# Angle between 1 and x

• Mar 26th 2010, 08:19 PM
dwsmith
Angle between 1 and x
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)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x
• Mar 26th 2010, 11:16 PM
Pulock2009
didnot understand
Quote:

Originally Posted by dwsmith
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)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x

• Mar 26th 2010, 11:17 PM
Drexel28
Quote:

Originally Posted by dwsmith
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)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x

$\displaystyle \left\langle 1,x\right\rangle=\int_0^1 x dx=\frac{1}{2}$
• Mar 27th 2010, 03:30 AM
HallsofIvy
Quote:

Originally Posted by dwsmith
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)$\displaystyle \int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x

You are thinking of C[0,1] as an inner product space with the standard inner product $\displaystyle <f, g>= \int_0^1 f(x)g(x)dx$.

Of course, in any inner product space, $\displaystyle <u, v>= |u||v|cos(\theta)$ where $\displaystyle \theta$ is the angle between u and v.

The particular calculation for the angle between u= 1 and v= x should be easy.
• Mar 27th 2010, 10:27 AM
dwsmith
After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?
• Mar 27th 2010, 11:07 AM
HallsofIvy
After taking what integral? The formula I gave you, $\displaystyle <u, v>= |u||v|cos(\theta)$ requires three integrals to determine <u, v>, |u|, and |v|.