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 $\displaystyle (g,h)=(x+1,x^2+1)=(1,x^2+1)$. I mean, how did they get "x+1" equal to "1"?

This is my own approach by the way:

$\displaystyle proj_{h(x)}g(x)= \frac{(g,h)}{(h,h)}h=$$\displaystyle \frac{\int^1_{-1}(x^2+1)(x+1)}{\int^1_{-1}(x^2+1)^2}(x^2+1) Since \displaystyle \int^1_{-1} (x^2+1)(x+1)dx =4 and \displaystyle \int^1_{-1}(x^2+1)^2 dx= 8, we have: \displaystyle \frac{4}{8}(x^2+1)=\frac{1}{4}(x^2+1) So, I've got 1/4, but it should've been 5/7 as the provided answer indicates. Could anyone please help? 2. \displaystyle \left<x+ 1, x^2+ 1\right>= \left<x, x^2+ 1\right>+ \left<1, x^2+ 1\right> \displaystyle \left<x, x^2+ 1\right>= \int_{-1}^1 x(x^2+ 1)dx But, as they say, x is an odd function and \displaystyle x^2+ 1 is an even function. Their product is an odd function, \displaystyle x^3+ x and so the integral is an even function: \displaystyle \int x^3+ x dx= \frac{1}{4}x^4+ \frac{1}{2}x^2 and that, evaluated at -1 and 1, gives the same thing: \displaystyle \left[\frac{1}{4}x^4+ \frac{1}{2}x^2\right]_{-1}^1=$$\displaystyle (\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 $\displaystyle \perp$- their inner product is 0. $\displaystyle \left<x+ 1, x^2+ 1\right>= \left<x, x^2+ 1\right>+ \left<1, x^2+ 1\right>= 0+ \left<1, x^2+ 1\right>$.

3. Originally Posted by demode
Since $\displaystyle \int^1_{-1} (x^2+1)(x+1)dx =4$ and $\displaystyle \int^1_{-1}(x^2+1)^2 dx= 8$, we have:
How did you get these? They are both wrong... showing the way would help us tell you where you went wrong.