text:
# The intercepts on the coordinate axes of the straight line with the given equation displaystyle{4}{x}-{3}{y}={12}.

Question
Alternate coordinate systems
The intercepts on the coordinate axes of the straight line with the given equation $$\displaystyle{4}{x}-{3}{y}={12}.$$

2021-02-26
Concept used:
If a straight line passes through the points (a, 0) (on the x-axis) and (0, b) (on the y-axis), we say that the x-intercept is a and the y-intercept is
b.
The x -intercept is found by setting $$\displaystyle{y}={0}$$ and solving for a, similarly, the y-intercept is found by setting $$\displaystyle{x}={0}$$ and solving for b.
Calculation:
For the given equation $$\displaystyle{4}{x}-{3}{y}={12}$$
$$\displaystyle{y}={0}\Rightarrow{4}{x}={12}\Rightarrow{x}-\text{intercept}={3},$$
$$\displaystyle{x}={0}\Rightarrow-{3}{y}={12}\Rightarrow{y}-\text{intercept}=-{4}$$
Conclusion:
For the given line, $$\displaystyle{x}-\text{intercept}={3}{\quad\text{and}\quad}{y}-\text{intercept}=-{4}$$

### Relevant Questions

To calculate: The intercepts on the coordinate axes of the straight line with the given equation $$\displaystyle{2}{y}-{4}={3}{x}$$

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$

The volume of the largest rectangular box which lies in the first octant with three faces in the coordinate planes and its one of the vertex in the plane $$\displaystyle{x}+{2}{y}+{3}{z}={6}$$ by using Lagrange multipliers.
A golf ball lies 2.00 m directly south of the hole on a levelgreen. On the first putt, the ball travels 3.00 m along astraight-line path at an angle of 5 degrees east of north; on thesecond putt, it travels a straight-line distance of 1.20 m at anangle of 6 degrees south of west.
What would be the displacement ofa third putt that would put the ball in the hole?
The answer in the back of the book is 1.3 m at 43 degreessouth of east. How do I get that?
For the equation (-1,2), $$y= \frac{1}{2}x - 3$$, write an equation in slope intercept form for the line that passes through the given point and is parallel to the graph of the given equation.
The coordinates of the point in the $$\displaystyle{x}^{{{p}{r}{i}{m}{e}}}\ {y}^{{{p}{r}{i}{m}{e}}}$$ - coordinate system with the given angle of rotation and the xy-coordinates.
The equivalent polar equation for the given rectangular - coordinate equation.
$$\displaystyle{y}=\ -{3}$$
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
b. From rectangular to polar:
$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}$$
$$\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}$$
$$\displaystyle\Rightarrow\ {\tan{\theta}}={1}$$
Thus, desired equivalent polar equation would be $$\displaystyle\theta={1}$$
$$\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}$$
a) $$\displaystyle{y}\ge{x}^{2},{z}\ge{0},$$
b) $$\displaystyle{x}\le{y}^{2},{0}\le{z}\le{2}.$$


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De Broglie wavelength and atom penetration

1. Sep 19, 2008

Bakery87

1. The problem statement, all variables and given/known data

Calculate the de broglie wavelength (DBW) of an electron with kinetic energy 60 GeV.

What percentage of an atom's diameter can it penetrate?

2. Relevant equations

DBW = h/p
p=mv

3. The attempt at a solution

Basically I have an electron traveling at the speed of light. I arrived at this from its kinetic energy (60 GeV) and by using the relativistic K-energy equation. So I get it's de broglie wavelength fairly easily (I have this part done).

The part I don't understand is the penetration. I guess I just need some guidance/equations. Any ideas?

2. Sep 19, 2008

Redbelly98

Staff Emeritus
Welcome to PF, Barkery87.

For an atom's diameter, they might mean take the diameter of the Bohr model for the hydrogen atom in its ground state. What percentage of that diameter is the deBroglie wavelength?

p.s.
Um, you didn't use the electron's rest mass to calculate p=mv, did you?

3. Sep 19, 2008

Bakery87

I used 0.511003 MeV/c^2

4. Sep 19, 2008

Redbelly98

Staff Emeritus
Momentum is calculated differently for relativistic motion. There should be a formula in your textbook or lecture notes, relating E, p, and m (the rest mass, sometimes called m0)

5. Sep 19, 2008

Bakery87

I did find something...

p = K/c

I'm still looking through my notes.

6. Sep 19, 2008

Redbelly98

Staff Emeritus
Actually, that's a valid approximation for extremely relativistic situations (like this one).

format: plain


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# Thermal conductivity and heat capacity

Using kinetic theory, we can derive an expression for the thermal conductivity of a gas to be
κ=nCmoleculeλ<v>/3
where n is the number density of the molecules in the gas, Cmolecule is the heat capacity of a single molcule (i.e the heat that must be given to each molecule to raise the temperature of the gas by unit temperature), λ is the mean free path and <v> is the mean speed of the molecules.

Now we can write nCmolecule=CV/V where CV is the heat capacity of the gas at constant volume and V is the total volume of the gas. Now I understand that nCmolecule=C/V where C is the heat capacity of the gas, and obviously because we have a gas we must have either C=CV or C=Cp because the gas must be held at either constant volume or constant pressure. However, I am not sure sure how to see why we have to consider the heat capacity at constant volume here - why can't it be constant pressure...

Bystander
Homework Helper
Gold Member
why we have to consider the heat capacity at constant volume here
n is the number density
Stick with the original constraints on the calculation.

Stick with the original constraints on the calculation.

What does this mean?

Bystander
Homework Helper
Gold Member
You started your calculation at constant V. Finish at constant V.

You started your calculation at constant V. Finish at constant V.

Ok, but I can't see why we started at constant V?

Bystander


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# diode number code

The battery it charges us 12 volt. ▶ Check our Supplier Directory, C = Transistor - audio frequency, low power, F = Transistor - high frequency, low power, R = Switching device, low power, e.g. Include your email address to get a message when this question is answered. Alibaba.com offers 623 diode number products. International Purchases: Our preferred method of … Exemple, I got a VIOLET-GREEN-GRAY-BLUE diode. Can you help in identifying this, I have a zener with 12.43 and 3 vertical dots. Below that, a 1N4148 signal diode uses a black ring to mark the cathode. The letters indicate the type and intended application of the device according to the following code. The top diode, a 1N4001 rectifier, has a grey ring near the cathode. There are many thousands of different types of diode, bipolar transistor and FET. Thyristor     This system has one drawback that the code will not show the function of the diode. Time and moneyare loss due to that we lack of knowledge of identify the right zener diode voltage. Schottky diode – Made from direct contact between silicon and metal. Semiconductor Device, Diode, Silicon, Voltage Regulator: 1N962B-1 through 1N992B-1, and 1N962BUR-1 through 1N992BUR-1, 1N962C-1 through 1N992C-1, and 1N962CUR-1 … These diodes can pass currents of up to 1 A, and they have peak inverse voltage (PIV) ratings that range from 50 to 1,000 V. I have a Diode that has failed, it is simply a almost spherical in shape, colour is creamy white with a black line at one end, there are four blue dots evenly spaced around the circumference of the white body. As a result it is essential that the different semiconductor devices are given different part numbers to distinguish them from each other. The DIODE dataset and the code is released using the MIT license. Unidirectional ElectroStatic Discharge (ESD) protection diode in a SOD882 leadless ultra small Surface Mounted Device (SMD) plastic package designed to protect one signal line from the damage caused by ESD and other transients. Power BJT: Introduction, Structure, Characteristics. Below that, a 1N4148 signal diode uses a black ring to mark the cathode. Take the resistance code and use the color number and put "1N" in from of the code you found and you got the zener part number! This makes Schottky diodes an excellent choice for reverse protection diodes. Exemple, I got a VIOLET-GREEN-GRAY-BLUE diode. On some occasions there may be a suffix letter that is added: This is useful to both manufacturers and users because when transistors are manufactured, there is a large spread in the levels of gain. (Not all manufacturers provide JEDEC numbers; many use their own commercial part numbers.) SMF5V0A to SMF58A www.vishay.com Vishay Semiconductors Rev. $0.20 per pack, or$0.02 per diode) SMD Zener Diodes Part No. The European System(ES) Diode Code Identification consists of an alphanumeric code as per below: The first naming method is for General purpose diodes and the second option is for special purpose diodes. The Pro-Electron numbering scheme to provide a standardised scheme for semiconductor numbering - in particular diodes, transistors and fild effect transistors was set up in 1966 at a meeting in Brussels, Belgium. In the early days of thermionic valve (vacuum tube) manufacture, each manufacturer gave a number to the types they manufactured. reads ‘the letter after the number means’……..hello . Joint Electron Device Engineering Council (JEDEC) These part numbers take the form: digit, letter, sequential number, [suffix] The letter is always 'N', and the first digit is 1 for diodes, 2 for transistors, 3 for four-leaded devices, and so forth. 1/2.     Return to Components menu . The output of the logic circuit is also located at that node. Warning Note: Please ensure the context of the component used in the electronic design is correct. I have T3D63 Diode and a 3D53 can you help with something that will work in stead? Never says what the numbers mean. Using the numbering scheme it can be seen that a transistor with the part number BC107 is a silicon low power audio transistor and a BBY10 is silicon variable capacitance diode for industrial or commercial use. For example parameters such as the transistor being intended for low frequency power, RF, etc can be determined. Note: Complete Technical Details can be found at the 1N4732A diode datasheet given at the end of this page. For example IN 4001, IN 4148. Please help me any knows, Hi, When choosing one of these di… How Zener Diode works as a Voltage Regulator? Silicon diode code in multisim. The Japanese Industrial Standards, JIS part numbering scheme for semiconductor devices is standardised under JIS-C-7012. Having industry standard numbering schemes for semiconductor devices has many advantages, not only for large scale manufacturers of electronic equipment right down to the hobbyist and student. Some of the special diodes areZener diode, Schottky diode,Varactor Diode, Tunnel diode, Gunn Diode, Light Emitting Diode(LED) etc. Pls,.help me to find a replacement for 3358r diode use for fire alarm switch series to the control panel with 24v supply thank you. Its blocking voltage varies from 50 volts (1N4001) to 1000 volts (1N4007). diode type number to marking code type number marking code package 1n821 1n821 sod68 (do34) 1n821a 1n821a sod68 (do34) 1n823 1n823 sod68 (do34) 1n823a 1n823a sod68 (do34) 1n825 1n825 sod68 (do34) 1n825a 1n825a sod68 (do34) 1n827 1n827 sod68 (do34) 1n827a 1n827a sod68 (do34) 1n829 1n829 sod68 (do34) 1n829a 1n829a sod68 (do34) 1n914 1n914 ph sod27 (do35) 1n4148 … Your email address will not be published. Tolerance I guess. Transistor Tutorial Includes: LEDs. This diode has 3 lines of numbers or letters: TCO/ /149 152C/ /PPC8/. This site uses Akismet to reduce spam. SMD type i search in the internet. Your email address will not be published. Simple digital logic gates, like the AND or the OR, can be built out of diodes. 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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Using the results of the above example on Example 20.3, find the drift velocity in a copper wire of twice the diameter and carrying 20.0 A.

The drift velocity in the copper wire is $1.08 \times {10^{ - 4}}\;{\rm{m/s}}$.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

• The gauge of the copper wire is a = 12.
• The diameter of the copper wire is $$d = 2 \times 2.053\;{\rm{mm}} \times \frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{mm}}}}{\rm{ = 4}}{\rm{.106 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}$$.
• The current carried by the copper wire is l = 20.0 A.

## Step 2: Significance of the drift velocity

The drift velocity is illustrated as the average velocity attained by a particle because of the involvement of an electric field. The drift velocity significantly contributes to the current’s change in an electric field.

## Step 3: Determination of the drift velocity

From the example 20.3, the number of copper atoms can be gathered that is$$n = 8.342 \times {10^{28}}\;{{\rm{m}}^{{\rm{ - 3}}}}$$ .

The equation of the magnitude of the current is expressed as:

${v_d} = \frac{I}{{nq\left( {\pi {{\left( {\frac{d}{2}} \right)}^2}} \right)}}$

Here, $$I$$is the magnitude of current, $${v_d}$$ is the drift velocity, $n$ is the number of the copper atoms, $q$ is the charge of one electron and $d$ is the diameter of the copper wire.

Substitute the values in the above equation.

\begin{aligned}{v_d} = \frac{{20.0\;{\rm{A}}}}{{\left( {8.342 \times {{10}^{28}}\;{{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {\left( {3.14} \right){{\left( {\frac{{{\rm{4}}{\rm{.106 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{2}} \right)}^2}} \right)}}\\ = \frac{{20.0\;{\rm{A}}}}{{\left( {1.33 \times {{10}^{10}}\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {\left( {3.14} \right){{\left( {{\rm{2}}{\rm{.106 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}} \right)}^2}} \right)}}\\ = \frac{{20.0\;{\rm{A}}}}{{\left( {1.33 \times {{10}^{10}}\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {\left( {3.14} \right)\left( {{\rm{4}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{{\rm{m}}^2}} \right)} \right)}}\\ = \frac{{20.0\;{\rm{A}}}}{{\left( {1.33 \times {{10}^{10}}\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {{\rm{1}}{\rm{.39 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{{\rm{m}}^2}} \right)}}\end{aligned}

Hence, further as:

\begin{aligned}{v_d} = \frac{{20.0\;{\rm{A}}}}{{\left( {184870\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 1}}}}} \right)}}\\ = 1.08 \times {10^{ - 4}}\;{\rm{A}} \cdot {\rm{m/C}}\\ = 1.08 \times {10^{ - 4}}\;{\rm{A}} \cdot {\rm{m/C}} \times \frac{{{\rm{1}}\;{\rm{C/s}}}}{{{\rm{1}}\;{\rm{A}}}}\\ = 1.08 \times {10^{ - 4}}\;{\rm{m/s}}\end{aligned}

Thus, the drift velocity in the copper wire is $1.08 \times {10^{ - 4}}\;{\rm{m/s}}$.


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What are the numbers that come next in these sequences: 3,3,6,9,15,24?

Apr 23, 2016

$39 , 63 , 102 , \ldots$

${a}_{n} = 3 {F}_{n} = \frac{3 \left({\phi}^{n} - {\left(- \phi\right)}^{- n}\right)}{\sqrt{5}}$

Explanation:

This is $3$ times the standard Fibonacci sequence.

Each term is the sum of the two previous terms, but starting with $3 , 3$, instead of $1 , 1$.

The standard Fibonnaci sequence starts:

$1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , \ldots$

The terms of the Fibonacci sequence can be defined iteratively as:

${F}_{1} = 1$

${F}_{2} = 1$

${F}_{n + 2} = {F}_{n} + {F}_{n + 1}$

The general term can also be expressed by a formula:

${F}_{n} = \frac{{\phi}^{n} - {\left(- \phi\right)}^{- n}}{\sqrt{5}}$

where $\phi = \frac{1}{2} + \frac{\sqrt{5}}{2} \approx 1.618033988$

So the formula for a term of our example sequence can be written:

${a}_{n} = 3 {F}_{n} = \frac{3 \left({\phi}^{n} - {\left(- \phi\right)}^{- n}\right)}{\sqrt{5}}$


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# Virtual Hydrogens

What's the difference between computational chemistry and cheminformatics? Computational chemists encode all the hydrogens. This tongue-in-cheek statement, whose origins I don't recall, isn't far off the mark. It stands to reason that if hydrogens are going to be deleted, there needs to be a good way to bring them back. This article proposes some language intended to simplify what has become an unnecessarily complex problem.

## Hydrogen Suppression

The practice of ignoring hydrogens to emphasize the carbon framework has a long tradition in organic chemistry. As far back as 1872, Kekulé himself used the convention when discussing alternative structures of benzene:

Over the next 100 years or so, authors experimented with hydrogen suppression on and off. By the 1970s, most chemists had standardized on the hydrogen-suppressed form now in common use. In particular, hydrogens are only added when the count is ambiguous or in certain special cases such as carbonyl carbons.

As the field of cheminformatics gathered steam the 1960s, it naturally gravitated toward hydrogen-suppressed representations. Given the severe resource constraints at the time, however, the earliest systems were based not on individual atoms, but on collections of them. The best-known example from this period is Wiswesser Line Notation (WLN)), which is based on functional groups.

Eventually it became feasible to represent each atom individually. But rather than elevating hydrogens to first-class status, the new systems being developed after the 1970s left hydrogens in the economy section, just like the structure drawings being published by organic chemists.

Somewhere along the line, the term "implicit hydrogen" was adopted. The Merriam-Webster dictionary defines "implicit" as: "capable of being understood from something else though unexpressed."

Unfortunately, the terms "implicit hydrogen" and "hydrogen suppression" have come to mean the same thing. They aren't. The resulting confusion has, I suspect, led to many mis-characterized chemical structures over the years.

## Implicit Hydrogens

An implicit hydrogen, as the name suggests, is a hydrogen atom whose presence can only be deduced by "something else." That "something else" is a set of conventions referred to valence rules or a valence model.

A valence model associates certain elements with one or more ideal hydrogen counts. For example, carbon has an ideal hydrogen count of four. Nitrogen has an ideal hydrogen count of three. Oxygen has an ideal hydrogen count of two, and so on. Valence models usually apply to a subset of the periodic table.

The implicit hydrogen count of any atom subject to a valence model can be determined by a simple procedure. First, determine the number of electrons donated by the atom to bonding. This value equals the sum of formal bond orders and is sometimes called "valence." Next, subtract the bonding electron count from the atom's ideal hydrogen count. The result is the atom's implicit hydrogen count. The implicit hydrogen count for an element not covered by a valence model is undefined.

Consider ethane with all hydrogens suppressed. We want to determine the implicit hydrogen count of a carbon atom. We note one single bond to a neighboring carbon, so the bonding electron count is one. Our valence model assigns carbon an ideal hydrogen count of four. Subtracting one for the bond to the neighboring carbon yields three hydrogens. This is the implicit hydrogen count for a carbon atom in ethane. The same procedure would yield two as the implicit hydrogen count for a carbon in ethene, and one for a carbon in ethyne.

To recap, an implicit hydrogen is one whose presence can only be deduced from a valence model. The implicit hydrogen count of an atom not covered by a valence model is undefined.

## Explicit Implicit Hydrogens

Clearly, there's more to hydrogen suppression than just implicit hydrogens and valence models. For example, organic chemists follow a modern drawing convention in which hydrogens attached to carbon are implicit, but the hydrogens attached to heteroatoms are explicit. The same phenomenon can be found in cheminformatics toolkits and exchange formats such as SMILES. Recall that the implicit hydrogen count of an atom not covered by a valence model is undefined. Or, consider a system in which valence models have been abolished altogether. In either case, brevity draws us toward hydrogen suppression, but no valence model applies.

Given the lack of an alternative, the phrase "explicit implicit hydrogen" has taken root to describe the atomic attribute representing a hydrogen count. The most obvious problem is that the terms "explicit" and "implicit" conflict. It's a linguistic pretzel. But the deeper problem is that the term "implicit hydrogen" carries the very specific association with a valence model. This seepage causes considerable problems to those implementing or using cheminformatics systems.

## Virtual Hydrogen

Can we do better? I think so. I'd like to propose the term virtual hydrogen.

A virtual hydrogen is a hydrogen atom encoded, not as a node and edge combination in a molecular graph, but within an atomic integer attribute. To convert an atomic hydrogen, delete it and its bond, and increment the virtual hydrogen count by one. To convert a virtual hydrogen, decrement the count and add an atomic hydrogen with bond. Multivalent hydrogens, as found in boranes for example, are not eligible for such conversion. Virtual and atomic hydrogen representations and both be present at the same atom.

Merriam-Webster defines "virtual" as: "being such in essence or effect through not formally recognized or admitted." The word "virtual" is also used extensively in computer science as reflected by such terms as: "virtual keyboard;" "virtual reality;" and "virtual memory." In these contexts, the word assumes a meaning similar to the Macmillan Dictionary definition: "almost the same as the thing that is mentioned." A virtual hydrogen is almost the same as an atomic hydrogen.

There's no linguistic or logical barrier to encoding "virtual hydrogens" explicitly. On the contrary, virtual hydrogens must be explicitly encoded.

The concept of virtual hydrogens allows us to talk about hydrogen suppression more precisely. For example, the mandatory hydrogen at pyrrole nitrogen (e.e., [nH]1cccc1) is virtual. Those hydrogens deduced to be present in propane (CCC) are implicit. Encoding a hydrogen as a standalone atom, as in [CH3][H] makes it atomic (note the corresponding deduction from the hosting carbon virtual hydrogen count). The V2000 Molfile format supports virtual hydrogens, but in a roundabout way. InChI eschews implicit hydrogens altogether in favor of virtual hydrogens encoded within a dedicated layer.

## Conclusion

Organic chemists discovered long ago that drawing every hydrogen atom leads to noise. Hydrogens can be safely suppressed, but only when the rules for interconversion are both simple and explicit. Two distinct forms of hydrogen suppression are possible, but poor terminology has led to their conflation. Implicit hydrogens are encoded through the use of a valence model. A second form of hydrogen suppression replaces atomic hydrogens and their bonds with a hydrogen count attribute. This latter form of hydrogen suppression requires an unambiguous name, and "explicit implicit hydrogens" just won't cut it. "Virtual hydrogen" avoids linguistic convolution while at the same time cleanly severing the link to a valence model.


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Mathematics Solutions Solutions for Class 8 Maths Chapter 12 Statistics are provided here with simple step-by-step explanations. These solutions for Statistics are extremely popular among Class 8 students for Maths Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

 No. of saplings (Scores) xi No. of students (frequncy) fi fi×xi 1 4 4 2 6 3 12 4 8 N = ∑fixi=

Mean $\overline{)x}$

=

=

∴ The average number of trees planted

 No. of saplings (Scores) xi No. of students (frequncy) fi fi×xi 1 4 4 2 6 = 2 $×$ 6 = 3 12 = 3 $×$ 12 = 4 8 = 4 $×$ 8 = N = ∑fixi=

Mean $\overline{)x}$ = $\frac{\overline{)\sum _{}{f}_{i}{x}_{x}}}{N}$

∴ The average number of trees planted

#### Question 2:

The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

 Electricity used (units) xi No. of Families (frequency) fi fi×xi 30 7 ....... 45 2 ....... 60 8 ....... 75 5 ....... 90 3 ....... N = ....... ∑fixi =.......
(1) How many families use 45 units electricity?
(2) State the score, the frequency of which is 5.
(3) Find N, and ∑fixi
(4) Find the mean of electricity used by each family in the month of May.

The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

 Electricity used (units) xi No. of Families (frequency) fi fi×xi 30 7 30 × 7 =  210 45 2 45 × 2 =  90 60 8 60 × 8 =  480 75 5 75 × 5 =  375 90 3 90 × 3 =  270 N =  25 ∑fixi =  1425

(1) 2 families use 45 units electricity.

(2) The score with a frequency of 5 is 75.

(3) N = 25, and ∑fixi = 1425

(4) Since,

$\mathrm{Mean}=\frac{\sum _{}{f}_{i}{x}_{i}}{N}\phantom{\rule{0ex}{0ex}}=\frac{1425}{25}\phantom{\rule{0ex}{0ex}}=57$

So, the mean of electricity used by each family in the month of May is 57.

#### Question 3:

The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2, 3, 5, 6, 4, 2. Prepare a frequency table and find the mean of members of 40 families.

 Number of members (xi) Number of families (fi) fixi 1 2 2 2 8 16 3 7 21 4 8 32 5 7 35 6 6 36 7 2 14 N = 40 $\sum _{}$fixi = 156

Since, mean = $\frac{\sum _{}{f}_{i}{x}_{i}}{N}=\frac{156}{40}=3.9$

So, the mean of members of 40 families is 3.9.

#### Question 4:

The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exibition is:
2 3 4 1 2 3 1 5 4 2 3 1 3 5 4 3 2 2 3 2. Prepare a frequency table and find the mean of the data.

The frequency table of the data is follows:

 Number of projects (xi) Frequency (fi) fixi 1 3 1 $×$ 3 = 3 2 6 2 $×$ 6 = 12 3 6 3 $×$ 6 = 18 4 3 4 $×$ 3 = 12 5 2 5 $×$ 2 = 10 N = 20 $\sum _{}{f}_{i}{x}_{i}$ = 55

Since, mean of the data = $\frac{\sum _{}{f}_{i}{x}_{i}}{N}=\frac{55}{20}=2.75$

Hence, the mean of the data is 2.75.

#### Question 1:

Observe the following graph and answer the question.

(1) State the type of the graph.
(2) How much is the savings of Vaishali in the month of April?
(3) How much is the total savings of Saroj in the months March and April?
(4) How much more is the total savings of Savita than the total savings of Megha?
(5) Whose savings in the month of April is the least?

(1) This is a bar graph.

(2) The savings of Vaishali in the month of April is Rs 600.

(3) The total savings of Saroj in the months March and April is Rs 800.

(4) Since, the total savings of Savita = Rs 1000, and

The total savings of Megha = Rs 500

Now, 1000 $-$ 500 = 500

So, the total savings of Savita is Rs 500 more than the total savings of Megha.

(5) The savings of Megha in the month of April is the least i.e. Rs 200.

#### Question 2:

The number of boys and girls, in std 5 to std 8 in a Z.P. school is given in the table. Draw a subdivided bar graph to show the data.
(Scale : On Y axis, 1 cm = 10 students)

 Standard 5th 6th 7th 8th Boys 34 26 21 25 Girls 17 14 14 20

We have,

 Standard 5th 6th 7th 8th Boys 34 26 21 25 Girls 17 14 14 20

The subdivided bar graph of the given data is as follows:

#### Question 3:

In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

 Town → Karjat Wadgoan Shivapur Khandala Year ↴ 2016 150 250 200 100 2017 200 300 250 150

We have,

 Town → Karjat Wadgoan Shivapur Khandala Year ↴ 2016 150 250 200 100 2017 200 300 250 150

The subdivided bar graph of the given data is as follows:

#### Question 4:

In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale : On Y axis : 1 cm = 500 students)

 Town → Paithan Yeola Shahapur Mean of Communication  ↴ cycle 3250 1500 1250 Bus and Auto 750 500 500 On foot 1000 1000 500

We have, the following table that shows the data of the transport means used by students in 8th standard for commutation between home and school is given.

 Town → Paithan Yeola Shahapur Mean of Communication  ↴ cycle 3250 1500 1250 Bus and Auto 750 500 500 On foot 1000 1000 500

The subdivided bar diagram of the given data is as follows:

#### Question 1:

Show the following information by percentage bar graph.

 Division of standard 8 A B C D Number of students securring grade A 45 33 10 15 Total number of students 60 55 40 75

We have,

The following information:

 Division of standard 8 A B C D Number of students securring grade A 45 33 10 15 Total number of students 60 55 40 75 Percentage of students securring grade A $\frac{45}{60}×100=75$ $\frac{33}{55}×100=60$ $\frac{10}{40}×100=25$ $\frac{15}{75}×100=20$

The percentage bar graph of the given data is as follows:

#### Question 2:

Observe the following graph and answer the questions.

(1) State the type of the bar graph.
(2) How much percent is the Tur production to total production in Ajita'a farm?
(3) Compare the production of Gram in the farms of Yash and Ravi and state whose percentage of production is more and by how much?
(4) Whose percentage production fo Tur is the least?
(5) State production percentages of Tur and gram in Sudha's farm

(1) This is a percentage bar graph.

(2) The Tur production is 60% of the total production in Ajita'a farm.

(3) The production of gram in the farm of Yash is 50%, whereas

The production of gram in the farm of Ravi is 30%.

Since, 50 $-$ 30 = 20

So, the production of Gram in the farms of Yash is more than that of Yash by 20%.

(4) The production fo Tur in the farm of Sudha is the least, i.e. 40%.

(5) The production percentages of Tur and gram in Sudha's farm are 40% and 60%, repectively.

#### Question 3:

The following data is collected in a survey of some students of 10th  standard from some schools. Draw the percentage bar graph of the data.

 School 1st 2nd 3rd 4th Inclination towards science stream 90 60 25 16 Inclination towards commerce stream 60 20 25 24

 School 1st 2nd 3rd 4th Inclination towards science stream 90 60 25 16 Inclination towards commerce stream 60 20 25 24 Total number of students 150 80 50 40 Percentage of students whose inclination towards science stream $\frac{90}{150}×100=60$ $\frac{60}{80}×100=75$ $\frac{25}{50}×100=50$ $\frac{16}{40}×100=40$ Percentage of students whose inclination towards commerce stream $100-60=40$ $100-75=25$ $100-50=50$ $100-40=60$


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# Kruskal–Wallis one-way analysis of variance

The Kruskal–Wallis one-way analysis of variance by ranks (named after William Kruskal and W. Allen Wallis) is a non-parametric method for testing whether samples originate from the same distribution. It is used for comparing more than two samples that are independent, or not related. The parametric equivalent of the Kruskal-Wallis test is the one-way analysis of variance (ANOVA). When the Kruskal-Wallis test leads to significant results, then at least one of the samples is different from the other samples. The test does not identify where the differences occur or how many differences actually occur. It is an extension of the Mann–Whitney U test to 3 or more groups. The Mann-Whitney would help analyze the specific sample pairs for significant differences.

Since it is a non-parametric method, the Kruskal–Wallis test does not assume a normal distribution of the residuals, unlike the analogous one-way analysis of variance. However, the test does assume an identically shaped and scaled distribution for each group, except for any difference in medians.

Kruskal–Wallis is also used when the examined groups are of unequal size (different number of participants).

## Method

1. Rank all data from all groups together; i.e., rank the data from 1 to N ignoring group membership. Assign any tied values the average of the ranks they would have received had they not been tied.
2. The test statistic is given by:
$K = (N-1)\frac{\sum_{i=1}^g n_i(\bar{r}_{i\cdot} - \bar{r})^2}{\sum_{i=1}^g\sum_{j=1}^{n_i}(r_{ij} - \bar{r})^2},$ where:
• $n_i$ is the number of observations in group $i$
• $r_{ij}$ is the rank (among all observations) of observation $j$ from group $i$
• $N$ is the total number of observations across all groups
• $\bar{r}_{i\cdot} = \frac{\sum_{j=1}^{n_i}{r_{ij}}}{n_i}$,
• $\bar{r} =\tfrac 12 (N+1)$ is the average of all the $r_{ij}$.
3. If the data contain no ties the denominator of the expression for $K$ is exactly $(N-1)N(N+1)/12$ and $\bar{r}=\tfrac{N+1}{2}$. Thus
\begin{align} K & = \frac{12}{N(N+1)}\sum_{i=1}^g n_i \left(\bar{r}_{i\cdot} - \frac{N+1}{2}\right)^2 \\ & = \frac{12}{N(N+1)}\sum_{i=1}^g n_i \bar{r}_{i\cdot }^2 -\ 3(N+1). \end{align}
The last formula only contains the squares of the average ranks.
4. A correction for ties if using the short-cut formula described in the previous point can be made by dividing $K$ by $1 - \frac{\sum_{i=1}^G (t_i^3 - t_i)}{N^3-N}$, where G is the number of groupings of different tied ranks, and ti is the number of tied values within group i that are tied at a particular value. This correction usually makes little difference in the value of K unless there are a large number of ties.
5. Finally, the p-value is approximated by $\Pr(\chi^2_{g-1} \ge K)$. If some $n_i$ values are small (i.e., less than 5) the probability distribution of K can be quite different from this chi-squared distribution. If a table of the chi-squared probability distribution is available, the critical value of chi-squared, $\chi^2_{\alpha: g-1}$, can be found by entering the table at g − 1 degrees of freedom and looking under the desired significance or alpha level. The null hypothesis of equal population medians would then be rejected if $K \ge \chi^2_{\alpha: g-1}$. Appropriate multiple comparisons would then be performed on the group medians.
6. If the statistic is not significant, then there is no evidence of differences between the samples. However, if the test is significant then a difference exists between at least two of the samples. Therefore, a researcher might use sample contrasts between individual sample pairs, or post hoc tests, to determine which of the sample pairs are significantly different. When performing multiple sample contrasts, the Type I error rate tends to become inflated.

## Exact probability tables

A large amount of computing resources is required to compute exact probabilities for the Kruskal-Wallis test. Existing software only provides exact probabilities for sample sizes less than about 30 participants. These software programs rely on asymptotic approximation for larger sample sizes. Exact probability values for larger sample sizes are available. Spurrier (2003) published exact probability tables for samples as large as 45 participants.[1] Meyer and Seaman (2006) produced exact probability distributions for samples as large as 105 participants.[2]


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# Abstracts

Anna Capietto, Università di Torino

Title: Existence and multiplicity of special solutions to a nonlinear eigenvalue problem in the half-line arising from the Dirac equation
Abstract: We provide the existence of a global continuum of solutions to a nonlinear PDE which have a special form. To this aim, we exploit the notion of partial wave subspace in relation with the well-known concept of rotation number.

Alberto Boscaggin, Università di Torino

Title: Periodic solutions of superlinear systems of ODEs: a symplectic approach
Abstract: We prove the existence of infinitely many nodal periodic solutions of  a weakly coupled superlinear system of ODEs. The proof relies on an elementary (higher dimensional) version of the Poincaré-Birkhoff fixed  point theorem, valid for exact symplectic maps with monotone twist. Joint work with Rafael Ortega (University of Granada).

Graham Hugh Cox, University of North Carolina at Chapel Hill

Title: Relating the Morse and Maslov index in an arbitrary domain
Abstract: In a recent paper of Deng and Jones, it was shown that the Morse index of an elliptic boundary value problem can be determined using methods from infinite-dimensional symplectic geometry. This idea extends the classic work of Smale, on higher-dimensional Morse theory, to more general boundary conditions, but has only been carried out for star-shaped domains. In this talk I will explain a generalization these results to domains that are diffeomorphic to a ball, and discuss some open problems for domains with nontrivial topology.

Walter Dambrosio, Università di Torino
Title:  On the boundedness of solutions to a nonlinear singular oscillator
Abstrac: We study a second order scalar equation of the form

$x''+V'(x)=p(t),$

where $p$ is a perodic function and $V$ is a singular potential. We give sufficient conditions on $V$ and $p$ ensuring that all solutions are bounded; we prove the existence of Aubry-Mather sets as well.

Francesca Dalbono, Università di Palermo
Title:  Nodal solutions for supercritical Laplace and p-Laplace equations
Abstract: We study asymptotic behaviour and nodal properties of the radial solutions to a superlinear p-Laplace equation. Our approach is based on the Fowler transformation combined with invariant manifold theory. (This is a joint work with Matteo Franca).

Russell Johnson, Università di Firenze

Title:  The nonautonomous version of the Yakubovich frequency theorem
Abstract: The Frequency Theorem is a basic result in linear-quadratic optimal control theory which has applications to problems concerning absolute stability, dissipative systems, and nonconvex optimization. It was formulated and proved by Yakubovich for control systems with time-periodic coefficients, and later extended to cover the general nonautonomous case when the coefficients  depend aperiodically on time. We will discuss the nonautonomous version of the Frequency Theorem, and indicate in particular how the rotation number for linear Hamiltonian ODEs can be used to verify its hypothesis. The results to be presented are contained in joint work with Roberta Fabbri and Carmen Nunez.

Chris Jones, University of North Carolina at Chapel Hill
Title:  Topological characterizations of the stability index for nonlinear waves
Abstract:TBA

Alessandro Portaluri, Università di Torino
Title:  Linear (in)stability for relative equilibria in singular Lagrangian systems
Abstract: Relative equilibria are maybe the simplest periodic orbit of Newtonian type singualr Lagrangian systems.  The goal of this talk is to  study  the linear and spectral (in)stability of relative equilibria for singular Lagrangian systems. By using symplectic and topological techniques we provide a sufficient condition for  a relative equilibrium to be linearly unstable as well as a necessary condition in order to be spectrally stable. (Joint with Vivina L. Barutello, R. Jadanza)

Nils Waterstraat, Politecnico di Torino
Title:  On the Morse index theorem in semi-Riemannian geometry
Abstrac: In the first part of our talk, we consider the Fredholm Lagrangian Grassmannian of a symplectic Hilbert space, the Maslov index and symplectic reductions.
Afterwards we give a brief recapitulation of the classical Morse index theorem from Riemannian geometry and its generalisation to geodesics in semi-Riemannian manifolds.
Finally, we introduce a new proof of the semi-Riemannian Morse index theorem, which is based on symplectic reductions. (Joint work with A. Portaluri)

Annunci


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Thread: Dimension of all complex valued continuos functions

1. Dimension of all complex valued continuos functions

Hi!

I'm trying to show that the set of all complex valued continuous functions defined on the real line is an infinite-dimensional complex vector space.

Well I don't know where to start.. My first intuitive idea is to show that if we take the Taylor series of the function only up to n finite terms this sum will not be equal to the function, only in the limit $n \to \infty$. (Thinking as the function as a vector written as a linear combination of the elements of the basis)

2. Take into account that the family:

$\mathcal{F}=\left\{{f_n:\mathbb{R}\rightarrow{\mat hbb{C}},\;f_n(x)=x^n,\;n\in \mathbb{N}}\right\}$

is linearly independent.

Regards.

3. Originally Posted by Ruun
Hi!

I'm trying to show that the set of all complex valued continuous functions defined on the real line is an infinite-dimensional complex vector space.

Well I don't know where to start.. My first intuitive idea is to show that if we take the Taylor series of the function only up to n finite terms this sum will not be equal to the function, only in the limit $n \to \infty$. (Thinking as the function as a vector written as a linear combination of the elements of the basis)

In fact, if $\dim_{\mathbb{C}}C\left[\mathbb{R},\mathbb{C}\right]=2^{\aleph_0}$. To see this we merely note that we may inject $C\left[\mathbb{R},\mathbb{C}\right]$ into $\mathbb{C}^{\mathbb{Q}}$ via the map $f\mapsto f_{\mid \mathbb{Q}}$. This is an injection since if two continuous maps into a metric space (in fact, a Hausdorff space) agree on a dense subset they must be the same function. But, note that $\text{card }\mathbb{C}^{\mathbb{Q}}=\left(2^{\aleph_0}\right) ^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$ so that $\text{card }C\left[\mathbb{R},\mathbb{C}\right]\leqslant 2^{\aleph_0}$ and in particular if $\mathcal{F}$ is a basis for $C\left[\mathbb{R},\mathbb{C}\right]$ then $\text{card }\mathcal{F}\leqslant 2^{\aleph_0}$. But, note that the family of maps $\left\{f_r: r\in\mathbb{R}^+\right\}$ given by $f_r:\mathbb{R}\to\mathbb{R}:x\mapsto x^r$ is linearly independent and has cardinality $2^{\aleph_0}$ and so $\text{card }\mathcal{F}\geqslant 2^{\aleph_0}$. It follows by the Schroeder-Bernstein theorem that $\text{card }\mathcal{F}=\dim_{\mathbb{C}}C\left[\mathbb{R},\mathbb{C}\right]=2^{\aleph_0}$

4. Thanks for even showing the dimension of the space.

I'm having some trouble with your notation, I'm a completely noob on real analysis. (My motivation of studying this is to study the Hilbert Spaces of Quantum Mechanics, wich are not quite good covered at school and I think that we are missing a huge point here).

First, let's see if I understood your post:

I think that the idea is that the theorem states (for our particular case) $\text{card }\mathcal{F}\leqslant 2^{\aleph_0} \mbox{ and} \text{ card }\mathcal{F}\geqslant 2^{\aleph_0} \mbox{ then } \text{card }\mathcal{F}=2^{\aleph_0}$ right? And as we know the cardinal of the basis we know the dimension.

Now the notation: I don't know if $\mathbb{C}^{\mathbb{Q}}$ stands for the cartesian product of $\mathbb{C}$ , $\mathbb{Q}$ times ("the number of rationals times").

Also I don't know wich set is $f_{\mid \mathbb{Q}}$.

I guess this is just above my math prerequisites, and I understand that an answer that I can understand I not just a post on a forum. (several weeks even months of study).

5. Originally Posted by Ruun
Thanks for even showing the dimension of the space.

I'm having some trouble with your notation, I'm a completely noob on real analysis. (My motivation of studying this is to study the Hilbert Spaces of Quantum Mechanics, wich are not quite good covered at school and I think that we are missing a huge point here).
Absolutely no worries! Notation can sometimes make math unreadable, and I in particular have a penchant for using "fancy looking notation" just to look "fancy". But anways

First, let's see if I understood your post:

I think that the idea is that the theorem states (for our particular case) $\text{card }\mathcal{F}\leqslant 2^{\aleph_0} \mbox{ and} \text{ card }\mathcal{F}\geqslant 2^{\aleph_0} \mbox{ then } \text{card }\mathcal{F}=2^{\aleph_0}$ right? And as we know the cardinal of the basis we know the dimension.
Correct. I am appealing to something called the Schroeder-Bernstein Theorem which in essence says that the relation $\leqslant$ on cardinal numbers (where $X\leqslant Y$ means "There exists an injection $f:X\to Y$" is antisymmetric). And then, right again. I'm appealing to the Dimension Theorem (which, by the way, is not trivial to prove. It requires something called the ultrafilter lemma) to conclude that since the cardinality of any basis must have cardinality $2^{\aleph_0}$ then $\dim_{\mathbb{C}}C[\mathbb{R},\mathbb{C}]=2^{\aleph_0}$

Now the notation: I don't know if $\mathbb{C}^{\mathbb{Q}}$ stands for the cartesian product of $\mathbb{C}$ , $\mathbb{Q}$ times ("the number of rationals times").
In essence, yes. But, with the definition of Cartesian product extended to the arbitrary case; namely that if $\left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ is an indexed class of sets then

$\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=\left\{f:\mat hcal{A}\to\bigcup_{\alpha\in\mathcal{A}}X_{\alpha} :f(\alpha)\in X_{\alpha}\text{ for all }\alpha\in\mathcal{A}\right\}$

So, without all the jargon really $\mathbb{C}^{\mathbb{Q}}$ is the set of all functions from $\mathbb{Q}$ into $\mathbb{C}$.

Also I don't know wich set is $f_{\mid \mathbb{Q}}$.
This is just the restriction of $f$ to $\mathbb{Q}$. Namely, for a function $f:\mathbb{R}\to\mathbb{C}$ we can create a new function, denoted $f\mid_{\mathbb{Q}}$ which maps $\mathbb{Q}\to\mathbb{C}$ and such that $f_{\mid\mathbb{Q}}(x)=f(x)$ for all $x\in\mathbb{Q}$. In other words, we just make the domain "smaller" and consider the function on that "smaller set". It turns out, like I said, that two continuous functions on $\mathbb{R}$ which agree on $\mathbb{Q}$ must be the same function (ask if you don't see why).


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CIS115 W6 Lab Solution

# CIS115 W6 Lab Solution

A
0 points

Week 6 iLab - Winner of the Chevy and Ford Racing Teams

TCO 5-Given a simple problem that requires iteration, create a solution algorithm that employs loops.

TCO 6-Given a simple problem that requires simple data structures, design, code, and test a solution algorithm that uses arrays.

TCO 8-Given a more complex problem, develop a complete solution that includes a comprehensive statement of the problem, complete program design, and program documentation.

Scenario

There are eight cars in each team called Chevy and Ford. One car from each team races its opponent on the drag strip. Read in the racing times for the eight Chevy cars and then read in the times for the eight Ford cars. Store the times into arrays called Chevy[ ] and Ford[ ]. Then list the winner of each pair, giving the...

CIS115 W6
Ash

A
0 points

#### Oh Snap! This Answer is Locked

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Excerpt from file: Name:_________________ Week 6 iLab Winner of the Chevy and Ford Racing Teams TCO 5Given a simple problem that requires iteration, create a solution algorithm that employs loops. TCO 6Given a simple problem that requires simple data structures, design, code, and test a solution algorithm that uses

Filename: cis115-w6-lab-solution-28.docx

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Print Length: 8 Pages/Slides

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## Energy System Benchmark Problems

Optimized grid design and operation is needed as modern society depends on vast quantities of electrical energy. The complexity of the electric grid presents difficult control problems that require powerful solvers and efficient formulations for tractable solutions.

The Gekko Optimization Suite is a machine learning and optimization package in Python for mixed-integer and differential algebraic equations and is capable of solving complex grid design and control problems. A series of non-dimensional benchmark cases are proposed for grid energy production. These include:

II: cogeneration
III: tri-generation
IV: energy storage with constant production
V: energy storage with load following
VI: energy storage with cogeneration

Individual case studies include ramp rate constraints, power production, and energy storage operation as design variables. Variables used in the benchmark problems are defined below.

 Symbol Description \Phi objective function d demand g generation r ramp rate e storage inventory ein, eout energy stored, recovered R renewable source sin, sout slack variables for storage switching \eta storage efficiency n number of generating units i subscript indicates product i

The first benchmark problem represents load following, a common scenario in grid systems. The optimizer seeks to match demand and supply with fluctuating demand dynamics. A single generator with ramping constraints attempts to respond to a single load with perfect foresight. The generation and demand match initially, but the generator must ramp in order to ensure this throughout the horizon while minimizing overproduction.

\begin{align} \min_{r} \ & \Phi = \int_{t=0}^{1} \left [ 1000 \max(0, d - g) + \max(0, g - d) \right ] \\ \mathrm{s.t.}\ & \frac{dg}{dt} = r \\ & d=\cos\left({2\pi t}\right) + 3 \label{eqn:benchmark1c}\\ & -1 \leq r \leq 1 \label{eqn:benchmark1d}\\ & g(0)=d(0)=4 \end{align}

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt

t = np.linspace(0,1,101)
m = GEKKO(remote=False); m.time=t

fe  = m.Param(np.cos(2*np.pi*t)+3)
de  = m.Var(fe[0])

e  = m.CV(0)
e.STATUS=1; e.SPHI=e.SPLO=0
e.WSPHI=1000; e.WSPLO=1
der = m.MV(0,lb=-1,ub=1); der.STATUS=1

m.Equations([de.dt()==der, e==fe-de])

m.options.IMODE=6; m.solve()

plt.plot(t,fe,'r-',label='Production')
plt.plot(t,de,'b:',label='Demand')
plt.plot(t,der,'k--',label='Ramp Rate')
plt.legend(); plt.grid(); plt.show()

#### Benchmark II: Cogeneration

In the second benchmark problem, one producer seeks to meet two objectives that are constraining at different times. Benchmark II enhances Benchmark I by replacing the generator with a cogeneration system n=2 that produces (1) electricity and (2) heat in response to electricity demand and a new heat demand profile, again with perfect foresight.

\begin{align} \min_{r} \ \ & \Phi = \sum_{i=1}^n \int_{t=0}^{1} \, \left [ 1000 \; \max(0, \, d_i - g_i) \vphantom{\int} + \max(0, \, g_i - d_i) \vphantom{\int} \right ] \\ \mathrm{s.t.}\ \ & \frac{dg_1}{dt} = r, \enspace g_2 = 2 g_1 \label{eqn:benchmark2b}\\ & d_1=\cos\left({2\pi t}\right) + 3 \\ & d_2=1.5\sin\left({2\pi t}\right) + 7 \\ & -1 \leq r \leq 1 \\ & g_1(0)=d_1(0)=4 \\ & g_2(0)=8, \; d_2(0)=7 \end{align}

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt

t = np.linspace(0,1,101)
m = GEKKO(remote=False); m.time=t

fe  = m.Param(np.cos(2*np.pi*t)+3)
fh  = m.Param(1.5*np.sin(2*np.pi*t)+7)
de  = m.Var(fe[0]); dh = m.Intermediate(de*2)

e  = m.CV(0); e.STATUS=1; e.SPHI=e.SPLO=0; e.WSPHI=1000; e.WSPLO=1
h  = m.CV(0); h.STATUS=1; h.SPHI=h.SPLO=0; h.WSPHI=1000; h.WSPLO=1
der = m.MV(0,lb=-1,ub=1); der.STATUS=1

m.Equations([de.dt()==der, e==fe-de, h==fh-dh])

m.options.IMODE=6; m.solve()

plt.figure(figsize=(8,5))
plt.subplot(3,1,1)
plt.plot(t,fe,'r-',label='Prod 1')
plt.plot(t,de,'b:',label='Demand 1')
plt.grid(); plt.legend()
plt.subplot(3,1,2)
plt.plot(t,fh,'r-',label='Prod 2')
plt.plot(t,dh,'b:',label='Demand 2')
plt.grid(); plt.legend()
plt.subplot(3,1,3)
plt.plot(t,der,'k--',label='Ramp Rate')
plt.grid(); plt.legend(); plt.xlabel('Time')
plt.show()

#### Benchmark III: Tri-generation

The third benchmark problem enhances the previous problems further still, creating a tri-generation system n=3 with two producers, three products, and three demand profiles. The primary producer is the same as the prior benchmark and is ramp-rate constrained to produce the two primary products (e.g., electricity and heat). An additional producer (e.g., a solid oxide electrolysis cell) uses these first two products to make a third product (e.g., hydrogen), thereby utilizing any excess system capacity and maximizing its production while avoiding supply shortages for products one and two. As before, the optimizer has perfect foresight.

\begin{align} \min_{r_1, r_3} \ \ & \Phi = \sum_{i=1}^n \int_{t=0}^{1} \left [ \vphantom{\int} 1000 \; \max(0, \, d_i - g_i) + \max(0, \, g_i - d_i) \vphantom{\int} \right ] - 0.1 \int_{t=0}^{1} g_3 \\ \mathrm{s.t.}\ \ & \frac{dg_1}{dt} = r_1, \; \frac{dg_3}{dt} = r_3 \\ & g_2 = 2 g_1 \\ & d_1=\cos\left({2\pi t}\right) + 3 + 2 g_3 \\ & d_2=1.5\sin\left({2\pi t}\right) + 7 + 3 g_3 \\ & d_3=\max\left[0,-0.2\sin\left(2\pi t\right) \right] \\ & -1 \leq r_1 \leq 1, \enspace -1 \leq r_3 \leq 1 \\ & g_1(0)=d_1(0)=4 \\ & g_2(0)=8, \; d_2(0)=7 \\ & g_3(0)=d_3(0)=0 \end{align}

import numpy as np
from gekko import GEKKO
import matplotlib.pyplot as plt

t = np.linspace(0,1,101)
m = GEKKO(remote=False); m.time=t

fe = m.Param(np.cos(2*np.pi*t)+3)
fh = m.Param(1.5*np.sin(2*np.pi*t)+7)
fz = m.Param(np.clip(-0.2*np.sin(2*np.pi*t),0,None))
de = m.Var(fe[0]); dh=m.Intermediate(de*2)
dz = m.Var(0,lb=0); dze=m.Var(0); dzh=m.Var(0)
te = m.Intermediate(fe+dze); th = m.Intermediate(fh+dzh)

e  = m.CV(0); e.STATUS=1; e.SPHI=e.SPLO=0; e.WSPHI=1000; e.WSPLO=1
h  = m.CV(0); h.STATUS=1; h.SPHI=h.SPLO=0; h.WSPHI=1000; h.WSPLO=1
z  = m.CV(0); z.STATUS=1; z.SPHI=z.SPLO=0; z.WSPHI=1000; z.WSPLO=1
der = m.MV(0,lb=-1,ub=1); der.STATUS=1; der.DCOST=0.0
dzr = m.MV(0,lb=-1,ub=1); dzr.STATUS=1; dzr.DCOST=0.0

m.Equations([de.dt()==der, e==te-de, h==th-dh])
m.Equations([dz.dt()==dzr, z==fz-dz, dze==dz*2, dzh==dz*3])
m.Maximize(dz)

m.options.IMODE=6; m.options.SOLVER=1; m.solve()

plt.figure(figsize=(8,7))
plt.subplot(4,1,1)
plt.plot(t,te,'r-',label='Demand 1')
plt.plot(t,de,'b:',label='Prod 1')
plt.legend(); plt.grid()
plt.subplot(4,1,2)
plt.plot(t,th,'r-',label='Demand 2')
plt.plot(t,dh,'b:',label='Prod 2')
plt.legend(); plt.grid()
plt.subplot(4,1,3)
plt.plot(t,fz,'r-',label='Demand 3')
plt.plot(t,dz,'b:',label='Prod 3')
plt.legend(); plt.grid()
plt.subplot(4,1,4)
plt.plot(t,der,'k:',label='Ramp Rate 1')
plt.plot(t,dzr,'k--',label='Ramp Rate 3')
plt.legend(); plt.grid(); plt.xlabel('Time')
plt.savefig('grid.png',dpi=600)
plt.show()

#### Benchmark IV: Constant Production with Storage

The fourth benchmark problem models a hybrid system with a single generator with constant production constraints coupled with energy storage that together must meet an oscillating electricity demand. The goal of the problem is to minimize the required power production and use energy storage to capture excess generation serve the oscillating energy demand while keeping the base-load generator production constant. As before, the model has perfect foresight. In order to prevent the energy storage from charging and discharging simultaneously without requiring mixed-integer variables, slack variables are used to control when the storage charges and discharges, allowing it to switch modes in a way that is both continuous and differentiable. This allows the modeling language to use automatic differentiation to provide exact gradients to the solver.

\begin{align} \min_g \ \ & g \\ \mathrm{s.t.} & \frac{de}{dt} = e_{in} - e_{out} \cdot \eta \\ & e_{in} = g - d + s_{in} \\ & e_{out} = d - g + s_{out} \\ & g - d = s_{out} - s_{in} \\ & s_{out}, \, s_{in} \geq 0, \enspace e_{out} \cdot e_{in} \leq 0 \\ & g + e_{out}/\eta - e_{in} \geq d \\ & e \geq 0, \enspace \eta = 0.7 \\ & d = 10 - 2 \sin(2 \, \pi \, t) \\ & e(0) = e(1) = 0 \end{align}

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt

m = GEKKO(remote=False)
m.time = np.linspace(0,1,101)

g = m.FV(); g.STATUS = 1 # production
s = m.Var(1e-2, lb=0)    # storage inventory
store = m.Var()          # store energy rate
s_in = m.Var(lb=0)       # store slack variable
recover = m.Var()        # recover energy rate
s_out = m.Var(lb=0)         # recover slack variable
eta = 0.7
d = m.Param(-2*np.sin(2*np.pi*m.time)+10)
m.periodic(s)
m.Equations([g + recover/eta - store >= d,
g - d == s_out - s_in,
store == g - d + s_in,
recover == d - g + s_out,
s.dt() == store - recover/eta,
store * recover <= 0])
m.Minimize(g)

m.options.SOLVER   = 1
m.options.IMODE    = 6
m.options.NODES    = 3
m.solve()

plt.figure(figsize=(6,3))
plt.subplot(2,1,1)
plt.plot(m.time,d,'r-',label='Demand')
plt.plot(m.time,g,'b:',label='Prod')
plt.legend(); plt.grid(); plt.xlim([0,1])

plt.subplot(2,1,2)
plt.plot(m.time,s,'k-',label='Storage')
plt.plot(m.time,store,'g--', label='Store Rate')
plt.plot(m.time,recover,'b:', label='Recover Rate')
plt.legend(); plt.grid(); plt.xlim([0,1])
plt.show()

#### Benchmark V: Load Following with Storage

The fifth benchmark combines energy storage with a load-following problem similar to Benchmark I. The first-half of the time horizon is nearly identical to Benchmark I, but now the excess energy can be stored. This allows the system to meet a higher demand in the second half of the time horizon without needing extremes in generation. The solver minimizes the ramping needs and operates more flexibly by storing and then recovering the overproduction caused by the ramping constraints. Energy storage allows this generator to meet the load without requiring significant overproduction.

\begin{align} \min_{r} \ \ & \Phi = \sum_{i=1}^n \int_{t=0}^{1} \, [ 1000 \, \max(0, \, \Psi) \vphantom{\int} + \, \max(0, \, -\Psi) \vphantom{\int} ] \\ & where \ \Psi = d - g - R + e_{out}/\eta - e_{in} \\ \mathrm{s.t.}\ \ & \frac{de}{dt} = e_{in} - e_{out} \cdot \eta \\ & e_{in} = g + R - d + s_{in} \\ & e_{out} = d - g - R + s_{out} \\ & g + R - d = s_{out} - s_{in} \\ & s_{out}, \, s_{in} \geq 0, \enspace e_{out} \cdot e_{in} \leq 0\\ & g + R + e_{out}/\eta - e_{in} \geq d \\ & e \geq 0, \enspace \eta = 0.85 \\ & d = 7 - 2 \sin(2 \, \pi \, t) \\ & R = \begin{cases} 3 + 3 \cos(4 \, \pi \, t) & \frac{1}{4} \leq t \leq \frac{3}{4} \\ 0 & \mbox{otherwise} \\ \end{cases} \\ & \frac{dg}{dt} = r, \enspace -1\leq r \leq 1 \\ & e(0) = e(1) = 0 \end{align}

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt

m = GEKKO(remote=False)
m.time = np.linspace(0,1,101)

# renewable energy source
renewable = 3*np.cos(np.pi*m.time/6*24)+3
num = len(m.time)
center = np.ones(num)
center[0:int(num/4)] = 0
center[-int(num/4):] = 0
renewable *= center
r = m.Param(renewable)

dg = m.MV(0, lb=-4, ub=4); dg.STATUS = 1
d = m.Param(-2*np.sin(2*np.pi*m.time)+7)
net = m.Intermediate(d-r)
g = m.Var(d[0])       # production
s = m.Var(0, lb=0)    # storage inventory
store = m.Var()       # store energy rate
s_in = m.Var(lb=0)    # store slack variable
recover = m.Var()     # recover energy rate
s_out = m.Var(lb=0)   # recover slack variable
m.periodic(s)
eta = 0.85            # storage efficiency
m.Minimize(g)

err = m.CV(0); err.STATUS = 1
err.SPHI = err.SPLO = 0
err.WSPHI = 1000; err.WSPLO = 1
m.Minimize(0.01*err**2)

m.Equations([g.dt() == dg,
err == d - g - r + recover/eta - store,
g + r - d == s_out - s_in,
store == g + r - d + s_in,
recover == d - g - r + s_out,
s.dt() == store - recover/eta,
store * recover <= 0])

m.options.SOLVER   = 1
m.options.IMODE    = 6
m.options.NODES    = 2
m.solve()

plt.figure(figsize=(7,5))
plt.subplot(3,1,1)
plt.plot(m.time,d,'r-',label='Demand')
plt.plot(m.time,g,'b:',label='Prod')
plt.plot(m.time,net,'k--',label='Net Demand')
plt.legend(); plt.grid(); plt.xlim([0,1])

plt.subplot(3,1,2)
plt.plot(m.time,r,'b-',label='Source')
plt.plot(m.time,dg,'k--',label='Ramp Rate')
plt.legend(); plt.grid(); plt.xlim([0,1])

plt.subplot(3,1,3)
plt.plot(m.time,s,'k-',label='Storage')
plt.plot(m.time,store,'g--', label='Store Rate')
plt.plot(m.time,recover,'b:', label='Recover Rate')
plt.xlim([0,1]); plt.xlabel('Time')
plt.legend(); plt.grid()
plt.show()

#### Benchmark VI: Cogeneration with Dual Storage

The sixth benchmark problem is a combination of Benchmarks II and V where the ramp rate of the generator is the manipulated variable but now must meet both electrical (1) and heat (2) demand with the use of both electrical and thermal storage. A renewable generation source (such as solar PV) is added to the system as an auxiliary electrical energy source that cannot be controlled. The objective is the same as in Benchmark IV, to minimize power production, but this time while meeting both the heat and power demands. The model again has perfect foresight. Slack variables are used in the same way as in Benchmarks IV and V.

\begin{align} \min_{r_1} \ \ & g_1 \\ \mathrm{s.t.}\ \ & \frac{de_i}{dt} = e_{in,i} - e_{out,i}\cdot \eta_i \\ & e_{in,i} = g_i + R_i - d_i + s_{in,i} \\ & e_{out,i} = d_i - g_i - R_i + s_{out,i} \\ & g_i + R_i - d_i = s_{out,i} - s_{in,i} \\ & s_{out,i}, \, s_{in,i} \geq 0, \enspace e_{out,i} \cdot e_{in,i} \leq 0 \\ & g_i + R_i + e_{out,i}/\eta_i - e_{in,i} \geq d_i \\ & e_i \geq 0, \enspace \eta_1 = 0.7, \enspace \eta_2 = 0.8 \\ & d_1 = 10 - 2 \sin(2 \, \pi \, t) \\ & d_2 = 15 + 1.5 \cos(2 \, \pi \, t) \\ & R_1 = \begin{cases} 2 + 2 \cos(4 \, \pi \, t) & \frac{1}{4} \leq t \leq \frac{3}{4} \\ 0 & \mbox{otherwise} \\ \end{cases} \\ & \frac{dg_1}{dt} = r_1, \enspace -3\leq r_1 \leq 3 \\ & R_2 = 0, \enspace g_2 = 1.5 \cdot g_1 \\ & e_1(0) = 0, \enspace e_2(0) = e_2(1) = 0.5 \end{align}

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt

m = GEKKO(remote=False)
t = np.linspace(0, 1, 101)
m.time = t

m.options.SOLVER = 3
m.options.IMODE = 6
m.options.NODES = 2
m.options.CV_TYPE = 1
m.options.MAX_ITER = 1000

p = m.SV(10)         # production (constant)
s = m.Var(0.1, lb=0) # storage inventory
stored = m.SV()      # store energy rate
recovery = m.SV()    # recover energy rate
vx = m.SV(lb=0)      # recover slack variable
vy = m.SV(lb=0)      # store slack variable

eps = 0.85 # Storage efficiency

d = m.MV((-20*np.sin(np.pi*t/12*24)+100)/10)
d_h = m.MV((15*np.cos(np.pi*t/12*24)+150)/10)

p_h_initial = m.Intermediate(p*1.5)

p_h = m.SV(p_h_initial)
s_h = m.Var(0.5,lb=0)
stored_h = m.SV()
recovery_h = m.SV()

#renewable energy source
renewable = (20*np.cos(np.pi*t/6*24)+20)/10
center = np.ones(len(t))
num = len(t)
center[0:int(num/4)] = 0
center[-int(num/4):] = 0
renewable *= center
r = m.Param(renewable)

r1 = m.MV(ub=3,lb=-3)
r1.STATUS=1

m.periodic(s_h)

zx = m.SV(lb=0)
zy = m.SV(lb=0)

eps_h = 0.8 # heat storage efficiency

net = m.Intermediate(d-r)
m.Equations([p + r + recovery/eps - stored >= d,
p + r - d == vx - vy,
stored == p + r - d + vy,
recovery == d - p - r + vx,
s.dt() == stored - recovery/eps,
p.dt() == r1,
stored * recovery <= 0,
p_h + recovery_h/eps_h - stored_h >= d_h,
p_h - d_h == zx - zy,
stored_h == p_h - d_h + zy,
recovery_h == d_h - p_h + zx,
s_h.dt() == stored_h - recovery_h/eps_h,
stored_h * recovery_h <= 0,
p_h == 1.5 * p])
m.Minimize(p)
m.solve()

# Plot solution
fig, axes = plt.subplots(5, 1, figsize=(5, 5.1), sharex=True)
axes = axes.ravel()

ax = axes[0]
ax.plot(t, d, 'r-', label='Demand 1 ($d_1$)')
ax.plot(t, p,'b:', label='Production 1 ($g_1$)',\
linewidth=2)
ax.plot(t, net, 'k--', label='Net ($d_1-R_1$)')

ax = axes[1]
ax.plot(t,r, 'b-',label='Source 1 ($R_1$)')
ax.plot(t,r1, 'k--', label='Ramp Rate ($r$)')

ax = axes[2]
ax.plot(t,s, 'k-', label='Storage 1 ($e_1$)')
ax.plot(t,stored,'g--',label='Stored ($e_{in,1}$)')
ax.plot(t,recovery,'b:',label='Recovered ($e_{out,1}$)',\
linewidth=2)

ax = axes[3]
ax.plot(t,d_h, 'r-', label='Demand 2 ($d_2$)')
ax.plot(t[1:], p_h.value[1:],'b:',\
label='Production 2 ($g_2$)',linewidth=2)

ax = axes[4]
ax.plot(t,s_h, 'k-', label='Storage 2 ($e_2$)')
ax.plot(t,stored_h,'g--',label='Stored ($e_{in,2}$)')
ax.plot(t[1:],recovery_h.value[1:],'b:',\
label='Recovered ($e_{out,2}$)',linewidth=2)
ax.set_xlabel('Time')

for ax in axes:
ax.legend(loc='center left',\
bbox_to_anchor=(1,0.5),frameon=False)
ax.grid()
ax.set_xlim(0, 1)
plt.tight_layout()
plt.savefig('grid_energy6.png', dpi=600,\
bbox_inches = 'tight')
plt.show()

#### Reference

Source code examples from the publication (under review):

• Gates, N.S., Hill, D.C., Billings, B.W., Powell, K.M., Hedengren, J.D., Benchmarks for Grid Energy Management with Python Gekko, 60th IEEE Conference on Decision and Control (CDC), Tutorial Session: Open Source Software for Control Applications, Austin, TX, Dec 13-15, 2021.


text:
# “Close” roots of polynomials and Pillai property

A sequence of integers $a(n)_{n \geq 0}$ has the Pillai property if there exists an integer $G \geq 2$ such that for all integers $k \geq G$ there exists an integer $n \geq 0$ such that none of the $k$ consecutive terms $a(n + 1), a(n + 2), \ldots, a(n + k)$ is relatively prime with all the others.

In [SS] it is proved that the sequence given by $a_n := f(n)$ has the Pillai property, where $f \in \mathbb{Z}[x]$ is any polynomial of degree at most $3$. (Actually, it is proved for degrees $2$ and $3$, the linear case was already proved by Evans [Eva].)

For $f(X) = a_k X^k + \cdots + a_1 X + a_0 \in \mathbb{Z}[X]$, $a_k \neq 0$, define $$\widetilde{f}(X) := a_k^{2k-2} \prod_{\substack{1 \leq i, j \leq k \\ i \neq j}}(X - (\alpha_i - \alpha_j)) ,$$ where $\alpha_1, \ldots, \alpha_j \in \mathbb{C}$ are the roots of $f$.

It seems that the only part of the proof where the hypothesis $\deg f = 2,3$ is used is the following elementary lemma:

Lemma 2.2. Suppose $f \in \mathbb{Z}[X]$ has degree $2$ or $3$, and that $p \mid \widetilde{f}(r)$ for some prime number $p \nmid 6a_k$ and some positive integer $r$. Then there exists an integer $n$ such that $$f(n) \equiv f(n + r) \equiv 0 \pmod p.$$

The authors showed that Lemma 2.2 is indeed false for quartic polynomials, by giving the examples:

• $f(X) = X^4 + 1$, since $7 \mid \widetilde{f}(3)$ but $f(X) \equiv 0 \pmod 7$ has no solutions.

• $f(X) = g(X)g(X + r)$ for a quadratic irreducible $g \in \mathbb{Z}[X]$ and a positive integer $r$, since $\widetilde{f}(r) = 0$ is divisible by all primes, but $f(X) \equiv 0 \pmod p$ has no solutions for infinitely many primes $p$.

My questions are:

1. Do the polynomials $f$ for which Lemma 2.2 fails (that is, $p \mid \widetilde{f}(r)$ for some prime $p \nmid 6a_k$ and some positive integers $r$, but $f(X) \equiv f(X + r) \equiv 0 \pmod p$ has no solutions) have a nice description? In particular, I am interested in irreducible $f$, since the second example $f(X)=g(X)g(X+r)$ is quite disappointing for the problem of Pillai property, being clear that one can always assume $f$ irreducible.

2. Can Lemma 2.2 be extended to higher degree polynomial by assuming further conditions than $p \mid \widetilde{f}(r)$? The hope of course is showing that also polynomials of degree greater than $3$ have the Pillai property.

[SS] Sanna and Szikszai, A coprimality condition on consecutive values of polynomials, Bull. London Math. Soc. (2017) (https://arxiv.org/abs/1704.01738)

[Eva] Evans, On $N$ consecutive integers in an arithmetic progression, Acta Sci. Math. 33 (1972), 295-296.


text:
Story of a mathematics professor

1. Jul 26, 2013

Alpharup

I have hard somewhere the story of mathematics professor(It maybe fake). The math professor was very strong in his concepts and had very good ability to solve math problems. Due to popularity among his peers and other mathematicians, his name spread throughout the entire city where he lived..He was awarded many prizes for his discoveries.

Once a common man approached him in his office.

Common man; I have heard that you are very strong in mathematics .....

Professor; Ya, may be ..

Common man; I think you can answer this question in one second? What is 1778565 multiplied by 167374?

Professor; How could...

(Common man interrupts)

Common man; I have heard that mathematicians can do this in one second. But you could not do
it. What sort of mathematician are you?

Professor; :grumpy:

This is how a common man views mathematicians.This is one of the common misconceptions regarding mathematicians... A common man thinks that one who does arithmetic proficiently is a mathematician..

Could you share some of the misconceptions regarding mathematicians and physicists which the general public has? If you have experienced the same, please share it...

2. Jul 26, 2013

WannabeNewton

Dude my relatives are exactly like that. Every time I meet a relative I haven't seen in a while and they ask me what subject I like most in university and I tell them pure math, they ask me to compute some insanely crazy arithmetic expression in my head. I seriously have no clue what goes on in their heads but I'm 100% sure they have no clue what pure math is. If I showed them my copy of Carothers or Spivak or w\e they would be asking where all the crazy computations are lol

3. Jul 26, 2013

Darth Frodo

I thought you liked physics more?

4. Jul 26, 2013

WannabeNewton

Nah. You can't beat pure math.

5. Jul 26, 2013

Alpharup

For a common person, arithmetic is so important. He should plan all his expenditures, income, etc... very quickly. The recommended math education in many countries is to multiply add, subtract, divide manually. So when math is being discussed, only 6374673+839874984 comes into their mind:rofl:

6. Jul 26, 2013

MathematicalPhysicist

Yes, my dad also thinks that I need to know how to compute a tip in my head in a heartbeat.

7. Jul 26, 2013

lisab

Staff Emeritus
At my work we identify projects by number. We have *lots* of projects every year, so there are *lots* of project numbers floating around in our communications to each other.

I can't remember a project number for longer than, oh, 5 seconds -- they're basically random numbers to me. I always have to look it up, to be sure. Once a coworker remarked to me, "You have to look up project numbers? I thought you were supposed to be good with math!" I was speechless .

8. Jul 26, 2013

Alpharup

Ha ha... Your story is worse than that of professor's:D

9. Jul 26, 2013

MarneMath

I have the same problem. One time I was eating lunch with my friend who is an accountant. We were eating with two people from human resources and an account executive. During lunch, I got a call about a work related problem, I did some dirty math on the back of the napkin to verify what I was being told and hung up. My friend the accountant said that he was terrible at math. One of other people looked at him and said "Then why are you an accountant!" The guy replied, "I can add subtract and find percentages but ask me how to solve a quadratic i'll ask my 14 yr old son!"

10. Jul 26, 2013

1MileCrash

Happens to me all the time. I find it happens the most with other college students, usually girls that I meet. They ask me to multiply two random double digit numbers when I tell them I am a math major. That's when I usually decide that her and I could never be friends or have any kind of meaningful relationship.

Also, very frequently when my dad brags about me to his friends, when I meet them, they immediately do the same thing.

11. Jul 26, 2013

Yanick

Happens to me with my father all the time. I think its common with Eastern Europeans that "being good at Math" means you can do arithmetic quickly and with fairly large numbers. I remember when I got my a non-A grade in Calculus 2 and a week or two later I couldn't do some kind of division or something in my head, my father immediately says, "No wonder you got that bad grade in Math!" I could only smile at the comment.

12. Jul 26, 2013

Staff: Mentor

You people meet a lot of weirdos.

13. Jul 26, 2013

Staff: Mentor

I have a similar story. One day a frind of the coach joined our cross country team for a few runs around our usual course. The coach told he was a ranger and he wanted to maintain his training preparedness. So we kept asking him about bears, wolves and what it was like to be a forest ranger. He just smiled and later we discovered he was a US Army Ranger on leave.

Being a programmer you get similar situations like hey you're a programmer can you fix my computer it's not working.

14. Jul 26, 2013

lisab

Staff Emeritus
Ranger....:rofl:

15. Jul 26, 2013

I_am_learning

I don't find it as much illogical as the maths case though. If you're a programmer you must be spending a lot of time with computers and you sure do know lots more about computers than others. Certainly you can't instantly solve every problem, but I believe computer programmers -- or anyone who spends lots of time with computers-- are better at fixing computer than those who don't. You may not like to, but that's another story.

16. Jul 27, 2013

fargoth

I know at least three programmers, who wouldn't know what to do if something went wrong with their computers, they would ask someone to fix it for them (if it's at home - their spouse, and if it's at work - maintenance).
They all have a masters degree in computer science and the least experienced one amongst them has more than six years of experience (the other two have more than 15 years).
Incidentally, all three are women... I think it has something to do with that - maybe it is not considered a feminine thing to fix computers?

Anyway, on topic - when I was studying physics, people used to ask me If i'm going to be a teacher... apparently they thought that's the only job you can get as a physicist.

17. Jul 27, 2013

Alpharup

When I was young, i too thought that the job of computer professional was to fix computers. Engineering and techical job are often confused. A mechanical engineer is thought to be mechanic, an electrical engineer is thought to be electrician, a civil engineer is thought to be one who mixes cement and concrete during construction. :)

18. Jul 27, 2013

Staff: Mentor

I got the same question a few times.
I think that is really problematic. A significant fraction of all jobs rely heavily on physics and/or physicists and too many are not aware of that.

Compared to that, a minor misconception:
When I say I work for the LHC (outside the physics community), the next question is often "so you work in Geneva / at CERN?". No, I do not. While I have been there multiple times, I did not even see the detectors yet.

19. Jul 27, 2013

I_am_learning

Pardon my little rambling,
People expect Electrical Engineers to be the electrician and ask for household electricity problem. Many (of my knowledge) Electrical Engineers regard themselves far superior than electrician and proudly declare that they don't know how they connect the washing machine to the utility, or similar trivial household problems as they are not electrician but EE.
My opinion is, you may not like to volunteer for such tasks, but being an Electrical Engineer you at least need to know how to. I admit one won't (shouldn't) know all the technicality, but basics?
I don't know much about mechanical Engineering and civil engineering though. Is it also filled with Engineers who are too knowledgeable to know the basics?

However, the OP isn't concerned with knowing the basics, but about doing the basics VERY FAST.

20. Jul 27, 2013

MathJakob

On the topic of solving math problems in your heads, can you be a half decent mathematician if you can't computer 87 * 54 in your head? or 756/6 for example? I hate mental math, in fact I have to write almost every problem down on paper before I can solve it.

Last edited: Jul 27, 2013
21. Jul 27, 2013

reenmachine

Personally , I find multiplications much easier to do in my head than divisions.At least with multiplications of two integers you know the answer is another integer.

Last edited: Jul 27, 2013
22. Jul 27, 2013

1MileCrash

lol, I can say with near certainty that not being able to do arithmetic in your head like that will have zero impact on your career as a mathematician.

Math quickly evolves out of computing things and into concepts and objects and properties and things like that. Any time I needed to do any type of arithmetic like that, I was in a class that allowed calculators, and any time I was in a class that did not allow calculators, I never had to do arithmetic.

Saying to a mathematician

"OK, you've proved this conjecture very nicely, but can you multiply 63*98 in your head right now?"

Is like saying to a electrical engineer "OK, I'm liking this circuit design. Now take this shovel out back and dig up some silicon."

23. Jul 27, 2013

fourier jr

I wish I could find the graph of how hard it is for someone doing something like functional analysis do teach calculus. maybe it was phd comics but I can't find it.

24. Jul 27, 2013

D H

Staff Emeritus

25. Jul 27, 2013

collinsmark

To this day I have yet to successfully come up with a description of what I do at my job, that most people can understand. I've tried several approaches of explanation. Still no success.

In particular, I've tried dozens of times to explain to my mom what I do at work (being my mom, she wants to herself so she can tell others, if for no other reason). I've tried and failed every time.

I've even tried to use analogous examples using more familiar systems. "Okay, suppose you have a bathroom scale that lacks calibration data." <blank stares> "Okay, that didn't work. Well, suppose you set the thermostat to your house to some temperature, but the furnace always makes the temperature 5 degrees hotter than the thermostat setting." <hesitant nods> "There's a dial on the inside of the thermostat box that can compensate for the difference by adding an offset to the temperature selection setting, thereby ..." <blank stares>

There is some truth to that, perhaps. But it's not necessarily out of arrogance. Sometimes it's just out of laziness.

This scenario has happened to me several times:

PERSON: Hey, collinsmark, I heard you have some sort of electrical degree! So do you wire houses?
ME: Well, It's actually a Master's degree in Electrical Engineering.
PERSON: Oh, so you wire bigger houses?
ME: Umm, no, it really doesn't ..."
PERSON: I've got some electrical wiring problem with my ceiling fan. The fan works, but there's no power to the light.
ME: My degree's emphasis was in communication systems theory and digital signal processing.
PERSON: <blank stare.> So will you fix my ceiling fan?
ME: <considers mentioning something about not having an electrician's license, but then realizes how that would open up a whole new line of questioning.>
ME: Okay, I'll fix your fan.

[Edit: Come to think of it, there's a whole swath of Electrical Engineering specific subject matter that really has nothing to do with electricity.]

Last edited: Jul 27, 2013


text:
chapter  5
34 Pages

## Combinatorial Problems in Algorithmic Differentiation

Conceptually, in Algorithmic (also known as Automatic) Differentiation (AD) we consider multivariate vector functions F : Rn → Rm mapping a vector of independent inputs x ∈ Rn onto a vector of dependent outputs y ∈ Rm as y = F (x). For the purpose of the current chapter the function F is assumed to be twice continuously differentiable in a neighborhood of all

arguments at which it is supposed to be evaluated. Hence, its Jacobian

∇F (x) = ( ∂yj ∂xi

∈ Rm×n

and Hessian

∇2F (x) = (

∂2yj ∂xi∂xk

∈ Rm×n×n

exist at all of these points. The Hessian ∇2F (x) ≡ H = (hj,i,k)j=1,...,mi,k=1,...,n is a symmetric 3-tensor, that is, hj,i,k = hj,k,i.


text:
## Thinking Mathematically (6th Edition)

$6\oplus 5=3$.
Consider the hour clock and find the value of given expression. Start with and move steps in a clockwise direction as shown. $6\oplus 5=3$. Also, from the table of the given clock, it can be seen that solution is .


text:
It looks like you have missed for the "rospack find MYROBOT_description". Things inside will be executed by bash first and output the result to replace the command. You should modify your command to this as the tutorial gives:
rosrun gazebo_ros spawn_model -file rospack find MYROBOT_description/urdf/MYROBOT.urdf -urdf -x 0 -y 0 -z 1 -model MYROBOT

And remember to source your project first so that bash can find out the package location. And I think you can try running $roscore and see if the "Unable to register with master node [http://localhost:11311]" error occurs again or not. Normally, roslaunch will run roscore automatically. It looks like you have missed  for the "rospack find MYROBOT_description". Things inside inside it will be executed by bash first and output the result to replace the command. You should modify your command to this as the tutorial gives: rosrun gazebo_ros spawn_model -file rospack find MYROBOT_description/urdf/MYROBOT.urdf -urdf -x 0 -y 0 -z 1 -model MYROBOT  And remember to source your project first so that bash can find out the package location. And I think you can try running$roscore and see if the "Unable to register with master node [http://localhost:11311]" error occurs again or not. Normally, roslaunch will run roscore automatically.


text:
# If I is the Unit Matrix of Order 2 X 2. Find the Matrix M Such That 5m + 3i = 4(2, -5),(0, -3) - Mathematics

Sum

If I is the unit matrix of order 2 x 2. Find the matrix M such that

5M + 3I  = 4[(2, -5),(0, -3)]

#### Solution

5M + 3I  = 4[(2, -5),(0, -3)]

5M = 4[(2, -5),(0, -3)] - 3[(1, 0),(0, 1)]

5M =  [(8, -20),(0, -12)] - [(3,0),(0, 3)]

5M = [(5, -20),(0, -15)]

M = 1/5[(5, -20), (0, -15)] = [(1, -4),(0, -3)]

Concept: Addition and Subtraction of Matrices
Is there an error in this question or solution?

#### APPEARS IN

Selina Concise Maths Class 10 ICSE
Chapter 9 Matrices
Exercise 9 (B) | Q 10.2 | Page 122


text:
# Atomic explanation of magnetic field [duplicate]

We all know how an electric current causes magnetic fields. But when a single electron is flowing through the wire, how is this electron creating the magnetic field or how are electrons responsible for magnetic field around a wire?

• The idea of "a single electron flowing through a wire" makes about as much sense as a single molecule of water flowing through a pipe. – Solomon Slow May 15 '15 at 2:27

I could give to you only an answer to some part of your question. The comment from gonenc points to an answer, there the first sentence is:

If you are not well-acquainted with special relativity, there is no way to truly explain this phenomenon.

There is a simple explanation how accelerated electrons create a magnetic field. It has to do with the intrinsic spin and the related magnetic dipole moment of the electrons. If you are not familiar with these phenomena of electrons please read Wikipedia, search in StackExchange or ask in separate questions.

Now, if you are familiar with the basics please read my explanations about how works the phenomenon of electromagnetic induction.

This is only the first step to answer your question. Now you could imagine how a accelerated charge - for example an electron in a coil - induces a magnetic field. How the magnetic field in a straight wire get induced I have explained in other posts.

But my posts were greeted with scepticism. I've no problem with that. What disturbs me is the general reaction that they are "wrong". My question is summarised as follows: Did any of these people actually follow my explanations and can anyone tell me where a mistake was made? (This sentences are adopted from a question of @G. 't Hooft.)

How are electrons responsible for magnetic field around a wire?

Via the "screw" nature of electromagnetism. Minkowski referred to this in Space and Time, as did Maxwell in On Physical Lines of Force: "a motion of translation along an axis cannot produce a rotation about that axis unless it meets with some special mechanism, like that of a screw". This why we have the right-hand rule which applies not just to electromagnetism, but to screw threads. Note section 11.10 of Jackson's Classical Electrodynamics where he says "one should properly speak of the electromagnetic field Fμν rather than E or B separately". Like Christoph said in the possible duplicate, the electron doesn't actually have an electric field, it has an electromagnetic field. And to envisage this field you take a tip from Maxwell's combination of curl and convergence:

Combine radial electric field lines with concentric magnetic field lines to give electromagnetic field lines like so:

The above is a flat depiction with no spin ½ dipole, but I think it's enough to get across this screw nature of the electromagnetism. It's also reminiscent of depictions of a vector field and a spinor, and something like the gravitomagnetic field, which is described as twisted space by NASA author Tony Phillips. The electromagnetic field isn't totally different, see Heaviside's 1893 paper. IMHO you get the gist of it by imagining you're walking into it with your arms outstretched: think of the electromagnetic field as a "twist" field, wherein if you were moving through it but didn't know it, you might say it was a "turn" field. That's kind of what a magnetic field is. We talk of curl aka rot which is short for rotor, and if you throw an electron through a magnetic field, it goes round in circles.

Anyway, in the current-in-the-wire you have in essence a column of electrons and a column of metal ions. The electrons all have a negative electromagnetic field, and the ions all have a positive electromagnetic field with the opposite chirality. If the electrons weren't moving, the opposite "twist fields" would cancel each other out$^*$. However the electrons are moving up the wire:

So it's like you're moving through one set of twist fields but not the other. And when you have motion relative to a twist field, you think of it as a turn field, and that's what a magnetic field is. So what you "see" is a magnetic field around the wire.

GNUFDL image by Jfmelero see Wikipedia

$*$ Actually they don't quite cancel each other out, but that's one for another day.


text:
# Why doesn't the $x^k$ function series converge in -1?

I'm studying power series and their radius of convergence, and my book says

$$\sum_{k\ge0} x^k$$

has a convergence radius of 1 and doesn't converge in 1 and -1, as opposite to

$$\sum_{k\ge1} \frac {1}{k} x^k$$

which has a convergence radius of 1 too, but converges in -1 as well and diverges in 1.

To my understanding, the latter converges in -1 because we keep adding and subtracting numbers and the sum doesn't ever get much big, while if we have 1 we have the $\sum_{k\ge1} \frac {1}{k}$ number series which is a known diverging one.

Doesn't the same apply to the former series? I mean, if we have $x=1$ it is pretty obvious why it would diverge, but with -1 wouldn't we keep adding and subtracting 1, thus never diverging?

Is it that the infinite jumping between 1 and 0 that the sum would do is not considered convergence, while not being divergence, because a limit does not exist, as opposite to the second series in which the added/subtracted numbers get infinitely smaller thus getting closer and closer to a fixed value?

• Is $s_n=(-1)^n$ a converging sequence?
– Surb
Apr 4, 2018 at 6:02
• This is what I was wondering. For sure it's not a diverging one, but a limit does not exist... So perhaps it's neither converging not diverging? Apr 4, 2018 at 6:03
• Definitely it is not converging, then what is diverging is another question.
– Surb
Apr 4, 2018 at 6:04
• Anyway, the radius of convergence describes convergence :). And the series, as you noticed, cleary does not converge at $-1$.
– Surb
Apr 4, 2018 at 6:06
• Ah Thanks, but I think the other answers answer already quite well your question.
– Surb
Apr 4, 2018 at 6:09

## 2 Answers

Using the $n$th term test, a requirement for a convergent series is:

$$\lim_{n\to\infty} a_n = 0$$

If not, then the series diverges. The limit, in this case, the doesn't exist as it oscillates between $1$ and $-1$ (and its partial sums $1$ and $0$), the series diverges:

$$\lim_{n\to\infty} (-1)^n = \text{DNE}$$

Divergence does not necessarily mean becoming really large, it just means it does not approach a value. Divergence essentially means the lack of convergence.

Diverging means not converging. You're right that the sum $1-1+1-1+\dots$ doesn't explode to infinity, but it does not converge since it's not getting closer to any particular value.