- #1

- 11

- 0

f(x) = sin(2x)*(x^2) (-pi < x < pi)

I got this answer, although it seems to be wrong.

f(x) = (-16/9)*sin(x) + (-1/8)*sin(2x) + sum((2*(-1)^n/(2-n)^2 - 2*(-1)^n/(2+n)^2)*sin(nx) , n, 3, 100)

- Thread starter pedro_bb7
- Start date

- #1

- 11

- 0

f(x) = sin(2x)*(x^2) (-pi < x < pi)

I got this answer, although it seems to be wrong.

f(x) = (-16/9)*sin(x) + (-1/8)*sin(2x) + sum((2*(-1)^n/(2-n)^2 - 2*(-1)^n/(2+n)^2)*sin(nx) , n, 3, 100)

- #2

- 11

- 0

Can you help me, please?

- #3

- 11

- 0

[PLAIN]http://img826.imageshack.us/img826/6708/semttulocz.png [Broken]

Last edited by a moderator:

- #4

- 22

- 0

What partCan you help me, please?

- #5

- 11

- 0

I did this exercise a lot of times, every time I get the same answer.

I can post the whole exercise to help me find what is wrong.

That Bn is kind hard to get right.

- #6

- 9,547

- 760

I think your b

I did this exercise a lot of times, every time I get the same answer.

I can post the whole exercise to help me find what is wrong.

That Bn is kind hard to get right.

[tex]b_2 = -\frac 1 8 +\frac {\pi^2} 3[/tex]

And remember when plotting your graph, that your formula for your original function only works on (-pi,pi), so look at the convergence there.

- #7

- 11

- 0

Thanks, the problem was at the B2.

Cya.

Cya.

- #8

- 11

- 0

I got this:

Bn = (2*(-1)^n)/(2-n)^2 - (2*(-2)^n)/(2+n)^2

B2 = (2/(2-2)^2) - 1/8

What did you make to get that: [tex]\frac {\pi^2} 3 [/tex]

Bn = (2*(-1)^n)/(2-n)^2 - (2*(-2)^n)/(2+n)^2

B2 = (2/(2-2)^2) - 1/8

What did you make to get that: [tex]\frac {\pi^2} 3 [/tex]

Last edited:

- #9

- 9,547

- 760

That formula doesn't work for n = 2 so you have to do bI got this:

Bn = (2*(-1)^n)/(2-n)^2 - (2*(-2)^n)/(2+n)^2

B2 = (2/(2-2)^2) - 1/8

What did you make to get that: (pi^2)/3

[tex]\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx[/tex]

- #10

- 11

- 0

Thanks, finally I got the right answer.That formula doesn't work for n = 2 so you have to do b_{2}separately. Just directly work out the integral

[tex]\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx[/tex]

You helped me a lot.

Last edited:

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 3

- Views
- 6K

- Last Post

- Replies
- 4

- Views
- 8K

- Replies
- 7

- Views
- 206

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 8

- Views
- 27K

- Last Post

- Replies
- 15

- Views
- 3K

- Last Post

- Replies
- 9

- Views
- 10K

- Last Post

- Replies
- 0

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 6K