# Homework Help: By choosing a suitable value of x, sum the series

1. Dec 9, 2011

### bubokribuck

The question:

Here's my thought so far:

This is pretty much what I have come up with. However, I don't know how to find a general solution for the sum of the series. Can someone help please. (the decimal part is where I'm stuck the most)

2. Dec 9, 2011

### SammyS

Staff Emeritus
$\displaystyle\frac{1}{1\cdot3}-\frac{3}{5\cdot7}+\frac{5}{9\cdot11}-\frac{7}{13\cdot15}+\dots$

What do we see?

The signs of the terms alternate.

Each successive numerator is the next odd natural number. If n is the index and starts at 0, then each numerator is (2n-1).

How are the two factors in each denominator related to each numerator ? (Consider what 2 times the numerator is for each.)

3. Dec 9, 2011

### Joffan

My guess is that those are not decimal points - they are multiplication dots.

Presumably you have noticed the alternating signs on the terms.

There is a simple relationship between the numerator and the two terms in the denominator in each term which will probably help.

4. Dec 9, 2011

### bubokribuck

Um... to be honest, I can't see how each denominator relates to each numerator, doesn't seem like there's a sequence.

5. Dec 9, 2011

### bubokribuck

Yes I have noticed the alternating signs, but couldn't find any relationship between the numerator and the denominator though :(

6. Dec 9, 2011

### Joffan

Well, calculate out the first ten terms, then come back and post them all.

$+\frac{1}{1\times3}$

$-\frac{3}{5\times7}$

$+\frac{5}{9\times11}$

$-\frac{7}{13\times15}$

$\ldots$

7. Dec 9, 2011

### Joffan

You might also put in the known values for a, b and c (1,1,3) in your formula for $n_{m}$ and simplify.

8. Dec 9, 2011

### SammyS

Staff Emeritus
Compare the factors in each denominator to 2 times the numerator !!

9. Dec 10, 2011

### spamiam

Just a small typo--I'm guess you meant the numerator is 2n+1.

bubokribuck: What is the first part of this problem (that you've cut off from the image you posted)? The statement of the problem ("By choosing a suitable value of x...") indicates that in previous parts of this problem you've found a series representation for some function, which you are supposed to use to solve this part.

10. Dec 10, 2011

### bubokribuck

Hi spamiam, here's the full question, but I don't like it has anything to do with the previous parts though.

11. Dec 10, 2011

### bubokribuck

Hi Joffan, here it is.

12. Dec 10, 2011

### bubokribuck

If the first term is n=1, then the numerator of the first term would be 2n-1 right? (as 2(1)-1=1)

13. Dec 10, 2011

### bubokribuck

Ah, I think I am getting it now!

Let N denotes the numerator, and D denotes the denominator, then D=(2N-1)(2N+1). Is this right? So for the nth term, it would be something like (-1)n+1$\frac{2n-1}{(4n-3)(4n-1)}$

Last edited: Dec 10, 2011
14. Dec 10, 2011

### spamiam

It definitely has to do with the previous part! What could you be choosing x for if not to plug into the Fourier series you found in an earlier part? So what did you get for the first 2 parts? In particular, what did you get for the Fourier series of the function?

You formula for the general term of the series looks correct.

15. Dec 10, 2011

### SammyS

Staff Emeritus
If the index, n, starts at zero, (like I said), then, yes, that should be (2n+1).

16. Dec 10, 2011

### bubokribuck

Hi spamiam, here's what I've got for the Fourier series:

And here's the sigma equation I've come up with for the last part of the question.

17. Dec 10, 2011

### spamiam

Hm, that's not what I get for the Fourier series. Could you show your calculations? Also, your series has a couple of oddities. For one thing, when n=1, you have terms of the form 1/0. Hopefully this doesn't pose a problem since these terms should cancel out. Which brings me to the second observation--you can simplify your sum since terms cancel out when n is odd.

18. Dec 10, 2011

### bubokribuck

Hi, the calculations are quite long, so I've attached a document. :)

#### Attached Files:

• ###### Find the Fourier series of f.zip
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19. Dec 11, 2011

### spamiam

Hey bubokribuck,

I (and I'm guessing this is the case with most other members from the 0 views your attachment has gotten) am wary of downloading zip files as a general rule. Could you upload it in another format like .doc, .pdf, or .jpg?

20. Dec 11, 2011

### bubokribuck

Hi, the doc exceeds the upload size limit, therefore I can only upload png. Thanks :)

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