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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 $f^n(x_0)$ ?

Thanks

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

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

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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 $$f(x)=\sum_{n=0}^{\infty}\frac{(x-a)^n}{(1-a)^{n+1}}$$

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Why would you think it is a pain?

Set $x_0=0$

$f^1(x)=-(1-x)^{-2}$

$f^2(x)=2(1-x)^{-3}$

$f^3(x)=-2*3(1-x)^{-4}$

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.

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  • $\begingroup$ 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

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