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This example comes from the book by Elias Wegert: Visual Complex Functions. Consider the function $f(z):=z^{1/2}$ for $z \in \mathbb{C}$ with $|z - 1| < 1.$ For these $z$, the function can be represented by a convergent power series

$$f(z) = \sum_{k=0}^{\infty} {1/2 \choose k} (z-1)^{k},$$

which means that $z^{1/2}$ is analytic in $D(1,1).$ Now, Theorem 3.3.1 in that book says that, given

$$f(z) = \sum_{k=0}^{\infty} a_k(z-z_0)^k, \quad \forall z \in D(z_0,r),$$

you can "rearrange" the latter series into

$$f(z) = \sum_{k=0}^{\infty} b_k (z-z_1)^k, \quad \forall z \in D(z_1,r_1),$$

where the $b_{k}$'s are given by $\sum_{n=k}^{\infty} {n \choose k} a_n (z_1-z_0)^{n-k},$ $k = 0, 1, \ldots, $ provided only that $|z_1 - z_0| < r$ and $r_1 := r - |z_1 - z_0|.$ Noting that $\rm{e}^{i \pi/4} \in D(1,1)$ we can try to apply the Theorem to the series defining $z^{1/2}$ with $z_1 = \rm{e}^{i \pi/4}$. This should give

$$f(z) = \sum_{k=0}^{\infty} \left(\sum_{n=k}^{\infty} {n \choose k} {1/2 \choose n} (\rm{e}^{i \pi/4} - 1)^{n-k} \right) (z-z_1)^k.$$

This in turn should be equal to

$$f(z) = \rm{e}^{i \pi/8} \sum_{k=0}^{\infty}\rm{e}^{-i k\pi/4} {1/2 \choose k} (z-z_1)^k,$$

the point being here that while the original series diverges at, say, $z = i,$ the last one is already convergent there (...). The only thing that completely eludes me is why (as it seems)

$$\sum_{n=k}^{\infty} {n \choose k} {1/2 \choose n} (\rm{e}^{i \pi/4} - 1)^{n-k} = \rm{e}^{i \pi/8}\rm{e}^{-i k\pi/4} {1/2 \choose k},$$

for $k = 0, 1 \ldots.$ I thought about induction on $k$ but was only able to do the base step $k=0.$ Could anybody help me? Thanks.

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Remember that the coefficients of the Taylor series of $f$ about $z_1$ are $\dfrac{f^{(k)}(z_1)}{k!}$. Writing $z^{1/2}$ for the square root, we have

$$z_1^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n}(z_1-1)^n,$$

which is the $k= 0$ case. Now differentiate the identity with respect to $z_1$,

$$\frac{1}{2}z_1^{1/2}z_1^{-1} = \sum_{n=1}^\infty \binom{n}{1}\binom{1/2}{n}(z_1-1)^{n-1}$$

is the case $k = 1$. Generally,

$$\frac{d}{dz_1} z_1^{1/2-k}\binom{1/2}{k} = (k+1)\cdot z_1^{1/2-(k+1)}\binom{1/2}{k+1},$$

and

$$\begin{align} \frac{d}{dz_1}\sum_{n=k}^\infty \binom{n}{k}\binom{1/2}{n}(z_1-1)^{n-k} &= \sum_{n=k+1}^\infty (n-k)\binom{n}{k}\binom{1/2}{n}(z_1-1)^{n-(k+1)}\\ &= \sum_{n=k+1}^\infty (k+1)\binom{n}{k+1}\binom{1/2}{n}(z_1-1)^{n-(k+1)}. \end{align}$$

Divide by $k+1$ to obtain the $k+1$ case from the $k$ case.

Nicer, in my opinion, is a direct exploitation of the series centered at $1$: Near $z_1$, where $\lvert z_1-1\rvert < 1$, we have

$$\begin{align} \sqrt{z} &= \sqrt{z_1}\cdot \sqrt{\frac{z}{z_1}}\\ &= \sqrt{z_1}\sqrt{1 + \left(\frac{z}{z_1}-1\right)}\\ &= \sqrt{z_1}\sum_{k=0}^\infty \binom{1/2}{k}\left(\frac{z}{z_1}-1\right)^k\\ &= \sqrt{z_1}\sum_{k=0}^\infty z_1^{-k}\binom{1/2}{k}(z-z_1)^k. \end{align}$$

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  • $\begingroup$ Well, that helped a lot, thank You. If I am right, you mean it like this: ok, we know that (under appropriate conditions) $f(z) = \sum_{n=0}^{\infty} b_n (z - z_1)^n,$ where the numbers $b_n$ are to be determined. Now, on one hand, $f^{(k)}(z_1) = k! \cdot b_k.$ On the other hand $f^{(k)}(z_1) = 1/2 \cdot (1/2-1) \cdot \ldots \cdot (1/2-k+1) z_1^{1/2-k},$ frome where $b_k = {1/2 \choose k} z_1^{1/2-k}.$ The other approach with $z = z_1 \cdot \frac{z}{z_1}$ seems to be, indeed, even much nicer (and shorter of course). $\endgroup$ – Jorge.Squared Jan 27 '14 at 16:29
  • $\begingroup$ Yes, that's the idea. And from writing $(z-1)^n = ((z-z_1)+(z_1-1))^n$, we have got the other representation $$b_k = \sum_{n=k}^\infty \binom{n}{k}\binom{1/2}{n}(z_1-1)^{n-k}.$$ Since the coefficients must be the same, the identity follows. $\endgroup$ – Daniel Fischer Jan 27 '14 at 16:35

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