Help(Using a fourier series to find the sum of second series

[Susanne217]

Homework Statement

I have found the complex Fourier series corresponding to following function f(x) = x \cdot (\pi -x) defined on the interval (0,\pi)

where I get that f(x) = \frac{\pi^2}{12} + \sum(\frac{-cos(n\pi)+1}{2n^2} + \frac{cos(n\cdot \pi)}{n^3\cdot \pi} \cdot i)

Then I suppose to use that series to find the sum of series \sum_{n=1}^{\infty} \frac{sin(2n-1)x}{(2n-1)^3} where x \in [-\pi,\pi]



Do I show the two series converge to the same point?

 

[LCKurtz]

First of all, your expression for f(x) doesn't have any x in it. But that's a bit irrelevant because I think you misunderstand what you need to do anyway. You are given a function defined on (0, π). And you are asked a question about a series that has only sine terms. Since the problem suggests the function you are given and the series you need to evaluate are related, what might be the connection? What if you extended your function x(π-x) from (0,π) to (-π,0) by taking the odd extension? What kind of terms would be in the resulting Fourier Series? Have you studied half-range expansions?

 

[Susanne217]

First of all, your expression for f(x) doesn't have any x in it. But that's a bit irrelevant because I think you misunderstand what you need to do anyway. You are given a function defined on (0, π). And you are asked a question about a series that has only sine terms. Since the problem suggests the function you are given and the series you need to evaluate are related, what might be the connection? What if you extended your function x(π-x) from (0,π) to (-π,0) by taking the odd extension? What kind of terms would be in the resulting Fourier Series? Have you studied half-range expansions?

maybe I am silly old girl, but am I suppose to deduce that from the the interval on which f(x) is defined??

Can you please give an example of a half range expression?

 

[LCKurtz]

You are given a function on (0,π). Presumably you are going to find the Fourier Series of a 2π periodic function. What are you going to use for f(x) on (-π, 0)? You are going to have to integrate f(x)cos(nx) and f(x)sin(nx) from -π to π to calculate your coefficients. So what formula are you going to use for f(x) on (-π, 0)?

Hint: What do you know about the Fourier coefficients for even and odd functions? What do you need to do with f(x) to have any chance of getting a series similar to what you are asked to sum? I'm asking about the a_n and b_n here, not the complex expansion.

 

[Susanne217]

You are given a function on (0,π). Presumably you are going to find the Fourier Series of a 2π periodic function. What are you going to use for f(x) on (-π, 0)? You are going to have to integrate f(x)cos(nx) and f(x)sin(nx) from -π to π to calculate your coefficients. So what formula are you going to use for f(x) on (-π, 0)?

Hint: What do you know about the Fourier coefficients for even and odd functions? What do you need to do with f(x) to have any chance of getting a series similar to what you are asked to sum? I'm asking about the a_n and b_n here, not the complex expansion.

If my assumption about my integral is semi correct then since it half a periode we are looking at its

\frac{1}{\pi}(\int_{0}^{\pi} (x \cdot (\pi-x) \cdot e^{-inx}) dx

anyway I found out that e^{-inx} = cos(nx)-i\cdot sin(nx) which also equals

(-1)^{-n)

does this integral then equal the series

\sum_{n=0}^{\infty}(-1)^{-n} \cdot (x\cdot (\pi-x))

 

[LCKurtz]

No, that integral does not equal that series. I'm afraid you are very confused. I have asked you several leading questions, the answers to which might help you understand, but you seem to ignore the questions. Let's start again with that last hint:

1. What do you know about the Fourier coefficients a_n and b_n in the case where f(x) is an even or odd function on (-π, π)?

2. In light of the answer to 1, what does f(x) need to be on (-π, 0) to have any chance of getting a series similar to what you are asked to sum?

Please think about the answers to these two questions. They are critical to understanding this problem. Then we can proceed.

 

[Susanne217]

No, that integral does not equal that series. I'm afraid you are very confused. I have asked you several leading questions, the answers to which might help you understand, but you seem to ignore the questions. Let's start again with that last hint:

1. What do you know about the Fourier coefficients a_n and b_n in the case where f(x) is an even or odd function on (-π, π)?

2. In light of the answer to 1, what does f(x) need to be on (-π, 0) to have any chance of getting a series similar to what you are asked to sum?

Please think about the answers to these two questions. They are critical to understanding this problem. Then we can proceed.

if n is odd and defined on the interval [-n,n] then the corresponding b_k in the Fourier series (according to my textbook)

b_k = \frac{2}{n} \int_{0}^{n} f(x) \cdot sin(\frac{k\cdot \pi \cdot x}{n}) dx

giving the corresponding Fourier series since f(x) is odd

\sum_{k=1}^{\infty} b_k \cdot sin(\frac{k\cdot \pi \cdot x}{n})

but what I don't understand here is how is this connect to f(x) is defined on a complex vector space. I thought since it stated that f(x) is defined on a complex vector space then its the same as the corresponding Fourier series is a complex Fourier seres?

have I misunderstood something here?

 

[LCKurtz]

"if n is odd and defined on the interval [-n,n] then..."

Not "if n is odd". Do you mean "if f(x) is odd"?

"have I misunderstood something here?"

Yes, I think so. You are given f(x) on (0,π), not a complex vector space. The fact that there is a complex form for the coefficients is irrelevant to the given question and is misleading you. Forget everything complex for the moment.

You have partially answered my first question by giving the formula for b_n if f(x) is odd. And what about the a_n in that case?

What about question 2?

And I just noticed your formula for b_k has typos. Check it.

 

[Susanne217]

"if n is odd and defined on the interval [-n,n] then..."

Not "if n is odd". Do you mean "if f(x) is odd"?

yes off cause I meant if f(x) is odd my mistake.

"have I misunderstood something here?"

Yes, I think so. You are given f(x) on (0,π), not a complex vector space. The fact that there is a complex form for the coefficients is irrelevant to the given question and is misleading you. Forget everything complex for the moment.

so everytime I get an assigment which states that f(x) is defined on the complex space. Then that has a trick question?

You have partially answered my first question by giving the formula for b_n if f(x) is odd. And what about the a_n in that case?

According to my textbook if f(x) is odd then the corresponding Fourier series doesn't contain and cos. I guess that's answer there.

What about question 2?
And I just noticed your formula for b_k has typos. Check it.

According to my textbook then f is odd

\int_{-n}^{n} f(x) dx = 0

and according to theory the left interval [-n,0] cancels out the right interval [0,n] and that's reason why since f is is odd we either deal if the left or right interval and not the whole interval [-n,n]

then to find the Fourier series which corresponds to f(x) = x \cdot (\pi - x) is and odd function then I find the

\begin{array}{cccccc}b_k = \frac{2}{\pi} \int_{0}^{\pi} (x \cdot (\pi - x)) \cdot sin (\frac{k \cdot \pi \cdot x}{\pi}) dx = - \frac{\pi \cdot x \cdot cos(kx)}{k} + \frac{k^2 \cdot x^2-2\cdot (cos(kx)}{k^3} - \frac{2x \cdot sin(kx)}{k^2} + \frac{\pi \cdot sin(kx)}{k^2}|_{x=0}^{\pi} = \frac{2((k\cdot \pi-2)\cdot cos(k\pi)-k\cdot cos(kx) \cdot \pi}{k^2}\end{array} ??

trying to simplify (hard)

cos(kx) = 1-k \sin^2(x)

then

b_k = \frac{2((k\cdot \pi-2)\cdot cos(k\pi)-k\cdot (1-k\cdot sin^2(x)) \cdot \pi}{k^2}

not sure how this connect to the series then being

\sum_{k=1}^{\infty} b_k \cdot sin(\frac{k \cdot \pi \cdot x}{\pi}) = \frac{k^2 \cdot \pi \cdot (sin^2 x) + 2 \cdot (k\pi-2) \cdot cos(k\pi)-k\cdot \pi) \cdot sin(kx) }{k^2} and how is suppose to lead to find the sum of

\sum_{n=1}^{\infty} \frac{sin(2n-1)\cdot x}{(2n-1)^3} ? series comparison test? Or is it something to do with convergence?

(Have I made a mistake my b_k? Cause that doesn't look like a simply expression which look any like the other series to me??

 

[LCKurtz]

You're getting there. Don't forget to put in your limits and simplify to get b_k.
Once you have b_k simplified, you are ready to write down your Fourier Series. Then it will be time to look at what you got and what the question asked and see if you can answer it.

 

[Susanne217]

You're getting there. Don't forget to put in your limits and simplify to get b_k.
Once you have b_k simplified, you are ready to write down your Fourier Series. Then it will be time to look at what you got and what the question asked and see if you can answer it.

Hi I have edited my previous post.

With some extra details. Hope you can point me in the right direction regarding these question:)

 

[LCKurtz]

Check your formula for b_k. I think you dropped a factor of 2 and have a sign wrong. And you will be able to simplify it additionally by looking at what you get when k is even vs. when k is odd.

Once you have b_k correct and simplified you need to write your Fourier series down, and put the b_k you have calculated in the series.

That needs to be done and correct. Then it will be time to look at what you have vs. the question you are trying to answer.

 

[Susanne217]

Changed my post again hope you maybe will look at it again then you have the time?

Is the first and the second series suppose to mirror each other??

I hope you can give me a hint on this, because my sill textbook hasn't any references to series that mirror each other :(

 

[LCKurtz]

It's getting hard to follow when you keep changing your post. Consider copying and pasting the relevant parts.

Your b_k's should only have k's and constants in their formula. No x's.

And when you do get them figured out, you substitute their values in the series for b_k, not on the right side of the equals like you have in the second equation from the bottom.

There is no point in trying to figure out the rest of the problem until you have your Fourier Series correct. You need to do two things to continue:

1. Work on your b_k until you can show that its value is:

b_k = \frac {4(1- cos(k\pi))}{\pi k^3}

2. Write down the series you get with these values for b_k.

 

[Susanne217]

It's getting hard to follow when you keep changing your post. Consider copying and pasting the relevant parts.

Your b_k's should only have k's and constants in their formula. No x's.

And when you do get them figured out, you substitute their values in the series for b_k, not on the right side of the equals like you have in the second equation from the bottom.

There is no point in trying to figure out the rest of the problem until you have your Fourier Series correct. You need to do two things to continue:

1. Work on your b_k until you can show that its value is:

b_k = \frac {4(1- cos(k\pi))}{\pi k^3}

2. Write down the series you get with these values for b_k.

I did the calculations have them on paper not here. But would simply like to how then I have this corresponding Fourier series for f(x). How do I use this to find the sum of the second series? Is there a theorem which I need to apply?

 

[LCKurtz]

When you get time, write down the series you get with those b_k and post it here. It will (it really will, eventually, trust me) come out very much like the series you are looking for. You won't need any theorems, just your eyeballs. There are a couple of things I may need to point out to you yet. You know the old saying, "if it was easy, it wouldn't be difficult".

 

[Susanne217]

When you get time, write down the series you get with those b_k and post it here. It will (it really will, eventually, trust me) come out very much like the series you are looking for. You won't need any theorems, just your eyeballs. There are a couple of things I may need to point out to you yet. You know the old saying, "if it was easy, it wouldn't be difficult".

Here is my calculations of the integral

\frac{2}{\pi} (x \cdot (\pi -x ) \cdot sin(kx) dx

we set x \cdot (pi -x ), du = \pi - 2x

and dv = sin(kx) , v = \frac{-cos(kx)}{k}

thus using integration by parts

x \cdot (pi -x ) \cdot \frac{-cos(kx)}{k} |_{0}^{\pi} - \int_{0}^{\pi} \frac{-cos(kx)}{k} \cdot (\pi - 2x) dx

yielding

\int_{0}^{\pi} \frac{-cos(kx)}{k} \cdot (\pi - 2x) dx

(since the left side is zero!)

choosing u =\pi - 2x, du = -2

dv = \frac{-cos(kx)}{k}, v = \frac{-sin(kx)}{k^2}

and again by integral by parts

(\pi - 2x) \cdot \frac{-sin(kx)}{k^2}|_{0}^{\pi} - \int_{0}^{\pi} \frac{2sin(kx)}{k^2} dx

yielding

\frac{sin(k\pi)\cdot \pi}{k^2} - \int_{0}^{\pi} 2 \cdot \frac{sin(kx)}{k} dx

finally result

b_k = \frac{sin(k\pi)\cdot \pi}{k^2} - \int_{0}^{\pi} 2 \cdot \frac{sin(kx)}{k} dx = \frac{-2cos(kx)}{k^3}|_{x=0}^{\pi} = \frac{-2(cos(kx)-1}{k^3}

thus

b_k = \frac{2}{\pi} \cdot (\frac{2cos(k \pi) + k \cdot sin(k\pi) \cdot \pi-2}{k^3}) = \frac{4 \cdot (cos(k\pi)-1)}{k^3\cdot \pi}

If I plug this into the formula for the odd function Fourier series I get

f(x) = \sum_{k=1}^{\infty} \frac{4 \cdot (cos(k\pi)-1)}{k^3\cdot \pi} \cdot sin(kx)

so how do I go about using this to find the sum of

\sum_{n=1}^{\infty} \frac{sin(2n-1)x}{(2n-1)^3} where x \in ]-\pi, \pi [

 

[LCKurtz]

OK, good. Now write down your Fourier series for f:

f(x) = your sum with your formulas in for the b_k.

Also write out the first 6 terms.

Oh I see you have already done that. Write out the first 6 terms now.

 

[LCKurtz]

Also, what do you notice about (cos(kπ) - 1) when k is even or odd?

 

[Susanne217]

OK, good. Now write down your Fourier series for f:

f(x) = your sum with your formulas in for the b_k.

Also write out the first 6 terms.

Oh I see you have already done that. Write out the first 6 terms now.

okay here is the first six terms

\frac{-8 \cdot sin(x)}{\pi} + 0 + \frac{-8 sin(3x)}{27 \pi} + 0 + \frac{-8 \cdot sin(5x)}{125 \pi} + 0

 

[LCKurtz]

Write out the first three terms of the series you were asked about and compare...

 

[Susanne217]

Also, what do you notice about (cos(kπ) - 1) when k is even or odd?

that for every odd k is produces -2 as a result?

 

[Susanne217]

Write out the first three terms of the series you were asked about and compare...

I get that for the second series first three terms

(sin(1) \cdot x) + \frac{sin(3)\cdot x}{27} + \frac{sin(5) \cdot x}{125}

I see that every odd term of my Fourier and the first three terms of the second series differ with a factor of \frac{-8}{\pi}??

 

[LCKurtz]

I think you are off by a minus sign in your b_k but you can chase that down later. So you should get 2 for even k instead of -2.

Do you know how to change the subscripts in your sum so it just gives the odd terms, which you can do since the even ones are all zero?

 

[LCKurtz]

You almost have it. Do you see where your equation

f(x) = your Fourier series

can have both sides multiplied by a constant to tell you what the value of your problem series is?

 

[Susanne217]

I think you are off by a minus sign in your b_k but you can chase that down later. So you should get 2 for even k instead of -2.

Do you know how to change the subscripts in your sum so it just gives the odd terms, which you can do since the even ones are all zero?

not sure how to change that? insteed of k = 1 is it then k+1? in the subscript?

 

[Susanne217]

You almost have it. Do you see where your equation

f(x) = your Fourier series

can have both sides multiplied by a constant to tell you what the value of your problem series is?

yes that constant looks to be \frac{4}{\pi}?

 

[LCKurtz]

Try writing the sum down as the sum for k = 1 to oo but inside the sum instead of writing k, write 2k-1 everywhere you have a k. When k goes 1,2,3,4... 2k-1 goes 1,3,5,7...

That will really make the two series look similar.

 

[LCKurtz]

8/π I think. So do you feel like you understand the solution now? What is your final answer to:

Your problem series = ?

 

[Susanne217]

Try writing the sum down as the sum for k = 1 to oo but inside the sum instead of writing k, write 2k-1 everywhere you have a k. When k goes 1,2,3,4... 2k-1 goes 1,3,5,7...

That will really make the two series look similar.

If I do that I get

\frac{4\cdot (2k\pi)+1) \cdot sin(2kx-x)}{((2k-1)^3) \cdot \pi}

do I need to get rid of this term here 4\cdot (2k\pi)+1)

is the sum of both series then \frac{8}{\pi}?

 

[LCKurtz]

You have the series

<br /> f(x) = \sum_{k=1}^{\infty} \frac{4 \cdot (cos(k\pi)-1)}{k^3\cdot \pi} \cdot sin(kx) <br />

and you know the numerator is 8 when k is odd and 0 when k is even. So rewrite it like this showing just the odd non-zero terms:

<br /> f(x) = \sum_{k=1}^{\infty} \frac{8}{(2k-1)^3\cdot \pi} \cdot sin((2k-1)x) <br />

<br /> = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

Solve it for your problem series and your answer is ?

 

[Susanne217]

You have the series

<br /> f(x) = \sum_{k=1}^{\infty} \frac{4 \cdot (cos(k\pi)-1)}{k^3\cdot \pi} \cdot sin(kx) <br />

and you know the numerator is 8 when k is odd and 0 when k is even. So rewrite it like this showing just the odd non-zero terms:

<br /> f(x) = \sum_{k=1}^{\infty} \frac{8}{(2k-1)^3\cdot \pi} \cdot sin((2k-1)x) <br />

<br /> = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

Solve it for your problem series and your answer is ?

which gives s_n = \lim_{k \rightarrow \infty} (\frac 8 \pi - \frac{1}{(2k-1)^3} \cdot sin((2k-1)x)) = \frac 8 \pi and this is my sum for both series?

 

[LCKurtz]

No no... You don't need any limit statement; that is already built into the infinite sum. You have an infinite sum (the "problem series") that was given to you at the beginning and you were asked to find what it added up to. You have worked a presumably related Fourier Series question for a particular f(x) that was given in your problem. You know that the Fourier Series you calculated gives the equation

f(x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

and it so happens that your problem series is in that equation. Solve for it. There's one more step after that.

 

[Susanne217]

No no... You don't need any limit statement; that is already built into the infinite sum. You have an infinite sum (the "problem series") that was given to you at the beginning and you were asked to find what it added up to. You have worked a presumably related Fourier Series question for a particular f(x) that was given in your problem. You know that the Fourier Series you calculated gives the equation

f(x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

and it so happens that your problem series is in that equation. Solve for it. There's one more step after that.

by this you mean finding the sum of series above and then showing that the difference between the sum of the two series is 8/pi?

do I try to rewrite this into a power series so find its sum?

if I take let's say f(x) = x^2 doesn't the Fourier series for that series look like the series for this one?

 

[LCKurtz]

by this you mean finding the sum of series above and then showing that the difference between the sum of the two series is 8/pi?

do I try to rewrite this into a power series so find its sum?

No. You are making this much more complicated than it is. I mean multiply both sides by π/8 to get the series on one side and what it equals on the other. That is, after all, what you want to know. You want to know the sum of the problem series, in other words, what it equals. You want a closed formula for its sum.

 

[Susanne217]

No. You are making this much more complicated than it is. I mean multiply both sides by π/8 to get the series on one side and what it equals on the other. That is, after all, what you want to know. You want to know the sum of the problem series, in other words, what it equals. You want a closed formula for its sum.

In other words by mulitiplying the second by 8/pi I (you and we) have then showed that find the the Fourier for the first series equal the second series and hence problem solved. And no need to find the sum?

 

[LCKurtz]

if I take let's say f(x) = x^2 doesn't the Fourier series for that series look like the series for this one?

You don't have the freedom to change f(x). It was given. And why would you expect the series for x² to be the same as this series which was for a different function?

 

[Susanne217]

You don't have the freedom to change f(x). It was given. And why would you expect the series for x² to be the same as this series which was for a different function?

Because I did a calculation which made this look a wee bit simular to the b_k of this series.

But then I thought. Never mind ;)

but do I still need to find sum of present the Fourier series?

 

[LCKurtz]

You aren't done with this problem until you have given an answer in this form:

the problem series = a closed form formula for a function of x

Much like if I asked for the sum of the series for |x| < 1

\sum_{n=0}^\infty x^n

and expected the answer

\frac 1 {1-x}

 

[Susanne217]

You aren't done with this problem until you have given an answer in this form:

the problem series = a closed form formula for a function of x

Much like if I asked for the sum of the series for |x| < 1

\sum_{n=0}^\infty x^n

and expected the answer

\frac 1 {1-x}

It look like a harmonic series (p-series) is that a correct assumption?

 

[LCKurtz]

"but do I still need to find sum of present the Fourier series?"

You have the sum of the Fourier series right here:

f(x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

You have a formula for f(x) given, at least on (0, π).

And if you will multiply both sides by π/8, you will have the sum of the problem series.

Once you "get" that, you are almost done.

 

[Susanne217]

"but do I still need to find sum of present the Fourier series?"

You have the sum of the Fourier series right here:

f(x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

You have a formula for f(x) given, at least on (0, π).

And if you will multiply both sides by π/8, you will have the sum of the problem series.

Once you "get" that, you are almost done.

i it true that this looks like a p-series?

 

[LCKurtz]

"is it true that this looks like a p-series?"

No. A p series has terms like 1/n^p and this has sin(nx) type terms.

 

[Susanne217]

"is it true that this looks like a p-series?"

No. A p series has terms like 1/n^p and this has sin(nx) type terms.

Sorry for my dumb questions. Is the series simular to this series which I can find in table and then use this to find the sum of the Fourier series?

 

[LCKurtz]

I am waiting for you to post the equation you get when you multiply that equation through by π/8 like I have been trying to get you to do. Write it with the series on the left side and the other on the right side.

 

[Susanne217]

I am waiting for you to post the equation you get when you multiply that equation through by π/8 like I have been trying to get you to do. Write it with the series on the left side and the other on the right side.

If I multiply by 8/pi

\frac{8sin(2k-1) \cdot x}{(2k-1)^3 \cdot \pi}

and I then write out terms for k = 1,...,6?

 

[LCKurtz]

I'm talking about this equation:

f(x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

 

[Susanne217]

I'm talking about this equation:

f(x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) <br />

so just to be clear what you are asking is I write out terms of this this as it is and then multiply by 8/pi and by this derivave the sum?

 

[LCKurtz]

Just leave the sum in there. Put it on the left so you have the form

sum = ..

Just multiply by pi/8 and do it. It's trivial...

 

[Susanne217]

Just leave the sum in there. Put it on the left so you have the form

sum = ..

Just multiply by pi/8 and do it. It's trivial...

\frac{(8\cdot sin(1) \cdot x)}{\pi} + \frac{8(sin(3))\cdot x}{27 \pi} + \cdots + \frac{8 \cdot sin(2k-1) \cdot x}{(2k-1)^3 \cdot \pi} = \frac{8}{\pi} \sum_{k=1}^{\infty} \frac{sin(2k-1)\cdot x}{(2k-1)^3}