# Inner product

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

## Answers and Replies

Homework Helper
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?

Mentor
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?

Homework Helper
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.

Homework Helper
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.

Homework Helper
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.