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Inner product

  • Thread starter Dustinsfl
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  • #1
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In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

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
 

Answers and Replies

  • #2
Dick
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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
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How would you find the angle between two ordinary vectors? Wouldn't you use something like this?

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

This idea can be generalized to any inner product space.
 
  • #5
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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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By definition, <1,1> is the integral from 0 to 1 of 1^2 which is 1/2.
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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By definition, <1,1> is the integral from 0 to 1 of 1^2 which is 1/2.
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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1 sorry. The definition of inner product on this space is (3)

<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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1 sorry. The definition of inner product on this space is (3)

<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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