- #1

Dustinsfl

- 2,285

- 5

Find the angle theta between 1 and x.

(3)[tex]\int_{0}^{1}f(x)g(x)dx[/tex]

Find the angle theta between 1 and x

I don't know what to do with polynomial inner product vector space

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Dustinsfl
- Start date

- #1

Dustinsfl

- 2,285

- 5

Find the angle theta between 1 and x.

(3)[tex]\int_{0}^{1}f(x)g(x)dx[/tex]

Find the angle theta between 1 and x

I don't know what to do with polynomial inner product vector space

- #2

Dick

Science Advisor

Homework Helper

- 26,263

- 619

Can't you compute the inner product of 1 and x given the definition?

- #3

Dustinsfl

- 2,285

- 5

Yes but how is that going to find the angle?

- #4

Mark44

Mentor

- 36,311

- 8,281

[itex]u \cdot v = |u| |v| cos(\theta)[/itex]

This idea can be generalized to any inner product space.

- #5

Dustinsfl

- 2,285

- 5

After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?

- #6

Dick

Science Advisor

Homework Helper

- 26,263

- 619

After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?

Didn't you see Mark44's suggestion? Set u=1 and v=x and then figure out u.v, |u|, |v| and then what cos(theta) is. Remember |v|=sqrt(v.v). It's a number, not a function.

- #7

Dustinsfl

- 2,285

- 5

And <x,x> is the integral from 0 to 1 of x^2 which is 1/3.

Now the equation is 1/2=1/6 cos theta so theta is arccos 3 which isn't pi/6.

- #8

Dick

Science Advisor

Homework Helper

- 26,263

- 619

And <x,x> is the integral from 0 to 1 of x^2 which is 1/3.

Now the equation is 1/2=1/6 cos theta so theta is arccos 3 which isn't pi/6.

The integral of 1*1 from 0 to 1 is 1/2??? And I already warned you that |x|=sqrt(<x,x>).

- #9

Dustinsfl

- 2,285

- 5

<u, v>=[tex]\int_{0}^{1}f(x)g(x)[/tex] which implies <x, x>=[tex]\int_{0}^{1}x^{2}=1/3[/tex].

So now we have 1/2=1/3 cos theta and now 3/2=cos theta which is greater than 1 so it doesn't exist.

- #10

Dick

Science Advisor

Homework Helper

- 26,263

- 619

<u, v>=[tex]\int_{0}^{1}f(x)g(x)[/tex] which implies <x, x>=[tex]\int_{0}^{1}x^{2}=1/3[/tex].

So now we have 1/2=1/3 cos theta and now 3/2=cos theta which is greater than 1 so it doesn't exist.

<x,x> is 1/3. That doesn't mean |x|=1/3.

- #11

Dustinsfl

- 2,285

- 5

Ok I understand now.

Share:

- Replies
- 5

- Views
- 271

- Replies
- 9

- Views
- 183

- Last Post

- Replies
- 15

- Views
- 451

- Replies
- 9

- Views
- 892

- Last Post

- Replies
- 9

- Views
- 878

- Last Post

- Replies
- 6

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 371

- Last Post

- Replies
- 16

- Views
- 567

- Replies
- 2

- Views
- 314

- Replies
- 5

- Views
- 2K