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Create a power series around $x=0$ and find its radius. What is $f(1)$? $$f(x)=\frac{1}{2+x}$$ Well according to me it is easy to see that the radius is $\left|x\right|<2$ and $f(1)=\frac{1}{3}$...right?

If I try to manipulate it to the geometric serie $f(x)=\frac{1}{1-x}$ I get $f(x)=\frac{1}{1-(-1-x)} =\frac{1}{2+x} $ which would be $\sum (-1-x)^k$ because $f(x)=\frac{1}{1-x}=\sum x^k $ but $\sum (-1-x)^k$ diverge when $x=1$. Why doesnt that work when $f(x)=\frac{1}{2}\frac{1}{1+x/2)}=\frac{1}{2}\sum (\frac{-x}{2})^k$ seems to do fine.$f(1) =1/3$

Do not use maclaurin to prove the series.

I am doing something fundamentally wrong here, please help me out.

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$\sum (-1-x)^k$ is the power series expansion of $f(x)$ around $x=-1$

$\frac12 \sum (\frac{-x}{2})^k$ is the power series expansion of $f(x)$ around $x=0$

Note that $f$ has a singularity at $x=-2$, therefore the former series has radius of convergence $1$ and can be used to compute the value of $f(x)$ in the interval $(-2,0)$ only (in particular not at $x=1$)

The later series has radius of convergence $2$ and can be used to compute values of $f(x)$ in the interval $(-2,2)$

So your problem with the first expansion is that by moving the base point to the left the radius of convergence becomes smaller since it can't go past the functions singularity and does no longer include 1.

Radii of convergence:

Radii of convergence

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you need to write $$\frac{1}{2+x} = \frac 12 \frac{1}{1 + x/2} = \frac 12 \left(1 - \frac x2 + \frac {x^2}4 - \frac {x^3} 8 \cdots \right), -2 < x <2.$$

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  • $\begingroup$ Sorry, I was just typing. Did not see that we have had same idea. $\endgroup$ – Frieder Mar 11 '15 at 12:06
  • $\begingroup$ @Frieder, that is quite alright. $\endgroup$ – abel Mar 11 '15 at 14:00
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I start with

$$\frac{1}{t+1}=\sum _{k=0}^{\infty } (-1)^k t^k$$

Now I substitude

$$t=\frac{x}{2}$$

It gives me

$$\frac{1}{\frac{x}{2}+1}=\sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k$$

Multiply with $\frac{1}{2}$ I get

$$\frac{1}{2 \left(\frac{x}{2}+1\right)}=\frac{1}{2} \sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k=\sum _{k=0}^{\infty } \frac{(-1)^k x^k}{2^{k+1}}$$

And this is

$$\frac{1}{x+2}=\sum _{k=0}^{\infty } \frac{(-1)^k x^k}{2^{k+1}}$$

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