How to obtain the Taylor expansion of any function?

enter image description here

Like the Taylor expansion of any function in the picture. How can I obtain the Taylor expansion of any function if I input it?

Series[Log[x + 1], {x, 0, 10}]

Can the output result display the same as in the picture, with each denominator written as a factorial form of a certain number?


1 Answer 1


The trial formuala for polynomials in a real variable x

p[a_List,x_Symbol,x0_ ] := a .  (x - x0) ^ Range[0,Length[a] 


a[[m]] == 1/m! D[p[a,x,x0,n],{x,m}] /.{x:>x0}] for all m

because the m differentiation steps xield a factor m(m-1)...1 == m!

All other summands vanish at x==x0 or disappear by differentiation of constants

D[ k(k1)... 1  a_k,x}==0  

the nonzeroes yield zero at x=x0

So one tries polynomials of any order m as approximation to a function f

S[f_,x_,x0_,m_]= Sum[1/n! D[f[x],{x,n}]/:x>x0 (x-x0)^n, {n,0,m}]

If these finite sums approximate the function as m->oo, that means

Abs[f[x]-S[f,x,x0,m] ]  < some positve number c for all m > a certain number M 

then this series is the new definition of f in some interval around x0.

There is a caveat: There exists known cases, where sum may converges but does not appriximate the function f in an open interval around x0.

So its a guess only and needs a proof of convergency.

That is the really difficult and constructive point of modern analysis after Euler, that ended the euclidean mathematics of equations and jump started the search for convergence criteria in function spaces: How to check convergency of series of finite approximations, if there exists no alternative way to find the value of an object at any point.

Example: By geometry we now that Cos[0]==1, cos[pi/2]==0. Show that

 1 == Sum[-(-1)^n (\[Pi]/2)^(2 n)/(2 n)!, {n, 1, \[Infinity]}]
  • 5
    $\begingroup$ This doesn’t really answer the question at hand. $\endgroup$ Apr 22, 2023 at 15:09
  • $\begingroup$ I like very much to formulate answers, that provoke downvotes. Then its rather probable, its an answer. Discussions on n-continuous nowhere n+1 differntiables are not an object of this forum, according to impressions during my short visit here. $\endgroup$
    – Roland F
    Apr 22, 2023 at 17:17
  • 4
    $\begingroup$ I'm missing the step where downvote implies answer. (I didn't downvote this, so maybe I'm simply not knowledgeable about that sort of thing.) $\endgroup$ Apr 22, 2023 at 17:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.