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

[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:

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

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:

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

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:

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

 

[Susanne217]

I'm talking about this equation:

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

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):

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

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):

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

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:

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

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:

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

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

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

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:

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

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:

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

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

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

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 $$(0,\pi)$$:

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

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 $$(0,\pi)$$:

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

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


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

 

[LCKurtz]

Yes. Typo there. I fixed it.

 

[Susanne217]

Yes. Typo there. I fixed it.

Please correct me if I am wrong but since its known that since f is odd

then

$$f(x) = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) = \left \{ \begin{array}{cl} \frac \pi 8 x(\pi-x) &0 < x < \pi \\ -\frac \pi 8 x(\pi-x) &-\pi < x < 0 \end{array} \right .$$

Because the LHS and the RHS cancel each other hence the definition. since the correct text says "find for every x on the interval [-pi,pi] the sum of the second series series.

Then

fourier series converges pointwise towards f on the whole interval [-pi,pi]

thus

$$(P_Nf)(x) = \sum_{k=-N}^{N} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) \rightarrow x \cdot (\pi-x)$$ for
$$N \rightarrow \infty$$

and direchlet condition isn't that its defined on the interval [-pi,pi] ?

And the sum is then zero. Hope I am correct now :)

 

[LCKurtz]

1. Draw a graph of your two piece function on $$(-\pi,\pi)$$. Is it odd?
2. Why are you suddenly changing the sum from -N to N?
3. Why are you now saying series equals $$x(\pi -x)$$? Is that what the above equation says?

 

[Susanne217]

1. Draw a graph of your two piece function on $$(-\pi,\pi)$$. Is it odd?
2. Why are you suddenly changing the sum from -N to N?
3. Why are you now saying series equals $$x(\pi -x)$$? Is that what the above equation says?

1) I don't know why I am saying that because if I plug-in (-pi) into f

I get $$-4\pi^3$$ but that not the same as f(pi) because f(pi) = 0.

for f to be odd then f(-pi) = -f(pi) = $$-4\pi^3$$

but if I plug in f(-pi) $$\neq f(pi)$$ then function is even either.

Am I missing something here?? (If i draw the graph is looks like a polynomial.

2) that was because you were referring to pointwise convergence and those "N" are used in the definition. Is that a mistake?

3) yes the original f(x) is $$x(\pi -x)$$ is that a problem ?

 

[LCKurtz]

1. So your guess of (-5/8)x(π - x) doesn't make an odd function extension for f(x). Your problem is to figure out what formula does make an odd extension.

2. You must have a theorem about the Dirichlet conditions. It will tell you under what conditions on f(x) the series converges to f(x) and you need to apply it to this problem. You need to make sure your f(x) satisfies these conditions. In particular, you haven't defined f(x) at 0 or π yet.

3. A couple of posts ago we had the formula for your problem series:

$$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) = \left \{ \begin{array}{cl} \frac \pi 8 x(\pi-x) &0 < x < \pi \\ -\frac \pi 8 x(\pi-x) &-\pi < x < 0 \end{array} \right .$$

The series on the left is the series you were asked to determine the sum for. The right side of that equation is going to be your answer to that question.

We know the first part of the definition for f(x) is given but you haven't got the second part correct yet.

[Edit] Shouldn't have been an f(x) on the left, fixed

 

[Susanne217]

1. So your guess of (-5/8)x(π - x) doesn't make an odd function extension for f(x). Your problem is to figure out what formula does make an odd extension.

Damnn it :(

I need to close this problem by tonight so here goes nothing.

Since the constant of 8/pi arrives then you integrate from 0 to pi then the left hand sum of the function that must the original f(x). I simply can't see anything else either that the left hand side zero.

2. You must have a theorem about the Dirichlet conditions. It will tell you under what conditions on f(x) the series converges to f(x) and you need to apply it to this problem. You need to make sure your f(x) satisfies these conditions. In particular, you haven't defined f(x) at 0 or π yet.

The theorem says that if the f is defined on entire R then its pointwise converges. If that the teorem you are referring to since its defined on [-pi,pi] which is not R. Then f is not pointwise convegent?

3. A couple of posts ago we had the formula for your problem series:

$$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \cdot sin((2k-1)x) = \left \{ \begin{array}{cl} \frac \pi 8 x(\pi-x) &0 < x < \pi \\ x (\pi-x) &-\pi < x < 0 \end{array} \right .$$

The series on the left is the series you were asked to determine the sum for. The right side of that equation is going to be your answer to that question.

We know the first part of the definition for f(x) is given but you haven't got the second part correct yet.

[Edit] Shouldn't have been an f(x) on the left, fixed

I hope I covered all the aspects now ?

 

[LCKurtz]

You need to figure out the correct formula on $$(-\pi,0)$$ to make an odd function. And what values at $$0,\, \pi,\, -\pi$$ make it continuous when you extend your odd function periodically to the whole line. Then the Dirichlet conditions will give you convergence of the series. What you have left to do is simple algebra. Draw a graph of what its odd extension should look like. Figure out its equation. Remember the definition of an odd function.

 

[Susanne217]

Is it true (to make a long story short) that if

I get the sum $$\frac 8 \pi \sum_{k=1}^{\infty} \frac{sin\{2k-1\}x}{(2k-1)^3}$$

and if $$k \rightarrow 0^+$$

then I get that $$\frac 8 \pi \sum_{k=1}^{\infty} \frac{sin\{2k-1\}x}{(2k-1)^3} = \frac 8 \pi \cdot sin(x)$$

and if $$n \rightarrow 0^+$$$$\sum_{n=1}^{\infty} \frac{sin\{2n-1\}x}{(2n-1)^3} = sin(x)$$

and thusly the sum of the second series is sin(x) ?

 

[LCKurtz]

No, nothing you have just written makes any sense. k is an index of summation and it doesn't make any sense to talk about k --> 0. Ditto for n. I fear that you haven't really understood much of what I have tried to help you with. Since your problem was apparently due to be finished yesterday, perhaps it is time for you to have a face-to-face meeting with your instructor about this problem.

 

[Susanne217]

No, nothing you have just written makes any sense. k is an index of summation and it doesn't make any sense to talk about k --> 0. Ditto for n. I fear that you haven't really understood much of what I have tried to help you with. Since your problem was apparently due to be finished yesterday, perhaps it is time for you to have a face-to-face meeting with your instructor about this problem.

One last time

The Fourier Convergence theorem

which states that the sum of any Fourier series

$$S_{f(x)} = \frac{1}{2} [f(x^+) + f(x^{-})]$$ (found it in my advanced Calculus Fourier book)

Can't I use this here?? If not why?

Sincerely Susanne.

I will like to thank you for you help. It difficult because having different textbook which explains the same things in different ways. Sorry about my stupidity.

p.p.s. Am me being a new newbie. Sorry