Questions tagged [series-expansion]

Questions on dealing with series data and constructing power series expansions of functions.

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Bug in SumConvergence

Bug introduced in 10.0.1 and fixed in 12.0.0 Version 11.2.0.0 on MacBook Pro: ...
Paul R.'s user avatar
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7 votes
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Design considerations behind `O` (a.k.a. BigOh, a.k.a. Landau Order)

This works without any warnings: O[Log[x]]. This raises a warning: O[x^2]. I have a few questions around this: Why is it a ...
Krastanov's user avatar
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6 votes
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How to revert behavior of SeriesData to pre 12.1

In Mathematica 12.0 and earlier, SeriesData[x, 0, {1/u + Log[x/y]}, 0, 3, 1] used to preserve its list of expressions in the form it was given. Now, in ...
QuantumDot's user avatar
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6 votes
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131 views

Getting around Series[Sinc] bug

Bug introduced in 8.0 or earlier and fixed in 10.4 For some reason, Series expansion of Sinc around ...
QuantumDot's user avatar
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5 votes
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MMA does not provide the correct asymptote for an integral function

Given is the function $$f(x)=\int_0^\infty \mathbb{exp}\left(-\frac{x^2}{2t^2}-t\right)\mathbb{d}t$$ Mathematica returns for the asymptotic behavior $x\to\infty$ using ...
granular_bastard's user avatar
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How to get an asymptotic of the real-valued branch of the inverse function?

Consider function $f:\mathbb R^+\to\mathbb R^+$, defined as $f(x) = x + x^2\left(1 + \log x\right)$. I need to find an asymptotic approximation of its inverse function $f^{\small(-1)}\!:\mathbb R^+\to\...
Vladimir Reshetnikov's user avatar
5 votes
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100 views

How to remove the smallest term from asymptotic expansion?

It is well-known that $e^{-1/x}\sim o(x^n)$ as $x\to 0^+$ for any $n\in\mathbb{N}$, thus if I do an asymptotic expansion for a function, say $f=1/(1-x)+e^{-1/x}$ as $x\to 0^+$, I expect to receive an ...
user142288's user avatar
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Issues with the series expansion of Nielsen generalized polylogarithms in Mathematica 11.3

Bug introduced in 11.3. Fixed in 12.0. In Mathematica 9, 10.3 and 11.0 I can easily expand PolyLog[2, 2, x] around $x=1$ using ...
vsht's user avatar
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5 votes
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135 views

Expansion of $E(i c \mid m)$ at $c\to\infty$?

Currently, I am using a Windows machine with Mathematica 8. I noticed a difference in a series expansion of the function EllipticE[] in comparison with a result ...
Kagaratsch's user avatar
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4 votes
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InverseSeries giving incorrect result

Somehow in Mathematica 13.2.0.0, InverSeries generates incorrect results. Let's look at the following two series that differs from each other by a constant number &...
mastrok's user avatar
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Calculate an n-order determinant by FindSequenceFunction

Calculate an n-order determinant: $\left|\begin{array}{cccccc}1 & 2 & 3 & \cdots & n-1 & n \\ n & 1 & 2 & \cdots & n-2 & n-1 \\ n-1 & n & 1 & \cdots ...
lotus2019's user avatar
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Series expansion of Lerch transcendent still buggy?

This series expansion of a Lerch transcendent seems fixed in V12. However, the following still fails: From the definition of a Lerch transcendent, ...
Roman's user avatar
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4 votes
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How to calculate the series of this function?

I have to calculate the series of the function F[r_] := 1 - a - a*r^(5 + n)/(r^(8 + n) + 1); for r->Infinity for generic ...
mattiav27's user avatar
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Another example where using FullSimplify gives different result than Simplify

I believe this question is very similar to Result of Series[expression] is different when I simplify the expression, however, due to my lack of Mathematica experience, I am reluctant to call it a bug. ...
Gabriel Nagaoka's user avatar
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75 views

Series of LegendreP[a,b,x] takes ages in Mathematica 11

My institution just upgraded to Mathematica 11.3 (from v10) and I'm experiencing a problem that is absent in Mathematica 10. Namely, ...
Paolo's user avatar
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Multivariate FindGeneratingFunction

For one dimensional problems sometimes I'm using the Mathematica function FindGeneratingFunction[seriescoefficient,x] which uses the coefficients of a powerseries ...
Ulrich Neumann's user avatar
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Change in behaviour of Series from 11.1

In 11.0, Series[(1 - Sqrt[1-4z])/2, {z, 1/4, 1}] gives 1/2 - I Sqrt[z-1/4] + O[z-1/4]^(3/2) as one would expect. From 11.1 ...
David Bevan's user avatar
4 votes
0 answers
129 views

Series expansion in Infinity issue with Zeta(s) function

With this code: Series[Zeta[s], {s, Infinity, #}] & /@ Range[10] // MatrixForm Series expansion for the Zeta(s) function ...
Mariusz Iwaniuk's user avatar
4 votes
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132 views

A series in powers of $(a-z)$ instead of $(z-a)$

Sometimes it is more convenient to find a series expansion (e.g., Taylor, Laurent, Puiseux, ...) in powers of $(a-z)$ than in powers of $(z-a)$. For instance, the command ...
Jára Cimrman's user avatar
4 votes
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Quirky behavior of Series[]

This is a "what's going on?" question about MMa behavior, not so much a "how to fix?" This code calculates a Taylor series for a two-term Gaussian Mixture: ...
Jerry Guern's user avatar
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Why does Mathematica not go to specified order in series?

For some reason Mathematica will not evaluate this asymptotic series to the requested order. Inputting: ...
DJBunk's user avatar
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1 answer
170 views

Proper treatment of roots and powers in Series?

I have the following problem in Mathematica 9 on Linux. I let Mathematica compute the Series expansion: ...
Kagaratsch's user avatar
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3 votes
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131 views

Cannot Understand nth Derivative of x/ArcTan[x]

The nth derivative of x/ArcTan[x]: f[x_, n_] = D[x/ArcTan[x], {x, n}] Evaluates to: I cannot get this general from to return ...
Josey Stevens's user avatar
3 votes
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487 views

Series expansion with fractional exponents

The following series expansion Series[1 + Sum[b[n] (x^(1/4))^n, {n, 1, 3}], {x, 0, 1}] gives terms up to O(x^{5/4}). I would ...
dpholmes's user avatar
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Power series raised to powers

Power series raised to powers: I am looking for a way to implement a replacement using this relation explained on wikipedia and in another post here: https://math.stackexchange.com/questions/1471438/...
Rupert Gillyweed's user avatar
3 votes
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203 views

Properties of the prime geometric series

§1. The well known geometric sum is given by g[x_] = Sum[x^n, {n, 0, \[Infinity]}] (* Out[659]= 1/(1 - x) *) Here we consider the "prime geometric series" in ...
Dr. Wolfgang Hintze's user avatar
3 votes
0 answers
221 views

Negative result of a integral of positive function

Let's try this: I have a function of two arguments, $e$ and $\omega$. First I integrate some function of $(e, \omega)$: ...
user16320's user avatar
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3 votes
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139 views

Error when simplifying a series expression

I have the following function which I call FF[q_,y_,u_] and this function is well known to have a reasonable Taylor expansion in all three variables. For example, ...
Benighted's user avatar
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3 votes
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78 views

Unexpected imaginary term in the asymptotic expansion of DawsonF

FullSimplify[Series[2 DawsonF[x], {x, ∞, 8}]] (* -I Exp[-x^2] √π + (1/x + 1/(2 x^3) + 3/(4 x^5) + 15/(8 x^7) + O[1/x]^9) *) What is the reason the term ...
Vladimir Reshetnikov's user avatar
3 votes
0 answers
454 views

Polynomial kernel expansion

I am trying to calculate the polynomial kernel expansion using Mathematica. I have tried the Expand and Simplify functions with ...
JC1's user avatar
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0 answers
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Discrepancy in the series expansion of $\log(z/(z-1))$ in Mathematica 5

When I calculate the series expansion of $\log\frac{z}{z-1}$ around $0$ in Mathematica 5.2, I obtain: ...
bsmile's user avatar
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3 votes
0 answers
228 views

Generating function for Newton series?

The function GeneratingFunction gives generating function for Taylor series. Is there a similar function for Newton's series? $$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k [f]\left (0\right)$$
Anixx's user avatar
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3 votes
0 answers
208 views

Expanding a Function in Series works, SeriesCoefficient Doesn't Work

Take the following definitions: ...
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3 votes
0 answers
219 views

Changing bounds of summation after differentiating symbolic sums

Suppose, I have a function written as Taylor-Maclaurin series f = Sum[c[n]*x^n, {n, 0, Infinity}] Now, I wish to differentiate this expression with respect to <...
SaF's user avatar
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2 votes
0 answers
69 views

Series expansion of an action to get quadratic order terms of the perturbation

I'm trying to solve this paper (Eqs 6-11). The action is defined as $S = \frac{1}{2 \pi \alpha'}\int_{\Sigma} d\tau d\sigma(\sqrt{-\det g_{ab}}+ B_{mn} \partial X^m \partial X^n)$. where, $\tau=t$, $\...
Entangled Quark's user avatar
2 votes
0 answers
189 views

How to approximate an exponential series?

Consider the following expression $$ y_j= \sum_{k=0}^{L} \frac{e^{-\sum_{i=-k}^k(k-|i|)x_{j+i}}-e^{-\sum_{i=-k}^k(k+1-|i|)x_{j+i}}}{\sum_{i=-k}^k x_{j+i}}\tag{1} $$ for $1\leq j \leq L$. Given smooth ...
sam wolfe's user avatar
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2 votes
0 answers
32 views

Can we approximate a matrix power series like NSum does?

Essentially, the following does not work, and I'm wondering if it can be made to: NSum[ MatrixPower[B,n], {n,0,∞}] (Here B is a ...
Good Boy's user avatar
  • 121
2 votes
0 answers
69 views

Asymptotic expansion for a function containing irrational exponents

I'm trying to find an asymptotic expansion of a function containing irrational exponents in the form of a power series (which also might contain irrational exponents). It works correctly if I request ...
Vladimir Reshetnikov's user avatar
2 votes
0 answers
122 views

Series expansion of PolyLog[2, 1/z]

There is a well known identity involving the Dilogarithm: $$ \mathrm{Li}_2(1/z) = - \mathrm{Li}_2(z) - \frac{\pi^2}{6} - \frac{1}{2} \log^2(-z) $$ As far as I understand it should be valid for all $z \...
AGim's user avatar
  • 133
2 votes
0 answers
64 views

Approximation of roots using Series

I am solving a fifth degree polynomial using Series. My equation looks like ...
Gaurav Maurya's user avatar
2 votes
0 answers
37 views

Multivariate Lagrange inversion

I know that the function InverseSeries (Reference here) provides an interface for the Lagrange inversion formula. However I can't find anything on the ...
Ada_a's user avatar
  • 21
2 votes
0 answers
55 views

Form a power series from a list

Say I have a list L = {0,1,2,4,6} how can I form a power series with coefficients in L x + 2x^2 + 4x^3 + 6x^4 ?
Bernoulli's user avatar
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2 votes
0 answers
160 views

Symbolic perturbation expansion for quantum mechanics using Hellmann-Feynman derivaties

I am interested in some quantum mechanical perturbation expansion for energies. Actually I want to implement these terms $E_n^{(k)}$. As is stated below one can do that using CAS. I would be ...
Raphael J.F. Berger's user avatar
2 votes
0 answers
85 views

Eliminating higher order trigonometric terms

I am interested in eliminating higher-order trigonometric terms from a long symbolic expression. Specifically I want to reproduce this simplification that is done (in a tutorial I am working through)...
Sonali Gera's user avatar
2 votes
0 answers
215 views

Finding symbolic series coefficients

Is there any way to get the symbolic form of a series? For simple series this seems possible e.g. SeriesCoefficent[Exp[x],{x,0,n}] tells me that the general ...
Rolfe Petschek's user avatar
2 votes
0 answers
63 views

Expansion of HypergeometricU

I try to get the first three terms of expansion of HypergeometricU in Mathematica. My input is ...
honeybadger's user avatar
2 votes
0 answers
132 views

Why do these sums return the same result when the results should be different?

Sum[1, {k, 1, Infinity}, Regularization -> Dirichlet] gives -1/2, which is right. ...
Anixx's user avatar
  • 3,575
2 votes
0 answers
348 views

Parallelizing / speeding up a replacement rule over large list

I have a list of differentiated expression of the form {f[0,0,0],(f^(a,b,c))[0,0,0],...} where I have left the function as a general ...
jpdomann's user avatar
  • 113
2 votes
0 answers
181 views

Formal Power Series and 0^0

I'm having the following problem: I define series = Exp[Sum[Subscript[J, n]/n t^n, {n, 1, \[Infinity]}]] and it's all fine. Also when I ask Mathematica ...
MaPo's user avatar
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2 votes
0 answers
105 views

How to perform series expansions of special functions with special arguments?

I want to determine the series expansion of PolyLog[n,z] for integer but arbitrary n near z=1...
Prahar's user avatar
  • 365