#Fourier Series Help needed
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Deam
This is what ive tried
And this is how he did it in class, but im not sure how he got to his final answer
I believe the second row is obtained after doing successive integration by parts
Which are all omitted
I will try to detail the steps below
\begin{align*}
\int_{0}^{1} (1 - t^2) \exp(-i\pi n t)dt &= \left[(1-t^2) \frac{\exp(-i\pi n t)}{-i \pi n} \right]_{0}^{1} - \int_{0}^{1} 2t \frac{\exp(-i\pi n t)}{-i \pi n} dt \\
&= \left[(1-t^2) \frac{\exp(-i\pi n t)}{-i \pi n} \right]_{0}^{1} - \left[2t \frac{\exp(-i\pi n t)}{(-i \pi n)^2} \right]_{0}^{1}\\ &+ \int_{0}^{1} 2 \frac{\exp(-i\pi n t)}{(-i \pi n)^2} \\
&= \left[(1-t^2) \frac{\exp(-i\pi n t)}{-i \pi n} \right]_{0}^{1} - \left[2t \frac{\exp(-i\pi n t)}{(-i \pi n)^2} \right]_{0}^{1} \\
+ \left[ 2 \frac{\exp(-i\pi n t)}{(-i \pi n)^3}\right]_{0}^{1}
\end{align*}
Rion
Putting it all together under one pair of brackets:
$$\int_{0}^{1} (1 - t^2) \exp(-i\pi n t)dt = \left[\exp(-i\pi n t) \left(\frac{1-t^2}{-i\pi n} - \frac{2t}{(-i \pi n)^2} + \frac{2}{(-i \pi n)^3} \right) \right]_{0}^{1 }$$
Rion
The rest just follows from simplifying all that
It seems that you did the integration by parts correctly here (it more or less matches the key), but you struggle on the simplification step
1/i can be turned into -i
(see that 1/i = i/i² = i/(-1) = -i)
and normally you should find what the key says
thank you, ill give it a shot, appreciate the help
you're welcome
+close
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