0
$\begingroup$

I have a question Let γ be the circular contour, positively oriented, with centre 0 and radius 8. Compute the following integral

$$\int \frac{5!\cos(z)}{(z-2\pi )^6}$$

I used the formula $$f^n(a)= \frac{n!}{2\pi i}\int \frac{f(z)}{(z-a)^{n+1}}$$

So I wrote $$f^n(a)=\frac{5!}{2\pi i} \int \frac{\cos z}{(z-2\pi)^6}$$

= $$\frac{5!}{2 \pi i} . \frac{2 \pi i}{5!}.f^5(2 \pi)=1$$

I'm new to Cauchy integrals so I'm not sure if I did this right help would be really appreciated.

$\endgroup$
  • $\begingroup$ Is $f^{(5)}(2\pi)=1$? $\endgroup$ – 민찬홍 Nov 18 '17 at 14:43
  • $\begingroup$ math.stackexchange.com/questions/2526074/… $\endgroup$ – Guy Fsone Nov 18 '17 at 14:52
  • $\begingroup$ I wrote f(z) as cosz and I f^5 (cos(z))=cosz so it would be cos 2pi if im not mistaken? $\endgroup$ – Sarah Angel Nov 18 '17 at 14:53
  • $\begingroup$ @SarahAngel No it is $-\sin z $. $\endgroup$ – 민찬홍 Nov 18 '17 at 15:02
0
$\begingroup$

There's a problem concerning the notation: it's $f^{(n)}$, not $f^n$. Besides, when you write:$$f^{(n)}(a)=\frac{5!}{2\pi i}\int\frac{\cos z}{(z-2\pi)^6},$$

  • $\mathrm dz$ is missing;
  • $\gamma$ is missing;
  • you don't tell us what $a$ is.

But the idea is correct.

$\endgroup$
  • $\begingroup$ and the parenthesis in the exponent means that many number of times differentiated, or am i tired? $\endgroup$ – mathreadler Nov 18 '17 at 15:40
  • $\begingroup$ @mathreadler Yes, that's what it means. $\endgroup$ – José Carlos Santos Nov 18 '17 at 15:41
  • $\begingroup$ I just wanted to clarify it as no one else had pointed it out. $\endgroup$ – mathreadler Nov 18 '17 at 15:42
  • $\begingroup$ That was all the information given to me in the question $\endgroup$ – Sarah Angel Nov 18 '17 at 16:03

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.