25
votes
Series sum approximation
Writing:
NSum[(-1)^n/Sqrt[Log[n]], {n, 2, Infinity}, Method -> "AlternatingSigns"]
I get:
0.690243
which is what you want.
In particular, directly from ...
- 4,462
18
votes
A more convenient Fourier series
For the reasons mentioned above, I wrote the following "shell" for these functions:
...
- 68.4k
15
votes
Accepted
How to convert this term to a Hypergeometric function?
For your first question, if we gather the factors into a single variable z, there's a simple hypergeometric function:
...
- 49.8k
14
votes
Help with Double Sum (lattice sum) over all integers m,n of 1/(a+m^2+n^2)
This kind of sum can be studied by integral transformation. Notice that $\int_0^1t^{x-1}dt=\frac{1}{x}$ if $\text{Re}(x)>0$:
...
- 49.8k
13
votes
How to find the perturbation of $x^2 − 1 = \epsilon x$?
Decide up to which power you would like to expand:
pow = 4;
Let's do one of the equations you mentioned as an example (bring all terms to one side and save as a ...
- 12.1k
13
votes
Accepted
How to get the Taylor series of implicit functions
You can use AsymptoticSolve for this purpose:
AsymptoticSolve[x+1/2y^2+1/2z+Sin[z]==0,{z,0},{{x,y},{0,0},4}]
{{z -> -((2 x)/...
- 132k
12
votes
Series vs Asymptotic in 12.1
Extended comment, I won't accept this as an answer.
Here are some cases I've found where Series might be a better choice than ...
- 1,018
12
votes
Accepted
11
votes
Accepted
How to represent $f(x) = (y-x)^k \log(y-x)$ as a summation of the form $f(x) = \sum\limits_{j=0}^\infty \cdots$?
Complete rewrite of answer
The expression to be expanded as a series is the argument of
ser = Series[(y - x)^k*Log[y - x], {x, 0, 5}]
I attempted to obtain the ...
- 62.1k
11
votes
Accepted
Why can’t mathematica find this residue?
You could use SeriesCoefficient instead:
SeriesCoefficient[(z+1)^2 Exp[3/z^2], {z, 0, -1}]
6
Addendum
Another possibility ...
- 132k
11
votes
Accepted
Legendre expansion of the Dirac delta function
As I understood you start from the completeness relation
$$\sum_{\ell=0}^\infty \frac{2\ell + 1}{2} P_\ell(x)P_\ell(y) = \delta(x-y)$$
and use that
$$
P_n(0) =
\begin{cases}
\frac{(-1)^{m}}{4^m} ...
- 19.7k
11
votes
Accepted
Getting terms and only evaluate specific parts of a series
times[{i_}] := i
times[{i__}] := Inactive[Times][i]
Sum[times@Table[2 i - 1, {i, n}]/times@Table[2 i, {i, n}], {n, 5}]
If you prefer ...
- 68.4k
11
votes
Accepted
How to get more terms with the Series[] expansion of InverseErf[x] around x=1?
This is not strictly speaking an answer, but I thought I provide code that can be used to generate the kind of expansion that OP mentions. Perhaps it is useful for other people here.
I use $y$ as an ...
- 12k
10
votes
Accepted
Using Fourier Series to acquire Nonlinear ODE Periodic Solutions
Direct solution of the last equation in the question also is feasible, because the Fourier series converges very rapidly. as will be seen below. The equation for a three term expansion can be written ...
- 62.1k
10
votes
Accepted
Plotting a Taylor Series of two-variable trigonometric function
Use Normal and Evaluate,e.g.:
...
- 64.8k
10
votes
Series expansion wrong
I think the behavior described is a bug, and I think it is related to the new enhanced support of Assumptions in Limit. In M11.1 ...
- 132k
10
votes
Accepted
How could this asymptotic expansion be obtained?
Exact expression for $\sigma_n$:
b[n_] = BesselJ[1, BesselJZero[0, n]]*BesselJZero[0, n]*StruveH[0, BesselJZero[0, n]];
σ[n_] = π/2*(-1)^n*(b[n + 1] - b[n]);
...
- 49.8k
10
votes
Accepted
Error Message when nothing should be evaluated
This is due to the special behavior of SetDelayed (:=) with regards to the first argument (see e.g. this question): The ...
- 34.4k
10
votes
How to convert this term to a Hypergeometric function?
Match up power series and solve for parameters for Hypergeometric2F1[a, b, c, d x]:
...
- 245k
10
votes
Accepted
How to "prepare" expression for Taylor expansion
Both options give the expected result same result as Ulrich shows with their method, however, it can be seen that this is not to the second order that OP indicates they desire expanding to.
...
- 3,362
10
votes
Accepted
Calculating relative error of Ramanujan formula for ellipse perimeter
We have to express a parameter $h=(a-b)^2/(a+b)^2$ in terms of the eccentricity of the ellipse $e = \sqrt{1-b^2/a^2}$.
Similarly we need comparing the second Ramanujan approximation for the ...
- 57.9k
9
votes
Series sum approximation
Using identity:
$$\int_0^{\infty } \frac{2 n^{-t^2}}{\sqrt{\pi }} \, dt=\frac{1}{\sqrt{\log (n)}}$$
then I have:
$$\sum _{n=2}^{\infty } \frac{(-1)^n}{\sqrt{\log (n)}}=\\\sum _{n=2}^{\infty } (-1)^n \...
- 16.4k
9
votes
Accepted
How can I get the minimum error term when manipulating Taylor series?
You could just apply Series to the expression of interest:
Series[(u[x+h] + u[x-h] - 2 u[x])/h^2, {h, 0, 3}] //TeXForm
$u''(x)+\frac{1}{12} h^2 u^{(4)}(x)+O\...
- 132k
9
votes
Accepted
Trying to get a Laurent expansion of a symbolic function
A standard approach is to use a partial fraction decomposition, rearranged appropriately for the region of interest, that is, so that the corresponding infinite sum will converge.
In this case the ...
- 60.3k
9
votes
Accepted
Removing higher order terms
You can use a variation of the idea I gave here:
Normal @ Series[
ss /. {f:u1|u2 -> (s f[#1,#2,#3,#4]&)},
{s, 0, 3}
] /. s->1
u2[x, y, z, t]^...
- 132k
9
votes
What series does Mathematica use for Hypergeometric1F1?
Recall that $(-n)_k=0$ for $k>n, n,k\in\mathbb N$. Thus, what you have is an appropriate truncation of the usual series for the Kummer function.
...
9
votes
Accepted
expand function as power series of another function
Try the following
Series[f[InverseFunction[g][y]],{y,0,10}]
- 19.7k
9
votes
How to "prepare" expression for Taylor expansion
Try
Normal[Series[(m^2 + M^2)/(m^2 - M^2)^2 /. m -> eps M , {eps, 0,3}] ] /. eps -> m/M
(*(3 m^2)/M^4 + 1/M^2*)
- 56.8k
9
votes
Zassenhaus formula in Mathematica
Here is my old implementation based on M.WEYRAUCH, D.SCHOLZ, COMPUTER PHYSICS COMMUNICATIONS, 180, (2009), 1558-1565
Returns 'unfolded' or 'folded' (in terms of commutators):
...
- 3,186
9
votes
Accepted
Why `AsymptoticSolve` doesn't work for a multivariate implicit function?
General remarks
It takes manual work sometimes when the starting value for the series cannot be easily determined. Also, for an exact solver like AsymptoticSolve (...
- 245k
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