Show that a certain complex power series satisfies $f'(w)=\frac{f(w)}{2w}$

Question

Suppose $$a_{n+1}=\left(\frac 1 2 - n\right)a_n$$ with $a_0=1$ defines a real sequence of numbers.

I am then given that the following complex power series is differentiable on $D(1,1)$: $$ f(w)= \sum_{n=0}^\infty \frac{a_n}{n!}(w-1)^n.$$ I know how to differentiate complex power series on the disk with radius $1$. Its derivative is given by: $$f'(w)=\sum_{n=1}^\infty n \frac{a_n}{n!}(w-1)^{n-1}.$$

$$f'(w)=\sum_{n=1}^\infty \frac{a_n}{(n-1)!}(w-1)^{n-1}.$$ An index shift gives us: $$f'(w)=\sum_{k=0}^\infty \frac{a_{k+1}}{k!}(w-1)^{k}.$$ Which almost is the original series again, apart from an incremented coefficient.

Now I am asked to show the following relation:

$f'(w)=\frac{f(w)}{2w}$

But I do not see how I can possibly get there from what I have. I tried binomial expansion and using the recursion, but by using the recursion I get two terms and I'm struggling to make it one single series again. Any insights/hints?

Answer 1

Observe that: $$ f'(w) =\sum_{n =1} ^\infty n \frac{a_n}{n!} (w-1)^{n-1} =\sum_{n=1} ^\infty \frac{a_n}{(n-1)!} (w-1)^{n-1}.$$ We are now asked to show: $$ f'(w) = \frac{1}{2w}f(w).$$ This is equivalent to: $$ 2w f'(w) = f(w).$$ Which can be written as: $$ 2((w-1) +1) f'(w) = f(w).$$ This is what we will prove instead. We fill in $f '(w)$: $$2wf'(w) = 2((w-1) +1)\sum_{n =1} ^\infty n \frac{a_n}{n!} (w-1)^{n-1}. $$ We work out the brackets: $$2wf'(w)= \sum_{n =1} ^\infty \frac{2 n a_n}{n!} (w-1)^{n} + \sum_{n =1} ^\infty \frac{2 a_n}{(n-1)!} (w-1)^{n-1}.$$ We apply an index shift in the second series: $$2wf'(w)= \sum_{n =1} ^\infty \frac{2 n a_n}{n!} (w-1)^{n} + \sum_{n =0} ^\infty \frac{2 a_{n+1}}{n!}(w-1)^{n}.$$ We combine the power series: $$2wf'(w)=\frac{2 a_1}{0!}+ \sum_{n=1}^\infty \frac{2 na_{n}+2 a_{n+1}}{n!} (w-1)^{n}.$$ Since $a_0=1$ we have $a_1=(1/2-0)a_0=1/2 \cdot 1=1/2.$ So $2a_1=1$, we also see that $a_{n+1}=(1/2-n)a_n \implies a_{n+1}+ n a_n= 1/2 a_n $ and thus $a_n=2na_n + 2a_{n+1}$. This is precisely what appears in our power series. We see that: $$2wf'(w)=\frac{2 a_1}{0!}+ \sum_{n=1}^\infty \frac{2 na_{n}+2 a_{n+1}}{n!} (w-1)^{n}=1+ \sum_{n=1}^\infty \frac{a_n}{n!} (w-1)^{n}, $$ $$2wf'(w) = \sum_{n=0}^\infty \frac{a_n}{n!} (w-1)^{n} = f(w) .$$ This then in turn is equivalent to out desired equality: $$ f'(w) = \frac{1}{2w} f(w). \quad \square $$

Answer 2

Let $g(z)=f(1+z)$. Then the identity to be shown is $$2 g'(z) = \frac{g(z)}{1+z}\text{.}$$

To show this identity, consider the the Chu-Vandermonde identity:

$$\binom{\alpha+\beta}{n}=\sum_{k=0}^n \binom{\alpha}{k}\binom{\beta}{n-k}\text{.}$$

Let $\alpha=\tfrac{1}{2}$, $\beta=-1$. Since

$$\binom{-1}{k}=(-1)^k$$ it follows that

$$\binom{-\tfrac{1}{2}}{n}=\sum_{k=0}^n\binom{\tfrac{1}{2}}{k}(-1)^{n-k}\text{.}$$

Then $$\sum_{n=0}^{\infty}\binom{-\tfrac{1}{2}}{n}z^n=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{\tfrac{1}{2}}{k}(-1)^{n-k}\right)z^n\text{,}$$ an equation which is equivalent to the required identity.