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Hi, please refer to the attached image.
I am having trouble when doing
Exercise 2
Here is what I did:
$ \int_{2}^{2}(f(x)\sin(\frac{m\pi x}{2}))dx = \sin(\frac{m\pi x}{2})a_{0} + \int_{2}^{2} \sum\limits_{n=1}^\infty (a_{n}(\cos(\frac{n\pi x}{2})\sin(\frac{m\pi x}{2})+b_{n}\sin(\frac{n\pi x}{2})\sin(\frac{m\pi x}{2}))$
and we have to show that $b_n$ is equal to $0$, for all $n$. However
is it not true, that since the terms in the sum are mutually orthogonal, we will no longer have anything within the sum when it is integrated from 2 to 2?
In which case, we will not have a $b_{n}$? As I type this, I realise... that this is what makes $b_{n}$ = 0, but in case I am wrong, could someone please point me to the right direction?
Do I even need to evaluate the left hand side integral or the integral for $a_0$ for exercise 2?
exercise 3
I am confused what to do after using the orthogonality again.
I am left with:
$ \int_{2}^{2} f(x) \cos(\frac{m \pi x}{2})dx $ = $\int_{2}^{2}a_0\cos(\frac{m \pi x}{2})dx$
how do i treat the f(x) ?
I am unsure how to get $a_n$ from this.
although, i know that since m is fixed, where m = n we will get a nonzero term. I feel like this has something to do with what I need.
Thanks
I am having trouble when doing
Exercise 2
Here is what I did:
$ \int_{2}^{2}(f(x)\sin(\frac{m\pi x}{2}))dx = \sin(\frac{m\pi x}{2})a_{0} + \int_{2}^{2} \sum\limits_{n=1}^\infty (a_{n}(\cos(\frac{n\pi x}{2})\sin(\frac{m\pi x}{2})+b_{n}\sin(\frac{n\pi x}{2})\sin(\frac{m\pi x}{2}))$
and we have to show that $b_n$ is equal to $0$, for all $n$. However
is it not true, that since the terms in the sum are mutually orthogonal, we will no longer have anything within the sum when it is integrated from 2 to 2?
In which case, we will not have a $b_{n}$? As I type this, I realise... that this is what makes $b_{n}$ = 0, but in case I am wrong, could someone please point me to the right direction?
Do I even need to evaluate the left hand side integral or the integral for $a_0$ for exercise 2?
exercise 3
I am confused what to do after using the orthogonality again.
I am left with:
$ \int_{2}^{2} f(x) \cos(\frac{m \pi x}{2})dx $ = $\int_{2}^{2}a_0\cos(\frac{m \pi x}{2})dx$
how do i treat the f(x) ?
I am unsure how to get $a_n$ from this.
although, i know that since m is fixed, where m = n we will get a nonzero term. I feel like this has something to do with what I need.
Thanks
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