# Tagged Questions

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

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### How to force Series[] to compute expansions by considering non commutative multiplication?

I wish to compute the Taylor series expansion of the following iteration method $x_{k+1}=x_k-f'(x_k)^{-1}f(x_k)$ up to four terms of error ($e_k=x_k-\alpha$). When this is a scalar iteration, I very ...
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### What is the formula for this numerical series?

I'm developing a questions game. My goal is that the score for each correct answer will increase as the user answers more questions. Initially there are 15 points for each correct answer. Every 4 ...
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### Evaluation of a triple sum does not finish in reasonable time

I'm trying to compute the following triple sum, but no result is produced within a reasonable amount of time. What to do? ...
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### Why does Mathematica fail to series expand this simple expression?

I wanted to expand the function $(x+2)^{x+2}$ around $x = -1$, that is, using Series[(x + 2)^(x + 2), {x, -1, 2}] and Mathematica returns the same expression. ...
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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}{... 1answer 483 views ### Erfi[z] expansion in Mathematica: is this a bug? Looking at the expansion: Series[Erfi[x], {x, Infinity, 1}] I obtain -I+E^x^2 (1/(Sqrt[\[Pi]] x)+O[1/x]^2) (note the ... 2answers 739 views ### How do I solve N simultaneous equations for N variables? I have a function: f[x_] := x + 31 x^3 + 5 x^25 Which I want to find an expansion for: ... 2answers 166 views ### How to get Series to recognize user-defined function has pole Edit for clarity: How does Mathematica's function Series know that Gamma[x] has a pole at ... 1answer 118 views ### Changing default behavior of Series to give slightly different SeriesData I am working with complicated expressions expr that contains the symbolic function f[x] which I know to have a pole at ... 1answer 303 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 ... 1answer 119 views ### Apparently contradictory output from Series and D I defined a function as: g[b_] := Integrate[f[n]Exp[-b DA (n.u)^2], n] Obviously D[g[b], b] /. b -> 0 gives ... 1answer 186 views ### List manipulation - various Update OK, Please forgive the messiness of this, but I am working with something like: ... 0answers 72 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 ... 1answer 124 views ### Peculiarities with Series and fractional exponents or bug? The documentation of the Series[] function states that it can handle "certain expansions involving negative powers, fractional powers, and logarithms." What are the ... 2answers 154 views ### Change all values of a series to positive values I would like to change all values of a series to positive values. I refer to my previous question, answered in the most part, very elegantly by RunnyKine. However, there remains a slight glitch that ... 2answers 138 views ### Solve of cubic and quartic equations too slow In my problem I have a third order algebraic equation for the variable sigma, all other letters are parameters. Here is it's right-hand side, left hand side is ... 0answers 60 views ### Best way to power series expand in multiple variables? [duplicate] A power series expansion of a multivariate function can be performed with the Series command, which performs each expansion consecutively. The results can be a bit ... 1answer 88 views ### Numerically solving a transcendental equation as a series Given a transcendental equation$f(x,y)=0\$, is there a way for Mathematica to automatically solve the equation as a series? I already know that I can use ...
I need to convert derivatives of g[x,v] wrt v into derivatives wrt x using this relation: $$\frac{dg[x,v]}{dv}=\frac{1}{2}\frac{d^{2}g}{dx^{2}}+\frac{1}{2}\left(\frac{dg}{dx}\right)^{2}$$ This code ...