Author Topic: Derivatives...the complex kind (Read 1837 times) Tweet Share

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?

Collin Li · « **Reply #1 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 $(1-i) = \sqrt{2}e^{-\frac{\pi}{4}i}$

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

$\therefore\; \frac{d^{18}}{dt^{18}}\left(e^{(1-i)t}\right) = 2^9e^{-\frac{\pi}{2}i}\cdot e^{(1-i)t}$

squance · « **Reply #2 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...

Collin Li · « **Reply #3 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.

squance · « **Reply #4 on:** September 04, 2008, 08:15:37 pm »

Oh...I see now
Thanks so much.

Collin Li · « **Reply #5 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

.

squance · « **Reply #6 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...)

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Author Topic: Derivatives...the complex kind (Read 1837 times) Tweet Share

squance

Derivatives...the complex kind

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Re: Derivatives...the complex kind

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