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

[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}

 

[LCKurtz]

What happened to the f(x) side of the equation? What's so difficult about multiplying both sides of an equation by something? And leave the sum in there. Don't expand it out.

Gotta run now. Will check back in later.

 

[Susanne217]

What happened to the f(x) side of the equation? What's so difficult about multiplying both sides of an equation by something? And leave the sum in there. Don't expand it out.

Gotta run now. Will check back in later.

its late as well but

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

and then dividing by 8/pi on both sides I get the formula

\frac{sin(2n-1)\cdot x}{(2n-1)^3}

I Hope this is the solution for the problems final question :) ?

 

[LCKurtz]

No. You have established a Fourier series for your given f(x):

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

where f(x) is the function you were given to begin with. This equation has a right-hand side and a left-hand side and an equals sign meaning the two sides are equal. The series on the right is equal to the function on the left which, I remind you, you were given a formula for.

All I am asking you to do is multiply both sides of this equation, as it stands, by π/8 and write it down. Don't expand it, don't change anything, and don't look at any other formulas. It will be convenient if when you write it down you put the summation of the left side of the equals and the rest on the right side.

 

[Susanne217]

No. You have established a Fourier series for your given f(x):

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

where f(x) is the function you were given to begin with. This equation has a right-hand side and a left-hand side and an equals sign meaning the two sides are equal. The series on the right is equal to the function on the left which, I remind you, you were given a formula for.

All I am asking you to do is multiply both sides of this equation, as it stands, by π/8 and write it down. Don't expand it, don't change anything, and don't look at any other formulas. It will be convenient if when you write it down you put the summation of the left side of the equals and the rest on the right side.

you mean like this \frac{8}{\pi} \cdot \pi \cdot (\pi - x) = \frac 8 \pi \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) ?

 

[LCKurtz]

What part of "don't expand it, don't change anything and don't look at any other formulas" don't you understand? And I don't think you multiplied both sides by pi/8.

 

[Susanne217]

What part of "don't expand it, don't change anything and don't look at any other formulas" don't you understand? And I don't think you multiplied both sides by pi/8.

I am sorry.

I have learned in Calculus that that if I need to find the sum of given series geometric, power etc.
Then my first step is to expand the series, look for a pattern and then go into partial sums and finally derive the sum from the series.

And you telling me that that that is not the procedure here? Sorry but I am confused

 

[LCKurtz]

I am sorry.

I have learned in Calculus that that if I need to find the sum of given series geometric, power etc.
Then my first step is to expand the series, look for a pattern and then go into partial sums and finally derive the sum from the series.

And you telling me that that that is not the procedure here? Sorry but I am confused

Yes, that is not the procedure here. You have the equation:

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

and if you would just multiply both sides of that equation by \pi/ 8, you would have your problem series on one side and what it is equal to on the other. Since I can't seem to get you to understand what I mean, it's like this:

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

or, as I have suggested, it reads better with the problem series on the left:

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

So if you want to know what the sum of your problem series is, you just look at the right hand side. You have a formula for f(x) on (0,\pi).

Once you understand that, the only remaining thing to figure out is what the formula for the sum of the series is on (-\pi,0).

 

[Susanne217]

Yes, that is not the procedure here. You have the equation:

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

and if you would just multiply both sides of that equation by \pi/ 8, you would have your problem series on one side and what it is equal to on the other. Since I can't seem to get you to understand what I mean, it's like this:

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

or, as I have suggested, it reads better with the problem series on the left:

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

So if you want to know what the sum of your problem series is, you just look at the right hand side. You have a formula for f(x) on (0,\pi).

Once you understand that, the only remaining thing to figure out is what the formula for the sum of the series is on (-\pi,0).

Hi again

So the trick is to just plug the two halfs of the interval on RHS and add them together? But since f(x) is an odd function the two halfs to the interval they cancel each other out?

If I plug the [-pi,pi] and into f(x) I get

-\pi\cdot (\pi - (-\pi)) + 0 \cdot (\pi + 0) + \pi \cdot (\pi+\pi) = 0

then since f(x) is odd then the sum I need to find is zero according to the definition?

 

[LCKurtz]

You need to know the formula for the sum of the series for all x. Here's what you know right now because you know the formula for f(x) on (0,\pi):

f(x) = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x)<br /> = \left \{<br /> \begin{array}{cl}<br /> \frac \pi 8 x(\pi-x) &amp;0 &lt; x &lt; \pi \\<br /> ?? &amp;-\pi &lt; x &lt; 0<br /> \end{array}<br /> \right .

What do you know about what f(x) needs to be on the negative interval? What is its formula? And, finally what about at the end points? Your text must have some theorem about convergence of the Fourier series. Does it mention the Dirichlet Conditions?

But your next step is to give the formula for f(x) on (-\pi,0)

 

[Susanne217]

You need to know the formula for the sum of the series for all x. Here's what you know right now because you know the formula for f(x) on (0,\pi):

f(x) = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x)<br /> = \left \{<br /> \begin{array}{cl}<br /> \frac \pi 8 x(1-x) &amp;0 &lt; x &lt; \pi \\<br /> ?? &amp;-\pi &lt; x &lt; 0<br /> \end{array}<br /> \right .

What do you know about what f(x) needs to be on the negative interval? What is its formula? And, finally what about at the end points? Your text must have some theorem about convergence of the Fourier series. Does it mention the Dirichlet Conditions?

But your next step is to give the formula for f(x) on (-\pi,0)

Isn't it suppose to be


f(x) = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x)<br /> = \left \{<br /> \begin{array}{cl}<br /> \frac \pi 8 x(\pi-x) &amp;0 &lt; x &lt; \pi \\<br /> ?? &amp;-\pi &lt; x &lt; 0<br /> \end{array}<br /> \right .