- #1

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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

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- Thread starter Dustinsfl
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- #1

- 699

- 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

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

- #3

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Yes but how is that going to find the angle?

- #4

Mark44

Mentor

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

This idea can be generalized to any inner product space.

- #5

- 699

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

- #6

Dick

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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

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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

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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

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<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

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<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

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Ok I understand now.

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