All Questions
Tagged with calculus-and-analysis series-expansion
144 questions
77
votes
6
answers
29k
views
Multivariable Taylor expansion does not work as expected
The basic multivariable Taylor expansion formula around a point is as follows:
$$
f(\mathbf r + \mathbf a) = f(\mathbf r) + (\mathbf a \cdot \nabla )f(\mathbf r) + \frac{1}{2!}(\mathbf a \cdot \nabla)...
- 17.4k
30
votes
2
answers
6k
views
How does Mathematica understand branchcuts of the complex logarithm?
Say I have the function $f(x) = x \tanh(\pi x) \log (x^2 +a^2)$ where $a$ is some positive real number. Then it seems to be me that Mathematica when given such a ...
- 1,191
16
votes
3
answers
2k
views
How can I obtain an asymptotic integral expansion at infinity?
I want to find the asymptotic expansion at $x \to \infty$ of the following function:
$$ I(x) \equiv \int_0^{\pi/2} e^{-xt^3 \cos(t)} dt.$$
To do this, I defined
...
- 7,753
15
votes
1
answer
318
views
Why do big-O terms disappear in definite integrals since Mathematica 9?
In Mathematica 8, when I computed the following input:
Integrate[Series[Cos[x], {x, 0, 2}], {x, 0, a}]
Mathematica returned an expression that had a O[a^4] in ...
- 753
13
votes
3
answers
544
views
Find asymptotics of $\sum\limits_{i=0}^{n/3} 2^i \binom{n-i-1}{\frac{2n}{3}-1}$
I have an expression
2^n / Sum[ 2^i Binomial[ n - i - 1, 2n/3 - 1], { i, 0, n/3}]
...
- 1,119
13
votes
1
answer
1k
views
Why is this infinite series wrongly computed by Mathematica?
Bug introduced in 7.0 and fixed in 10.0
Could you let me know if Mathematica (newer versions) is able to correctly compute this one?
...
- 1,415
13
votes
2
answers
865
views
How to compute the residue of $e^{z-\frac{1}{z}}$ at z=0?
I've tried the following but it didn't work:
Residue[Exp[z - 1/z], {z, 0}]
not even this:
Residue[Exp[1/z], {z, 0}]
...
- 295
13
votes
2
answers
348
views
Differentiation and series expansion of dot product - inconsistent results
Bug introduced in 9.0 or earlier and persisting through 11.3 or later
Bug resolved in 12.0
As of 12.0, we have an unevaluated result - inconsistent with the differentiation result, but not invalid.
...
- 17.1k
13
votes
1
answer
390
views
Series with ArcTan gives wrong symbolic answer in Wolfram Language
Bug introduced after 9 and persisting through 13.1. Resolved in 13.2
Recently, I have found a very bad problem with Wolfram Language. It gives the wrong answer for a quite simple expression!
When ...
- 231
12
votes
3
answers
8k
views
Laurent series expansion
Can someone share how to find the Laurent series expansion of
$$f(z)=\frac{1}{(z^2-1)(z^2-4)}$$
centered at $0$ on the annulus $1<|z|<2$?
- 15k
12
votes
2
answers
498
views
Why do I get a wrong result from SeriesCoefficient?
Bug introduced in 7.0.1 or earlier and fixed in 13.3
Consider the following code:
func[x_] = Sin[x^3]/(x - 1/3);
c[n_] = SeriesCoefficient[func[x], {x, 0, n}]
<...
- 7,182
12
votes
1
answer
540
views
Dirichlet coefficients as limits: wrong
Perhaps I should have included the word "bug" in my question. Here we go
There is a bug in this Limit
Limit[3^s (-1 - 2^-s + Zeta[s]), s -> ∞]
(* 0 *)
which ...
- 13.1k
12
votes
1
answer
1k
views
Contour Integration along a contour containing two branch points
I need to compute following contour integrations:
$$f(u)=\oint_\alpha dz \sqrt{z^3+z+u} \qquad ; \qquad g(u)=\oint_\beta dz \sqrt{z^3+z+u}$$
In which $\alpha$ and $\beta$ are two contours in ...
- 221
11
votes
2
answers
1k
views
Series expansion of an inverse
I have to find the series expansion of the inverse function of : $\arctan\left(\frac{\ln(1+x)}{1+x}\right)$
How do I find out the series expansion of any inverse ?
Note: The inverse of a function $f$...
- 2,549
11
votes
2
answers
2k
views
Find closed form expression for series expansion coefficients [duplicate]
Is there a built-in function that will find a general expression for the coefficient of the series expansion of a function?
Series will only give the explicit ...
- 236k
11
votes
2
answers
305
views
Is it possible to circumvent a bug inside SeriesCoefficient?
Bug introduced in 9.0.1 or earlier and fixed in 13.3.
As far as I can tell, there seems to be a bug in SeriesCoefficient:
...
- 559
10
votes
1
answer
842
views
Expansion for Modified Bessel Function Around Infinity
I'm somewhat new to Mathematica, and I don't understand why I'm getting inconsistent series expansions for the modified Bessel Function of first kind near $x=\infty$.
First problem:
I get different ...
- 101
10
votes
1
answer
247
views
How to predict the degree of the first series coefficient?
Given an expression f that is a function of x and a number x0, what is the least integer <...
- 245k
9
votes
2
answers
877
views
Analytical approximation of indefinite integral on a given interval to a given precision
I'm looking for an analytical approximation of
$\int_a^b e^{-x^2}\mathrm{erf}(x+c) dx$
that would be accurate to precision $\varepsilon$ for $a,b,c$ within a certain range. How do I ask Mathematica ...
- 767
9
votes
1
answer
464
views
Calculating relative error of Ramanujan formula for ellipse perimeter
On this page, they present the Ramajujan's second formula for the perimeter of an ellipse:
$$P \approx \pi (a+b) \left(1+ \frac{3 h}{\sqrt{4-3 h}+10}\right),$$
where $h=(a-b)^2/(a+b)^2$. They expand ...
- 179
8
votes
1
answer
2k
views
Series expansion for small ratio of variables
I have a messy expression of variables that I would like to simplify under the assumption that certain ratios of the variables are small. For example, consider $\sqrt{x+y}\;$ expanded for small $\...
- 183
6
votes
2
answers
2k
views
Why can’t mathematica find this residue?
I am trying to find the residue of the function $$f(z)=(z+1)^2e^{3/z^2}$$
at $z=0$.
I tried the following in Mathematica
Residue[(z+1)^2*Exp[3/z^2],{z,0}]
which ...
- 409
6
votes
1
answer
3k
views
How to expand a function into a power series with negative powers?
Is there any way to expand this expression
a+b(1-Exp[-T/(b c)]/(z-Exp[-T/ (b c)])
(where a, ...
- 71
6
votes
3
answers
417
views
Neglecting higher order terms in a Lagrangian
I have a lagrangian which is modified by variable change. I want to neglect all the 4th order and higher terms in the new lagrangian. The code being used is given below:
...
- 2,963
6
votes
1
answer
221
views
Getting terms and only evaluate specific parts of a series
How to write the first five terms of this series in the following form by MMA code?
$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot \cdots \cdot 2 n}= \frac{1}{2}+\frac{...
- 2,425
6
votes
1
answer
699
views
Find exponential generating function from the first few terms
The function FindGeneratingFunction computes the ordinary generating function of a sequence, given a sufficient number of initial terms. I have two questions in ...
- 163
6
votes
2
answers
273
views
Problem with `Derivative`/`Series` and `InverseFunction`
Here is the Mathematica code I'm asking about:
...
- 502
6
votes
1
answer
425
views
Is it possible to find a function from first few terms in the expansion
Is it possible to find a function if first few terms of the expansion is known. For example if I have this series
$f(x)=\frac{k^3 x^2}{6}-\frac{k^5 x^4}{120}+\frac{k^7
x^6}{5040}-\frac{k^9 x^8}{...
- 16k
6
votes
3
answers
196
views
Extracting a logarithmic divergence of an expression using Series
Consider the following expression:
...
- 1,553
6
votes
1
answer
233
views
Using Integrate and then Series seem to produce a wrong result
Bug introduced in 11.1.0.0 or earlier and persisting through 14.0 or later
Run this:
...
5
votes
6
answers
429
views
How to find the derivative value at $(\pi,0)$ for this implicit function $n$ times?
I am trying to take the implicit derivative at $\sin(x+y)+\sin(x)=y$ and substitute $x=\pi$ and $y=0$ at least 6-7 times since I need to find the Taylor series for this function.
Since I barely ...
- 121
5
votes
2
answers
327
views
Calculating the n-th term of the series expansion of a special function [closed]
I am trying to calculate the $n^{\text{th}}$ term of the following polynomial:
$$\, _2F_1\left(-n,n+3;\frac{3}{2};x\right)$$
To do this I calculate:
...
- 65
5
votes
1
answer
2k
views
Asymptotic expansion
I wanted to expand a function of $x$ about $x=\infty$ and see its coefficients as a function of the parameters $m,n,q,y$ and I wrote this - but it didn't work!
It gave me back a very complicated ...
- 1,191
5
votes
2
answers
214
views
Assumptions for FourierSeries
I want to calculate the Fourier series of the following function.
$u(t)=\left\{\begin{array}{lc}0, & -\frac{T}{2} \leqslant t<-\frac{\tau}{2} \\ h, & -\frac{\tau}{2} \leqslant t<\frac{\...
- 2,425
5
votes
3
answers
125
views
Why earlier terms generated from AsymptoticDSolveValue change when increasing the order?
I was comparing my hand solution with Mathematica's. I noticed the particular solution terms generated for a Frobenius series solution change depending on the order. This should not happen. As when ...
- 151k
5
votes
2
answers
917
views
Fast way to the Taylor series expansion coefficients of multivariable function?
Is there a fast method to get the coefficients of a Taylor series expansion of the function $f(x_1,x_2,...,x_d)$ with maximal summed partial derivative up to $n$, where $d,n$ can be relatively large (...
- 117
5
votes
2
answers
754
views
Expansion of the Meijer G Function
I'm trying to do the integral
Integrate[ B^2*BesselK[0, ko*ρ]^2*2 π*ρ, {ρ, a, ∞}]
which I figured should be relatively simple as the integral of a Bessel ...
- 105
5
votes
2
answers
340
views
Series expansion of a certain infinite product
I wanted to find Series[Product[Cos[t/n], {n, 1, Infinity}], {t, 0, 5}] but failed.
Then, I tried ...
- 3,896
5
votes
1
answer
134
views
Why the coefficient function is very fast
When looking for the coefficients of an desired series, I found that the Coefficient function is very fast compared to other functions and methods.
In the following summary, we find the different ...
5
votes
4
answers
463
views
`Series` with non-analytic terms
I have a function which has a smooth non-analytic term, a simplified version of which is
$$
f(\lambda) = \frac{1}{1 + k \lambda + e^{-1/\lambda}}.
$$
I want a series expansion of this, in Mathematica, ...
- 235
5
votes
1
answer
207
views
How to Approximate at Non-differentiable Point (forced Series Expansion around Branch Cut)
I need a numerical approximation around some functions at $x = 0^{-}$ from the left side, where $x = 0$ is unfortunately the right end of the domain (in the reals) so the functions are not ...
4
votes
1
answer
1k
views
How to get the Taylor series of implicit functions
Given that the equation $x+\frac{1}{2} y^{2} +\frac{1}{2} z+\sin (z)=0$ can determine an implicit function $z(x,y)$ at {0, 0}, I now need to expand the implicit function $z(x,y)$ to a fourth-order ...
4
votes
1
answer
409
views
Taylor series of functions with defined derivatives?
Suppose that I have the following equations
$$\dot{x}(t) = p(t),$$
$$\dot{p}(t) = -V'(x(t)).$$
I am trying to compute the Taylor series of $p(t)$ at $t=0$.
Here are the codes I use:
...
4
votes
1
answer
2k
views
Series approximation to integral
I would like to approximate the integral
$$
\int_0^\infty dy\,\frac{1}{\sqrt{2\pi y\sigma^2}}\exp\left(-\frac{(x-y)^2}{2y\sigma^2}\right)f(y),
$$
as a series expansion in the limit $\sigma\rightarrow ...
- 145
4
votes
1
answer
368
views
Finding periodic ODE solution via small parameter method
I want to figure out how to find periodic solutions of ODE via the small parameter method. I will provide couple of examples of what I mean.
Consider equation $\ddot x + 3x = 2 \sin t + \mu \dot x^2$....
- 293
4
votes
2
answers
147
views
4
votes
1
answer
157
views
How to expand Lie characters?
The following involves characters of affine Lie algebras, and I will be using as reference the book on CFT by Francesco et at (here are some screenshots if useful). But hopefully the post will be self-...
4
votes
2
answers
2k
views
Radius of Convergence when using Series[]
I want to write function expand[f] which gives the Taylor series expansion of $f(x)$ up to $O(x^4)$ in $\TeX$ form, as well as return the radius of convergence.
I ...
- 450
4
votes
1
answer
546
views
Problem with series expansion and integrate
I have the following very simple code. A power series in a-parameter, function f[x], is integrated with respect to x, and the ...
- 101
4
votes
1
answer
144
views
Workaround for Series messing up Inactive[Integrate]
Bug introduced in 10.0
Series does not correctly construct expansions of inactive Integrate:
...
- 19.8k