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# NAG CL Interfacec02abc (poly_real_fpml)

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## 1Purpose

c02abc finds all the roots of a real polynomial equation, using a fourth-order convergent modification of Laguerre's method.

## 2Specification

 #include
 void c02abc (const double a[], Integer n, Integer itmax, Nag_Root_Polish polish, Complex z[], double berr[], double cond[], Integer conv[], NagError *fail)
The function may be called by the names: c02abc or nag_zeros_poly_real_fpml.

## 3Description

c02abc attempts to find all the roots of the $n$th degree real polynomial equation
 $P(z) = a0zn + a1zn-1 + a2zn-2 + ⋯ + an-1 z + an = 0 .$
The roots are located using a modified form of Laguerre's method, as implemented by Cameron (2018).
c02abc is a wrapper around the corresponding complex routine c02aac.
The relative backward error of a root approximation ${z}_{j}$ is given by
 $η(zj) = |P(zj)| α(zj) ,$
where $\alpha$ is the perturbed polynomial
 $α(z) = ∑ k=0 n (3.8k+1) |an-k| |z|k .$
A root approximation is deemed to have converged if $\eta \left({z}_{j}\right)\le 2\text{ }×$ machine precision, at which point updates of that root cease. If the stopping criterion holds, then the computed root is the exact root of a polynomial whose coefficients are no more perturbed than the floating-point computation of $P\left({z}_{j}\right)$.
The condition number of each root is also computed, as a measure of sensitivity to changes in the coefficients of the polynomial. It is given by
 $κ(zj) = α(zj) |zj| |P′(zj)| .$
Root approximations can be further refined with optional 'polishing' processes. A simple polishing process is provided that carries out a single iteration of Newton's method, which proves quick yet often effective. Alternatively, a compensated polishing process from Cameron and O'Neill (2019) can be applied. This iterative method combines the implicit deflation of the modified Laguerre method, with the accuracy of evaluating polynomials and their derivatives using the compensated Horner's method from Graillat et al. (2005). Compensated polishing yields approximations with a limiting accuracy as if computed in twice the working precision.
It is recommended that you read Section 9.1 for advice on selecting an appropriate polishing process.

## 4References

Bini D A (1996) Numerical computation of polynomial zeros by means of Aberth's method Numerical Algorithms 13 179–200 Springer US
Cameron T R (2018) An effective implementation of a modified Laguerre method for the roots of a polynomial Numerical Algorithms Springer US https://doi.org/10.1007/s11075-018-0641-9
Cameron T R and O'Neill A (2019) On a compensated polishing technique for polynomial root solvers To be published
Graillat S, Louvet N, and Langlois P (2005) Compensated Horner scheme Technical Report Université de Perpignan Via Domitia
Petković M, Ilić S, and Tričković S (1997) A family of simultaneous zero-finding methods Computers & Mathematics with Applications (Volume 34) 10 49–59 https://doi.org/10.1016/S0898-1221(97)00206-X
Wilkinson J H (1959) The evaluation of the zeros of ill-conditioned polynomials. Part I Numerische Mathematik (Volume 1) 1 150–166 Springer-Verlag

## 5Arguments

1: $\mathbf{a}\left[{\mathbf{n}}+1\right]$const double Input
On entry: ${\mathbf{a}}\left[\mathit{i}\right]$ must contain the coefficient of ${z}^{n-\mathit{i}}$, for $\mathit{i}=0,1,\dots ,n$.
Constraint: ${\mathbf{a}}\left[0\right]\ne 0.0$.
2: $\mathbf{n}$Integer Input
On entry: $n$, the degree of the polynomial.
Constraint: ${\mathbf{n}}\ge 1$.
3: $\mathbf{itmax}$Integer Input
On entry: the maximum number of iterations to be performed.
Suggested value: ${\mathbf{itmax}}=30$.
Constraint: ${\mathbf{itmax}}\ge 1$.
4: $\mathbf{polish}$Nag_Root_Polish Input
On entry: specifies the polishing technique used to refine root approximations.
${\mathbf{polish}}=\mathrm{Nag_Root_Polish_None}$
No polishing.
${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Simple}$
Single iteration of Newton's method.
${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Compensated}$
Iterative refinement using the compensated Horner's method.
Suggested value: ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Simple}$.
Constraint: ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_None}$, $\mathrm{Nag_Root_Polish_Simple}$ or $\mathrm{Nag_Root_Polish_Compensated}$.
5: $\mathbf{z}\left[{\mathbf{n}}\right]$Complex Output
On exit: ${\mathbf{z}}\left[\mathit{j}-1\right]$ holds the approximation of the root ${z}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,n$.
6: $\mathbf{berr}\left[{\mathbf{n}}\right]$double Output
On exit: ${\mathbf{berr}}\left[\mathit{j}-1\right]$ holds the relative backward error, $\eta \left({z}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,n$.
7: $\mathbf{cond}\left[{\mathbf{n}}\right]$double Output
On exit: ${\mathbf{cond}}\left[\mathit{j}-1\right]$ holds the condition number, $\kappa \left({z}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,n$.
8: $\mathbf{conv}\left[{\mathbf{n}}\right]$Integer Output
On exit: ${\mathbf{conv}}\left[\mathit{j}-1\right]$ indicates the convergence status of the root approximation, ${z}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,n$.
${\mathbf{conv}}\left[j-1\right]\ge 0$
Successfully converged after ${\mathbf{conv}}\left[j-1\right]$ iterations.
${\mathbf{conv}}\left[j-1\right]=-1$
Failed to converge after itmax iterations.
${\mathbf{conv}}\left[j-1\right]=-2$
Overflow was encountered.
Note: if ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Compensated}$, conv refers to convergence in the compensated polishing process.
9: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).

## 6Error Indicators and Warnings

NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value.
On entry, ${\mathbf{itmax}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{itmax}}\ge 1$.
NE_CONVERGENCE
Convergence has failed for at least one root approximation.
Check conv and consider increasing itmax.
NE_INT
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 1$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_OVERFLOW
c02abc encountered overflow during at least one root approximation.
Check conv and consider scaling the polynomial (see Section 9.2).
NE_ZERO_COEFF
On entry, the real variable ${\mathbf{a}}\left[0\right]=0.0$.

## 7Accuracy

All roots are evaluated as accurately as possible, but because of the inherent nature of the problem complete accuracy cannot be guaranteed.

## 8Parallelism and Performance

c02abc is not threaded in any implementation.

### 9.1Selecting a Polishing Process

The choice of polishing technique ultimately depends on two factors: how well conditioned the problem is, and a preference between run time and accuracy. For a detailed analysis of the polishing techniques, see Cameron and O'Neill (2019).
Well-conditioned Problems
Simple polishing is effective in reducing the error in approximations of well-conditioned roots, doing so with a negligible increase in run time. Compensated polishing has comparable accuracy, but it is approximately ten times slower than when using simple polishing.
Simple polishing (${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Simple}$) is recommended for well-conditioned problems.
Ill-conditioned Problems
There is a dramatic difference in accuracy between the two polishing techniques for ill-conditioned polynomials. Unpolished approximations are inaccurate and simple polishing often proves ineffective. However, compensated polishing is able to reduce errors by several orders of magnitude.
Compensated polishing (${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Compensated}$) is highly recommended for ill-conditioned problems.

### 9.2Scaling the Polynomial

c02abc attempts to avoid overflow conditions where possible. However, if the function fails with ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_OVERFLOW, such conditions could not be avoided for the given polynomial. Use conv to identify the roots for which overflow occurred, as other approximations may still have succeeded.
Extremely large and/or small coefficients are likely to be the cause of overflow failures. In such cases, you are recommended to scale the independent variable $\left(z\right)$ so that the disparity between the largest and smallest coefficient in magnitude is reduced. That is, use the function to locate the zeros of the polynomial $sP\left(cz\right)$ for some suitable values of $c$ and $s$. For example, if the original polynomial was $P\left(z\right)={2}^{-100}i+{2}^{100}{z}^{20}$, then choosing $c={2}^{-10}$ and $s={2}^{100}$, for instance, would yield the scaled polynomial $i+{z}^{20}$, which is well-behaved relative to overflow and has zeros which are ${2}^{10}$ times those of $P\left(z\right)$.

## 10Example

The example program for c02abc demonstrates two problems, given in the functions ex1_basic and ex2_polishing. (Note that by default, the second example is switched off because the results may be machine dependent. Edit the program in the obvious way to switch it on.)
Example 1: Basic Problem
This example finds the roots of the polynomial
 $a0 z6 + a1 z5 + a2 z4 + a3 z3 + a4 z2 + a5 z+a6 = 0 ,$
where ${a}_{0}=3.5$, ${a}_{1}=1.0$, ${a}_{2}=-7.0$, ${a}_{3}=12.5$, ${a}_{4}=2.5$, ${a}_{5}=-17.0$ and ${a}_{6}=32.5$.
Example 2: Polishing Processes
This example finds the roots of a polynomial of the form
 $(z-1) (z-2) ⋯ (z-n) = 0 ,$
first proposed by Wilkinson (1959) as an example of polynomials with ill-conditioned roots, that are sensitive to small changes in the coefficients. A polishing mode is demonstrated with $n=10$, and the maximum forward and relative backward errors of the approximations displayed.

### 10.1Program Text

Program Text (c02abce.c)

### 10.2Program Data

Program Data (c02abce.d)

### 10.3Program Results

Program Results (c02abce.r)


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# Isometry in $\mathbb{R}^n$

I'm trying to prove that if $f\colon\mathbb{R}^n \to \mathbb{R}^n$ is a $\mathcal{C}^1$ mapping such that $f'(x)$ is a (linear) isometry for every $x \in \mathbb{R}^n$, then $f$ is an isometry. By an application of inverse mapping theorem and mean value theorem, we have that $|f(x) - f(y)| = |x-y|$ as long as $x$ and $y$ are sufficiently close. How to extend this to the whole space?

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I'm not really sure what you intend with $f'$ here, but do you really mean $f'(x)$ to be a map? Presumably evaluating a derivative at a point would result in an element of the field, but not a map. Please take some time to clarify these two points :) –  rschwieb May 28 '12 at 1:55
@rschwieb: For maps $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$, the derivative at a point is defined to be a linear map satisfying certain properties. When viewed this way, for a function $f:\mathbb{R}\rightarrow\mathbb{R}$, the "usual" derivative at a point $x_0$ should be thought of as "multiplication by $f'(x_0)$." –  Jason DeVito May 28 '12 at 1:59
I suspect he means that $f'(x)$ is the differential of $f$ at the point $x$. For example, in one variable the functions equal to their own derivatives in this sense are really the linear functions $f(x) = mx$, not $f(x) = e^x$ (whose derivative at $x_0$ is the map $x \mapsto e^{x_0}x$ on the tangent space at $x_0$). Maybe one can use some kind of connectedness argument to show a local formula for $f$ as an orthogonal map composed with a translation has to work globally. –  KCd May 28 '12 at 2:00
His formulation is correct. The derivative is a linear transformation. This is the approach taken in Spivak's Calculus on Manifolds. In this case $f': \mathbb{R}^n \rightarrow L(\mathbb{R}^n)$. –  ncmathsadist May 28 '12 at 2:02
OK just checking :) I wasn't able to guess all of this. Thanks! –  rschwieb May 28 '12 at 2:07

Edit

Here's a much simpler argument. Let $X\subseteq\mathbb{R}^n\times\mathbb{R}^n$ with $$X = \{(x,y)\in\mathbb{R}^n\times \mathbb{R}^n|d(x,y) = d(f(x),f(y)) \}.$$ Note that $(x,x)\in X$ for any $x$, so $X\neq \emptyset$. Your observation is equivalent to the statement that $X$ is open. To see $X$ is closed, note that if $g:\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}$ with $g(x,y) = d(x,y) - d(f(x),g(y))$, then $g$ is continuous since $d$ and $f$ are, and $X = g^{-1}(0)$. Hence, $X$ is a nonempty clopen subset of $\mathbb{R}^n\times\mathbb{R}^n$, so $X = \mathbb{R}^n\times\mathbb{R}^n$, so $f$ is an isometry.

(End edit)

Here's one approach, borrowed from Riemannian geometry.

Let $\gamma:\mathbb{R}\rightarrow\mathbb{R}^n$ be any straight line paramaterized by arclength, meaning that $\|\gamma(t)-\gamma(s)\| = |t-s|$ for any $t$ and $s$. We will show that $f\circ \gamma$ is also a straight line parametrized by arclength.

Believing this for a second, for $x$ and $y$ in $\mathbb{R}^n$, if $\gamma$ is chosen to be the line going through $x$ and $y$ with $\gamma(t) = x$ and $\gamma(s) = y$, then we have \begin{align*} d(x,y) &= \|\gamma(t)-\gamma(s)\|\\\ &= |t-s|\\\ &= \|f(\gamma(t))-f(\gamma(s))\| \\\ &= d(f(x),f(y)), \end{align*} establishing what we want.

So, why is $f\circ \gamma$ a straight line parameterized by arclength? This follows from your observation that $\|f(x)-f(y)\| = \|x-y\|$ for $x$ and $y$ close together. More specifically, looking at the point $\gamma(t)$, we know that for $s$ close to $t$ (and therefore $\gamma(s)$ close to $\gamma(t)$), that $\|f(\gamma(t)) - f(\gamma(s))\| = \|\gamma(t)-\gamma(s)\|$. This implies that for the line segment of points near $\gamma(t)$, that $f($segment$)$ is another line segment, parametrized by arclength.

Since $\mathbb{R}$ is connected, so is $f(\gamma(\mathbb{R}^n))$. This implies that $f(\gamma(\mathbb{R}^n))$ is a union of line segments end point to end point where each segment is parameterized by arclength. This is a straight line iff there are no corners. But if $f(\gamma(t_0))$ is a corner, then $f$ brings points near $\gamma(t_0)$, but on either side of it, closer together, contradicting your earlier observation that locally $f$ preserves distance.

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format: plain


text:
dcpo
attitudes
measurement
Author

Frederick Solt

Published

January 12, 2020

• Frederick Solt. 2020. “Modeling Dynamic Comparative Public Opinion.” SocArXiv.

• ## Abstract

The study of public opinion in comparative context has been hampered by data that is sparse, that is, unavailable for many countries and years; incomparable, i.e., ostensibly addressing the same issue but generated by different survey items; or, most often, both. Questions of representation and of policy feedback on public opinion, for example, cannot be explored fully from a cross-national perspective without comparable time-series data for many countries that span their respective times of policy adoption. Recent works (Claassen 2019; Caughey, O’Grady, and Warshaw 2019) have introduced a latent variable approach to the study of comparative public opinion that maximizes the information gleaned from available surveys to overcome issues of sparse and incomparable data and allow comparativists to examine the dynamics of public opinion. This paper advances this field of research by presenting a new model and software for estimating latent variables of public opinion from cross-national survey data that yield superior fit and more quantities of theoretical interest than previous works allow.

## BibTeX citation

@misc{Solt2020,
author={Solt, Frederick},
title={Modeling Dynamic Comparative Public Opinion},
publisher={SocArXiv},
year={2020},
month={Jan},
url={osf.io/preprints/socarxiv/d5n9p},
doi={10.31235/osf.io/d5n9p}
}


text:
# de Moivre identity, proof of

To prove the de Moivre identity, we will first prove by induction on $n$ that the identity holds for all natural numbers.

For the case $n=0$, observe that

 $\cos(0\theta)+i\sin(0\theta)=1+i0=\left(\cos(\theta)+i\sin(\theta)\right)^{0}.$

Assume that the identity holds for a certain value of $n$:

 $\cos(n\theta)+i\sin(n\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{n}.$

Multiply both sides of this identity by $\cos(\theta)+i\sin(\theta)$ and expand the left side to obtain

 $\displaystyle\cos(\theta)\cos(n\theta)-\sin(\theta)\sin(n\theta)+i\cos(\theta)% \sin(n\theta)+i\sin(\theta)\cos(n\theta)$ $\displaystyle=\left(\cos(\theta)+i\sin(\theta)\right)\left(\cos(n\theta)+i\sin% (n\theta)\right)$ $\displaystyle=\left(\cos(\theta)+i\sin(\theta)\right)^{n+1}.$

By the angle sum identities,

 $\displaystyle\cos(\theta)\cos(n\theta)-\sin(\theta)\sin(n\theta)$ $\displaystyle=\cos(n\theta+\theta)$ $\displaystyle\cos(\theta)\sin(n\theta)+\sin(\theta)\cos(n\theta)$ $\displaystyle=\sin(n\theta+\theta)$

Therefore,

 $\cos((n+1)\theta)+i\sin((n+1)\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{% n+1}.$

Hence by induction de Moivre’s identity holds for all natural $n$.

Now let $-n$ be any negative integer. Then using the fact that $\cos$ is an even and $\sin$ an odd function, we obtain that

 $\displaystyle\cos(-n\theta)+i\sin(-n\theta)$ $\displaystyle=\cos(n\theta)-i\sin(n\theta)$ $\displaystyle=\frac{\cos(n\theta)-i\sin(n\theta)}{\cos^{2}(n\theta)+\sin^{2}(n% \theta)}$ $\displaystyle=\frac{1}{\cos(n\theta)+i\sin(n\theta)}\cdot\frac{\cos(n\theta)-i% \sin(n\theta)}{\cos(n\theta)-i\sin(n\theta)}$ $\displaystyle=\frac{1}{\cos(n\theta)+i\sin(n\theta)},$

the denominator of which is $\left(\cos(n\theta)+i\sin(n\theta)\right)^{n}$. Hence

 $\cos(-n\theta)+i\sin(-n\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{-n}.$
Title de Moivre identity, proof of DeMoivreIdentityProofOf 2013-03-22 14:34:08 2013-03-22 14:34:08 rspuzio (6075) rspuzio (6075) 10 rspuzio (6075) Proof msc 12E10 proof of de Moivre’s formula proof of de Moivre’s theorem AngleSumIdentity


text:
Measuring cpu load of interrupt routine

I have an isr that is updating a display at a fixed frequency. I would like to tune my routine to minimize overhead and keep as much cpu time as possible open for other processing, but I don't have any good way to collect metrics to determine my cpu load.

I could look at the assembly and analyze the routine, but I don't have the patience or the ability to do that accurately. I don't feel like I need terribly fine grained results either, just a simple percentage of cpu time occupied by the isr.

I could set a pin high only when the isr is active and measure it externally. That has a minimum of overhead in the code, but I do not know what to measure it with. I don't have an oscilloscope or anything like that. Is there a simple ic or an easy way to use another micro to measure duty cycle? I have heard of dedicated frequency counter chips, but is there anything for duty cycle?

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Just a half-baked idea, you might be able to make use of timers like so (pseudo code):

int main(void)
{

/* ... init timers and uart here, enable your interrupts ... */

start_timer0();
while (!timer1Started()){}
stop_timer1();

uart_puts("Idle ticks: %d, ISR ticks: %d", timer0_value, timer1_value);

}

and in your display ISR...

ISR_display()
{
stop_timer0();
start_timer1();

/* ... your ISR routine ... */
}

I've made a few assumption here. 1 - that you're not using your timers for anything else, and that, 2 - the overhead of starting and stopping a timer is minimal (typically done with a single register write). EDIT: and a 3rd assumption, you can capture all of this before a timer overflow occurs, but maybe you can account for that as well.

There will be some context switch overhead that you won't be able to catch, and this also adds two additional operations in your ISR (be sure to use macros for your start_timer/stop_timer to eliminate function call overhead). If you can get the total number of cycles used for start+stop timer macros, then you can subtract those ticks from the timer1_value to get the ISR ticks value a little more accurately. Your final calculation for % of CPU time used would simply be:

$$Usage_{cpu} = (\frac{Ticks_{isr}}{Ticks_{isr} + Ticks_{idle}}) * 100$$

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+1 for a solution requiring no external hardware. Probably a more cumbersome way to do it, but it fits in my constraints. I just need to learn about timers now. Thanks. –  CMP Nov 7 '11 at 19:44
I used this on an AVR for a similar project. It works very well. –  drxzcl Nov 7 '11 at 22:26

Set an output pin when you enter the ISR and clear it before returning from it. Filter the output with an RC filter. The voltage across the capacitor should give you the ISR duty cycle.
For instance, if your power supply is 3.3V and you measure 33mV, then you spend 1% of the time in the ISR.

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This is exactly what we do, except for the RC filter part. We set an output pin on entering the ISR, then clearing it on return. Looking at this on an o-scope will give all sorts of useful info. –  user3624 Nov 7 '11 at 18:03
@David - I also just watch the pulses on a scope, but OP says he doesn't have one. –  stevenvh Nov 7 '11 at 18:05
It looks like I need to start saving for a good scope. –  CMP Nov 7 '11 at 19:40
@stevenvh, exactly what I was going to suggest when I saw the absence of a scope. This is why those embedded guys love a few extra digital out pins. –  Kortuk Nov 7 '11 at 20:03
Filtering to meter in absence of a scope a nice idea. Turn LED on/off and measure effective brightness :-).(Very hard to see even quite large variations - but cute.) –  Russell McMahon Nov 7 '11 at 22:14
show 4 more comments

The easiest way to do this is to run the code in the simulator and measure the cycles taken by the interrupt routine. The Microchip MPLAB simulator, for example, has a handy stopwatch feature that is very useful for this purpose.

Failing that, raising a pin at the start of the interrupt and lowering it at the end can help. The easiest way to look at that is with a oscilloscope. If you are doing microcontroller and electronics projects, you should get one anyway.

Without a scope, you could simply low pass filter the pin voltage, then measure it with a voltmeter. That voltage divided by the processor power voltage will give you the fraction of the time the pin is high. For example, if the processor is running on 3.3V and the low pass filtered pin voltage is 800mV, then the pin is high 800mV / 3.3V = 24% of the time.

Since you are apparently using a compiler, you need to derate this answer somewhat. The compiler is likely adding some interrupt entry code before your code runs and interrupt exit code after your code runs. The true interrupt time will extend a few cycles on either side of the pulse. Of course if you care about interrupt timing, you shouldn't be using a compiler in the first place.

-


text:
# Prime Problem Solving

What is the largest prime number that is less than 220?

It seems like a pretty simple question. Maybe you think it’s something similar to “What is the capital of Missouri?” And if what you’re expecting is a snap answer, those are very similar questions. They both leave two possible responses: a memorized answer or a guess. However, the ability to memorize lists of facts isn’t a great indicator or intelligence. That makes it a less than ideal test question as the test maker wants to gauge your ability to succeed academically at the next level. You won’t be asked about the capital of Missouri (Jefferson City) on your test, but you may be asked the question I posed above. Why?

Well unlike asking about a capital city, there’s a logical way to go about solving the problem above, even if you haven’t memorized a list of primes. Let’s break it down.

Looking for primes is looking for something that isn’t there. The way to find a number is a prime is to find that it doesn’t have any factors other than 1 and itself. While checking all possible factors may seem like a daunting task, let’s take our current problem and see a shortcut.

The nice thing about factors is that they come in pairs: a larger factor and a smaller factor (or two equally sized factors in the case of a square). Since we have no need to find all factors of the number simply finding one member of a pair is sufficient. If you take the square root of a number, you know that the smaller member of the pair must be less than the square root.

In the case of 220, we should know that 225 is 15 squared, so any number less than 220 that has factors will have a factor less than 15. So that leaves 15 numbers to check, right? Not really?

15- Don’t have to check because it isn’t prime. If 15 is a factor, 3 and 5 will also be factors.

14- Like above, if 14 is a factor, 2 will also be a factor, so no need to check.

13- Prime, so CHECK

12- Not prime, don’t check

11- Prime, so CHECK

10- Not prime, don’t check

9- Not prime, don’t check

8- Not prime, don’t check

7- Prime, so CHECK

6- Not prime, don’t check

5- Prime, so check

4- Not prime, don’t check

3- Prime, so CHECK

2- Prime, so CHECK

1- One will always be a factor, so no need to check

So really we only need to check the possible prime factors less than 15 with our answer choices:

A) 219

B) 217

C) 211

D) 209

E) 201

A- Since the digits add to a multiple of 3, we know this number is a multiple of 3, not prime

B- This one’s less obvious, but when we go to check whether 7 is a factor we see that 30*7=210, and 217 is one more 7 so 217 is not prime.

C- After checking our 6 possible factors we find that 211 is prime and since it is the greatest of our remaining answer choices, it is the correct answer.

Success on your test isn’t based on memorization. It’s based on finding logical ways to break down big problems.

# Trick or Treat

The town of Halloweenville welcomes hundreds of trick-or-treaters every Halloween. There are 1000 houses in the town, with address numbers 1-1000. All of the houses are black except for houses with the following characteristics: Prime numbered houses have an orange stripe on them. Houses whose address number end in 7 have an orange door. Houses whose address number have an integer as a square root have white ghosts painted on them. If a trick-or-treater picks a house at random, what is the probability that that house has both an orange stripe and white ghosts?

A) 0

B) 1/1000

C) 1/500

D) 1/100

E) It cannot be determined

Trick! Sometimes there are problems that require more logic than math. This is one of those problems. Orange-striped houses have prime numbers as addresses. Houses with white ghosts are square numbers. Any square number (other than one) MUST have a factor other than 1 and itself, so there cannot be a house with both features.

Three things are meant to throw you off. First, it’s a word problem. Second, there’s irrelevant information thrown in there. Third, the question asks for a probability which throws many students off.

Keep your cool and analyze what the question is asking for, and you’ll avoid these tricks and grab a nice treat: a correct answer.

# Snap a Picture, Get a Result

There’s a new app out there that allows you to take a picture of an equation with your smartphone, and the app will solve it. Before you start grabbing your phone to download it, here are three important things to consider.

1. It’s a Tool, Not a Magic Cure- Much like a calculator, this app is a tool that can be very useful when used correctly and under the right circumstances. Need to figure out whether the giant TV you want to buy will fit on the wall space you have? Sure, jot the equation and solve away.Trying to figure out how many jelly beans are in the jar to win a prize at the local fair? Write out an equation for your volume estimates and snap the picture. Trying to learn how to solve quadratic equations? Stop right there. If you don’t have the proper underlying knowledge first, you aren’t just taking a shortcut, you’re potentially setting yourself up for disaster.

2. Technology isn’t Perfect- Neither are we! The problem with that combination is that we often expect technology to be perfect and that can lead to bad consequences. If you attempted to type 9 + 2 into your calculator and the result came up 101 you wouldn’t take that answer as truth. You’d recognize that the sort of answer you should get should be around 10, and you might even be able to work backward to find that the mistake you made was entering 99 + 2 on accident. However, if you’re snapping a picture of an equation you don’t really understand and the software interprets the numbers wrong, or you don’t take the picture correctly, you may end up proceeding off a bad result!

3. You’re Never Going to Be Able to Use That on Your Test- We want to test the processor that’s mounted above your shoulders, not carried in your pocket. Using a shortcut when time is of the essence makes sense, but knowing that you don’t need the shortcut when it comes right down to it is even better.

# An Exponent Problem

Consider the following question:

$x=2^{y}-(8^{7}-8^{5})$

Which of the following values of y produces an x that is closest to 0?

A) 24

B) 21

C) 20

D) 16

E) 14

Here’s a question designed for a calculator, you might say. But let’s set the calculator down for a moment and break this down. Realize that a problem like this is designed to be solvable. Let’s look at it step-by-step.

Step 1: Ask yourself how could this be possible? It’s a useful question to ask whenever you encounter a difficult problem, whether inside the test prep world or outside, but it’s an especially useful question to ask in a situation such as this where you know there is some possible and not extremely difficult solution. Here, the answer to the question is that this must be true:

$2^{y}-2^{y}=0$

That’s our path toward a solution.

Step 2: Make the information you have look as much like your projected solution as possible. How do we make a base of 8 into a base of 2?

$8^{7}=(2^{3})^{7}=2^{21}$

So, we can start to break down the second piece of the equation into:

$8^{7}-8^{5}=2^{21}-2^{15}$

Step 3: Analyze. Now we must find what value of $2^{y}$ is closest to the value in the equation above. Again, you might be tempted to pull out a calculator, but resist the urge. $2^{21}-2^{15}$ appears to be an ugly jumble of numbers and symbols. What does it mean? Think about what powers of 2 mean. It means $2^{21}$ is twice as big as $2^{20}$ which is in turn twice as big as $2^{19}$ and so on. $2^{15}$ is only 1/64th as big as $2^{21}$ so subtracting it out doesn’t move us very close to $2^{20}$$2^{21}$ is still the value of y that gets us closest to 0, and B is your correct answer.


text:
# Math Help - Logarithms - help please!

1. ## Logarithms - help please!

Log3x – log3 (x-5) = log3 A
Express A in terms of x.
Hence, solve the equation
Log3 x – log3 (x-5) = 1

Thank you so much.

2. every single 3 is a base, it just didn't show up.

3. Originally Posted by kayteee92
Log3x – log3 (x-5) = log3 A
Express A in terms of x.
Hence, solve the equation
Log3 x – log3 (x-5) = 1

Thank you so much.
Recall that $\log_a X - \log_a Y = \log_a \frac XY$

So that,

$\log_3 x - \log_3 (x - 5) = \log_3 A$

$\Rightarrow \log_3 \frac x{x - 5} = \log_3 A$

Now what?

4. oh okay, so next i did

A = x/x-5

and i crossed multiplied to get x = A(x-5)

but for the next part of the problem, i replaced this for x and now i am stuck.

5. Originally Posted by kayteee92
oh okay, so next i did

A = x/(x-5)
stop there. that is A in terms of x.

but for the next part of the problem, i replaced this for x and now i am stuck.


text:
# Neutron decay at neutron star

1. Jan 20, 2013

### Mordred

On decays I have only a preliminary understanding. I was looking up the half life of a neutron for another reason.
The value I found for neutron decay for a free neutron was 10.3 seconds. Hope thats reasonably accurate.
Anyways knowing that neutrons are stable in a nucleus held together by the strong force.
In the case of neutron stars which are comprised mainly of neutrons which is said to be held by gravity not the strong force.
Is the local gravity allowing neutron stability or does the neutrons have the above mentioned half life?

2. Jan 20, 2013

### snorkack

Wrong order of magnitude. 10 minutes, not seconds.
In neutron stars, it is electron chemical potential that makes neutrons stable. Gravity as such is irrelevant.

3. Jan 20, 2013

### Mordred

Lol I meant minutes not sure why I typed seconds. Anyways thanks for the answer.


text:
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Problem 106

When heated, the DNA double helix separates into two random coil single strands. When cooled, the random coils reform the double helix: double helix $\Longrightarrow 2$ random coils.
(a) What is the sign of $\Delta S$ for the forward process? Why?
(b) Energy must be added to break $\mathrm{H}$ bonds and overcome dispersion forces between the strands. What is the sign of $\Delta G$ for the forward process when $T \Delta S$ is smaller than $\Delta H ?$
(c) Write an expression for $T$ in terms of $\Delta H$ and $\Delta S$ when the reaction is at equilibrium. (This temperature is called the melting temperature of the nucleic acid.)

Check back soon!

## Discussion

You must be signed in to discuss.

## Video Transcript

were given a generic reversible reaction to model. How adding heat to the double helix structure of DNA causes the hydrogen bonds that form that structure to break and therefore caused the two strands to form two different random coils. Report A. We want to determine whether the sign of the change in entropy, where the forward direction of this reaction is positive or negative, and we want to explain how we know. So remember that the total entropy of a given system be correlated to the amount of disorder or randomness in that system, so that an increase in randomness or disorder, is equal to an increase in entropy. If we examine the the reaction that were given, we see that the word random even appears on on the product side. And when we go from left side to the right side in the forward direction, we start with one molecule of DNA double helix, two strands bonded together to two separate strands. So we increase the number of molecules in therefore increase molecular complexity and randomness. And that means that to go to go in the forward direction for this reaction, the increase entropy and that's why the change is going to be positive. Since we increase entropy, we have a higher ah, higher value for the entropy on the products than on the reactant. In the change in entropy is the entropy of the products, minus the entropy of the reactant. And if we have a higher value for the entropy of the products than the reactant stand, when we subtract them, Delta s comes out to be positive. And so that's the explanation as to how we know that the change in entropy Delta s for this. This generic reaction will be positive. And part B. We want to know, based on the equation, Delta G equals Delta H minus Tito s. We're told that this product of the temperature in the change in entropy is is smaller than the value for Delta H. And so because of that, we say we can plug in, we can plug in some numbers in order to qualitatively see whether Delta G will be positive or negative. We know that Delta H is greater than t Delta s. We're subtracting t Delta s from Delta H. So if Delta H were to equal to and Tito s would equal one. When we subtract them, we get a positive value of one. So since we know that Delta H is greater than T Delta s for any values we plug in for those the result meaning Delta G will always be positive, meaning that react, that reaction will always be non spontaneous under those conditions. So that comes out to a positive value. We're Delta G, meaning it's greater than zero. Finally, for part C, we want to arriving expression for the temperature. He in terms of Delta H and Delta s. When this reaction is at equilibrium member that when when a system is an equilibrium, a Delta G equals zero. When that's the case, we can get rid of this term Delta G instead of equal to zero and move over the product of Tea Delta s to the other side so that we can read out this equation now, as Delta H equals T Delta s. And now we just need to isolate T by dividing over Delta s and our final equation for he at equilibrium is still to each divided by Delta s


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# Integer solution to a square root equation

Given the following function: $$f(x) = \sqrt{36 x^2+16416 x +61084}$$ where x is an integer and the function also generates an integer value, is there an algorithm to determine its integer solutions?

-
What do you mean by "solution"? Do you mean the x for which $f(x) =0$? –  DavidButlerUofA Jul 16 '14 at 20:05
Your question is a little unclear: are you asking for solutions to "$f(x) = 0$"? Or "$f(x)$ is an integer"? Or something else? –  Hurkyl Jul 16 '14 at 20:06
The term $72*x*228$ also hints at a typo. –  hardmath Jul 16 '14 at 20:07
Sorry , I can not figure out what to do with the 228 –  bobbym Jul 16 '14 at 20:14
Sorry but the function f(x) generates an integer value. Also 72*x*228 = 16416*x. –  Nominal Jul 16 '14 at 21:00

Given

$$f(x) = \sqrt{ 36 x^2 + 16416 x + 61084}.$$

For what integer $x$ is $f(x)$ also an integer?

We can write

$$f(x) = \sqrt{ \Big( 6 x + 1368 \Big)^2 - 1810340 },$$

therefore

$$\Big( 6 x + 1368 \Big)^2 - 1810340 = \Big( 6 x + 1368 - p \Big)^2,$$

so

$$p \left\{ 2 \Big( 6 x + 1368 \Big) - p \right\} = 1810340.$$

Thus

$$x = \left( \left( \frac{1810340}{p} + p \right) \Big/ 2 - 1368 \right) \Big/ 6.$$

Note the symmetry

$$\frac{1810340}{p} + p = \frac{1810340}{1810340/p} + 1810340/p.$$

Using

$$1810340 = 2^2 \times 5 \times 7 \times 67 \times 193,$$

we can use limited values for $p$, like $p=2 \times 5 \times 7$, which is equivalent with $p= 2 \times 67 \times 193$.

We obtain the following results:

$$\begin{array}{ccccc|c|c} 2 & 5 & 7 & 67 & 193 & p & x\\ \hline \times &&&&& 2 &75203\\ \times & \times &&&& 10 &14859\\ \times && \times &&& 14 &10549\\ \times & \times & \times &&& 70 &1933\\ \times &&& \times && 134 &909\\ \times &&&& \times & 386 &195\\ \times & \times && \times && 670 &53\\ \times & \times &&& \times & 1930 &11\\ \end{array}$$

-

You can write $f(x) = \sqrt{(6(x-228))^2 - 1810340}$.

Now, $f(x)$ is an integer if and only if $(6(x-228))^2 - 1810340$ is a square. Write $$k^2=(6(x-228))^2 - 1810340$$ so $$1810340 = (6(x-228))^2 - k^2 = (6x-6*228+k)(6x-6*228-k)$$

Now you factorize $1810340 = 2^2 * 5* 7* 67* 193$.

So you have to solve $48$ linear systems of the form $$\left\{ \begin{matrix} 6x-6*228&+k &=& a \\ 6x-6*228&-k &=& b \end{matrix} \right.$$

For all possible values of $ab = 1810340$.

-
$f(6195)$ is in fact not an integer (I just plugged it into Wolfram). One potential issue I see with your proof is that $x$ and $k$ are not independent, so you cannot simply solve the last system of equations. –  Divergent Queries Jul 16 '14 at 21:18
No, I did a worse mistake. I made a mistake in completing the square. In fact, if you compute $(3*6195-228)^2-36713$ you get exactly $18356^2$. Ok, sorry, I made a mistake but I'm not erasing my answer. –  Crostul Jul 16 '14 at 21:37
That's fine. Mind if I edit it, though? –  Divergent Queries Jul 16 '14 at 21:39
I am editing right now. –  Crostul Jul 16 '14 at 21:41
It should be $\sqrt{(6(x\color{red}{+}228))^2 - 1810340}$... –  johannesvalks Jul 16 '14 at 21:57


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# Formally: Let f be a function defined on...

• Lab Report
• 4

This preview shows page 1 - 2 out of 4 pages.

MA103 Lab Report 6 Prep Page 1 MA103 Lab Report 6 Prep – Existance Theorems; Curve Sketching Maximum and Minimum Function Values Formally: Let f be a function defined on interval I . Then f ( c ), c I , is the absolute maximum value of f if f ( c ) f ( x ) for all x I , and f ( d ), d I , is the absolute minimum value of f if f ( d ) f ( x ) for all x I . ii) Evaluate f at all critical values in step 1 and evaluate f ( a ) and f ( b ).


text:
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Anu, Vinu and Manu can complete a work alone in $15 \; \text{days}, 12 \; \text{days}$ and $20 \; \text{days},$ respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is

1. $6$
2. $5$
3. $8$
4. $7$

We can draw a table for better understanding.

$$\begin{array} {llll} & \text{Anu} & \text{Vinu} & \text{Manu} \\\hline \text{Time :} & 15\;\text{days} & 12 \;\text{days} & 20 \;\text{days} \\ \text{Total work :} & \text{LCM(15,12,20)} & = & 60\;\text{units} \\ \text{Efficiency :} & 4\;\text{units/day} & 5\;\text{units/day} & 3 \;\text{units/day} \end{array}$$

Now,

• First day: Vinu  + Anu $= 5+4 =9\;\text{units}$
• Second day: Vinu  + Manu $= 5+3 =8\;\text{units}$
• Third day: Vinu  + Anu $= 5+4 =9\;\text{units}$
• Fourth day: Vinu  + Manu $= 5+3 =8\;\text{units}$
• Fifth day: Vinu  + Anu $= 5+4 =9\;\text{units}$
• Sixth day: Vinu  + Manu $= 5+3 =8\;\text{units}$
• Seven day: Vinu  + Anu $= 5+4 =9\;\text{units}$

Total work $= 60\; \text{units}$

$\text{(Or)}$

In odd days, Vinu + Anu $=5+4 = 9\;\frac{\text{units}}{\text{day}}$

In even days, Vinu + Manu $=5+3 = 8\;\frac{\text{units}}{\text{day}}$

• $2$ days $\longrightarrow 17\;\text{units}$
• $6$ days $\longrightarrow 51\;\text{units}$

In the Seventh day (Vinu + Anu) $= 5+4 = 9\;\frac{\text{units}}{\text{day}}$

• $7$ days $\longrightarrow 60\;\text{units}$

$\therefore$ The number of days needed to complete the work is $7$ days.

Correct Answer $:\text{D}$

10.3k points

1 vote
1
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text:
Lemma 39.21.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. The category of groupoid schemes cartesian over $(U, R, s, t, c)$ is equivalent to the category of pairs $(V, \varphi )$ where $V$ is a scheme over $U$ and

$\varphi : V \times _{U, t} R \longrightarrow R \times _{s, U} V$

is an isomorphism over $R$ such that $e^*\varphi = \text{id}_ V$ and such that

$c^*\varphi = \text{pr}_1^*\varphi \circ \text{pr}_0^*\varphi$

as morphisms of schemes over $R \times _{s, U, t} R$.

Proof. The pullback notation in the lemma signifies base change. The displayed formula makes sense because

$(R \times _{s, U, t} R) \times _{\text{pr}_1, R, \text{pr}_1} (V \times _{U, t} R) = (R \times _{s, U, t} R) \times _{\text{pr}_0, R, \text{pr}_0} (R \times _{s, U} V)$

as schemes over $R \times _{s, U, t} R$.

Given $(V, \varphi )$ we set $U' = V$ and $R' = V \times _{U, t} R$. We set $t' : R' \to U'$ equal to the projection $V \times _{U, t} R \to V$. We set $s'$ equal to $\varphi$ followed by the projection $R \times _{s, U} V \to V$. We set $c'$ equal to the composition

\begin{align*} R' \times _{s', U', t'} R' & \xrightarrow {\varphi , 1} (R \times _{s, U} V) \times _ V (V \times _{U, t} R) \\ & \xrightarrow {} R \times _{s, U} V \times _{U, t} R \\ & \xrightarrow {\varphi ^{-1}, 1} V \times _{U, t} (R \times _{s, U, t} R) \\ & \xrightarrow {1, c} V \times _{U, t} R = R' \end{align*}

A computation, which we omit shows that we obtain a groupoid scheme over $(U, R, s, t, c)$. It is clear that this groupoid scheme is cartesian over $(U, R, s, t, c)$.

Conversely, given $f : (U', R', s', t', c') \to (U, R, s, t, c)$ cartesian then the morphisms

$U' \times _{U, t} R \xleftarrow {t', f} R' \xrightarrow {f, s'} R \times _{s, U} U'$

are isomorphisms and we can set $V = U'$ and $\varphi$ equal to the composition $(f, s') \circ (t', f)^{-1}$. We omit the proof that $\varphi$ satisfies the conditions in the lemma. We omit the proof that these constructions are mutually inverse. $\square$

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In this post, we will discover the three different ways you could use the delete specifier in C++. We are going to see how

• you can disallow an object from being copied
• you can limit what kind of implicit conversions you allow for a function call
• you can limit what kind of template instantiations you allow

## How to disallow copying/moving for a class?

The first question to answer is why would you need such a feature? You might not want a class to be copied or moved, so you want to keep related special functions unreachable for the caller.

In order to achieve this, there is a legacy and a modern option.

The legacy option is to declare them as private or protected and the modern one (since C++11) is that you explicitly delete them.

class NonCopyable {
public:
NonCopyable() {/*...*/}
// ...
private:
NonCopyable(const NonCopyable&); //not defined
NonCopyable& operator=(const NonCopyable&); //not defined
};

Before C++11 there was no other option than declaring the unneeded special functions private and not implementing them. As such one could disallow copying objects (there was no move semantics available back in time). The lack of implementation/definition helps against accidental usages in member functions, friends, or when you ignore the access specifiers. It doesn’t cause a compile-time failure, you’ll face a problem at linking time.

Since C++11 you can simply mark them deleted by declaring them as = delete;

class NonCopyable {
public:
NonCopyable() {/*...*/}
NonCopyable(const NonCopyable&) = delete;
NonCopyable& operator=(const NonCopyable&) = delete;
// ...
private:
// ...
};

The C++11 way is a better approach because

• it’s more explicit than having the functions in the private section which might only be a mistake by the developer
• in case you try to make a copy, you’ll already get an error at compilation time

It’s worth to note that deleted functions should be declared as public, not private. In case, you make them private some compilers might only complain about that you call a private function, not that a deleted one.

## How to disallow implicit conversions for function calls?

You have a function taking integer numbers. Whole numbers. Let’s say it takes as a parameter how many people can sit in a car. It might be 2, there are some strange three-seaters, for some luxury cars it’s 4 and for the vast majority, it’s 5. It’s not 4.9. It’s not 5.1 or not even 5 and a half. It’s 5. We don’t traffic body parts.

How can you enforce that you only receive whole numbers as a parameter?

Obviously, you’ll take an integer parameter. It might be int, even unsigned or simply a short. There are a lot of options. You probably even document that the numberOfSeats parameter should be an integral number.

Great!

So what happens if the client call still passes a float?

#include <iostream>

void foo(int numberOfSeats) {
std::cout << "Number of seats: " << numberOfSeats << std::endl;
// ...
}

int main() {
foo(5.6f);
}
/*
Number of seats: 5
*/

The floating-point parameter is accepted and narrowed down into an integer. You cannot even say that it’s rounded, it’s implicitly converted, narrowed down into an integer.

You might say that this is fine and in certain situation it probably is. But in others, this behaviour is simply not acceptable.

What can you do in such cases to avoid this problem?

You might handle it on the caller side, but

• if foo is often used, it’d tedious to do the checks at each call and code reviews are not reliable enough,
• if foo is part of an API used by the external world, it’s out of your control.

As we have seen in the previous section, since C++11, we can use the delete specifier in order to restrict certain types from being copied or moved. But = delete can be used for more. It can be applied to any functions, member or standalone.

If you don’t want to allow implicit conversions from floating-point numbers, you can simply delete foo’s overloaded version with a float:

#include <iostream>

void foo(int numberOfSeats) {
std::cout << "Number of seats: " << numberOfSeats << std::endl;
// ...
}

void foo(double) = delete;

int main() {
// foo(5);
foo(5.6f);
}

/*
main.cpp: In function 'int main()':
main.cpp:12:13: error: use of deleted function 'void foo(double)'
12 |     foo(5.6f);
|             ^
main.cpp:8:6: note: declared here
8 | void foo(double) = delete;
|      ^~~
*/

Et voila! - as the French would say. That’s it. By deleting some overloads of a function, you can forbid implicit conversions from certain types. Now, you are in complete control of the type of parameters your users can pass through your API.

## How to disallow certain instantiations of a template

This kind approach also works with templates, you can disallow the instantiations of your templated function with certain types:

template <typename T>
void bar(T param) { /*..*/ }

If you call this function, let’s say with an integer, it will compile just fine:

bar<int>(42);

However, you can delete the instantiation with int, and then you receive a similar error message compared to the previous one:

#include <iostream>

template <typename T>
void bar(T param) { /*..*/ }

template <>
void bar<int>(int) = delete;

int main() {
bar<int>(5);
}
/*
main.cpp: In function ‘int main()’:
main.cpp:10:15: error: use of deleted function ‘void bar(T) [with T = int]’
10 |     bar<int>(5);
|               ^
main.cpp:7:6: note: declared here
7 | void bar<int>(int) = delete;
|      ^~~~~~~~
*/

Just keep in mind, that T and const T are different types and if you delete one, you should consider deleting the other too. This is only valid for the templates, not when you delete function overloads.

## Conclusion

Today we saw 3 ways how to use the delete specifier that is available for us since C++11. We can make classes non-copyable and/or non-movable with its help, but we can also disallow implicit conversions for function parameters and we can even disallow template instantiations for any type. It’s a great tool to create a tight, strict API that is difficult to misuse.


text:
# Laboratoire de Mécanique des Fluides et d’Acoustique - UMR 5509

LMFA - UMR 5509
Laboratoire de Mécanique des Fluides et d’Acoustique
Lyon
France

## Nos partenaires

Alexandre Delache

## Isotropie rétablie en turbulence avec effet de stratification et de rotation : role de l’échelle d’Ozmidov et de Hopfinger

Jeudi 16 juin 2016 13h, UCBL Bât Omega

In oceanic, atmospheric or engineering flows, turbulence can be strongly affected by stratification and rotation. In contrast with isotropic turbulence, anisotropic structures emerge in stably stratified turbulence or in rotating turbulence :
quasi-horizontal structures appear to be organized in vertically sheared layers. These structures are a mix between internal wave and turbulence.
quasi-vertical structures appear as column of Taylor. These structures are a mix between inertial wave and turbulence.

Introducing the turbulent dissipation rate eps , two length scales can be obtained, namely the Ozmidov length scale [1] $L_o = \sqrt{\epsilon/N^3}$ and the Hopfinger length scale [2-3] $L_h = \sqrt{\epsilon/f^3}$. Lo and Lh compare the relative effects of inertia and of the buoyancy force or of the Coriolis force respectively, and thus quantify the rise of anisotropy in different scale ranges : at large scales (larger than Lo or Lh ) the anisotropy due to strong stratification or strong rotation is dominant, whereas at small scales (smaller than Lo or Lh ), universal 3D isotropic characteristics of turbulence appear to be restored.

To confirm directly the role of these two scales, we performed numerical simulations at high resolution (2048³ points) in freely decaying turbulence at four different stratification rates and six rotating rates. We confirm the role played by Lo and Lh by considering the angular energy spectra.

[1] R.V. Ozmidov, Izvestia Acad. Sci. USSR, Atmosphere and Ocean Physics, 1965,N 8
[2] Mory, M. & Hopfinger, Proceedings of a Workshop Held at INRIA, Sophia-Antipolis, Fance, December 10–14, 1984, Springer Berlin Heidelberg, E. Frisch, U. ; Keller, J. B. ; Papanicolaou, G. C. & Pironneau, O. (Eds.), 1985, 218-236
[3] O. Zeman,Phys. Fluids 6, 3221 (1994).


text:
1. ## power problem

hi this is my first post over here i wish i can get help with this

well i got this problem in my homework and i cant find a way to solve it
well the problem is
find the last digit of 7^1000000 and 11^1000000

so if plz if anyone has idea about how to solve it that would be great

thx

2. Originally Posted by spywx
hi this is my first post over here i wish i can get help with this

well i got this problem in my homework and i cant find a way to solve it
well the problem is
find the last digit of 7^1000000 and 11^1000000

so if plz if anyone has idea about how to solve it that would be great

thx
$7^{100000}=49^{50000}=(40+9)^{50000}$

So the last digit of $7^{100000}$ is the same as the last digit of $9^{50000}$.

Now repeat the process:

$9^{50000}=81^{25000}=(80+1)^{25000}$

So the last digit of $9^{50000}$ is the same as the last digit of $1^{25000}$ which is $1$.

So the last digit of $7^{100000}$ is $1$.

RonL

3. Hello, spywx!

Welcome aboard!

Find the last digit of: $7^{1,000,000}\text{ and }11^{1,000,000}$
Did you try anything?

I would try consecutive powers of 7 and see what happens . . .

$\begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline
7^1 & \to & 7 \\
7^2 & \to & 9 \\
7^3 & \to & 3 \\
7^4 & \to & 1\\
\vdots & & \vdots \end{array}$

We see that $7^4$ ends in 1.

We have: . $7^{1,000,000} \;=\;(7^4)^{250,000} \:\to\:1^{250,000} \:\to\:1$

Try consecutive powers of 11 . . .

$\begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline
11^1 & \to & 1 \\
11^2 & \to & 1 \\
11^3 & \to & 1 \\
& \text{Hey!} \end{array}$

4. Originally Posted by Soroban
Hello, spywx!

Welcome aboard!

Did you try anything?

I would try consecutive powers of 7 and see what happens . . .

$\begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline
7^1 & \to & 7 \\
7^2 & \to & 9 \\
7^3 & \to & 3 \\
7^4 & \to & 1\\
\vdots & & \vdots \end{array}$

We see that $7^4$ ends in 1.

We have: . $7^{1,000,000} \;=\;(7^4)^{250,000} \:\to\:1^{250,000} \:\to\:1$

Try consecutive powers of 11 . . .

$\begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline
11^1 & \to & 1 \\
11^2 & \to & 1 \\
11^3 & \to & 1 \\
& \text{Hey!} \end{array}$

no i did nt try this way b4 i was using totally different method, but your seems to be reasonable
so lets take another example i have 9^1000000 == (9^4)^250000

9^4 ==6561 last digit 1 then (9^4)^250000 last digit also must be 1

thx man

5. Originally Posted by spywx
no i did nt try this way b4 i was using totally different method, but your seems to be reasonable
so lets take another example i have 9^1000000 == (9^4)^250000

9^4 ==6561 last digit 1 then (9^4)^250000 last digit also must be 1

thx man
So you came on a forum asking for help on Prof. Boklan's question huh?

ah the power of google XD

6. Originally Posted by ramelan
So you came on a forum asking for help on Prof. Boklan's question huh?

ah the power of google XD
x 2 LOL

7. Originally Posted by ramelan
So you came on a forum asking for help on Prof. Boklan's question huh?

ah the power of google XD


text:
# Should Dirac's delta function be regarded as a subclass of the Gaussian distribution?

In Wikidata it is possible to link probability distributions (like everything else) in an ontology, e.g., that the t-distribution is a subclass of the noncentral t-distribution, see, e.g.,

https://angryloki.github.io/wikidata-graph-builder/?property=P279&item=Q209675&iterations=3&limit=3

There are various limiting cases, e.g., when the degrees of freedom in the t-distribution goes to infinity or when the variance approaches zero for the normal distribution (Gaussian distribution). In the latter case the distribution will go towards Dirac's delta function.

I note that on the English Wikipedia the variance parameter is presently stated as larger than zero, so with a strict interpretation one would not say that the Dirac's delta function is a subclass of the normal distribution. However, to me it seems quite ok, as I would say that the exponential distribution is a superclass of the Dirac's delta function.

Are there any problems with stating that Dirac's delta function is a subclass of the Gaussian distribution?

• IF dirac delta is a subclass of gaussian then its kurtosis must be 3, right? – Aksakal Sep 7 '16 at 22:57
• I guess that if we regard the Dirac delta as a subclass of several probability distributions, then the kurtosis is inconsistent for the Dirac Delta. It speaks against regarding the Dirac delta as a subclass of any of these distributions. – Finn Årup Nielsen Sep 8 '16 at 8:46
• In probability context delta is described as a generalized function. It's not am ordinary function – Aksakal Sep 8 '16 at 11:16

Dirac's delta is regarded as a Gaussian distribution when it is convenient to do so, and not so regarded when this viewpoint requires us to make exceptions.

For example, $(X_1, X_2, \ldots, X_n)$ are said to enjoy a multivariate Gaussian distribution if $\sum_i a_iX_i$ is a Gaussian random variable for all choices of real numbers $a_1, a_2, \ldots, a_n$. (Note: this is a standard definition in "advanced" statistics). Since one choice is $a_1=a_2=\cdots=a_n=0$, the standard definition treats the constant $0$ (a degenerate random variable) as a Gaussian random variable (with mean and variance $0$). On the other hand, we ignore our regard for the Dirac delta as a Gaussian distribution when we are considering something like

"The cumulative probability distribution function (CDF) of a zero-mean Gaussian random variable with standard deviation $\sigma$ is $$F_X(x) = P\{X \leq x\} = \Phi\left(\frac{x}{\sigma}\right)$$ where $\Phi(\cdot)$ is the CDF of a standard Gaussian random variable."

Note that this statement is almost right but not quite right if we regard the Dirac delta as the limiting case of a sequence of zero-mean Gaussian random variables whose standard deviation approaches $0$ (and hence as a Gaussian random variable). The CDF of the Dirac delta has value $1$ for $x \geq 0$ whereas $$\lim_{\sigma\to 0}\Phi\left(\frac{x}{\sigma}\right) = \begin{cases} 0, & x < 0,\\ \frac 12, & x = 0,\\ 1, & x > 0.\end{cases}$$ But, lots of people will tell you that regarding a Dirac delta as a Gaussian distribution is sheer nonsense since their book says that the variance of a Gaussian random variable must be a positive number (and some of them will down-vote this answer to show their displeasure). There was a very vigorous and illuminating discussion of this point a few years ago on stats.SE but unfortunately it was only in the comments on an answer (by @Macro, I believe) and not as individual answers, and I cannot find it again.

• +1. I'm not sure there's a problem concerning the CDF, because I believe the limiting value of a sequence of CDFs at any jump of the limit doesn't matter. There are two ways to see that. One is to note that your limiting formula is not a valid CDF (it is not cadlag). Another is to note that you obtain a Dirac distribution at $0$ when you let $(\mu,\sigma)\to(0,0)$ simultaneously, but you can contrive to have the limiting value of $\Phi_{\mu,\sigma}(0)$ be anything between $0$ and $1$ (or not to have a limit at all). – whuber Feb 16 '16 at 20:30
• The conversation you reference happened in the comments of this answer, though I sincerely hope that to most readers the discussion will not appear too vigorous. (+1) – cardinal Feb 17 '16 at 0:50
• @cardinal Deep knowledge of our community. Well done! – Matthew Drury Feb 17 '16 at 1:21

The delta functions fit into a mathematical theory of distributions (which is quite distinct from the theory of probability distributions, terminology here could not be more confusing).

Essentially, distributions are generalized functions. They cannot be evaluated like a function can, but then can be integrated. More precisely, a distribution $D$ is defined as follows

Let $T$ be the collection of test functions. A test function $\theta$ is a true, honest to god function, smooth, with compact support. A distribution is a linear mapping $D: T \rightarrow \mathbb{R}$

An honest function $f$ determines a distribution by the integration operator

$$T(\theta) = \int_{-\infty}^{+\infty} f(x)\theta(x) dx$$

There are distributions that are not associated to true functions, the dirac operator is one of them

$$\delta(\theta) = \theta(0)$$

In this sense, you can consider the dirac a limiting case of the normal distributions. If $N_t$ is the family of pdf's of normal distributions with mean zero and variance $t$, then for any test function $\theta$

$$\theta(0) = \lim_{t \rightarrow 0} \int_{-\infty}^{+\infty} N_t(x) \theta(x) dx$$

This is probably more commonly expressed as

$$\theta(0) = \int_{-\infty}^{+\infty} \delta(x) \theta(x) dx = \lim_{t \rightarrow 0} \int_{-\infty}^{+\infty} N_t(x) \theta(x) dx$$

which a mathematician would consider an abuse of notation, because the expression $\delta(x)$ does not actually make any sense. But then again, who am I to criticize Dirac, who is the best.

Of course, whether this makes the dirac a member of the family of normal distributions is a cultural question. Here I'm just giving a reason why it may make sense to consider it so.

• Whilst I agree with your statements, I think this implies the opposite. A delta function is not a subset of gaussians. Just as a limit of continuous functions need not be a continuous function. – seanv507 Feb 16 '16 at 22:38
• @seanv507 I did my best to not state a conclusion either way! – Matthew Drury Feb 17 '16 at 1:24
• I thought distributions are very much like probability distributions, with a Dirac delta (probability) distribution indicating a deterministic variable... – Mehrdad Feb 17 '16 at 6:37
• If you don't write the limits of the integrals they might be confused indefinite integrals. Also, this sentence doesn't make sense: "A test function θ is a true, honest to god function, smooth, with compact support". – jkabrg Sep 7 '16 at 20:09
• @jkabrg Why doesn't it make sense? Since I wrote it, it's hard for me to see it not making sense. – Matthew Drury Sep 7 '16 at 20:25

No. It's not a subclass of normal distribution.

I think the confusion comes from one of the representations of Dirac function. Remember that it's defined as follows:

$$\int_{-\infty}^\infty\delta(x)dx=1$$ $$\delta(x)=0,\forall x\ne 0$$

It's defined as an integral, which is great but sometime you need to operationalize it by a function representation rather than an integral. So, people came up with all kinds of alternatives, one of them looks like Gaussian density: $$\delta(x)=\lim_{\sigma\to 0} \frac{e^{\frac{-x^2}{2\sigma^2}}}{\sqrt{2\pi}\sigma}$$

However, this is not the only representation, e.g. there's this one: $$\delta(x)=\frac{1}{2\pi}\sum_{k=-\infty}^\infty e^{ikx}, \forall x\in (-\pi,\pi)$$

Hence, it's best to think of Dirac function in terms of its integral definition, and take the function representations, such as Gaussian, as tools of convenience.

UPDATE To @whuber's point, a better even example is this representation of Dirac's delta:

$$\delta(x)=\lim_{\sigma\to 0} \frac{e^{-\frac{|x|}{\sigma}}}{2\sigma}$$

Does this look like Laplacian distribution to you? Shouldn't we consider then Dirac's delta as a subclass of Laplacian distribution?

• At some point in this answer you seem to switch from discussing distributions to discussing "functions." The question refers explicitly to "probability distributions." Those are not generally given by density functions, but can always be given by their distribution function. The distribution of an atom--the "Dirac delta"--fits beautifully in with all the other Gaussian distributions as a limiting case. (In Matthew Drury's setup, it is defined as that limit!) Your argument seems similar to claiming that, say, circles are not ellipses. Enforcing such exceptions doesn't seem constructive. – whuber Sep 7 '16 at 18:19
• @whuber, what's "distribution of an atom"? – Aksakal Sep 7 '16 at 18:35
• An "atom" is a lump of probability at a single point. Equivalently, it is the distribution of any random variable that is constant almost everywhere. – whuber Sep 7 '16 at 18:36
• @whuber, Oh, I was thinking of a physical atom. No, my point is that Dirac's delta is not a subclass of Gaussian, because it can be represented also by Laplacian like distros – Aksakal Sep 7 '16 at 18:45
• Re: your point about Laplace distributions. Just as a square is both a rectangle and a rhombus, and the Uniform$(0,1)$ distribution is both a special case of a Uniform$(0,\theta)$ distribution and a Beta$(\alpha,\beta)$ distribution, a distribution can belong to multiple named families of distributions. The delta distributions in fact belong to every location-scale family and at least one delta distribution belongs to every scale family. Geometrically, families are curves in a space of distributions; a given distribution is a point; and (obviously) any point may belong to many curves. – whuber Sep 7 '16 at 19:21


text:
# Probability space

#### Did you know...

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The definition of the probability space is the foundation of probability theory. It was introduced by Kolmogorov in the 1930s. For an algebraic alternative to Kolmogorov's approach, see algebra of random variables.

## Definition

A probability space $(\Omega, \mathcal F, P)$ is a measure space with a measure P that satisfies the probability axioms.

The sample space $\Omega,$ is a nonempty set whose elements are known as outcomes or states of nature and are often given the symbol $\omega.$The set of all the possible outcomes of an experiment is known as the sample space of the experiment.

### Events

The second item, $\mathcal F$, is a σ-algebra of subsets of $\Omega$. Its elements are called events, which are sets of outcomes for which one can ask a probability.

Because $\mathcal F$ is a σ-algebra, it contains $\Omega$; also, the complement of any event is an event, and the union of any (finite or countably infinite) sequence of events is an event.

Usually, the events are the Lebesgue-measurable or Borel-measurable sets of real numbers.

### Probability measure

The probability measure $P$ is a function from $\mathcal F$ to the real numbers that assigns to each event a probability between 0 and 1. It must satisfy the probability axioms.

Because $P$ is a function defined on $\mathcal F$ and not on $\Omega$, the set of events is not required to be the complete power set of the sample space; that is, not every set of outcomes is necessarily an event.

When more than one measure is under discussion, probability measures are often written in blackboard bold to distinguish them. When there is only one probability measure under discussion, it is often denoted by Pr, meaning "probability of".

## Related concepts

### Probability distribution

Any probability distribution defines a probability measure.

### Random variables

A random variable X is a measurable function from the sample space $\Omega$; to another measurable space called the state space.

If X is a real-valued random variable, then the notation ${\scriptstyle\Pr(X \geq 60)}$ is shorthand for ${\scriptstyle\Pr(\{ \omega \in \Omega \mid X(\omega) \geq 60 \})}$, assuming that ${\scriptstyle X \geq 60}$ is an event.

### Conditional probability

Kolmogorov's definition of probability spaces gives rise to the natural concept of conditional probability. Every set $A$ with non-zero probability (that is, P(A) > 0 ) defines another probability measure

$P(B \vert A) = {P(B \cap A) \over P(A)}$

on the space. This is usually read as the "probability of B given A".

### Independence

Two events, A and B are said to be independent if P(AB)=P(A)P(B).

Two random variables, X and Y, are said to be independent if any event defined in terms of X is independent of any event defined in terms of Y. Formally, they generate independent σ-algebras, where two σ-algebras G and H, which are subsets of F are said to be independent if any element of G is independent of any element of H.

The concept of independence is where probability theory departs from measure theory.

### Mutual exclusivity

Two events, A and B are said to be mutually exclusive or disjoint if P(AB)=0. (This is weaker than AB=, which is the definition of disjoint for sets).

If A and B are disjoint events, then P(AB)=P(A)+P(B). This extends to a (finite or countably infinite) sequence of events. However, the probability of the union of an uncountable set of events is not the sum of their probabilities. For example, if Z is a normally distributed random variable, then P(Z=x) is 0 for any x, but P(Z is real)=1.

The event AB is referred to as A AND B, and the event AB as A OR B.

## Examples

### First example

If the space concerns one flip of a fair coin, then the outcomes are heads and tails:

$\Omega = \{H,T\}$

The events are

• {T}: tails,
• {}: neither heads nor tails, and

So, $F=\{\{H\},\{T\},\{\},\{H,T\}\}.$

There is a fifty percent chance of tossing either heads or tail: P({H}) = P({T}) = 0.5. The chance of tossing neither is zero: P({})=0, and the chance of tossing one or the other is one: P({H,T})=1.

### Second example

If 100 voters are to be drawn randomly from among all voters in California and asked whom they will vote for governor, then the set of all sequences of 100 Californian votes would be the sample space Ω.

The set of all sequences of 100 Californian voters in which at least 60 will vote for Schwarzenegger is identified with the "event" that at least 60 of the 100 chosen voters will so vote.

Then, $\mathcal F$ contains: (1) the set of all sequences of 100 where at least 60 vote for Schwarzenegger; (2) the set of all sequences of 100 where fewer than 60 vote for Schwarzenegger (the converse of (1)); (3) the sample space Ω as above; and (4) the empty set.

An example of a random variable is the number of voters who will vote for Schwarzenegger in the sample of 100.


text:
# Find Antiderivatives

dfklajsdfald
Member warned that an effort must be shown

## Homework Statement

find the anti-derivative of xe ^x so its x to the power of e to the power of x

## The Attempt at a Solution

i have 0 idea where to even start. this was a question on my quiz today

Homework Helper
Hi dfk,

No homework. Still an attempt on your part required $$F'(x) = x^{e^x} \ \Rightarrow \ F(x) = ??$$

dfklajsdfald
well i know that the derivative of ex is ex so the anti derivative of ex is ex and then my attempt would be exx but i don't know if that's right or even a good start

dfklajsdfald
Hi dfk,

No homework. Still an attempt on your part required $$F'(x) = x^{e^x} \ \Rightarrow \ F(x) = ??$$
actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it

Mentor
actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it
So is ##\frac d{dx}(x^{e^x + 1}) = x^{e^x}##?

For a power function, such as ##x^n##, the derivative is ##nx^{n - 1}##, but is that what you have here?

Homework Helper
Gold Member
Do you confirm that the formula in #2 is the one you meant?

Then you don't need to ask us whether a proposed solution is correct, surely you can differentiate?

If then you find it wrong, It may at least suggest something else to try. Or maybe some other direction to look.

dfklajsdfald
Do you confirm that the formula in #2 is the one you meant?

Then you don't need to ask us whether a proposed solution is correct, surely you can differentiate?

If then you find it wrong, It may at least suggest something else to try. Or maybe some other direction to look.

i don't know what ur asking, the question is find the antiderivative of the function x to the power of e to the power of x as stated at the start. I am not asking whether or not a solution is correct I am asking how abouts would i go to try and get the antiderivative because I've tried multiple ways (one which i have stated in a different reply) and can't seem to get there. I am only asking for guidance not for an answer of whether or not my solution is correct.

dfklajsdfald
So is ##\frac d{dx}(x^{e^x + 1}) = x^{e^x}##?

For a power function, such as ##x^n##, the derivative is ##nx^{n - 1}##, but is that what you have here?
what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i do xn+1 divided by the power instead? so wouldn't it be xe^x +1 all divided by ex(the plus one is added to ex not to the entire function I am just having trouble showing it up on here properly)

Mentor
what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i do xn+1 divided by the power instead?
That's not how it works -- you are applying the power rule for a function for which it doesn't apply.
Power rule: ##\frac d{dx} x^n = nx^{n - 1}##
Here the exponent, n, is a constant. For your function the exponent, ##e^x + 1## is not a constant.

By your reasoning, ##\frac d{dx} x^x =xx^{x - 1} = x^x##, which isn't true. The derivative of ##x^x## is actually ##x^x(\ln x + 1)##.
dfklajsdfald said:
so wouldn't it be xe^x +1 all divided by ex(the plus one is added to ex not to the entire function I am just having trouble showing it up on here properly)

cnh1995
Homework Helper
Dearly Missed
what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i do xn+1 divided by the power instead? so wouldn't it be xe^x +1 all divided by ex(the plus one is added to ex not to the entire function I am just having trouble showing it up on here properly)

To get the derivative, write ##x^{e^x}= e^{f(x)}##, where ##f(x) = \ln(x) e^x##, and use the chain rule.

Homework Helper
Gold Member
i don't know what ur asking, the question is find the antiderivative of the function x to the power of e to the power of x as stated at the start. I am not asking whether or not a solution is correct I am asking how abouts would i go to try and get the antiderivative because I've tried multiple ways (one which i have stated in a different reply) and can't seem to get there. I am only asking for guidance not for an answer of whether or not my solution is correct.

I thought and rereading #3 it still looks to me you were asking whether an answer was correct. We do often get to ask that question for integrals and have to point out that checking is the easy part - and also necessary. It seems to me you have now got some guidance to work on.

dfklajsdfald
To get the derivative, write ##x^{e^x}= e^{f(x)}##, where ##f(x) = \ln(x) e^x##, and use the chain rule.
ya but the question is asking for antiderivative

Mentor
ya but the question is asking for antiderivative
You claimed earlier that the antiderivative was ##x^{e^x + 1}##. If this is correct, you should be able to differentiate this function to get your original integrand. That's what we're helping you with.

On the other hand, if you differentiate ##x^{e^x + 1}##, and DON'T get ##x^{e^x}##, then that tells you that your purported antiderivative is incorrect.

dfklajsdfald
You claimed earlier that the antiderivative was ##x^{e^x + 1}##. If this is correct, you should be able to differentiate this function to get your original integrand. That's what we're helping you with.

On the other hand, if you differentiate ##x^{e^x + 1}##, and DON'T get ##x^{e^x}##, then that tells you that your purported antiderivative is incorrect.
ohh okay that makes more sense. and i wasnt claiming that that's what the antiderivative was. i was saying that was how i was trying to solve the antiderviative but was getting the wrong answer so i needed guidance to go abouts finding the antiderivative correctly. you asked what my initial attempt was which is why i responded with that, i know its wrong but u wanted to know what i did so that's why i showed that

Mentor
ohh okay that makes more sense. and i wasnt claiming that that's what the antiderivative was.
In post #4 you said:
dfklajsdfald said:
actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it
It looks to me like you were claiming that ##x^{e^x + 1}## is the antiderivative of ##x^{e^x}##.

dfklajsdfald said:
i was saying that was how i was trying to solve the antiderviative but was getting the wrong answer so i needed guidance to go abouts finding the antiderivative correctly. you asked what my initial attempt was which is why i responded with that, i know its wrong but u wanted to know what i did so that's why i showed that

Please don't use "text speak" here at PF. In the rules for this site (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/ ), under General Posting Guidelines, it says"
SMS messaging shorthand ("text-message-speak"), such as using "u" for "you", "please" for "please", or "wanna" for "want to" is not acceptable.

dfklajsdfald
In post #4 you said:

It looks to me like you were claiming that ##x^{e^x + 1}## is the antiderivative of ##x^{e^x}##.

Please don't use "text speak" here at PF. In the rules for this site (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/ ), under General Posting Guidelines, it says"
i said my "attempt" i never claimed anything?? but anyway nevermind i'll figure it out on my own i just wanted some help and guidance not an argument or a forum of whose wrong whose right i'll delete this shortly


text:
## Chemistry: Principles and Practice (3rd Edition)

LiCl will dissolve in water. CO$_2$ will remain a gas. In water, an ionic compound like LiCl will dissociate to for Li$^+$ and Cl$^-$ ions. Molecular compounds like CO$_2$ will not dissociate and will remain as a gas in solution.
LiCl will dissolve in water. CO$_2$ will remain a gas. In water, an ionic compound like LiCl will dissociate to for Li$^+$ and Cl$^-$ ions. Molecular compounds like CO$_2$ will not dissociate and will remain as a gas in solution.


text:
PL EN

Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
Czasopismo

## Acta Arithmetica

1994 | 66 | 3 | 221-228
Tytuł artykułu

### On the positivity of the number of t-core partitions

Autorzy
Treść / Zawartość
Warianty tytułu
Języki publikacji
EN
Abstrakty
EN
A partition of a positive integer n is a nonincreasing sequence of positive integers with sum $n.$ Here we define a special class of partitions. \de{1.} Let $t ≥ 1$ be a positive integer. Any partition of n whose Ferrers graph have no hook numbers divisible by t is known as a t- core partition} of $n.$ \vskip 4pt plus 2pt The hooks are important in the representation theory of finite symmetric groups and the theory of cranks associated with Ramanujan's congruences for the ordinary partition function [3,$\,$4,$\,$6]. If $t≥ 1$ and $n ≥ 0$, then we define $c_t(n)$ to be the number of partitions of n that are t-core partitions. The arithmetic of $c_t(n)$ is studied in [3,$\,$4]. The power series generating function for $c_t(n)$ is given by the infinite product: ∑_{n=0}^{∞} c_t(n)q^n= \prod_{n=1}^{∞
Słowa kluczowe
Czasopismo
Rocznik
Tom
Numer
Strony
221-228
Opis fizyczny
Daty
wydano
1994
otrzymano
1993-04-06
poprawiono
1993-10-12
Twórcy
autor
• Department of Mathematics, the University of Georgia, Athens, Georgia 30602, U.S.A.
Bibliografia
• [1] G. Andrews, EΥPHKA! num = Δ + Δ + Δ, J. Number Theory 23 (1986), 285-293.
• [2] P. Deligne, La conjecture de Weil. I, Publ. Math. I.H.E.S. 43 (1974), 273-307.
• [3] F. Garvan, Some congruence properties for partitions that are t-cores, Proc. London Math. Soc. 66 (1993), 449-478.
• [4] F. Garvan, D. Kim and D. Stanton, Cranks and t-cores, Invent. Math. 101 (1990), 1-17.
• [5] B. Jones, The Arithmetic Theory of Quadratic Forms, Carus Math. Monographs 10, Wiley, 1950.
• [6] A. A. Klyachko, Modular forms and representations of symmetric groups, Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 116 (1982), 74-85 (in Russian).
• [7] K. Ono, Congruences on the Fourier coefficients of modular forms on Γ₀(N), Ph.D. Thesis, The University of California, Los Angeles, 1993.
• [8] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Forms, Publ. Math. Soc. Japan 11, Princeton Univ. Press, 1971.
Typ dokumentu
Bibliografia
Identyfikatory


text:
# 5.5.10. Broyden Algorithm¶

algorithm('Broyden', secant=False, initial=False, count=10)

Create a Broyden algorithm for general unsymmetric systems which performs successive rank-one updates of the tangent at the first iteration of the current time step.

 secant (bool) Flag to indicate to use secant stiffness. (optional) initial (bool) Flag to indicate to use initial stiffness.(optional) count (int) Number of iterations. (optional)