Angle Between 1 and x in C[0,1] Using Inner Product (3)

[Dustinsfl]

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)\int_{0}^{1}f(x)g(x)dx

Find the angle theta between 1 and x

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

 

[Dick]

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

 

[Dustinsfl]

Yes but how is that going to find the angle?

 

[Mark44]

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

u \cdot v = |u| |v| cos(\theta)

This idea can be generalized to any inner product space.

 

[Dustinsfl]

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

 

[Dick]

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.

 

[Dustinsfl]

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.

 

[Dick]

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>).

 

[Dustinsfl]

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

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


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.

 

[Dick]

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

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


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.

 

[Dustinsfl]

Ok I understand now.