Proof of double derivative.

**cupid** · February 28th 2011, 06:03 AM

Question - In Pic

My progress is that i found out that $f''(x) = \frac{[f'(x)]^2}{f(x)}$

but i couldn't get to the solution

**NOX Andrew** · February 28th 2011, 06:21 AM

If you use the identity:

$\cos{x} + i\sin{x} = e^{ix}$ ,

then

$y = e^{ix} \cdot e^{3ix} \cdot ... \cdot e^{\left(2n-1\right)ix}$

$= e^{ix + 3ix + ... + \left(2n-1\right)ix}$

$= e^{ix\left(1 + 3 + ... + 2n-1\right)$

$= e^{n^2ix}$

**cupid** · February 28th 2011, 06:34 AM

Thanks a lot Andrew
that made the question soo simple
thanks a lot

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