Angle between 1 and x

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

Find the angle theta between 1 and x
March 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) $\int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x

didnot get your question. could you please elaborate???
March 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) $\int_{0}^{1}f(x)g(x)dx$

Find the angle theta between 1 and x

$\left\langle 1,x\right\rangle=\int_0^1 x dx=\frac{1}{2}$
March 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) $\int_{0}^{1}f(x)g(x)dx$

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 $<f, g>= \int_0^1 f(x)g(x)dx$ .

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

The particular calculation for the angle between u= 1 and v= x should be easy.
March 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?
March 27th 2010, 11:07 AM
HallsofIvy

After taking what integral? The formula I gave you, $<u, v>= |u||v|cos(\theta)$ requires three integrals to determine <u, v>, |u|, and |v|.