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# New paper on neural networks

Michael Buice and I have a new paper in Frontiers in Computational Neuroscience as well as on the arXiv (the arXiv version has fewer typos at this point). This paper partially completes the series of papers Michael and I have written about developing generalized activity equations that include the effects of correlations for spiking neural networks. It combines two separate formalisms we have pursued over the past several years. The first was a way to compute finite size effects in a network of coupled deterministic oscillators (e.g. see here, herehere and here).  The second was to derive a set of generalized Wilson-Cowan equations that includes correlation dynamics (e.g. see here, here, and here ). Although both formalisms utilize path integrals, they are actually conceptually quite different. The first formalism adapted kinetic theory of plasmas to coupled dynamical systems. The second used ideas from field theory (i.e. a two-particle irreducible effective action) to compute self-consistent moment hierarchies for a stochastic system. This paper merges the two ideas to generate generalized activity equations for a set of deterministic spiking neurons.

# Paper on compressed sensing and genomics

New paper on the arXiv. The next step after the completion of the Human Genome Project, was the search for genes associated with diseases such as autism or diabetes. However, after spending hundreds of millions of dollars, we find that there are very few common variants of genes with large effects. This doesn’t mean that there aren’t genes with large effect. The growth hormone gene definitely has a large effect on height. It just means that variations of genes that are common among people have small effects on the phenotype. Given the results of Fisher, Wright, Haldane and colleagues, this was probably expected as the most likely scenario and recent results measuring narrow-sense heritability directly from genetic markers (e.g. see this) confirms this view.

Current GWAS microarrays consider about a million or two markers and this is increasing rapidly. Narrow-sense heritability refers to the additive or linear genetic variance, which means the phenotype is given by the linear model $y= Z\beta + \eta$, where $y$ is the phenotype vector, $Z$ is the genotype matrix, $\beta$ are all the genetic effects we want to recover, and $\eta$ are all the nonadditive components including environmental effects. This is a classic linear regression problem. The problem comes when the number of coefficients $\beta$ far exceeds the number of people in your sample, which is the case for genomics. Compressed sensing is a field of high dimensional statistics that addresses this specific problem. People such as David Donoho, Emmanuel Candes and Terence Tao have proven under fairly general conditions that if the number of nonzero coefficients are sparse compared to the number samples, then the effects can be completely recovered using L1 penalized optimization algorithms such as the lasso or approximate message passing. In this paper, we show that these ideas can be applied to genomics.

Here is Steve Hsu’s summary of the paper

Application of compressed sensing to genome wide association studies and genomic selection

(Submitted on 8 Oct 2013)

We show that the signal-processing paradigm known as compressed sensing (CS) is applicable to genome-wide association studies (GWAS) and genomic selection (GS). The aim of GWAS is to isolate trait-associated loci, whereas GS attempts to predict the phenotypic values of new individuals on the basis of training data. CS addresses a problem common to both endeavors, namely that the number of genotyped markers often greatly exceeds the sample size. We show using CS methods and theory that all loci of nonzero effect can be identified (selected) using an efficient algorithm, provided that they are sufficiently few in number (sparse) relative to sample size. For heritability h2 = 1, there is a sharp phase transition to complete selection as the sample size is increased. For heritability values less than one, complete selection can still occur although the transition is smoothed. The transition boundary is only weakly dependent on the total number of genotyped markers. The crossing of a transition boundary provides an objective means to determine when true effects are being recovered. For h2 = 0.5, we find that a sample size that is thirty times the number of nonzero loci is sufficient for good recovery.

 Comments: Main paper (27 pages, 4 figures) and Supplement (5 figures) combined Subjects: Genomics (q-bio.GN); Applications (stat.AP) Cite as: arXiv:1310.2264 [q-bio.GN] (or arXiv:1310.2264v1 [q-bio.GN] for this version)

# Happiness and divisive inhibition

The Wait But Why blog has an amusing post on why Generation Y yuppies (GYPSYS) are unhappy, which I found through the blog of Michigan economist  Miles Kimball. In short, it is because their expectations exceed reality and they are entitled. What caught my eye was that they defined happiness as “Reality-Expectations”. The key point being that this is a subtractive expression. My college friend Peter Lee, now Professor and Director of the University Manchester X-Ray imaging facility, used to define happiness as “desires fulfilled beyond expectations”. I always interpreted this as a divisive quantity, meaning “Reality/Expectations”.

Now, the definition does have implications if we actually try to use it as a model for how happiness would change with some quantity like money. For example, consider the model where reality and expectations are both proportional to money. Then happiness = a*money – b*money. As long as b is less than a, then money always buys happiness, but if a is less than b then more money brings more unhappiness. However, if we consider the divisive model of happiness then happiness = a*money/ b*money = a/b and happiness doesn’t depend on money at all.

However, the main reason I bring this up is because it is analogous to the two possible ways to model inhibition (or adaptation) in neuroscience. The neurons in the brain generally interact with each other through two types of synapses – excitatory and inhibitory. Excitatory synapses generally depolarize a neuron and make its potential get closer to threshold whereas inhibitory neurons hyperpolarize the neuron and make it farther from threshold (although there are ways this can be violated). For neurons receiving stationary asynchronous inputs, we can consider the firing rate to be some function of the excitatory E and inhibitory I inputs. In subtractive inhibition, the firing rate would have the abstract form f(E-I) whereas for divisive inhibition it would have the form f(E)/(I+C), where f is some thresholded gain function (i.e. zero below threshold, positive above threshold) and C is a constant to prevent the firing rate from reaching infinity. There are some critical differences between subtractive and divisive inhibition. Divisive inhibition works by reducing the gain of the neuron, i.e. it makes the slope of the gain function shallower while subtractive inhibition makes the threshold effectively higher. These properties have great computational significance, which I will get into in a future post.

# The cost of the shutdown and sequester

People may be wondering how the US government shutdown is affecting the NIH. I can’t speak for the rest of the institutes but I was instructed to not come to work and to not use my NIH email account or NIH resources. Two new fellows, who were supposed to begin on Oct 1, now have to wait and they will not be compensated for the missed time even if Congress does decides to give back pay to the furloughed employees. I really was hoping for them to start in August or September but that was pushed back because of the Sequester (have people forgotten about that?), which cut our budgets severely. In fact, because of the Sequester, I wasn’t able to hire one fellow because the salary requirements for their seniority exceeded my budget. We were just starting to get some really interesting psychophysics results on ambiguous stimuli but that had to be put on hold because we couldn’t immediately replace fellow Phyllis Thangaraj, who was running the experiments and left this summer to start her MD/PhD degree at Columbia. Now it will be delayed even further. I have several papers in the revision process that have also been delayed by the shutdown. All travel has been cancelled and I heard that people at conferences were ordered to return immediately, including those who were on planes on Oct 1. My quadrennial external review this week has now been postponed. All the flights for the committee and ad hoc members have to be cancelled and we now have to find another date where 20 or more people can agree on. All NIH seminars and the yearly NIH research festival has been cancelled. I was supposed to review an external NIH research proposal this week and that has been postponed indefinitely along with all other submitted proposals awaiting review. Academic labs, students and postdocs depending on their NIH grants this fiscal year will be without funding until the government is reopened. Personally, I will probably come out of this reasonably intact. However, I do worry how this will affect young people, who are the future.

# Heritability and additive genetic variance

Most people have an intuitive notion of heritability being the genetic component of why close relatives tend to resemble each other more than strangers. More technically, heritability is the fraction of the variance of a trait within a population that is due to genetic factors. This is the pedagogical post on heritability that I promised in a previous post on estimating heritability from genome wide association studies (GWAS).

One of the most important facts about uncertainty and something that everyone should know but often doesn’t is that when you add two imprecise quantities together, while the average of the sum is the sum of the averages of the individual quantities, the total error (i.e. standard deviation) is not the sum of the standard deviations but the square root of the sum of the square of the standard deviations or variances. In other words, when you add two uncorrelated noisy variables, the variance of the sum is the sum of the variances. Hence, the error grows as the square root of the number of quantities you add and not linearly as it had been assumed for centuries. There is a great article in the American Scientist from 2007 called The Most Dangerous Equation giving a history of some calamities that resulted from not knowing about how variances sum. The variance of a trait can thus be expressed as the sum of the genetic variance and environmental variance, where environment just means everything that is not correlated to genetics. The heritability is the ratio of the genetic variance to the trait variance.

# Government shutdown

As of today, I am officially furloughed without pay since the NIH is officially closed and nonessential employees like myself are barred from working without pay by the Antideficiency Act of 1884. However, given that blogging is not considered an official duty, I can continue to post to Scientific Clearing House. Those who are not up on American politics may be wondering why the US government has shutdown. The reason is that the US fiscal year begins on Oct 1 and according to the the US Constitution, only Congress can appropriate funds for the functioning of government and they did not pass a budget for the new fiscal year by midnight of September 30. Actually, Congress has not passed a budget on time in recent years but has passed Continuing Resolutions that to keep the government going. So why have they not passed a budget or a CR this year? Well, currently the US government is divided with the Democratic party controlling the Senate and Presidency and the Republican party controlling the House of Representatives. All three entities must agree for a law to pass. Three years ago, the Democrats controlled the Congress, which includes both the House and Senate, and passed the Affordable Care Act, also known as Obamacare, which the President signed into law.   The Republicans took control of the House in 2011 and have been trying to repeal the ACA ever since but have been stopped by the Senate. This year they decided to try a new tactic, which was to pass a budget that withholds funding for the ACA. The Senate did not agree, passed a budget with the ACA and sent it back to the House, which then took out funding for the ACA again with some modifications and sent it back. This went on back and forth without converging to an agreement and thus we are closed today.


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# Meeting Details

Title: Black-Scholes type formulas for European options Probability and Financial Mathematics Seminar Manfred Denker, PSU Mathematics The classical Cox-Ross-Rubinstein model assumes that the option price of a risky asset jumps up by a factor of $u$ or drops down by a factor $d$ in one time unit. We take a more general viewpoint that one has several possibilities of this type, and derive the option price as a function of the jump sizes and the probability of occurrences of these hedging options. This is done as in the classical case as a space time approximation of a continuous limit model. The advantage of such a model is that the error bounds of the option price are better to control.

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# R^2>=Sin2Theta, Solved Find The Area Of The Region That Is Above The X Ax Chegg Com

R^2>=Sin2Theta. The graphs are different too. Find the center and radius. How do you graph #r=(12)/(3+4 cos theta)#? Theta can be 45 degrees or it can be 315 degrees. Degree) 2sin theta and sin2 theta are different.we can also prove this by putting other values.but if theta becomes zero degree they become same. ** use x in place of theta. If two polar curves have different equations, can they be the same curve? A2 = integral{(1/2)sin(2*t)dt} evaluated from t = 0 to pi/8. Compute answers using wolfram's breakthrough technology & knowledgebase, relied on by millions of students & … How do you graph #x^2 + y^2 = 16# and #r = 4#? After that, i computed for the area of the two full petals at the bottom. You subtract when one is inside of the other, when two curves are like that , you add them. Right)+{1 \over r\sin \theta }{\partial a_{\varphi } \over \partial \varphi }}. Transform the polar equation r=2sin(theta) into a rectangular equation. See all questions in more with polar curves.

R^2>=Sin2Theta Indeed recently has been sought by consumers around us, perhaps one of you personally. People now are accustomed to using the internet in gadgets to see image and video information for inspiration, and according to the title of the article I will discuss about R^2>=Sin2Theta.

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## An angle of a right triangle has a cota… sec(theta) = _____.

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# Lookup table logarithmic.

Status
Not open for further replies.

#### dr pepper

##### Well-Known Member
I'm having issues with pic16 lookup tables, I'm working in asm.

I have a pot wired to an analogue in on a pic16f88, its a linear pot.
I want to select either a linear function or log function as per the old fashioned log volume pots.
Linear is no problem I can just use the value direct from the atod module, to convert lin to log I was thinking of using a lookup table, my maths are a bit rusty, anyone know where I can find a 256 element lookup table for the purpose?

Also I found with the dt function if I have more than 1 line I get errors from the assembler, can I have a 256 element table that isnt all on one line, I'm sure I've done it before but I cant find that program.

#### Nigel Goodwin

##### Super Moderator
Funnily enough that's near enough the exact same question I asked when I first joined here

I was programming a PIC to work like an LM3915/4, switch-able between log and linear.

No one was prepared to help, so I wrote a small Delphi program that generated a log table of the switching points for each LED - I then used that in the program, along with a linear table, selected by a switched pin.

Unfortunately, both the Delphi program and the PIC source have long since been lost

As for a table longer than 256 words, check my tutorials (specifically the 8x8 LED matrix one).

#### dr pepper

##### Well-Known Member
This is my routine that contains a 255 byte lookup table, the table is assembled at 0x200 using the org directive (0x1fe).
The table is all on one line, can I put the table on several lines and use 'dt' on each line?
My code doesnt run properly, that might be an issue elsewhere.
Is crossing the page boundary a nono just after assing to pcl, as I understand it pclath is added to the high byte so this should work, it does on mpsim.
I could lose the last 2 elements in the table wouldnt make much difference.
The table is in decimal, I did use R=dec directive at the start of the code.

Code:
;
;-------------------------------lookup conversion to log table--------------------------
;
logcon    movlw    d'2'
movwf    pclath
;
org        0x1fe
;
;
dt    0,0,0,0,0,1,1,1,1,1,2,2,2,2,3,3,3,4,4,5,5,5,6,6,7,7,8,9,9,10,10,11,12,12,13,14,14,15,16,17,17,18,19,20,21,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,40,41,42,43,44,45,47,48,49,50,52,53,54,55,57,58,59,61,62,63,65,66,67,69,70,72,73,74,76,77,79,80,82,83,85,86,88,89,90,92,93,95,97,98,100,101,103,104,106,107,109,110,112,113,115,117,118,120,121,123,124,126,128,129,131,132,134,135,137,138,140,142,143,145,146,148,149,151,152,154,155,157,158,160,162,163,165,166,167,169,170,172,173,175,176,178,179,181,182,183,185,186,188,189,190,192,193,194,196,197,198,200,201,202,203,205,206,207,208,210,211,212,213,214,215,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,234,235,236,237,238,238,239,240,241,241,242,243,243,244,245,245,246,246,247,248,248,249,249,250,250,250,251,251,252,252,252,253,253,253,253,254,254,254,254,254,255,255,255,255,255,255,255,255,255,255,255,255,255
;
end

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#### Nigel Goodwin

##### Super Moderator
Like I said, check my tutorials for long tables, specifically tutorial 13.4

Last edited:

#### dr pepper

##### Well-Known Member
Can someone please explain how I search tutorials, I'm having a useless moment.

#### large_ghostman

##### Well-Known Member
Can someone please explain how I search tutorials, I'm having a useless moment.

WELCOME to my world!

#### dr pepper

##### Well-Known Member
This is more difficult than I thought.
I bodged it.
Reducing the table to 64 entries and dividing the atod result by 4 to suit means I can nick someone elses code which has a 64 byte long log lookup.
Heres my altered code, I still dont seem to be able to get dt to work for more than 1 line of data entries:
LG, your own world can be a good thing, mines nice and quiet.

Code:
;
;-------------------------------lookup conversion to log table--------------------------
;
logcon        org    0x200            ;shift assy to 0x200
;
movlw    d'2'            ;setup pclath to correct page
movwf    pclath
;
movf    adresh,w        ;transfer atod result to temp and divide by 4
movwf    temp
bcf    status,c
rrf    temp,f
bcf    status,c
rrf    temp,f
;
;
dt    0,  2,  3,  4,  4,  5,  6,  6,  7,  8     ;64 logarithmic table entries
9, 10, 11, 12, 13, 15, 16, 17, 19, 20
22, 24, 26, 28, 30, 32, 34, 37, 39, 42
45, 48, 51, 54, 57, 61, 65, 68, 72, 77
81, 86, 91, 96,101,107,112,118,125,131
138,145,152,160,168,176,185,194,203,213
223,234,244,256
;
end

#### Nigel Goodwin

##### Super Moderator
Heres my altered code, I still dont seem to be able to get dt to work for more than 1 line of data entries:
Then use a dt on each line.

#### ChrisP58

##### Well-Known Member
I used Excel to generate a sine wave lookup table. After doing all of the number crunching and scaling, I converted the output to Hex values and then formatted it into three columns that I just copied and pasted into MPLAB.

The three columns were:
1: empty space
2: The text "retlw"
3: The value of that line in Hex.

The result looked like:
Code:
    org        0x0200
SineTable:
movlw    high SineTable     ; get high order part of the beginning of the table
movwf    PCLATH
movf     pointer,w
nop
retlw    0x00    ; beginning of paste from excel
retlw    0x03
retlw    0x06
retlw    0x09
retlw    0x0C
retlw    0x10
retlw    0x13
retlw    0x16
retlw    0x19
retlw    0x1C
retlw    0x20

etc. for 250 lines

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#### dr pepper

##### Well-Known Member
Yes figured out the dt in each line, what confused me there was in the past I have use dt for text strings and a different string on each line, but this is data.

I'm not very good with excel, having to use it at work puts me off using it at home too.

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text:
# Proofing the derivatives of e^x from the limit approach

1. May 3, 2012

### ibelive

I was searching for the proof of $$\frac{d}{dx} e^x = e^x$$.
and I found one in yahoo knowledge saying that

$$\frac{d}{dx} e^x = \lim_{Δx\to 0} \frac {e^x(e^{Δx}-1)} {Δx}$$

$$= \lim_{Δx\to 0} \frac {e^x [\lim_{n\to\infty} (1+ \frac{1}{n})^{n(Δx)}-1]} {Δx}$$

Let $$h= \frac {1}{n}$$, So that $$n = \frac {1}{h}$$

That makes
$$\lim_{Δx\to 0} \frac {e^x * [\lim_{h\to 0} (1+ h)^{\frac {1}{h}(Δx)}-1]} {Δx}$$

And it said variable h of the power $$\frac{1}{h}$$ and that of the limit $$\lim_{h\to 0}$$ of $$e^{Δx} = \lim_{h\to 0} (1+ h)^{\frac {1}{h}}$$ can be replaced by $$Δx$$, to make the whole formula becomes
$$= \lim_{Δx\to 0} \frac {e^x * [\lim_{Δx\to 0} (1+ Δx)^{\frac {1}{Δx} * {Δx}}-1]} {Δx}$$

because both variables are going to zero and that they can "replace" each other.

Of course, the calculation can still come to the answer that $$\frac{d}{dx} e^x = e^x$$.

But I wanna ask if it is true that, two different variables(neither of it is a function of another) approaching to the same value can be "replaced" by each other, or in other words, equal?
If that's true, what are the situations that this kind of "replacements" can't be done.
Also, if that's not true,is there any other way to prove that the derivatives of e^x =e^x without the use of any logarithms?

Last edited: May 3, 2012
2. May 3, 2012

### hamsterman

Not in general, I think. For example, $\lim \limits_{x \rightarrow 0} \frac { \lim \limits_{y \rightarrow 0} y } {x}$ is undefined. If you picked y = x, you'd get 1. You could, however, pick y = 2x as that too satisfies the arrow.
I'm not sure why it was done here.

Anyway, e^x can be derived if you just apply l'Hopital to the definition of derivative (your first line).

3. May 3, 2012

### tiny-tim

welcome to pf!

hi ibelive! welcome to pf!

(try using the X2 button just above the Reply box )

d/dx ( ∑ xn/n! ) = … ?

4. May 3, 2012

### dextercioby

Yes, it's a tricky subject. I asked myself the same question on a different thread a few days ago. Anyway, you cannot use the series expansion (MacLaurin), unless you define e^x by the infinite sum.

5. May 3, 2012

i do!

6. May 3, 2012

### lavinia

If you define the exponential as the inverse function of the integral of 1/x from 1 to x, everything follows immediately from the Chain Rule.

the equation is

x = $\int^{exp(x)}_{1}$1/xdx

Differentiating wrsp to x shows that exp(x) is its own derivative. This though is using the logarithm without giving it a name.

If you could show somehow that the derivative of exp is 1 at 0 then the rest follows instantly from the Newton quotient.

i.e. exp'(x) = exp(x)lim h -> 0 (exp(h) - 1)/h

So exp'(x) = exp(x)exp'(0)

Last edited: May 3, 2012
7. May 3, 2012

### dextercioby

This is interesting. But how would you prove it ?

8. May 3, 2012

### HallsofIvy

Staff Emeritus
That's not quite correct. What you say is correct for
$$\lim_{(x,y)\to (0, 0)} \frac{y}{x}$$

But what you wrote,
$$\lim_{x\to 0}\frac{\lim_{y\to 0} y}{x}= 0$$
specifically takes the limit as y goes to 0 first, then takes the limit, as x goes to 0, of 0/x which is 0 for all non-zero x and so goes to 0.

But applying L'Hopital to the first line requires taking the derivative of $e^x$ doesn't it?

Last edited: May 3, 2012
9. May 3, 2012

### HallsofIvy

Staff Emeritus
Are you refering to Lavinia's post?

First define
$$ln(x)= \int_1^x \frac{1}{x} dx$$
Since 1/x is defined and continuous for all x except 0, it follows that this ln(x) exists, is continous and differentiable for all positive x. The derivative is, from the Fundamental Theorem of Calculus, 1/x. Since that derivative is positive, ln(x) is an increasing function.

By appropriate substitutions one can show that, for x and y both positive, ln(1/x)= -ln(x), ln(xy)= ln(x)+ ln(y), and, for x positive, $ln(x^y)= yln(x)$.

In particular, since ln(x) is differentiable for all positive x, it is, in particular, on the closed interval [1, 2] and we can apply the mean value theorem on that interval- there exist c in [1, 2] such that
$$\frac{ln(2)- ln(1)}{2- 1}= ln(2)= 1/c$$
And since $1\le c\le 2$, $1/2\le 1/c\le 1$. That is, $ln(x)\ge 1/2$. From that it follows that, for X any positive number, $ln(2^{2X})= (2X)ln(2)\ge (2X)(1/2)= X$. That is, ln(x) is unbounded above and, since it is increasing, $\lim_{x\to\infty} ln(x)= \infty$. From ln(1/x)= -ln(x), we have $\lim_{x\to 0} ln(x)= -\infty$.

That tells us that ln(x) maps the set of all positive numbers one-to-one and onto the set of all real numbers. That, in turn, means that it has an inverse function which we can call "exp(x)", a differentiable function from the set of all real numbers to the set of all positive numbers.

If y= exp(x), then x= ln(y) and so dx/dy= 1/y from which it follows that $dy/dx= y= e^x$.

Of course, I've just "called" that exp(x) and it is true that I have not shown that this is an "exponential". What we need is this:

Suppose y= exp(x). Then x= ln(y). If x is not 0, we can divide on both sides and have $1= (1/x)ln(y)= ln(y^{1/x})$. Going back to the exponential form, $y^{1/x}= exp(1)$ and so $y= exp(x)= (ln(1))^x$. If we define e= ln(1), we have $exp(x)= e^x$.

10. May 3, 2012

### lavinia

Well all of the exponential functions f(x) = exp(ax) satisfy the relations

f(0) = 1 and f(x + y) =f(x)f(y).

11. May 3, 2012

### HallsofIvy

Staff Emeritus
Yes. You need to define e as f(1).

12. May 3, 2012

### lavinia

This is all fine but more simply if you differentiate the equation

x = $\int^{exp(x)}_{1}$1/xdx

you get 1 on the left side and by the Chain Rule you get exp'(x)/exp(x) on the right.

13. May 3, 2012

### HallsofIvy

Staff Emeritus
Yes, but you still need to show that "exp(x)" really is a number to the x power.

14. May 3, 2012

### lavinia

I don't see this. Defining e as f(1) just gives different definitions of e as f varies. Unless you know e beforehand by some other definition. If you did then how would the proof go?

15. May 3, 2012

### lavinia

Differentiating both sides gives you exp(x) = exp'(x)

I guess you are saying that you still need to show that exp(x + y) = exp(x)exp(y).

The same method works I think.

The derivative of f(x) = $\int^{exp(x)exp(y)}_{1}$1/xdx is again by the Chain Rule,

[1/exp(x)exp(y)]exp(x)exp(y) = 1 so f(x) = x + c since both functions have the same derivative. Setting x to zero shows that c = y.

But x + y = $\int^{exp(x + y)}_{1}$1/xdx

Last edited: May 3, 2012
16. May 3, 2012

### lavinia

The equation f'(x) = f'(0)f(x) shows that f is infinitely differentiable.

The Taylor series looks like 1 + xf'(0) + .... 1/n!x$^{n}$f'(0)$^{n}$ + .....

This equation shows that each choice of f'(0) determines the function, f. Choosing f'(0) = 1 can be defined as the exponential function.

One must then shows that the Taylor series converges for all x but this is not hard.

17. Nov 4, 2012

### Hetware

Define:
$\ln x \equiv \int_1^x \frac{1}{t} \, dt$
$\ln e\equiv 1$

By the fundamental theorem of calculus:

$\frac{d}{dx}\ln x=\frac{1}{x}$

By the chain rule:

$\frac{d}{dx}\ln x^n=\frac{1}{x^n}nx^{n-1}=\frac{n}{x}=\frac{d}{dx}(n\ln x)$

Two functions with equal derivatives over a domain differ by a constant over that domain:

$\ln x^n=n\ln x+C \text{ for all } x>0$

Set $x=1$

$\ln 1^n=n\ln 1=0\Longrightarrow C=0$

So:

$\ln x^n=n\ln x \text{ for all } x>0$

From above we have:

$\ln e^x=x \ln e=x$

So:

$\frac{d}{dx}\ln e^x=\frac{d}{dx}x=1=\frac{1}{e^x}\frac{d}{dx}e^x \Longrightarrow e^x=\frac{d}{dx}e^x$

q.e.d.

Technically this only applies to rational numbers. We have to throw in some irrational spackle to make it apply to real numbers.

I started out to prove this using limits, and realized that those proof are summed up in the fundamental theorem of calculus.

18. Nov 8, 2012

### Gloyn

Hetware, maybe it's lame question, but where did the last line come from? I mean, the thing

1=e^x d(e^x)/dx

19. Nov 8, 2012

### lurflurf

^
1=x'=(log(e^x))'=(e^-x)(e^x)'

20. Nov 8, 2012

### Gloyn

Thanks, don't know how I could miss that :).


text:
# What's the most rewarding path?

Get from the top-left to the bottom-right using only right and down moves. Pick up as much gold as possible. There is only one maximum.

• If there were a lateral tag, I'd think the numbers 6 and 9 are "G"olds (as the relation between the numbers and golds is not mentioned). – WhatsUp Nov 11 '19 at 14:58

It is a 10x10 map and you need to go east 9 times and south 9 times. There are total $${18\choose9}$$ = 48620 paths which are impractical to be done with pen and paper in a reasonable amount of time.
But if you observe

If I know the best answer (max gold) that can be collected until the cell on my north and cell on my west, I should perhaps always choose greedily because those two answers are optimal.

This can be done pretty quickly with just 10x10=100 calculations.

109 (Assuming my unverified calculations are correct)

Proof of work

Method

Write the cumulative sum of the first row and column (Blue)
For every other cell, write the sum of that cell and maximum of the top cell and cell to the left and mark the source with an arrow.
The last cell contains the answer. Use arrows to reach the source recursively until you reach the top cell.

• This method always finds the optimal answer, even in cases where the greedy algorithm does not. It's a pity that for the given grid the greedy algorithm also mostly works, apart from the ambiguity at the cell with score 100. – Jaap Scherphuis Nov 11 '19 at 7:17
• It's called dynamic programming (done by hand). We used to have this sort of problems in math contests for children. – WhatsUp Nov 11 '19 at 14:53

@Mohit Jain's solution is nice and provably correct. It also shows you the best way to reach any square from the top left corner.

I happened to solve the puzzle in a way that's pretty much identical, but in reverse.

Both methods are equally valid, but somehow I felt that finding the optimum path from any square to the bottom right might be a more interesting "side effect" of solving the puzzle, so I decided to post my solution too, so here it is:

Start at

the bottom right, and mark the best possible score for its neighbours.

Then continue in the same manner throughout.

Like this:

and going on,

And then you make an error because single digit addition is hard, and after backtracking, you finally reach the top left corner (again):

And now you have the optimal score for every starting square, and to find the best path to the bottom right you can always just go to the neighbour with the better score.

At the risk of flogging a dead horse, I wanted to solve it using Dijkstra's algorithm, since an answer implied that it required a computer to do so.

First, we change it to a minimize problem. We know that all solutions use 19 steps, so we simply change the values to $$10-x$$ and then minimize the result:

Initialize the top left to a value of 1. Everywhere else is value infinity. Then we just check the lowest, unvisited nodes and move on one step.

The above is after a handful of steps. Visited nodes have been circled and need not be looked at again. The unvisited nodes have the total value that it took to get there. We choose the one with the lowest value (21 in this case) for the next step. In this case, it goes to the right and that node gets a value of 23.

We then do the same again for the unvisited nodes of 22. Then 23, marking them as visited (circled) in each case:

Here we've gone as far as 30. Note that we take the lowest total value if necessary. In this particular problem, it was not really necessary (there was one (literal) corner case, where I counted the corner two steps as one):

Here we've finished it:

The minimum value is 81. To convert back to the maximum value, we undo our transformation ($$190-81=109$$). Phew!

To figure out the path, we work backwards from the last node. It has total value of 81 and a value of 8, so it must have come from a 73. There's only one node with 73. It has a value of 7, so must have come from a 66. Again only one node does it. Repeat all the way:

Local maximum solution:

Consider we should always choose larger value while pick either right or down move as next step(if both values are same, join the next 2 steps into consideration), the local maximum from start to end is: $$9\downarrow6\to8\downarrow8\downarrow3\to7\to9\to4\to6\downarrow7\downarrow3\downarrow6\to7\downarrow8\to5\downarrow4\to4\to3\downarrow2 = 109$$

And consdier from end back to the start, the local maximum is: $$9\downarrow6\to8\downarrow8\downarrow3\to7\to9\downarrow3\downarrow5\to8\to3\downarrow6\to7\downarrow8\to5\downarrow4\to4\to3\downarrow2 = 108$$

But there should exist an optimal maximum, still verifying the algorithm :P
Remark: This solution could be solved by Dijkstra's algorithm, but this may need computer to easy-solved, however this puzzle requires no-computer.

• I also approached the same way, and found the same solution of 109, but only reason I did not post it as an answer is : this approach does not necessarily yield the maximum solution. For example, consider the 3X3 table with the rows as : [ 1, 2, 1] , [0, 3, 4], [9, 2, 3]. The current approach says the optimum path is 1 + 2 + 3 + 4 + 3 = 13 whereas the actual optimum is 1 + 0 + 9 + 2 + 3 = 15 – SamRoy Nov 11 '19 at 6:11
• @SamRoy Yes, so it's only local but not optimal solution, and Dijkstra's algo. could find out the optimal but handy work may be lousy if no computer used... – Conifers Nov 11 '19 at 6:20
• I also wrote a computer program which gives 109 as the maximum reward, and the same path you mentioned. But I guess there is no general rule to find out the optimal solution by analytical approach only – SamRoy Nov 11 '19 at 6:22
• For a reference, it's not a Dijkstra algorithm but Dynamic Programming approach instead (shown by @MohitJain) -- In your local optimum one, it's Greedy approach. The Dijkstra is to find the shortest path given the lengths of each segments of path (edges). – athin Nov 11 '19 at 6:45
• @athin we could turn the value as the weighted edge and find the maximum path, may like the Dijkstra's but a twisted(Just thought in mind, need transformation proofs.) – Conifers Nov 11 '19 at 6:50


text:
Prismatization seminar

## Fall 2022

The goal of this seminar is to understand Drinfeld's prismatization paper with an eye towards F-gauges, as well as some related papers and applications.

### Time and place

Thursdays 11:30 AM - 1 PM, 528 Mathematics

### References

[1] Bhargav Bhatt and Jacob Lurie. "Absolute prismatic cohomology." arXiv preprint arXiv:2201.06120 (2022).

[2] Bhargav Bhatt and Jacob Lurie. "The prismatization of p-adic formal schemes." arXiv preprint arXiv:2201.06124 (2022).

[3] Bhargav Bhatt and Peter Scholze. "Prisms and prismatic cohomology." arXiv preprint arXiv:1905.08229 (2019).

[4] Vladimir Drinfeld. "Prismatization." arXiv preprint arXiv:2005.04746 (2020).

### Schedule

 Date Speaker Topic References Notes September 8 Avi Zeff Introduction, organization and overview [1, 2, 3, 4] PDF September 15 Avi Zeff Overview of prismatic cohomology [3] PDF September 22 Ivan Zelich Perfectoid rings and THH Perfectoid rings and THH Cyclic homology offers, as shown by Connes, a way to refine de Rham cohomology to the non-commutative setting. To unpack what this means, we will delve into the classical theory of Hochschild cohomology with an emphasis on its relation to perfect rings. The modern view point on this theory, afforded by recent developments in homotopy theory, is to consider topological variants, where the theory can be generalised to more homotopically enriched categories. It is possible, in this setting, to perform analogous computations for perfect rings in characteristic p, like in the classical case. If one is to take the analogy that perfectoid rings are characteristic 0 analogues of perfect rings in characteristic p seriously, then one would expect that similar computations can be undertaken for perfectoid rings that would yield similar results. We will attempt to survey such results, and in particular, outline how perfectoid rings can arise as certain Thom spectra. September 29 Ivan Zelich Perfectoid rings and THH, continued October 6 Avi Zeff An introduction to the stack $$\Sigma$$ [4] PDF October 13 Avi Zeff WCart: Bhatt-Lurie's perspective [1] PDF October 20 Avi Zeff $$\Sigma'$$: construction and properties [4] PDF October 27 Ivan Zelich Syntomic cohomology and $$\Sigma''$$ Syntomic cohomology and $$\Sigma''$$ Motivated by proving certain comparison theorems between crystalline and etale cohomology in p-adic hodge theory, Fontain and Messing defined their "syntomic cohomology" as a way to transfer between both worlds. As we've understood throughout this seminar, prismatic cohomology serves as a "universal" p-adic cohomology theory refining all known p-adic cohomology theories. One may ask whether there is a notion of syntomic cohomology that may compute the prismatic cohomology, in the spirit of Fontain-Messing's approach to p-adic Hodge theory. We will discuss a few interpretations of this question, and hopefully be able to compare the modern treatments with the classical treatments satisfactorily. November 3 Not meeting November 10 Avi Zeff Syntomization: syntomic cohomology in terms of $$\Sigma''$$ and $$X^{\Delta''}$$ PDF November 17 Ivan Zelich The quasisyntomic site December 1 Avi Zeff Application of syntomic duality: unramified $$p$$-primary class field theory for function fields PDF


text:
# Status of Grothendieck's conjecture on homomorphisms of abelian schemes

In [1] Grothendieck posits the following:

Conjecture. Let $S$ be a reduced connected scheme, locally of finite type over Spec($\mathbf{Z}$) or a field $k$, $A$ and $B$ two abelian schemes over $S$, $l$ a prime number, $u_l: T_l(A) \rightarrow T_l(B)$ a homomorphism, and suppose that there exists a point $s\in S$ such that $u_{ls}$ comes from a homormophism $u_s: A_s \rightarrow B_s$ of abelian schemes over $k(s)$. Then there exists an integer $n>0$ and a homomorphism $v: A \rightarrow B$ such that $T_l(v)=n u_l$.

Next Grothendieck remarks that the conjecture follows from Tate's conjectures on algebraic cycles. He then proceeds to a give a proof, using the Serre-Tate theorem, for the case of a characteristic zero base field for which the statement holds with $n=1$.

Question: What is the current status of this conjecture? If not resolved, are there any partial results for the positive characteristic case?

Here is what I've spotted in the literature so far, all characteristic zero:

Deligne in the second of three papers on Hodge theory, proves results about homomorphisms of abelian schemes over schemes $S$ of finite type over $\mathbf{C}$. For instance:

Proposition (Deligne 4.4.12 [2]) : Let $f: X \rightarrow S$ be an abelian scheme over a smooth scheme $S$. The following conditions are equivalent:

$(i)$ For every abelian scheme $g: Y \rightarrow S$, we have $$\mathrm{Hom}_S(X,Y) \cong \mathrm{Hom}_S(R_1 f_* \mathbf{Z}, R_1 g_* \mathbf{Z} )$$

$(ii)$ The condition $(i)$ is verified for $X=Y$, and the centre $Z$ of $\mathrm{End}_S(X)\otimes \mathbf{Q}$ does not admit a complex place $\rho: Z \rightarrow \mathbf{C}$ such that the direct factor $R_1 f_*\mathbf{Q} \otimes_{Z,\rho} \mathbf{C}$ of $R_1 f_* \mathbf{C}$ is of pure Hodge type (-1,0).

Along similar lines, S.G. Tankeev in a 1976 paper [3] proves if $S$ is a connected smooth curve over $\mathbf{C}$, and $\pi_i: X_i \rightarrow S, i=1,2$ are two abelian schemes, then under some natural conditions, the canonical homomorphism $$\mathrm{Hom}_S(X_1,X_2) \rightarrow \mathrm{Hom}(R_1 \pi_{1*} \mathbf{Z}, R_1 \pi_{2*} \mathbf{Z})$$ is an isomorphism. In a paper the following year [4] Tankeev proves a Tate module variant of the above, of the form $$\mathrm{Hom}_k(I_1,I_2)\otimes_\mathbf{Z} \mathbf{Z}_l \cong \mathrm{Hom}_G(T_l(I_1), T_l(I_2)),$$ where $G=\mathrm{Gal}(\overline{k}/k)$, and $I_1, I_2$ are abelian varieties over $k(t)$ whose Néron models admit compactifications with certain properties.

Are there other well-known results along the same lines? I would especially like to know about the characteristic $p$ case, as well any results that actually spread a homomorphism of abelian varities over a point of S to maps of abelian schemes over S, as in the conjecture.

[1] A. Grothendieck Un théorème sur les homomorphismes de schémas abéliens Invent. Math. 2 59-78 (1966)

[2] P. Deligne Théorie de Hodge: II Publications mathématiques de l'I.H.E.S, tome 40 (1971) p.5-57

[3] S.G. Tankeev On homomorphisms of abelian schemes Math. USSR Izvestija Vol. 10 (1976), No. 4

[4] S.G. Tankeev On homomorphisms of abelian schemes II Math. USSR Izvestija Vol. 11 (1977), No. 6

-
Suppose that $S$ is a normal variety over $\mathbb{F}_p$ with generic point $\eta\to S$. Then, by Zarhin, there exists a homomorphism of the generic fibers $u_{\eta}:A_{\eta}\to B_{\eta}$ giving rise to $u_{\ell}$ (one doesn't even need the existence of $u_s$ for this). Now, by Proposition 2.7 in Chai-Faltings 'Degeneration of abelian varieties', $u_{\eta}$ extends over all of $S$. The idea basically is to use the Neronian property of abelian schemes to show that the closure of the graph of $u_{\eta}$ in $A\times B$ maps isomorphically onto $A$. –  Keerthi Madapusi Pera Mar 1 '13 at 14:39
Thanks, this is great! Though I'm somewhat confused about the application of Zarhin to obtain $u_\eta$. Isn't $Hom_G(T_l(A), T_l(B))$ only isomorphic to $Hom(A,B)$ up to tensoring with $\mathbf{Z}_l$? –  Zavosh Mar 1 '13 at 19:25
Ah, that's right! I was being cavalier as usual. I think the existence of $u_s$ should force it to actually lie in $Hom(A_{\eta},B_{\eta})$. –  Keerthi Madapusi Pera Mar 1 '13 at 20:45
More precisely, what my comment shows is that we have $Hom(A_{\eta},B_{\eta})=Hom(A,B)$. So, by Zarhin, $u_{\ell}$ arises from an element of $Hom(A,B)\otimes\mathbb{Z}_{\ell}$. To check that some multiple of it actually lies in $Hom(A,B)$, it suffices to check that, for some point $s\in S$, the image in $Hom(A_s,B_s)\otimes\mathbb{Z}_{\ell}$ lies in $Hom(A_s,B_s)$. –  Keerthi Madapusi Pera Mar 1 '13 at 20:50
That sounds right. If you would like to enter it as an answer I would accept it. –  Zavosh Mar 2 '13 at 0:17

This is an expansion of my comments. Grothendieck's conjecture is now a theorem, except perhaps when $S$ is in characteristic $p$ and $\ell=p$. The explanation that follows is presumably what Grothendieck had in mind in his comment regarding the Tate conjecture.
First, we make some reductions: We can certainly assume that $S$ is integral. In fact, we can assume that $S$ is normal. This follows from Proposition 1.2 of Grothendieck's article.
Finally, we can assume that the generic point $\eta$ of $S$ is the spectrum of a finitely generated field. This can be shown as in 2.2 of Grothendieck's article.
In this scenario, by Proposition 2.7 in Chai-Faltings, 'Degeneration of Abelian varieties', the natural restriction map $Hom(A,B)\to Hom(A_{\eta},B_{\eta})$ is a bijection. Suppose now that $\eta$ is the spectrum of a finitely generated field. Then, by the Tate conjecture for homomorphisms of abelian varieties over finitely generated fields (proved by Zarhin in finite characteristic, and Faltings in characteristic $0$), the natural map of $\mathbb{Z}_{\ell}$-modules: $$Hom(A,B)\otimes\mathbb{Z}_{\ell}\to Hom_G(T_{\ell}(A_{\eta}),T_{\ell}(B_{\eta}))$$ is an isomorphism. Here, on the right hand side, $G$ is the absolute Galois group of the spectrum of $\eta$. In particular, $u_{\ell}$ gives rise to an element of $Hom(A,B)\otimes\mathbb{Z}_{\ell}$.
Now, suppose that the specialization of $u_{\ell}:T_{\ell}(A)\to T_{\ell}(B)$ over a point $s\in S$ arises from an honest homomorphism $u_s:A_s\to B_s$. Then we find that, for some $n\in\mathbb{Z}_{\geq 1}$, $nu_{\ell}$ arises from a homomorphism $u':A\to B$. Essentially, the map $$Hom(A,B)_{\mathbb{Q}}\to Hom(A_s,B_s)_{\mathbb{Q}}$$ identifies the left hand side with a vector sub-space of the right hand side. In fact, if $S$ is of characteristic $p$, we can take $n$ to be a power of $p$, since the specialization map will have saturated image after inverting $p$.


text:
# Our solar system

1. Apr 15, 2006

### wolram

I had a thought about the possition of the sun in our SS, and wondered if
the suns possition varied by any significant amout due to the gravitational
pull of the planets, all the simulations i can find have the sun static.

2. Apr 15, 2006

### Garth

The Sun orbits around the Centre of Mass of the Solar system, just as all the planets do.

However, in the two body Newtonian problem you can transform the orbital motion around the CoM into an orbital motion, to a non-inertial frame of reference, around either of the masses. Normally the Sun is taken as the origin of the coordinate system, and the planet performs a Keplerian ellipse around it.

Garth

Last edited: Apr 15, 2006
3. Apr 15, 2006

### wolram

Hi Garth, i imagine the center of mass is not far from the sun, so the sun would wobble , rather than actually orbit, but is this a chaotic wobble, due
to having so many planets to pull on it. or is it more than a wobble.

4. Apr 15, 2006

### mubashirmansoor

Wolram, your imagination is abseloutely correct and its just a wobble and not more.

5. Apr 15, 2006

### Labguy

With the Sun being 99.8% of the mass of the entire solar system and Jupiter being most of the rest, the center of mass would be very near the core and well below the photosphere, It might even be in the core (however defined) and the wobble would be very small. Even that small wobble effected by Jupiter would often be offset some by any other solar system masses opposite Jupiter at the time and I don't know if the wobble would even be detectable...

6. Apr 15, 2006

### Kurdt

Staff Emeritus
The wobble of stars is one of the main tools used in detecting extra solar planets. This can be done by astrometry (carefully measuring the position of a star over a long period of time) or by doppler shifting of the stars spectra. I did see a website a while back that demonstrated the suns wobble with a video and it is rather erratic and mainly due to Jupiter. I can't find it at the minute but if i do I'll post it.

7. Apr 15, 2006

### Labguy

I was going to mention that (extra solar planets) but I think I remember that most found by this method are much larger than Jupiter and quite a few are much closer to thier primary than Jupiter is to the Sun.

But, a quick search shows me that I was wrong and that the barycenter of the Earth-Jupiter system is a bit outside the Sun's photosphere as described by the heretical "Wikepedia" source:
and more, including the animations, at:
http://en.wikipedia.org/wiki/Center_of_mass
and more stuff at:
http://sprott.physics.wisc.edu/lectures/seasons/sld016.htm

Guess I better stick to stellar stuff and leave the solar system stuff to Janus..

Last edited: Apr 15, 2006
8. Apr 15, 2006

### Garth

Last edited: Apr 15, 2006
9. Apr 15, 2006

### tony873004

The solar system barycenter is often outside the photosphere of the Sun, but not outside the corona.

Here is an image of the Sun locked to the center of the screen:
http://orbitsimulator.com/BA/sbc1.GIF

Unlocking the Sun exposes the solar system barycenter which is now locked to the center of the screen. The Sun spends nearly a decade in this image circling the solar system barycenter:
http://orbitsimulator.com/BA/sbc2.GIF

Editing the Sun and setting its size to 0 while retaining its mass allows me to zoom in on the solar system barycenter and observe the Sun's path around it:
http://orbitsimulator.com/BA/sbc4.GIF

The center of the Sun strays near its maximum barycenter distance of about 1.3 million kilometers on Dec. 3, 2022.This can be corroborated by querying the JPL Ephemeris system which gives the following data for Dec. 3, 2022.
$$SOE 2459916.500000000 = A.D. 2022-Dec-03 00:00:00.0000 (CT) 1.357796530286104E+06 -5.340761458326381E+04 -3.117957726829686E+04 -9.802716335062548E-04 1.565705785929941E-02 -1.077297045285839E-04 4.533817068543462E+00 1.359204163100999E+06 -1.592002604514373E-03$$EOE

The boldfaced, italicized number represents the Sun / solar system barycenter distance (km) on that date.

In the simulation, editing Jupiter, and setting its mass to 0 demonstrates that Jupiter is responsible for the majority of wobble.
http://orbitsimulator.com/BA/sbc5.GIF

Saturn is the next strongest perturber. Additionally, setting its mass to 0 shows:
http://orbitsimulator.com/BA/sbc6.GIF

The next strongest perturber is Neptune. Additionally, setting its mass to 0 causes the Sun's motion around the solar system barycenter to trace a circle around its barycenter. Uranus is responsible for this circle. The Sun's period around the barycenter and Uranus' period around the barycenter match.
http://orbitsimulator.com/BA/sbc7.GIF

Zooming in exposes the effects of the smaller planets on the Uranus-induced circle.
http://orbitsimulator.com/BA/sbc8.GIF

Setting Uranus' mass to 0 eliminates the Uranus-induced wobble. The Sun's center now appears to rest on the solar system barycenter.
http://orbitsimulator.com/BA/sbc9.GIF

But zooming in further exposes the influences of the remaining planets on the solar system barycenter: The Earth/Moon system is responsible for the majority of the wobble
http://orbitsimulator.com/BA/sbc10.GIF

Setting the Earth/Moon mass to 0 leaves Venus as the most significant perturber. It has the following influence:
http://orbitsimulator.com/BA/sbc11.GIF

Setting Venus' mass to 0 leaves Mercury, Mars, and Pluto as the sole perturbers. They cause the center of the Sun to trace the following path around the solar system barycenter:
http://orbitsimulator.com/BA/sbc12.GIF

Zooming in for a clearer view, the effects of Mercury and Mars are seen. Pluto's effects is simply pulling this pattern off-center. If left to run for a quarter century of sim time, the entire pattern would trace a circle around the center of the screen:
http://orbitsimulator.com/BA/sbc13.GIF

Setting Mars, which is now the most significant perturber, to 0 shows Mercury's influence causing the center of the Sun to trace circles around the solar system barycenter:
http://orbitsimulator.com/BA/sbc14.GIF

Letting this simulation run for half of a Pluto orbit exposes Pluto's influence on the solar system barycenter.
http://orbitsimulator.com/BA/sbc15.GIF

10. Apr 15, 2006

### wolram

Thankyou everyone, the links you provided etc have been most interesting,
i imagine the seasoned space pilot would have to know his orbits.


text:
# All Questions

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### How to prove that a function is not pseudorandom?

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### Finding Private Key $d$ using RSA [duplicate]

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### Proof of RSA security dependent on public key exponent

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text:
Monday Math 77

Find .

Using , we see

Next, we try to eliminate the ex term in the numerator. By noting that , we do so as follows:
.
For the second integral, we make the u-substitution u=2x to get

and renaming the variable back to x, we get
.

Now, we showed here that . Thus, our integral is
.

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One Response to “Monday Math 77”

1. Monday Math 78 « Twisted One 151’s Weblog Says:

[…] parts, with and ; then and Now, using and , integration by parts tells us: . Now, we proved here that , so the above is just this for the case n=2, so since and (see here), then our integral is […]


text:
# 22.9 Magnetic fields produced by currents: ampere’s law  (Page 3/12)

 Page 3 / 12

## Magnetic field produced by a current-carrying solenoid

A solenoid    is a long coil of wire (with many turns or loops, as opposed to a flat loop). Because of its shape, the field inside a solenoid can be very uniform, and also very strong. The field just outside the coils is nearly zero. [link] shows how the field looks and how its direction is given by RHR-2.

The magnetic field inside of a current-carrying solenoid is very uniform in direction and magnitude. Only near the ends does it begin to weaken and change direction. The field outside has similar complexities to flat loops and bar magnets, but the magnetic field strength inside a solenoid    is simply

$B={\mu }_{0}\text{nI}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left(\text{inside a solenoid}\right),$

where $n$ is the number of loops per unit length of the solenoid $\left(n=N/l$ , with $N$ being the number of loops and $l$ the length). Note that $B$ is the field strength anywhere in the uniform region of the interior and not just at the center. Large uniform fields spread over a large volume are possible with solenoids, as [link] implies.

## Calculating field strength inside a solenoid

What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current?

Strategy

To find the field strength inside a solenoid, we use $B={\mu }_{0}\text{nI}$ . First, we note the number of loops per unit length is

$n=\frac{N}{l}=\frac{\text{2000}}{2.00 m}=\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{-1}=\text{10}\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{-1}\text{.}$

Solution

Substituting known values gives

$\begin{array}{lll}B& =& {\mu }_{0}\text{nI}=\left(4\pi ×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}T\cdot \text{m/A}\right)\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{m}^{-1}\right)\left(\text{1600 A}\right)\\ & =& 2\text{.01 T.}\end{array}$

Discussion

This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000 loops squeezed into a meter’s length would produce significant heating. Higher currents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields.

There are interesting variations of the flat coil and solenoid. For example, the toroidal coil used to confine the reactive particles in tokamaks is much like a solenoid bent into a circle. The field inside a toroid is very strong but circular. Charged particles travel in circles, following the field lines, and collide with one another, perhaps inducing fusion. But the charged particles do not cross field lines and escape the toroid. A whole range of coil shapes are used to produce all sorts of magnetic field shapes. Adding ferromagnetic materials produces greater field strengths and can have a significant effect on the shape of the field. Ferromagnetic materials tend to trap magnetic fields (the field lines bend into the ferromagnetic material, leaving weaker fields outside it) and are used as shields for devices that are adversely affected by magnetic fields, including the Earth’s magnetic field.

## Phet explorations: generator

Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light.

## Section summary

• The strength of the magnetic field created by current in a long straight wire is given by
$B=\frac{{\mu }_{0}I}{2\mathrm{\pi r}}\left(\text{long straight wire}\right),$
where $I$ is the current, $r$ is the shortest distance to the wire, and the constant ${\mu }_{0}=4\pi \phantom{\rule{0.15em}{0ex}}×\phantom{\rule{0.15em}{0ex}}{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}$ is the permeability of free space.
• The direction of the magnetic field created by a long straight wire is given by right hand rule 2 (RHR-2): Point the thumb of the right hand in the direction of current, and the fingers curl in the direction of the magnetic field loops created by it.
• The magnetic field created by current following any path is the sum (or integral) of the fields due to segments along the path (magnitude and direction as for a straight wire), resulting in a general relationship between current and field known as Ampere’s law.
• The magnetic field strength at the center of a circular loop is given by
$B=\frac{{\mu }_{0}I}{2R}\text{}\left(\text{at center of loop}\right),$
where $R$ is the radius of the loop. This equation becomes $B={\mu }_{0}\text{nI}/\left(2R\right)$ for a flat coil of $N$ loops. RHR-2 gives the direction of the field about the loop. A long coil is called a solenoid.
• The magnetic field strength inside a solenoid is
$B={\mu }_{0}\text{nI}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left(\text{inside a solenoid}\right),$
where $n$ is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction.

## Conceptual questions

Make a drawing and use RHR-2 to find the direction of the magnetic field of a current loop in a motor (such as in [link] ). Then show that the direction of the torque on the loop is the same as produced by like poles repelling and unlike poles attracting.

Why is there no 2nd harmonic in the classical electron orbit?
how to reform magnet after been demagneted
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
what is the error during taking work done of a body..
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
Julius
great Mudang
Kossi
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
ancient greek language physis = nature
isidor
what is phyacs
if i am going to start studying physics where should i start?
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir


text:
Definition 88.8.1. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of categories cofibered in groupoids over $\mathcal{C}_\Lambda$. We say $\varphi$ is smooth if it satisfies the following condition: Let $B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda$. Let $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{G}(B)), x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$, and $y \to \varphi (x)$ be a morphism lying over $B \to A$. Then there exists $x' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(B))$, a morphism $x' \to x$ lying over $B \to A$, and a morphism $\varphi (x') \to y$ lying over $\text{id}: B \to B$, such that the diagram

$\xymatrix{ \varphi (x') \ar[r] \ar[dr] & y \ar[d] \\ & \varphi (x) }$

commutes.

There are also:

• 2 comment(s) on Section 88.8: Smooth morphisms

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text:
# Changes between Version 30 and Version 31 of WRF4GTutorial2

Ignore:
Timestamp:
Oct 2, 2012 1:55:03 PM (10 years ago)
Comment:

--

### Legend:

Unmodified
 v30 * MN SLRUM (only for Red Española de Supercomputación) In order to add new resources, you need to edit the section Computing Resources in the [wiki:WRF4Gframework4g_conf framework4g.conf] file which is located under $WRF4G_LOCATION/etc directory. The file has to contain one resource per line, with the format: In order to add new resources, you need to edit Computing Resources section in the [wiki:WRF4Gframework4g_conf framework4g.conf] file which is located under $WRF4G_LOCATION/etc directory. The file has to contain one resource per line with the format: {{{ SGE_cluster        local://localhost?LRMS_TYPE=sge;PROJECT=l.project }}} If you configure a remote computing resource through ssh protocol, you need to put your private keys into your [[http://en.wikipedia.org/wiki/Ssh-agent|ssh-agent]], and it will handle your authentication thereafter (see [wiki:DRM4G#AppendixA| Appendix A]) or set up SSH login without ''password'' (see [wiki:DRM4G#AppendixB| Appendix B]). If you want to configure a remote computing resource through ssh protocol, you need to put your private keys into your [[http://en.wikipedia.org/wiki/Ssh-agent|ssh-agent]], and it will handle your authentication thereafter (see [wiki:DRM4G#AppendixA| Appendix A]) or set up SSH login without ''password'' (see [wiki:DRM4G#AppendixB| Appendix B]). {{{ remote_PBS_cluster ssh://user@meteo1.macc.unican.es?LRMS_TYPE=pbs;QUEUE_NAME=short }}} [[NoteBox(warn, After modifying this file\, in order to make changes effective\, users will have to execute: || wrf4g_framework reload)]] [[NoteBox(warn, After modifying Computing Resources section\, in order to make changes effective\, you have to execute: || wrf4g_framework reload)]]