I've been stuck for days now and I just can't solve it. Can someone please solve this exercise or help me with them? Here's what is given:
Let $a>0$. (Definition:) A function $f: [−a,a] \rightarrow \mathbb{R}$ is called even if $f(x) = f(−x)$ for all $x \in [−a, a].$ In my case $f(x) :=|x| \cdot cos(x)$ and we have $n + 1$ nodes on the x-axis defined as $x_j := -a + 2a\frac{j}{n}$ for $j \in \{0,...,n\}$.
The polynomial that interpolates $f$ is $p(x) = \sum_{i=0}^n c_i x^i$ and we know that for all odd $i$ the coefficient $c_i$ satisfies $c_i = 0$ (I already proved that).
In the following, (1) means the property of $c_i$ above ($c_i = 0$ for all odd i) and $\tilde{c}_i$ means the computer calculated value of the coefficient $c_i$:
Thanks a lot for any help!
The number of even-degree terms in the polynomial of degree $n$ is the same as the number of non-negative nodes. So you get the interpolation polynomial $p(x)=q(x^2)$ from using standard interpolation for $q$ on the points $(x_j^2,y_j)$, $j=ceil(n/2),...,n$.