always regular).

There is a parenthesis mismatch in 58.

It’s not hard to show that 365 implies 170, but not vice versa.

108 seems a bit odd, since $\omega_2$ is itself never the countable union of countable sets (it could be the countable union of countable unions of countable sets), so it’s power set certainly isn’t either.

And that’s all from the image appearing on this post… :)

1. Thank you so much for the observations, I agree with all of it! I think Form 108 has a typo, and instead of “denumerable union of denumerable sets” it should be “denumerable union of sets of cardinality $\aleph_\alpha$” (given that this form is taken from Gitik’s “Regular cardinals in models of ZF”, see page 67, question 6).

All this from one featured image, exactly! :) Just imagine what will come out while staring at the wallpaper version of the full diagram! :D


text:
# Secretary of State Kissinger engaged inbetween Israel and Arab nations.to reach a cease-fireA. shuttle diplomacyB. embargoesC. Big Stick diplomacyD.

###### Question:

Secretary of State Kissinger engaged in between Israel and Arab nations.
to reach a cease-fire
A. shuttle diplomacy
B. embargoes
C. Big Stick diplomacy
D. boycotts

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Multiply this please  $Multiply this please $...


text: The proof of Taylor's theorem in its full generality may be short but is not very illuminating. Likewise, the uniform limit of a sequence of (real) differentiable functions may fail to be differentiable, or may be differentiable but with a derivative which is not the limit of the derivatives of the members of the sequence. The insight into this property comes from geometric algebra, where objects beyond scalars and vectors (such as planar bivectors and volumetric trivectors) are considered, and a proper generalization of Stokes' theorem. Watson. This is significant because one can then prove Cauchy's integral formula for these functions, and from that deduce these functions are infinitely differentiable. Let f be holomorphic in an open set that contains an open disc Dand its boundary circle C, equipped with the positive (counterclockwise) orientation. a surface for which $Q$( see (5)) … Let Dbe a domain in C and suppose that f2A(D):If 1; 2 are continuously deformable into each other closed curves, then Z 1 f(z)dz= Z 2 Also, we show that an analytic function has derivatives of all orders and may be represented by a power series. &= \frac{(k+1)! Now we are in position to prove the Deformation Invariance Theorem. But inside this closed disk of radius R, f is a continuous function and therefore cannot go off to infinity. &= \frac{k! An illustration of a 3.5" floppy disk. Cauchy-Goursay theorem, Cauchy’s integral formula. More generally, γ\gammaγ is the boundary of any region whose interior contains aaa. Let D be a disc in C and suppose that f is a complex-valued C1 function on the closure of D. Then[3] (Hörmander 1966, Theorem 1.2.1). \end{aligned}f(k+1)(a)=dadf(k)(a)=2πik!∫γdad(z−a)k+1f(z)dz=2πik!∫γ(z−a)k+2(k+1)f(z)dz=2πi(k+1)!∫γ(z−a)k+2f(z)dz.. Cauchys integral formula Theorem 15.1 (Cauchy’s Integral formula). Sign up to read all wikis and quizzes in math, science, and engineering topics. denotes the principal value. https://brilliant.org/wiki/cauchy-integral-formula/. Tema 6- Parte 2 Integrales Complejas Teorema de Cauchy Goursat - Duration: 54:55. We prove the Cauchy integral formula which gives the value of an analytic function in a disk in terms of the values on the boundary. If ˆC is an open subset, and T ˆ is a is completely contained in U. We can use a combination of a Möbius transformation and the Stieltjes inversion formula to construct the holomorphic function from the real part on the boundary. Similarly, one can use the Cauchy differentiation formula to evaluate less straightforward integrals: Compute ∫Cz+1z4+2z3 dz,\displaystyle \int_{C} \frac{z+1}{z^4 + 2z^3} \, dz, ∫Cz4+2z3z+1dz, where CCC is the circle of radius 111 centered at the origin. A direct corollary of the Cauchy integral formula is the following (((using the above definitions of fff and γ):\gamma):γ): f(n)(a)=n!2πi∫γf(z)(z−a)n+1 dz.f^{(n)}(a) = \frac{n! ... Complex Integration And Cauchys Theorem by Watson,G.N. ) Suppose the differentiation formula holds for n=kn=kn=k. Then for any aaa in the disk bounded by γ\gammaγ. Cauchy’s theorem is a big theorem which we will use almost daily from here on out. Fortunately, a very natural derivation based only on the fundamental theorem of calculus (and a little bit of multi-variable perspective) is all one would … Then f(z) = 1 2ˇi Z C f( ) z d for every z2D. Theorem 6.4 (Cauchy’s Theorem for a Triangle) Let f:D → C be a holo-morphic function deﬁned over an open set D in C, and let T be a closed triangle contained in D. Then Z ∂T f(z)dz = 0. }{2\pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz.f(n)(a)=2πin!∫γ(z−a)n+1f(z)dz. a Theorem 1.4. Here, contour means a piecewise smooth map . The key technical result we need is Goursat’s theorem. While Cauchy’s theorem is indeed elegant, its importance lies in applications. Then, f(z) = X1 n=0 a n(z z 0)n; 7 TAYLOR AND LAURENT SERIES 5 where the series converges on any disk jz z 0j


text:
How to put epsilon transitions in jflap

Transitions jflap epsilon

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J le must match the alphabet used in the student’s submitted le. It relies on the assumption that shear and shuffle are decoupled and requires some peculiar magnetic order, different from that of \alpha and \epsilon phases as found in Density-Functional Theory. The UTM feeds w to M and simulates M's entire operations against w until M halts. that students know whether to draw transitions to a dead (trap) state, or to completely leave how these transitions out of the diagram. Add a new start state if necessary with this transition to how to put epsilon transitions in jflap the original intended start state ε, ε -> $In how to put epsilon transitions in jflap JFLAP, instead of adding a how to put epsilon transitions in jflap new start state, that step is essentially done automatically. Finally, we put how to put epsilon transitions in jflap forward an original mechanism for this transition, how to put epsilon transitions in jflap based on. JFLAP History • 1990 - Started as NPDA • 1991 – LL, LR parser • 1991 – FLAP how to put epsilon transitions in jflap • 1996 – JFLAP • Thanks to Many Students! Sa v e this F Ainthe le lab1b. For instance, the machine above accept \epsilon. И как это работает? (Use the collapse state tool to remove nonfinal, noninitial how to put epsilon transitions in jflap status. JFLAP was made available to students to help them visualize the different concepts how to put epsilon transitions in jflap and construct solutions. Using JFLAP (instructions are b elo w), construct nite automata for the follo wing languages. In addition, these informal statements must be coded as invariant predicates used to validate the roles of the state. A path in the Burgers description of the bcc/hcp transition different from those previously considered is proposed. If, in every situation, at most one such transition action is possible, then the how to put epsilon transitions in jflap automaton is called a deterministic pushdown automaton (DPDA). ) The default corresponds to: python gridworld. The constraint is formulated as ht. Notice we have more than one transition out of a how to put epsilon transitions in jflap state, we can have epsilon-transitions, and we don't need to have a transition from every alphabet symbol from a state. The original title says: title="Wait, forgot to escape a space. F A (c) = f a; b g, L= w. ② Simulate automaton. In order to understand how this works, enable jflap the Equation View, and look at the implementation of the Dirichlet condition (in this case, a prescribed temperature):. The$ character is used to represent the empty string symbol (epsilon) but should not be listed in the alphabet. Enabling the Equation View. 2 Grading: We will check that you only changed. JFLAP denotes the start state by marking it with a triangle. 19209e-07 Note that the above result varies from machine to machine.

6 |- When do FA exist: Balanced parantheses Parity mod Is legal ASCII 1 Is legal 10. (Noise refers to how often an agent ends up in an unintended successor state when they perform an action. F is a set of final state/states of Q (F ⊆ Q). For example, creating one transition using 1,2 is wrong, so you have to create two transitions: one for 1 and another for 2. Rodger * Computer Science Department * Duke University * Aug * Copyright how to put epsilon transitions in jflap (c. In other words, ε-closure for a state can be obtained how to put epsilon transitions in jflap by union operation of the ε-closure of the states which can be. JFLAP itself allows missing transitions in its de nition of DFAs. how to put epsilon transitions in jflap Put together: Given an input symbol, current state, and stack symbol, the automaton can follow a transition to another state, and optionally manipulate (push jflap or pop) the stack.

For the travel industry, the new destination is contactless. how to put epsilon transitions in jflap Stack Exchange network consists of 176 Q&A how communities including Stack Overflow, the largest, most trusted online community for developers to learn, share. Procopiuc, Cavalcante, how to put epsilon transitions in jflap Hung, Grammond, Geer, Karweit, Hardekopf, Bressler, Reading, Finley. 0 instead of lambda by changing the preferences:. The “(S)tate Expander” button, if pressed, will fill how out all transitions and create any new DFA states those how to put epsilon transitions in jflap jflap transitions require when you click on an existing DFA state. We can read one symbol at a time. A PDA machine configuration (p, w, y) can be. puts forth a systematic development strategy in how to put epsilon transitions in jflap the form of a design recipe 2, 3.

JFLAP file from class implementing the TM described in Lecture 28. 2 12c (Converting NFA to regular expression) Use JFLAP to construct the regular expression, but note that as stated in the JFLAP Activities. The use of JFLAP was not compulsory. jffl File Input Test View Convert Help Editor Automaton Size JFLAP : (RG2DFA. jffl File Input Test View Convert Help Editor. • The newest in the FA family. – Applying the transition function will give, not 1 state, but 0 or more states. Turn in the JFLAP regular expression file.

Answer: a Explanation: A PDA is a finite machine which has an additional stack storage. TRABAJANDO CON JFLAP 4. The Equation View for the Temperature node. how to put epsilon transitions in jflap The initial state of our new machine is q0,q1,q2 the epsilon closure of q0 The final state(s) of our new machine is the new state(s) that contain a state symbol that was a final state in the original. Способ использования JFLAP для создания Pushdown how to put epsilon transitions in jflap Automaton.

COSC 3340: Introduction to Theory of how to put epsilon transitions in jflap Computation University of Houston Dr. JFLAP converts from a regular expression R to how to put epsilon transitions in jflap an NFA in a slightly different way than the text. (a) = f a; b g, L= a n b p j n> how to put epsilon transitions in jflap 0;p > g.

In the RHS side, Q is the subset of 2 Q which indicates Q is how to put epsilon transitions in jflap contained in 2 Q or Q is a part of 2 Q, however, the how to put epsilon transitions in jflap reverse isn’t true. pdf file, the nfa must meet the requirements of the. . Exercise 4: Epsilon transitions and non-determinism – background for obligatory assignment Make an NFA in JFLAP corresponding to: Try it with various inputs, e. • Introducing. .

In general, if several actions are possible, then the automaton is called a. &0183;&32;Animation of the temperature distribution how to put epsilon transitions in jflap as the prescribed how to put epsilon transitions in jflap temperature spot travels along the bar. Verma Lecture 14 PDA’s and CFG’s For every CFG G there is a PDA M such that L(G) = L(M) For every PDA M there is a CFG G such that L(M) = jflap L(G) CFG PDA Given CFG G = (V,, R, how to put epsilon transitions in jflap S) Let PDA M = (Q,, V $,, qstart, qaccept) Q = qstart, qloop,qaccept contains transitions for the form ((qstart,, ), (qloop, S$)) For. Remember you must feed it input including a left angle bracket as endmarker) June 22 - Prelim 2, covering how to put epsilon transitions in jflap all the material on CFLs up to and including the 6/20 class June 23 - More examples and. Also, the lion share of Randall’s irony is always put into titles and you for some reason ruined it.

How restaurant how to put epsilon transitions in jflap marketers can deliver engaging contactless experiences and gain guest affinity Novem Read more. Hence, given a symbol, a next-state can be found by either following a transition with that symbol how to put epsilon transitions in jflap or by rst doing how to put epsilon transitions in jflap any number of epsilon-transitions and then a transition with the. (JFLAP is a software package for. However, the alphabet used in the \answer key". Pushdown Automata CPSC 388 Ellen Walker Hiram College PDA = FA + Stack A PDA Transaction Read the next symbol from the input Pop a symbol from the stack Push a symbol on the stack Enter a new state Epsilon in PDA Read epsilon No character is read, input location stays the same Push or pop epsilon The stack remains unchanged Format of a Transition how to put epsilon transitions in jflap Items before how ; are “input”, items after. δ is the transition function where δ: Q &215; ∑ → Q.

Project Description We are going to design and implement a "Universal Turing Machine (UTM)" whose input is the definition of an arbitrary TM called M and how to put epsilon transitions in jflap an arbitrary input string of M, called w. Generate random DFA Generate random NFA Generate random eNFA. Exactly one state of a finite automaton is designated as the start state.

Instead how to put epsilon transitions in jflap of building up the machine from small pieces, as the text does on pages 80-81, JFLAP "expands" a simple machine that has a regular expression transition R from a start how to put epsilon transitions in jflap state to a final state into a true NFA with single input symbol transitions and lambda transitions. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute. 25, You can see the greek letter epsilon in JFLAP 6. jff) File Input how to put epsilon transitions in jflap Test View Convert Help Editor Automaton Size how to put epsilon transitions in jflap q JFLAP (RG2DFA. Pushdown automata are computational models—theoretical computer-like machines—that can do more than a finite state machine, but less than a Turing machine. Alternatively, if one wants to see the whole DFA right away, one can click on the “Complete” button. Answer jflap to Construct a Turing machine that subtracts one ternary integer (base 3) from jflap another.

This is a fantastic article, always struggled to. We say the NFA accepts w roughly if there is some sequence of transitions beginning with the start how to put epsilon transitions in jflap state, that processes each character of w, and ends in an accept state. Epsilon transitions The last example is a bit more complex and introduces something we how did not talk about yet:. If the next symbol of string w is a and the pointer is on state p, move the pointer to δ(p, a). For each left put in the symbol as a transition from one state to the next non-terminal Then, put in the recursion which are the loops. How to use one or more of JFLAP, paper, the Understanding Computation FSA implementation to build finite automata; The differences between deterministic and nondeterministic finite automata (from a formal perspective as well as intuitively) How to determine whether a finite automaton accepts a particular string or a particular language; Somewhere between today and next class (more likely the.

Exercise 5 a) As jflap probably not everybody completed this during group last week how (ex 4a), we repeat it here: Work. &0183;&32;The Epsilon S Launch Vehicle, which will be developed based on the current Enhanced Epsilon, is aimed at strengthening Epsilon’s international competitiveness in the satellite launch market by achieving synergistic effects with the H3 Launch Vehicle, a large-size liquid-propellant rocket currently under development. automaton pushdown-automaton jflap. &0183;&32;Transition state table for the above NFA.

Instead of \$, JFLAP uses Z 2. JFLAP – Regular Languages •Create –DFA and jflap NFA. To run how to put epsilon transitions in jflap JFLAP, just double-click on the JFLAP7. F A (b) = f a; b g,L = w 2 j has an ev en n um ber of a 's and an o dd n b 's g. Pushdown automata are nondeterministic finite state machines augmented with additional memory in the form of a stack, which is why jflap the term “pushdown” is used, as elements are pushed down onto the stack. The initial state is denoted by. /* * JFLAP - Formal Languages and Automata Package * * * Susan H. STATES 0 1 epsilon; A B, C A B; B – B C; C C C – Epsilon (∈) – closure : Epsilon closure for a given state X is a set of states which can be reached from the states X with only (null) or ε moves including the state X itself.

Do you see how they differ? Generate a FSM to see a valid example. Now, put a pointer to the start state q and read the input string w from left to right and move the pointer according to the transition function, δ.

May 27 - Nondeterminism; epsilon-transitions (Lectures 5 and 6, and additional handout about computation trees for NFAs). jar file name and be patient, the startup is a little slow.

How to put epsilon transitions in jflap

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text:
# 9.78 StringTable¶

This domain provides a table type in which the keys are known to be strings so special techniques can be used. Other than performance, the type StringTable(S) should behave exactly the same way as Table(String,S). See TableXmpPage for general information about tables.

This creates a new table whose keys are strings.

t: StringTable(Integer) := table()

 table()

Type: StringTable Integer

The value associated with each string key is the number of characters in the string.

for s in split("My name is Ian Watt.",char " ")
repeat
t.s := #s

Void

for key in keys t repeat output [key, t.key]

["Ian",3]
["My",2]
["Watt.",5]
["name",4]
["is",2]

Void


text:
1. Jun 7, 2009

### hippopviolin

Hello, everyone

I am learning special relativity in high school.. I understand the time dilation thing...but i have trouble to understand the length contration..

I am getting confused why

Lm=ts*V and Ls=tm*V

2. Jun 7, 2009

### diazona

I'm not sure what you're referring to by Lm, ts, etc. But in general, length contraction is a very similar effect to time dilation, so if you understand one, it should not be hard to understand the other. The difference is that whereas time intervals appear longer when you are moving relative to the clock, length intervals appear shorter when you are moving relative to the yardstick (or whatever length-measuring device there is).

Maybe you can explain a little more what you're confused about?

3. Jun 7, 2009

### feynmann

The problem really is this, how do you measure the length of a moving object, like meter stick?
After you measure one end, the other end has moved, so its position changed.
The answer is that you must measure both ends at the "same time". That's the key.
But at the same time, relative to who? Since two events happen at the same time relative to one frame, do not happen at the same time relative to another frame, if there is relative motion between them

4. Jun 7, 2009

### hippopviolin

Thanks to Diazona and Feynmann respond to my question

Ls is the proper length, and Ts is the proper time

now, I think I understand the formula..

however, i still dont really get one concept..

My textbook says,

when you look at the length of the spaceship as proper length, then the distance between Planet A and Planet B will shrink.

If you look at the distance between Planet A and Planet B as proper length, then the length of the spaceship will shrink..

I understand the first idea, and I dont get the second one

Thank you for helping

5. Jun 7, 2009

### feynmann

Since special relativity is really "relative", so its symmetric. You see my length contract and I see your length contracted. Nothing wrong with it. How could it be? after all, how could my length is shorter than yours and your length is shorted than mine? The answer is "it's about "time". Einstein's great insight is "time slow down when moving".

Last edited by a moderator: May 4, 2017
6. Jun 7, 2009

### JesseM

When you write Lm=ts*V and Ls=tm*V, what are Lm, tm and V?

Normally, if Ls is the length of an object in its own rest frame, and Lv is its length in a frame where it's moving at speed v, then the relation between the two lengths is:

$$L_v = L_s * \sqrt{1 - v^2/c^2}$$

Likewise, if Ts is the proper time between two readings on a clock as measured by the clock itself, and Tv is the time between those same readings in a frame where the clock is moving at speed v (where it is ticking slower, so it takes longer to go from the first reading to the second), the relation is:

$$T_v = \frac{T_s}{\sqrt{1 - v^2/c^2}}$$

Probably the first means that in the rest frame of the spaceship, the distance from planet A to planet B is less than it is in the planet's own rest frame, and likewise the second would mean that in the rest frame of the planets, the length of the spaceship would be less than it is in the ship's own rest frame.

7. Jun 7, 2009

### hippopviolin

Thank you JesseM
"Probably the first means that in the rest frame of the spaceship, the distance from planet A to planet B is less than it is in the planet's own rest frame, and likewise the second would mean that in the rest frame of the planets, the length of the spaceship would be less than it is in the ship's own rest frame."

This helps a lot..

What I confused about the formula is why Ls is calculated by Tv instead of Ts..

why should we use the moving time to calculate the proper length

8. Jun 7, 2009

### JesseM

The proper length Ls isn't related to time in the formulas I posted, it's related to Lv, the length in the frame where the object is moving at speed v. Why would you think length is calculated using time? Does it say that in your textbook?

9. Jun 7, 2009

### hippopviolin

Yeah..My textbook use the time dilation relationship to derive the length contraction formula.

They mutiply both sides by v, which is the speed of the spaceship..

then V*Ts becomes Lv, and V*Tv becomes Ls

10. Jun 7, 2009

### JesseM

But that's just a step in a longer derivation, right? Without seeing the context of what assumptions they're making in the derivation I can't really make sense of that...what's the name of the textbook? Maybe we can find the page on google books.

11. Jun 7, 2009

### hippopviolin

haha thank you for helping me

It is Thomson Nelson Physics 12. By Hirsch, Martindale, Sewart and Barry

It is in page 574

12. Jun 7, 2009

### JesseM

Excellent, turns out I could actually download that section from http://access.mmhs.ca/docs/Science/MMHS%20Web%20folder/physics12/Physics%20CD/physics12/a40.html [Broken]. The key is that Lm = v*ts is not meant to be a general relationship that would hold in any situation, but only in the case of the specific situation they describe on p. 573, where a ship is traveling between two planets, and in the ship's frame the distance between the planets is Lm, while the time between passing each planet as measured by the ship's clock is ts. So naturally in this frame if the second planet is a distance Lm away at the moment the ship is next to the first planet, and the ship remains at rest while the second planet rushes towards it, reaching the stationary ship a time ts later, then since that planet covered a distance of Lm in a time of ts the planet's speed must be v = Lm/ts, just based on the fact that in any frame speed is defined as distance/time (in terms of that frame's coordinates). Likewise, in the frame of the planets it takes the ship a time of tm to get from one planet to the other, and the distance between the planets is Ls in this frame, so the speed of the ship in this frame must be v = Ls/tm. And obviously if you take this equation and multiply both sides by tm you get Ls = v*tm, likewise the first equation can be rearranged to give Lm = v*ts.

Last edited by a moderator: May 4, 2017
13. Jun 8, 2009

### hippopviolin

Thank YOu Jesse M, i think I get it now.

but..ah..I am not sure how do you know which one is suppose to be the proper length.

In this scenario, is it true that the captain finds the distance separating the two events to be Lm is because of she thinks the planets are moving, and she is stationary?

If that is true, so basically the objects that are moving will seem to be shorten to the observers who are stationary?

14. Jun 8, 2009

### feynmann

If you have to ask this kind of question, that means you know the math equation but you do not understant its physical meaning. I guess you must know the so-called Lorentz transformation by now. He had similar question about time, since there are two "time" in the his equations. One is "time", one is "tau". He did not understand it, so he called one "time" and the other "local time". Lorentz think local time is just a mathematical trick that greatly eased the computational burden of finding new solutions of Maxwell's equations. Einstein cleared things up in one stroke. We now call one the "coordinate time' and the other "proper time", or (Eigen-Zeit) in German, it means its own time.
No, it is "shorten" if you measure it with ruler, Not just "seem to be shorten" to the observers who are stationary.

http://www.pitt.edu/~jdnorton/Goodies/rel_of_sim/index.html

Last edited: Jun 8, 2009
15. Jun 8, 2009

### JesseM

Proper length is the distance between the two endpoints of an object (or two separate objects with space between them like the planets) in the frame where the endpoints are at rest.
Right.
Well, there's no absolute truth about who's "moving" and who's "stationary" since all speeds are relative, but objects that are moving relative to a given frame of reference will shorten to observers who are stationary relative to that same frame. This explains why if your ship is moving relative to my ship, and both ships have the same proper length, then in my rest frame your ship will shorten relative to mine, while in your rest frame my ship will shorten relative to yours.

16. Jun 8, 2009

### hippopviolin

oh yes..that makes perfect sense now..thank you so much


text:
MathSciNet bibliographic data MR602900 55P20 Cohen, F. R.; May, J. P.; Taylor, L. R. $K({\bf Z},\,0)$$K({\bf Z},\,0)$ and $K(Z\sb{2},\,0)$$K(Z\sb{2},\,0)$ as Thom spectra. Illinois J. Math. 25 (1981), no. 1, 99–106. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.