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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 ...
Artes's user avatar
  • 57.5k
9 votes
Accepted

Series expansion not working with $\sqrt{1-x^d}$

Perhaps this?: Asymptotic[Sqrt[(1 - u)], {u, 0, 2}] /. u -> x^d (* 1 - x^d/2 - x^(2 d)/8 *) If we examine the output of ...
Michael E2's user avatar
  • 237k
6 votes
Accepted

Why does Mathematica get the infinite sum of this representation of the taylor series of $\cos^2(x)$ wrong?

Expanding on the comment by @Goofy and after reading the docs we see that: "ParallelFirstToSucceed" try each method in parallel until one succeeds ...
bmf's user avatar
  • 16.1k
5 votes

Discrepancy with Hurwitz Zeta function

HurwitzZeta When calculating a sum one can choose an option GenerateConditions -> True to get appropriate conditions on parameters for convergence of a series <...
Artes's user avatar
  • 57.5k
5 votes

Evaluating series expansion is very slow

One can try breaking the expression into pieces: ...
b.gates.you.know.what's user avatar
5 votes

Evaluating series expansion is very slow

For alternative algorithm and for speed up you can may use: NSeries: ...
Mariusz Iwaniuk's user avatar
5 votes
Accepted

First argument -h is not a valid variable

h is a local to Series, you cannot use it as a functionargument. Try ...
Ulrich Neumann's user avatar
5 votes

Find Generalized Series with Symbolic Variable

With SeriesCoefficient, the order and expansion point can be symbolic. ...
Bob Hanlon's user avatar
  • 160k
5 votes

Can Mathematica simplify an ordinary differential equation (ODE) by assuming a power series solution and obtain the recurrence relation?

Yes, Mathematica is able to do that. ...
Ulrich Neumann's user avatar
4 votes

Asymptotic integral expansion at infinity

Solving this problem analytically is difficult. However, it is possible to get exact values of the integral In in symbolic form for each ...
Vaclav Kotesovec's user avatar
4 votes

Find Generalized Series with Symbolic Variable

If I understand correctly what you want the following is your friend ...
bmf's user avatar
  • 16.1k
4 votes
Accepted

A simple series expansion which seems to be wrong

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Bob Hanlon's user avatar
  • 160k
4 votes

Series expansion of a given function

I am not sure why this is happening, but here's a quick fix. Do a replacement $\xi_3 \rightarrow \tfrac{1}{x_3}$ and then expand around $\infty$. ...
bmf's user avatar
  • 16.1k
4 votes
Accepted

Getting coefficients list of a series expansion at a point different from 0

One way is to do the following: expand to whatever order you want, and subsequently substitute $x-1$ by $y$. Do CoefficientList in the ...
bmf's user avatar
  • 16.1k
3 votes
Accepted

How to convert DifferenceRoot into a special function?

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Bob Hanlon's user avatar
  • 160k
3 votes

Set the value of a parameter in a Series expression

For others struggling I figured it out. Just use Normal@Series when doing the expansion. Then you can use $/.e\rightarrow 1$ without issue.
Matthew Ward's user avatar
3 votes
Accepted

How to find the asymptotic envelope of a function?

The Series command of 14.0 on Windows 10 produces the required answer for BesselJ[n,x] ...
user64494's user avatar
  • 27k
3 votes
Accepted

How to invert a series with two variables, where the series is expanded in the other variable?

You can use AsymptoticSolve for this purpose: ...
Carl Woll's user avatar
  • 131k
3 votes

Series expansion of a given function

Use Assuming. ...
Jie Zhu's user avatar
  • 1,471
2 votes
Accepted

How to obtain a list of pairs of exponents in a double series expansion?

Using CoefficientRules, Normal and SortBy: ...
E. Chan-López's user avatar
2 votes
Accepted

Comparing two power series and extracting their coefficients

In sum2, the powers of x and y are less than or equal to 6, whereas in ...
march's user avatar
  • 23.9k
2 votes

Laurent series expansion

Another way is to use partial fraction decomposition: Apart[1/((z^2 - 1) (z^2 - 4))] $$\frac{1}{\left(z^2-1\right) \left(z^2-4\right)}=-\frac{1}{6 (z-1)}+\frac{1}{...
Gappy Hilmore's user avatar
2 votes

First argument -h is not a valid variable

Or s[h_, n_ : 4] := Normal[f[t] + O[t, x]^(n + 1)] /. t -> x + h s[h] + s[-h]
cvgmt's user avatar
  • 76.6k
2 votes

Why does Mathematica get the infinite sum of this representation of the taylor series of $\cos^2(x)$ wrong?

Vs 6 gets the correct result, but Wolfram $\alpha$ decides, too, that $$\sum_{m,n=0}^\infty \frac{ (-1)^{n+m} x^{2m+2n} }{(2n)!(2m)!} \quad \ne \quad \sum_{m=0}^\infty\left(\sum_{n=0}^\infty \frac{ ...
Roland F's user avatar
  • 3,887
2 votes
Accepted

Asymptotic solution of a system of ODEs

A simple approach is to introduce the ansatz $$ f(u)=\sum_{k\ge0}\frac{f_k}{u^k},\qquad a(u)=\sum_{k\ge0}\frac{a_k}{u^k} $$ which you plug into the ODE and solve for the coefficients. I find \begin{...
AccidentalFourierTransform's user avatar
2 votes

BUG: Why is Series[] getting this expression wrong?

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Bob Hanlon's user avatar
  • 160k
2 votes
Accepted

Expanding polynomials using valuation

Maybe: Min[CoefficientRules[#[[2]], x][[All, 1, 1]]] + #[[1, 1]] \[Lambda]0 & /@ CoefficientRules[poly, \[Lambda]] (* {2 \[Lambda]0, 1} *) Or: ...
Goofy's user avatar
  • 3,277
2 votes
Accepted

Solving algebraic equations perturbatively (using function series)

...
2 votes

Series expansion not working with $\sqrt{1-x^d}$

Since the expansion is at an analytic point of the function, the power series is a Taylor expansion: $$ S(x)=\sum_{n=0}^{\infty} \frac{\frac{d^n}{dx}(1-x^d)^{1/2}}{n!}x^n $$ However when $d$ is ...
josh's user avatar
  • 2,432
1 vote
Accepted

Finding constant term in product expression

You may use "Expand" to get the expanded form of your polynomial. And then "Cases" can pick the constant terms. here is an example. First we create some large polynomial: ...
Daniel Huber's user avatar

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