# Inner product

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

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

Yes but how is that going to find the angle?

Mark44
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.

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

Dick
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.

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
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>).

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
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.

Ok I understand now.