Projection of Polynomials

**demode** · April 17th 2010, 02:45 AM

The following is a worked example:

In the very first line of working, I don't understand how they've got $(g,h)=(x+1,x^2+1)=(1,x^2+1)$ . I mean, how did they get "x+1" equal to "1"?

This is my own approach by the way:

$proj_{h(x)}g(x)= \frac{(g,h)}{(h,h)}h=$ $\frac{\int^1_{-1}(x^2+1)(x+1)}{\int^1_{-1}(x^2+1)^2}(x^2+1)$

Since $\int^1_{-1} (x^2+1)(x+1)dx =4$ and $\int^1_{-1}(x^2+1)^2 dx= 8$ , we have:

$\frac{4}{8}(x^2+1)=\frac{1}{4}(x^2+1)$

So, I've got 1/4, but it should've been 5/7 as the provided answer indicates. Could anyone please help?

**HallsofIvy** · April 17th 2010, 05:18 AM

$\left<x+ 1, x^2+ 1\right>= \left<x, x^2+ 1\right>+ \left<1, x^2+ 1\right>$

$\left<x, x^2+ 1\right>= \int_{-1}^1 x(x^2+ 1)dx$

But, as they say, x is an odd function and $x^2+ 1$ is an even function. Their product is an odd function, $x^3+ x$ and so the integral is an even function: $\int x^3+ x dx= \frac{1}{4}x^4+ \frac{1}{2}x^2$ and that, evaluated at -1 and 1, gives the same thing: $\left[\frac{1}{4}x^4+ \frac{1}{2}x^2\right]_{-1}^1=$ $(\frac{1}{4}+ \frac{1}{2})- (\frac{1}{4}+ \frac{1}{2})= 0$ .

That is what is meant by saying that an odd function and an even function are $\perp$ - their inner product is 0. $\left<x+ 1, x^2+ 1\right>= \left<x, x^2+ 1\right>+ \left<1, x^2+ 1\right>= 0+ \left<1, x^2+ 1\right>$ .

**Defunkt** · April 17th 2010, 06:04 AM

Originally Posted by demode

Since $\int^1_{-1} (x^2+1)(x+1)dx =4$ and $\int^1_{-1}(x^2+1)^2 dx= 8$ , we have:

How did you get these? They are both wrong... showing the way would help us tell you where you went wrong.

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