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$f(z)= log(2+z)$ for $z_{0}=0$; $g(z) = (z-2)^{-1}$for $z_{0}=1$

It is possible just to make derivatives and hope. But there is a faster way by using known taylor series.


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You know the series for $\log(1-z)$ and $\dfrac{1}{1-z}$, yes? – J. M. Nov 3 '11 at 2:08
…………………….yes……. – VVV Nov 3 '11 at 2:12
There is a theorem: If you know that your function $f$ has a representation $f(z)=\sum_{k=0}^\infty a_k z^k$ which converges in a disk of positive radius, then this representation is in fact the Taylor series of $f$; i.e., $a_k=f^{(k)}(0)/k!$ for all $k\geq0$. – Christian Blatter Nov 3 '11 at 9:47
How to use these answers if $z_{0}$ is not 0. Since the series known are usually $z_{0}=0$ ? – VVV Nov 3 '11 at 16:43
up vote 0 down vote accepted

I'll do $g(z)=\frac{1}{z-2}$. We have

$$ \frac{1}{z-2}=-\frac{1}{1-(z-1)}. $$


$$ \frac{1}{1-x}=\sum_{n=0}^\infty x^n $$

we set $x=1-z$ to get

$$ -\frac{1}{1-(z-1)}=\sum_{n=0}^\infty -(z-1)^n. $$

You can do $\log(2+z)$ similarly.

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What about $z_{0}=1$ I your representation takes $z_{0}=0$???? – VVV Nov 3 '11 at 16:38
@VVV: I'm not sure what you are saying. But, the series above is centered at $z_0=1$ just like you asked. It has a radius of convergence of $1$ and so $z=0$ can't be plugged in. – Joe Johnson 126 Nov 6 '11 at 12:52

We look first at the logarithm question, which is more interesting, and then at the reciprocal question, which is far more important, but easy.

The logarithm: The following is an idea that works for $\log(2+x)$, where $|x|$ is small.

Try to express $2+x$ as $\frac{1+u}{1-u}$. Then $\log(2+x)=\log(1+u)-\log(1-u).$$

We calculate $u$. From the equation $2+x=\frac{1+u}{1-u}$, we obtain, after a little manipulation, $$u=\frac{1+x}{3+x}.$$ Now we can expand $\log(1+u)$ and $\log(1-u)$ as usual in powers of $u$, that is, of $\frac{1+x}{3+x}$. If $x$ is close to $0$, we get moderately rapidly convergent series.

Take for example $x=0$. We get $\log 2$ as the difference between two fairly rapidly convergent series. The ordinary series for $\log(1+x)$ does converge when $x=1$, but glacially slowly. We can also use the idea with $\log(3+x)$, and obtain $\log 3$ as the difference between two fairly quickly convergent series. Note that in this case the ordinary series does not converge.

Tricks of this type go back to Napier's first computations of logarithms. Euler also used this idea.

The reciprocal: Less interesting, but practically much more useful, is the problem of expanding $\frac{1}{x-2}$, and similar expressions. Note that $x-2=-2(1-x/2)$. Thus $$\frac{1}{x-2}=-\frac{1}{2}\frac{1}{1-x/2}.$$ We know the expansion of $\frac{1}{1-u}$ in powers of $u$. Substitute $x/2$ for every occurrence of $u$.

Comment: The explicit numerical examples for the logarithm used real $x$, but nothing changes if instead we use complex $x$, except that we have to be careful about the branch of the logarithm. In the case of the reciprocal, whether the variable is complex or real makes no difference.

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