I am confused about the point $x_0$ in Taylor Series Expansion.

$f(x) = 1/(1-x)$ and $\sum_n^\infty x^n$ at $x_0=0$

so I thought that if $x_0$=2 I don't need to go through all the process solving for $f^n(x_0)$

and I have a shortcut for $x_0=2$ like :

$$f(x) = 1/(1-(x-2))$$ and series for it is$$\sum_n^\infty (x-2)^n$$ but it is not true.

My exercise was to find $f(x)$ for $\sum_n^\infty((x-1)/2)^n$. So I see here that $x_0=-1$ and

immediately I thought of this:

$$1/(1-(x-1)/2) = \sum_n^\infty((x-1)/2)^n$$

so I just got lucky with this example.

Question : Isn't there a shortcut to find the series of a function like $1/(1-x)$ where $x_0=const$ without going through all the pain of finding the derivative for $f^n(x_0)$ ?


  • $\begingroup$ $\sum_n^\infty((x-1)/2)^n$ is the sum of terms in geometric progression, isn't it ? Then .... $\endgroup$ Jan 3 '15 at 14:35
  • $\begingroup$ I don't understand , I am saying that using my bad logic I got lucky only with this example $\endgroup$
    – Oleg
    Jan 3 '15 at 14:43

You didn't get lucky in the second example, you just didn't do it properly in the first.

Let's see.

You have $f(x)=\frac{1}{1-x}=\sum x^n$. You want to obtain $f(x)=\sum a_n(x-2)^n$.

Then $\sum x^n=\sum a_n(x-2)^n$. Put $x=y+2$.

So $f(y+2)=\sum (y+2)^n=\sum a_ny^n$.

Therefore, to obtain the coefficients $a_n$ we just need to expand $f(y+2)$ at $y_0=0$.

But $f(y+2)=\frac{1}{1-(y+2)}=-\frac{1}{1-(-y)}=-\sum (-y)^n=\sum(-1)^{n+1}y^n$. Therefore $a_n=(-1)^{n+1}$. Therefore

$$f(x)=\sum (-1)^{n+1}(x-2)^n.$$

Summarizing: To expand $f(x)$ at $x_0$ (assuming that what we know is to expand things at $0$) is to expand $$f(x+x_0)=\sum a_nx^n$$ at $0$ and take those coefficients to get

$$f(x)=\sum a_n(x-x_0)^n.$$


Consider the general case where you need the Taylor expansion of $$f(x)=\frac{1}{1-x}$$ built at $x=a$. The derivatives are easy to compute and you arrive at $$f(x)=\sum_{n=0}^{\infty}\frac{(x-a)^n}{(1-a)^{n+1}}$$


Why would you think it is a pain?

Set $x_0=0$




So, as you can see, $f^n(0)=(-1)^{n}*n!$

And even if you don't want to set $x_0=0$, for any $x_0$, you'd have $f^n(x_0)=(-1)^n*n!*(1-x_0)^{n-1}$

For other cases, I am not an expert at this, but there are some helpful "tricks".

Assume that the expression for the derivative is easy to obtain, so that we have the coefficients $c_0',c_1',\ldots$ then we have $(f')^{(k)}(0)=c_k'*k!$ and thus $f^{k+1}(0)=c'_k*k!$ which would make it easy to get the coefficients for $f$

After simplification, you'd get $f(x)=f(0)+\frac{c_0'}{1}x+\frac{c_1'}{2}x^2+\ldots$

This would be helpful in the case of getting the taylor series for $arctan(x)$ for example.

  • $\begingroup$ for this function it is not a pain , but for others might be , I wanted to know if I can short it out :) $\endgroup$
    – Oleg
    Jan 3 '15 at 14:40
  • $\begingroup$ I'll edit my answer. $\endgroup$
    – Hasan Saad
    Jan 3 '15 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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