1. ## Projection of Polynomials

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:

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

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

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

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. $\left= \left+ \left<1, x^2+ 1\right>= 0+ \left<1, x^2+ 1\right>$.

3. 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.