Assume that $zeta(3)=n$, we can calculate that $$n^2=sum_{i,jinmathbb{Z^+}}i^{-3}j^{-3}$$. It is also known that $$begin{align}n^2&=2l+frac{pi^6}{945}\n&=sqrt{2l+frac{pi^6}{945}}end{align}$$For when $$l=sum_{i<j}frac{1}{i^3j^3}$$This reminds me of the Basel Problem for which we use the sine function and it’s Taylor series to find the value of $sum_{i<j}frac 1{i^2j^2}$. I assume that there will be a function such that $$f(x)=xprod_{i=1}^{infty}bigg(1-frac {x^3}{i^3pi^3}bigg)$$, What will be the intuition of finding this function? If it is a trigonometric function with zeroes for when $x^3=k^3pi^3$ for all positive integer $k$, or $(x-kpi)(x^2+pi^2k^2+pi kx)=0$, does that imply that the value may be in complex form?

# Tag: constant

## co.combinatorics – Explicit constant in Green/Tao’s version of Freiman’s Theorem?

Green and Tao’s version of Freiman’s theorem over finite fields (doi:10.1017/S0963548309009821) is as follows:

If $A$ is a set in $mathbb{F}_2^n$ for which $|A+A| leqslant K|A|$, then $A$ is contained in a subspace of size $2^{2K+O(sqrt{K}log(K))}$.

Does anybody know of a version of this bound with an **explicit** constant for the error term, or just an explicit exponent with more or less the same order of magnitude? For example, replacing the $O$ term by another multiple of $K$ might be sufficient for my purposes. (I would like to apply the bound to some specific sets $A$ and values of $n$. Roughly speaking, my $n$ range from about $50$ to $100$, and my $K$ range from about $50$ to $1000$, in case that helps.)

There is an earlier result of Ruzsa and Green, in which the upper bound is given explicitly as $K^22^{2K^2-2}$, but this is far too large for what I need it for.

I suppose one might be able to infer an explicit bound by reading Green and Tao’s proof with enough care, but of course it would be easier not to have to do this.

## macos – Constant password input and resets on Mac

I got a new MacBook Pro from my job and was pretty excited, but now I try to hold myself from tossing it out the window!

It constantly asks me to enter passwords for installing things, keeps signing out of apps so I need to re-enter again and again on a daily basis, and once every 2 days it wants me to reset my password when I’m logging in after sleep.

I tried disabling the password from the system preferences but the “Require password after sleep” is greyed out! I really struggle to understand the thinking behind this system…

Any Mac experts that can help?

## homology cohomology – Why are Simplicial Chains assumed Constant?

Just curious and wondering if I am missing something obvious:

Let K be a simplicial complex. Why are Simplicial chains with coefficients in a group G described as $$ g sigma_i ^m $$

for g in G and $ sigma_i^m $ an m-simplex in the m-th chain group $C_m(K,G)$, i.e., chains seem

to be assumed to be constant in every simplex?

## integration – How could I express this divergence nullifying, constant, C, without its sums?

Note before reading: each sum here implies its limit.

Question: how in tarnation can the constant $C$ be expressed without the sums and in terms of $A$ and $p$?

$$C=sum_{n_{L}=0}^{A}left(sum_{n_{G}=n_{L}}^{A}left(left(-1right)^{n_{G}+n_{L}+1}cdotfrac{p^{n_{G}}}{n_{L}!}right)right)$$

Context:

If one wanted to make an integration formula for $x^p$, where $pinBbb C$, specifically by expanding the expression into the Taylor series, the steps I took to solve it are below:

- Since: $x^p$ is identical to $sum^infty_{n=0}(frac {ln^n(x)*p^n}{n!})$

The integral on both sides would be:

$$int x^pdx=int sum^infty_{n=0}biggr(frac {ln^n(x)*p^n}{n!}biggr)dx$$

Notice that each term in the sum is in the form $c*ln^n(x)$ where $c$ is a constant with respect to $x$: this is important for the second step.

- To get rid of the integral, one would need an integration formula for $ln^n(x)$. Using the sum formula for$int ln^n(x)dx$, one would get that

$$int ln^n(x)dx=sum^n_{j=0}biggr((-1)^{j+n}xln^j(x)frac{n!}{j!}biggr)$$.

- Plugging this into the original equation, it becomes

$$int_{ }^{ }x^{p}dx=sum_{n=0}^{infty}left(frac{p^{n}}{n!}cdotleft(sum_{j=0}^{n}left(left(-1right)^{j+n}cdot xln^{j}left(xright)cdotfrac{n!}{j!}right)right)right)$$

Which, after simplifying, becomes

$$int_{ }^{ }x^{p}dx=sum_{n=0}^{infty}left(sum_{j=0}^{n}left(left(-1right)^{j+n}cdot xln^{j}left(xright)cdotfrac{p^{n}}{j!}right)right)$$

- Notice that the only problem about this series is, despite its derivative being correct, it is divergent! OH NO!!!

After noticing that an identical series can be expressed as

$$lim_{Ato infty} sum_{n_{L}=0}^{A}left(sum_{n_{G}=n_{L}}^{A}left(left(-1right)^{n_{G}+n_{L}}cdotfrac{xleft(lnleft(xright)right)^{n_{L}}cdot p^{n_{G}}}{n_{L}!}right)right)$$

I noticed that in order to complete the formula, I must find constant $C$ that will,∀$Ain Bbb R$ and ∀$pin Bbb C$, satisfy the following equation true when $x=0$ (for simplicity):

$$0=sum_{n_{L}=0}^{A}left(sum_{n_{G}=n_{L}}^{A}left(left(-1right)^{n_{G}+n_{L}}cdotfrac{xleft(lnleft(xright)right)^{n_{L}}cdot p^{n_{G}}}{n_{L}!}right)right)+C$$

Treating the product $xln^{n_L}(x)$ as its limit as $x$ approaches $0$, which is $1$, we get that $C$ must equal the following series:

$$C=sum_{n_{L}=0}^{A}left(sum_{n_{G}=n_{L}}^{A}left(left(-1right)^{n_{G}+n_{L}+1}cdotfrac{p^{n_{G}}}{n_{L}!}right)right)$$

Which lead me to my question, which I emphasize: how, in the world, in tarnation can the constant $C$ be expressed without the sums and in terms of $A$ and $p$?

Please provide your method of solving; thank you!

## Mathematica can’t simplify asymptotic expressions containing constant symbols

I want to calculate simple asymptotic expressions involving positive constant symbols ($a > 0$), such as

$$lim_{xtoinfty} operatorname{sech}(a x) sim 2 e^{-a x}$$

Surprisingly, the `Asymptotic`

function of Mathematica can’t calculate this limit.

The code

Assuming(a > 0, Asymptotic(Sech(a x), x -> Infinity))

returns

Sech(a x)

while

Asymptotic(Sech(3 x), x -> Infinity)

correctly returns

2 E^(-3 x)

How can I get Mathematica to evaluate this asymptotic limit correctly?

Edit:

One hack is to replace $a$ with $pi$, then calculate the asymptotic limit, then convert $pi$ back to $a$.

Asymptotic(Sech(a x) /. a -> π, x -> Infinity) /. π -> a

returns the desired limit

2 E^(-a x)

## sg.symplectic geometry – Constant holomorphic strips in Lagrangian floer cohomology

Let $(P,omega)$ be a symplectic manifold , when defining the lagrangian floer cohomology of a pair $(L_0,L_1)$ of lagrangian submanifolds with $Sigma_{L}geq 3$ we will want to look at the space $mathcal{M}_{J}(x,y)$ ,where $x,yin L_0cap L_1$, of $J$-holomorphic strips connecting intersection points and satisfying a sobolev condition, namely that $uin W^{k,2}(mathbb{R}times (0,1), P)$ for $k>1$. Now if we had $x=y$ will the constant strip be an element of $mathcal{M}_{J}(x,x)$? I would say no since we want the paths in this space to satisfy a sobolev in condition ,and in particular the constant path will not satisfy it , but I wanted to make sure this reasoning is correct.

Any enlightment is appreciated , thanks in advance.

## units – How do I get a numerical value for a physical constant in Mathematica 12.1

I’ve been working on some optics problems and regularly need to use constants like the charge of an electron, vacuum permittivity, etc. I’d like to be able to call these constants in Mathematica as decimal values without having to go in and define them every time I start a new notebook. So far I’ve tried using Quantity and UnitConvert on constant names ported over from older versions of Mathematica, but thus far I have been unsuccessful in calling constants completely divorced from their units. Is there any way to do this without having to include a bunch of definitions at the top of the notebook?

## asymptotics – Solution to a recurrence changing by a constant factor

## java – How should an abstract class require a value that will be constant for each subclass?

Let’s say I have an `abstract class A`

exposing a method called `getE()`

which returns an object of type `E`

which is an `enum`

. The value returned by `getE()`

will be defined per subclass. Put differently, every object of a certain subclass of `A`

will always return the same value if `getE()`

is called.

This is how I have implemented that system right now:

```
public enum E {
VAL_1, VAL_2
}
public abstract class A {
private final E e;
protected A(E e) {
this.e = e;
}
public E getE() {
return e;
}
}
public class A1 extends A {
public A1() {
super(E.VAL_1);
}
}
public class A2 extends A {
public A2() {
super(E.VAL_2);
}
}
```

But this wastes memory, as every instance of `A`

has to keep a reference to `E`

, and also clutters `A`

‘s constructor.

Another solution I have thought of is this one:

```
public abstract class A {
public abstract E getE();
}
public class A1 extends A {
@Override
public E getE() {
return E.VAL_1;
}
}
public class A2 extends A {
@Override
public E getE() {
return E.VAL_2;
}
}
```

But adding new methods to a class probably is memory-intenisve as well and might cause confusion.

Which of these ways do you think is better or are there other alternatives?

I have already asked this question on Code Review and Stack Overflow and got redirected to here.