2
$\begingroup$

In proving $$f^{(n)}(w) = \frac{n!}{\rho^n \pi} \int_{0}^{2\pi} \Re ( f(w+\rho e^{i\theta})) e^{-in\theta} d\theta$$ where $f$ is analytic in $D(0;R)$, $0<r<R$, and let $w$ be an arbitary point in $D(0;r)$, I have some confusion.

The first idea is the Cauchy Integral Formula for derivatives, where we have $$f^{(n)}(w) = \frac{n!}{2\pi i} \int_{\partial D(w;\rho)} \frac{f(z)}{(z-w)^{n+1}} dz.$$

Plug in $ z = w + \rho e^{i\theta}$ we have $$f^{(n)}(w) = \frac{n!}{2\pi i} \int_{0}^{2\pi} \frac{f(w+\rho e^{i\theta})}{\rho^n e^{in\theta}} d\theta = \frac{n!}{2 \rho^n \pi i} \int_{0}^{2\pi} \frac{\Re(f(w+\rho e^{i\theta}))e^{i\theta}}{e^{in\theta}} d\theta,$$ and that's where I can go no further. I didn't know how to cancel the $2i$ in the denominator, or did the integrand has wrong power?

What is the trick?

$\endgroup$
0
2
$\begingroup$

There are two problems in your calculation. First, as Kavi said, with the parametrization $\gamma(t) = w+ \rho e^{i\theta}$ we have $\gamma'(t) = i\rho e^{i\theta}$, and therefore $$\tag{1} f^{(n)}(w) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{(z-w)^{n+1}} \, dz = \frac{n!}{2 \pi \rho^n} \int_0^{2 \pi} f( w+ \rho e^{i\theta}) e^{-in\theta} \, d\theta \, . $$

The second problem is that you cannot simply replace $f( w+ \rho e^{i\theta})$ by its real part in the integral. Instead we use Cauchy's integral theorem $$ \tag{2} 0 = \frac{n!}{2\pi i} \int_\gamma f(z)(z-w)^{n-1} \, dz = \frac{n!\rho^n}{2 \pi } \int_0^{2 \pi} f( w+ \rho e^{i\theta}) e^{in\theta} \, d\theta $$ for $n \ge 1$. Dividing this by $\rho^{2n}$ and taking the complex conjugate gives $$\tag{3} 0 = \frac{n!}{2 \pi \rho^n} \int_0^{2 \pi} \overline{f( w+ \rho e^{i\theta})} e^{-in\theta} \, d\theta \, . $$ Now we add $(1)$ and $(3)$ and note that $$ f( w+ \rho e^{i\theta}) + \overline{f( w+ \rho e^{i\theta})} = 2 \operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) \, , $$ so that the factor $2$ cancels in the result: $$ f^{(n)}(w) = \frac{n!}{ \pi \rho^n} \int_0^{2 \pi} \operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) e^{-in\theta} \, d\theta \, . $$

Remark: We have assumed that $n \ge 1$. The formula does not hold for $n=0$.

$\endgroup$
2
  • $\begingroup$ My initial thought is to rewrite $ f((w+ρe^{i\theta}) $ as $ R(f(w+ρe^{i\theta}))e^{i\theta}$, since we can rewrite a complex number as a product of its real part and its argument. Is this a valid apporach? $\endgroup$ – Dinoman May 11 at 11:32
  • 1
    $\begingroup$ @Dinoman: That does not work. What you perhaps mean is that a complex number is the product of its absolute value and $e^{i\theta}$ where $\theta$ is the argument. Also, the argument of $f(w+ρe^{i\theta})$ is not $\theta$ or $e^{i\theta}$. $\endgroup$ – Martin R May 11 at 11:35
3
$\begingroup$

With $ z = w + \rho e^{i\theta}$ we have $f^{n}(w) = \frac{n!}{2\pi i} \int_{0}^{2\pi} \frac{f(w+\rho e^{i\theta})}{\rho^{n+1} e^{i(n+1)\theta}} i\rho e^{i\theta} d\theta$ = $ \frac{n!}{2 \rho^n \pi } \int_{0}^{2\pi} {\{f(w+\rho e^{i\theta})\}e^{-in\theta}} d\theta$.

$\endgroup$
3
  • $\begingroup$ I think I have replaced it and it got cancelled off. The original denominator is to the power (n+1) $\endgroup$ – Dinoman May 11 at 9:18
  • $\begingroup$ How does $i$ still remain in the final expression? @Dinoman $\endgroup$ – Kavi Rama Murthy May 11 at 9:24
  • $\begingroup$ You made several mistakes in your calculations. How did you get $(z-w)^{n}$ instead of $(z-w)^{n+1}$ in the denominator? Please check each step carefully. $\endgroup$ – Kavi Rama Murthy May 11 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.