# Determining coefficients

1. Nov 7, 2014

### MathewsMD

For the question attached in the file, how exactly does one go about finding a solution? Problem 28 says that if n does not equal m, then $\int_{-1} ^{1} {P_n}{P_m} = 0$

With that statement, I've tried treating this as a Taylor series (centred at 0, arbitrarily) and then trying to find a general term that would allow me to isolate for $a_k$ but that was unsuccessful.

Since $P_k (x)$ and $f(x)$ have similar orders, can I assume $\int_{-1} ^{1} {P_k (x)}{f(x)} = 0$? This doesn't really help, unless I want the trivial solution for $a_k = 0$

I also tried substituting the $a_k$ term into the sum to show the two sides are equivalent, but end up with $f(x) = \sum [{P_k} \int_{-1} ^{1}f(x){P_k}(x)]$ but I can't quite do anything with this either.
I just don't quite see how the above statement from Problem 28 is useful in this problem, and am uncertain on how to proceed. Any hints or explanations would be greatly appreciated!

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2. Nov 7, 2014

### Simon Bridge

It may be easier to see if I adjust the notation a tad:
You have (1) $f(x)=\sum_{n=0}^N a_nP_n(x)$, and (2) $\int_{-1}^1 P_nP_m\;dx = 0:n\neq m$ (wat happens for n=m?)
Your task is to find the kth coefficient in (1) exploiting the property in (2). (Do you know anything else about P?)
Note: $\int_{-1}^1 P_k(x)f(x)\;dx \neq 0$

3. Nov 9, 2014

### MathewsMD

Do you have any relatively simple proofs for when m = n? I have searched a few and couldn't quite get one with the similar strategy used for $m \neq n$ and the ones for when m = n seem a little longer than what I was expecting.

Also, if I just multiply by $P_n (x)$, can I just integrate both sides then and then include $P_n (x)$ with the sum terms? Then, uses the the property of it equaling for 2/(2n+1) for m = n, this does seem to make sense.

4. Nov 10, 2014

### MathewsMD

How exactly does multiplying a sum by a function, and then integrating, work? When you multiply 2 functions, and then integrate, is this analogous to taking the sum of the the two products? If one of the functions is a sum itself, can the initial sum of terms be taken out (e.g. $f(x)=\sum_{n=0}^N a_nP_n(x)$ and $g(x) = h(x)f(x)$ ; $\int_{-1}^1 g(x) = h(x)f(x) = \sum_{k=0}^K h(x)a_nP_n(x)$ OR is it $\int_{-1}^1 g(x) = h(x)f(x) = \sum_{k=0}^K h(x)\sum_{n=0}^Na_nP_n(x) = \sum_{k=0}^K \sum_{n=0}^N h(x)a_nP_n(x)$
OR none of the above)?

I'm trying to manipulate the equation, but am a little uncertain on the valid operations regarding sum and integral combinations.

5. Nov 10, 2014

### Simon Bridge

Well is $f=2x^2$ and $g=3x^3$ then $h=fg=6x^5$ and $\int h\;dx = \int fg\;dx = x^6+c$... that's how it works.
You don't need analogies - just do it: concentrate on what is in front of you.

Note: if $f = \sum_n g_n$ then $\int f\;dx = \int \sum g_n\;dx = \sum \int g_n\;dx$

6. Nov 10, 2014

### MathewsMD

Ok, thank you for the feedback! I did this:

Multiply both sides by $P_n (x)$ to get:

$P_n (x) f(x)= P_n (x) \sum_{n=0}^N a_nP_n(x)$ Then I integrated both sides wrt x, from -1 to 1.

$\int _{-1} ^1 P_n (x) f(x)= \int _{-1} ^1 P_n (x) dx \sum_{n=0}^N a_nP_n(x)$

$\int _{-1} ^1 P_n (x) f(x)= \sum_{n=0}^N a_n \int _{-1} ^1 P_n (x)P_n(x) dx$

$\int _{-1} ^1 P_n (x) f(x)= \sum_{n=0}^N [\frac {2}{2n+1}a_n]$

But I want to find value for the general $a_n$ term, not for the sum of n terms of a up to n. How exactly do I cancel out the sum to find the general term? Have I done anything incorrect so far?

7. Nov 10, 2014

### Simon Bridge

You want:
$$\int_{-1}^1 P_k(x)f(x)\;dx = \sum_{n=0}^N a_n \int_{-1}^1 P_k(x)P_n(x)\;dx$$ ... which is why I asked about what happens when n=k.

8. Nov 10, 2014

### MathewsMD

Hmmm..okay. I know that $\int_{-1}^1 P_k(x)P_n(x)dx = \frac {2}{2n+1}$ for $k = n$, but from here (I may just be missing something extremely obvious) I am a little lost. I can't seem to separate the $\sum_{n=0}^N \frac {2}{2n+1}a_n$ and isolate for simply $a_n$ in $\int_{-1}^1 P_k(x)f(x)\;dx = \sum_{n=0}^N \frac {2}{2n+1}a_n$ . I just don't quite see how to break up the series and find the general expression for an arbitrary a term. Any hints?

9. Nov 10, 2014

### Simon Bridge

... you've got it back to front.
In the equation it is n that varies and k that is fixed. What happens when n=k in the sum?

$$\int_{-1}^1 P_kf(x)\;dx \neq \sum_{n=0}^N \frac{2}{2n+1}a_n$$ ... the step right before that one was wrong.

Go back to the last equation I posted and get rid of the summation symbol on the RHS by expanding it explicitly into each term.
What happens to all the terms where $n\neq k$?

10. Nov 10, 2014

### MathewsMD

Ahh yes, sorry. I had just mixed up my notation.

11. Nov 10, 2014

### MathewsMD

Would the correct steps be:

$\int _{-1} ^1 P_k (x) f(x)= \sum_{n=0}^N a_n \int _{-1} ^1 P_n (x)P_k(x) dx$

Then for k = n

$\int _{-1} ^1 P_k (x) f(x)= a_k \frac{2}{2k+1}$ which is the answer since the integral terms become 0 when $m \neq n$. Thank you so much for the help!

12. Nov 10, 2014

### Simon Bridge

No, we do not need to consider that for $n\neq k$ that $a_n=0$ ... why would you need to?
Anyway - that would invalidate the proof.

By picking the value of k, you are choosing to evaluate the coefficient of a single term in the sum.
The point of premultiplying by $P_k$ was to isolate that term.

Last edited: Nov 10, 2014
13. Nov 10, 2014

### MathewsMD

Yes. Sorry, I saw what you were saying now. Thank you! I just want to clarify one step still, since I don't quite grasp how exactly integrals and sums are still combined.

For this step:

$\int _{-1} ^1 P_k (x) f(x)\ dx= \int _{-1} ^1 P_k (x) \sum_{n=0}^N a_nP_n(x)\ dx$

to

$\int _{-1} ^1 P_k (x) f(x)\ dx= \sum_{n=0}^N a_n \int _{-1} ^1 P_k (x)P_n(x)\ dx$

How exactly is it allowed for me to push the integral in the sum? Is it because the $a_n$ terms are considered constant by integration wrt to x, and the $P_k (x)$ function is independent of n? Is what I did here valid?