## Task 5 - 301245

Determine a polynomial

$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n} \quad\left(a_{0}, a_{1}, \ldots, a_{n} \text { real } ; a_{n} \neq 0\right)
$$

that satisfies the conditions

$$
f(-4)=0, f(-3)=1, f(-2)=0, f(-1)=-1, f(0)=0, f(1)=1, f(2)=0, f(3)=-1, f(4)=0
$$

and has the lowest possible degree $n$.
Due to $f(-4)=f(-2)=f(0)=f(2)=f(4)=0$, $f(x)$ is divisible by

$$
g(x):=(x+4)(x+2) x(x-2)(x-4)=\left(x^{2}-16\right)\left(x^{2}-4\right) x=x^{5}-20 x^{3}+64 x
$$

so there exists a polynomial $h(x)$ with $f(x)=g(x) \cdot h(x)$. Given $g(-3)=105, g(-1)=-45$, $g(1)=45$ and $g(3)=-105$, we are now looking for a polynomial $h(x)$ of the lowest possible degree with $h( \pm 3)=\frac{1}{105}$ and $h( \pm 1)=\frac{1}{45}$.

Consider the polynomial $h_{2}(x):=315 \cdot h(x)-3$, which has the same degree as $h$. Then $h_{2}( \pm 3)=3-3=0$ and $h_{2}( \pm 1)=7-3=4$.

Clearly, $h_{2}$ cannot be of degree 0 or 1, as it would then have to be constantly zero due to $h_{2}( \pm 3)=0$, which contradicts the second condition on $h_{2}$. However, 

$$
h_{2}(x)=-\frac{1}{2} \cdot\left(x^{2}-9\right)=-\frac{1}{2} x^{2}+\frac{9}{2}
$$

obviously satisfies all conditions and can thus be chosen as the polynomial of the smallest degree. Therefore,

$$
h(x)=\frac{h_{2}(x)+3}{315}=-\frac{1}{630} x^{2}+\frac{1}{42}
$$

and

$f(x)=g(x) \cdot h(x)=-\frac{1}{630} x^{7}+\left(\frac{2}{63}+\frac{1}{42}\right) x^{5}-\left(\frac{32}{315}+\frac{10}{21}\right) x^{3}+\frac{32}{21}=-\frac{1}{630} x^{7}+\frac{1}{18} x^{5}-\frac{26}{45} x^{3}+\frac{32}{21} x$