numerics – Problem evaluating numerical value of meijerG function at some parameter?

MejerG function could not be evaluate in mathematica ?
I need to evaluate this sum:

Where x = 33.6614 and $$k,l,s$$ are non-negative index of the three sum. For example $$k,l,s=0,1,2,3,…$$.

Furthermore, $$M$$ is a positive integer that is $$M=1,2,3,4…$$

For the case of $$s = 0,k = 0,l = 0,M = 4$$, I enter my meijerG function in this form

``````MeijerG({{2}, {}}, {{0, 2, 3, 4 + 2}, {}}, 33.6614)
``````

Mathematica said that:

``````"MeijerG({{2},{}},{{0,2,3,6},{}},33.6614) does not exist. Arguments are not consistent"
``````

Why does this happen and is there anyway to work around this ?

Thank you very much !

P/S:

For more information, the big meijerG function that I am trying to evaluate come from the evaluation of this integral

``````Integrate((-z/(Gamma))^S* (1/z)^L *Exp(-(a + c)/(z*(CapitalOmega)))*
MeijerG({{1}, {}}, {{k + 1, k + 2, k + M + 1}, {}}, z*e), {z,
0, (Infinity)},
Assumptions ->
Element({a, c, (CapitalOmega), (Gamma), e}, PositiveReals))
``````

simplifying expressions – How to prevent TrigReduce auto evaluating Sin and Cos

Mathematica automatically evaluates `Sin` and `Cos` when `TrigReduce` is applied. For example, `TrigReduce(Cos(2*t)*Cos(2*t))` results in `1/2 (1 + Cos(4 t))`. In my application I would like `TrigReduce` to keep the `Sin` and `Cos` as is. I’d like to have `1/2 (Cos(0 t) + Cos(4 t))` as the output of the preceding example. I tried to use `Inactivate`, but that seems to prevent `TrigReduce` from working as well.

simplifying expressions – How to prevent TrigReduce auto evaluating Sin/Cos

Mathematica automatically evaluates Sin/Cos when `TrigReduce`. For example, `TrigReduce(Cos(2*t)*Cos(2*t))` results in `1/2 (1 + Cos(4 t))`. My application would like `TrigReduce` to keep the Sin/Cos as is. For example, I’d like to have `1/2 (Cos(0 t) + Cos(4 t))` as the output. I tried to use `Inactivate`, but that seems to prevent `TrigReduce` from working as well.

calculus – Evaluating \$lim_{xto 0}frac{cos(sin x)-(1+x^2)^{frac{-1}{2}}}{x^4}\$

$$L=lim_{xto 0}frac{cos(sin x)-(1+x^2)^{frac{-1}{2}}}{x^4}$$
To Evaluate the limit I used few terms of taylor series:

$$cos(sin x)=1-frac{sin^2 x}2+cdots$$

$$(1+x^2)^{frac{-1}2}=1-frac12x^2+cdots$$

$$L=lim_{xto 0}frac{1-frac{sin^2 x}{2}-1+frac{x^2}{2}}{x^4}=lim_{xto0}frac{x^2-sin^2 x}{2x^4}$$

$$sin^2 x=(x-frac{x^3}{6}+O(x^4))^2=x^2-frac{x^4}{3}+O(x^5)$$

So we have:$$lim_{xto 0}frac{x^2-x^2+frac{x^4}{3}}{2x^4}=frac{1}{6}$$

But the right answer is $$frac{-1}6$$. Why my final answer is wrong?

summation – Algorithm used by Mathematica for evaluating partial sums

Today, while using mathematica, I entered the command `Sum(1/Factorial(n), {n, 0, x})` and found that:
$$sum_{xgeq ngeq0}frac{1}{n!}=frac{eGamma(x+1,1)}{Gamma(x+1)}$$
Where $$Gamma(x,y)$$ is the Incomplete Gamma function.
It is well known that
$$sum_{xgeq0}frac{1}{n!}=e$$
But which algorithm(or formula) did Mathematica use? If the algorithm is too complex for human use, how can we prove the first equality?
Another result it gave was
$$sum_{mgeq ngeq 1}frac{1}{n,n!}=frac{(mathrm{Ei}(1)-gamma)m(m+1)!+(mathrm{Ei}(1)-gamma)(m+1)!-_2F_2(1,m+1;m+2,m+2;1)}{(m+1)(m+1)!}$$
Where $$mathrm{Ei}$$ is the Exponential integral and $$F$$ is the Generalized hypergeometric function. This looks very hard to evaluate by hand, is there any formula or algorithm to evaluate this? I shouldn’t say this on MSE, but I don’t know how to do this, I can’t do even a step.

real analysis – Evaluating \$sum _{k=1}^{infty }frac{H_{2k}}{k^3:4^k}binom{2k}{k}\$.

As the solutions here, I tried similar ways.

I tried using:
$$sum _{k=1}^{infty }frac{x^{2k}}{k:4^k}binom{2k}{k}=-2ln left(1+sqrt{1-x^2}right)+2ln left(2right)$$
$$-2sum _{k=1}^{infty }frac{1}{k:4^k}binom{2k}{k}int _0^1x^{2k-1}ln left(1-xright):dx=4int _0^1frac{ln left(1-xright)ln left(1+sqrt{1-x^2}right)}{x}:dx$$
$$-4ln left(2right)int _0^1frac{ln left(1-xright)}{x}:dx$$
$$sum _{k=1}^{infty }frac{H_{2k}}{k^2:4^k}binom{2k}{k}=4int _0^1frac{ln left(1-xright)ln left(1+sqrt{1-x^2}right)}{x}:dx+4ln left(2right)zeta left(2right)$$

But I can’t get the right manipulations so that I end up with:
$$sum _{k=1}^{infty }frac{H_{2k}}{k^3:4^k}binom{2k}{k}$$

I think I might have another way to go, I’ll edit as soon as i advance on it.

np complete – How can you modify a SUBSET-SUM instance so evaluating a set outputs either 0 or 1?

An SUBSET-SUM instance is a list of $$n$$ integers $${ a_1, a_2,… a_n}$$ and a target $$W$$. To evaluate a set is to find the sum of a selection of numbers in the set. However, I want to know, is it possible to modify the instance, or SUBSET-SUM in general, where evaluating a set outputs $$0$$ if the sum equals $$W$$, and $$1$$ otherwise?

Bonus: Give a list of other NP-complete problems (other than 3SAT, where you evaluate a formula that either outputs $$0$$ or $$1$$ depending on the set of binary variables being passed into it), where evaluating an analogous instance outputs $$0$$ if it satisfies some objective related to the problem and $$1$$ otherwise.

Evaluating in PARI/GP the Fourier coefficients of the j -invariant?

Which PARI-commands evaluates the q-expansion, i.e. the Fourier coefficents of the modular j-invariant?

Note. ellj(x) computes single values of j; and mfcoefs(-,-) computes coefficents. But mfcoefs(ellj,8) is not accepted.

Evaluating \$int _0^1frac{ln left(1-xright)ln left(1-x^4right)}{x}:dx\$

How can i evaluate $$int _0^1frac{ln left(1-xright)ln left(1-x^4right)}{x}:dx$$
WolframAlpha offers no closed form and it numerically approximates it to $$1.07801$$.

I’d appreciate any hints.

calculus and analysis – Problem with DiracDelta evaluating to zero

I am having an issue (in Mathematica 11.2.0.0) where the following integral evaluates to zero:

`Integrate(E^(-(1/4) (2 π q + a t)^2 σx2^2) Sqrt(π) σx2 DiracDelta(5 kL - π (p - q)), {q, -∞, ∞}, Assumptions -> {kL ∈ Reals, a ∈ Reals, σx2 ∈ Reals, t ∈ Reals, p ∈ Reals})`

When all of the variables are real, as I’ve tried to make it assume, then the Dirac delta should make this integral equivalent to evaluating the exponential at $$q=(-5k_L+ppi)/pi$$, which would be a nonzero result. But when I try to evaluate this (as the very first thing evaluated in the kernel, so I know no other assumptions are acting), I get zero.

Of course I can do this by hand, but the actual integral I want to evaluate has over fifty terms of this form, and all but one evaluates to zero individually, when none of them should. (The one that survives has the same exponential, but the Dirac delta is simply $$delta(p-q)$$.) So it’d be nice to figure out what the problem is with Mathematica.

What could be going wrong here?

EDIT:

This integral (without the $$a t$$ term in the exponential) works fine:

`Integrate(E^(-(1/4) (2 π q)^2 σx2^2) Sqrt(π) σx2 DiracDelta(5 kL - π (p - q)), {q, -∞, ∞}, Assumptions -> {kL ∈ Reals, a ∈ Reals, σx2 ∈ Reals, t ∈ Reals, p ∈ Reals})`

I can in fact replace $$a t$$ with any number, and it evaluates correctly. But any variable other than $$q$$ there results in zero. What is this apparent madness?