Proving a quadratic form is well defined

Question

Let $V$ be the space of polynomials with real coeffiscients and degree at most $n$, for each polynomial $p$ we define: $$\Phi(p)=\sum_{k=0}^\infty p(k)p(-k)e^{-k}$$ Show that $\Phi$ is well defined and is a quadratic form in $V$.

This was part of an exam of a second course in linear algebra.
First thing I'm not sure about is: what does it mean to prove $\phi$ is well defined? Does it mean I have to prove the series converge? That seems like a calculus task (and I haven't prepared my calculus exam yet), in that case do you thik there is some way to prove it using just linear algebra? So far this is my attempt:

Consider the basis $\{1,x,x^2,\ldots,x^n\}$, we have for any natural number m that $\Phi(x^m)$ converges (by the ratio test). Now, if there were a bilinear form $\phi$ associated to to $\Phi$, then we would have (using the polarization formula) that for any i,j in $\{0,1,\ldots,n\}$ $$\phi(x^i,x^j)=\frac{1}{2}[\sum_{k=0}^\infty [(k^i+k^j)((-k)^i+(-k)^j)-k^i(-k)^i-k^j(-k)^j]e^{-1}]$$ which again can be proven to converge

This is all I could do but it's not enough because I need to prove $\phi$ is linear in both entries (clearly it is symmetric, so it would suffice proving linearity in one entry) and I can't figure out how to do it with this formula.

Any hint or suggestion is welcome.

Answer 1

HINT: In general, the symmetric bilinear form $b$ associated to a quadratic form $q$ is given by $$b(x,y):=\frac{1}{2}(q(x+y)-q(x)-q(y)).$$

---

It is then a matter of checking the definition; for this (possible) quadratic form $\Phi$ you have \begin{eqnarray*} \varphi(p,q)&:=&\frac12\Big(\Phi(p+q,p+q)-\Phi(p)-\Phi(q)\Big)\\ &=&\frac12\sum_{k=0}^{\infty}(p+q)(k)(p+q)(-k)e^{-k}-\sum_{k=0}^{\infty}p(k)p(-k)e^{-k}-\sum_{k=0}^{\infty}q(k)q(-k)e^{-k}\\ &=&\frac12\sum_{k=0}^{\infty}\Big((p+q)(k)(p+q)(-k)-p(k)p(-k)-q(k)q(-k)\Big)e^{-k}\\ &=&\frac12\sum_{k=0}^{\infty}\Big(p(k)q(-k)+q(k)p(-k)\Big)e^{-k}. \end{eqnarray*} Now it is not hard to verify that $\varphi$ is bilinear, and that its associated quadratic form is $\Phi$. Do note that the rearranging of the series above requires the absolute convergence of these series.

Answer 2

You're correct that the well definedness issue comes down to convergence. The way to see convergence is to note that you're multiply something that decays exponentially by something that increases (in absolute value) at a polynomial rate. You want to have the intuition that such a sum will converge be fairly automatic; if you want a quick formal proof you could use the "integral test."