ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: squance on September 04, 2008, 07:56:24 pm

Title: Derivatives...the complex kind
Post by: squance on September 04, 2008, 07:56:24 pm: I can't seem to do this question from my green calculus book.

Find the derivative with respect to the real variable t, using the complex exponential:

the 18th derivative of e^(1-i)t

Can someone hlep me please?
Title: Re: Derivatives...the complex kind
Post by: Collin Li on September 04, 2008, 08:04:24 pm: $\frac{d^{18}}{dt^{18}}\left(e^{(1-i)t}\right) = (1-i)^{18}e^{(1-i)t}$

Since $(1-i) = \sqrt{2}e^{-\frac{\pi}{4}i}$

$\implies (1-i)^{18} = 2^9e^{-\frac{18\pi}{4}i} = 2^9e^{-\frac{\pi}{2}i}$

$\therefore\; \frac{d^{18}}{dt^{18}}\left(e^{(1-i)t}\right) = 2^9e^{-\frac{\pi}{2}i}\cdot e^{(1-i)t}$
Title: Re: Derivatives...the complex kind
Post by: squance on September 04, 2008, 08:10:36 pm: Is that it? wow....
The answer in the book says -512ie^(1-i)t...

im not sure...
Title: Re: Derivatives...the complex kind
Post by: Collin Li on September 04, 2008, 08:13:43 pm: Yep, that's right.

$2^9 = 512$ , and

$e^{-\frac{\pi}{2}} = -i$

So our answers are equivalent. That answer looks nicer.
Title: Re: Derivatives...the complex kind
Post by: squance on September 04, 2008, 08:15:37 pm: Oh...I see now
Thanks so much. :)
Title: Re: Derivatives...the complex kind
Post by: Collin Li on September 04, 2008, 08:18:06 pm: No problem!

Quote from: squance on September 04, 2008, 08:10:36 pm
Is that it? wow....

You see where it comes from right? If I differentiate $e^{at}$ , I get $ae^{at}$ (first derivative). Do it again I'd get $a^2e^{at}$ (second derivative)... doesn't take much imagination to see what happens at the eighteenth derivative now ;).
Title: Re: Derivatives...the complex kind
Post by: squance on September 04, 2008, 08:21:21 pm: Yep i get it now :)
thanks again :)

(sometimes im just a bit slow with understanding maths...esepcially anything to do with complex numbers...they are too complex...)