1
$\begingroup$

I would like to find a polynomial that is parallel to $f(x)=x^2$ that isn't just a scalar multiple of $f$. Will such functions exist?

Define the inner product as $\langle f,g \rangle = \int_{-1}^1 fg$.

1.) Am I correct that, besides taking the cross product, I need to check when

$$\frac{\langle x^2, g \rangle}{\Vert x^2 \Vert \Vert g \Vert} = \pm 1$$

for $\cos$ of 0 or $\pi$? I found something online that said parallel requires $\langle f,g \rangle = 1$, but I think that is wrong.

2.) I have been trying $g(x) = x+k$, for some constant $k$.

I get $k=\pm i\sqrt{2}$. For $h(x)=k$, I get no solutions. Are there other options or any mistakes here?

$\endgroup$
  • $\begingroup$ For unit vectors, $\langle f, g\rangle = 1$ implies parallel. The condition $\langle f, g\rangle = -1$ is sometimes called "anti-parallel." $\endgroup$ – Michael Lee Dec 2 '17 at 17:51
  • $\begingroup$ What is your definition of "parallel"? I would define it as "scalar multiple", so the answer would be no by definition... $\endgroup$ – Eric Wofsey Dec 2 '17 at 21:36
1
$\begingroup$

If we are just considering the inner product over the space of polynomials, we have the following:

We want $\langle f, g \rangle=\int_{-1}^{1}fg=1$, if we want polynomials parallel to $x^2$, we obtain $\int_{-1}^{1}x^2g=1$. If we want $g$ to be a monic polynomial, we simply compute $\int_{-1}^{1}x^2x^n=\int_{-1}^{1}x^{n+2}.$ When we integrate, we simply get $\frac{x^{n+3}}{n+3}$ and we can set up the equation $\frac{1}{n+3}-\frac{(-1)^{n+3}}{n+3}=1$. However, this has no integer solution.

We can tweak this by considering $g=cx^n$ where $n\neq 2, c\neq 1$. We obtain $\int_{-1}^{1}cx^2x^n=1$, factoring out $c$, and combing terms, we now have:

$c\int_{-1}^{1}x^{n+2}=1$, which reduces to $c(\frac{1}{n+3}-\frac{(-1)^{n+3}}{n+3})=1$.

Consider the case when $n$ is even, (of the form $n=2k$) we have $c(\frac{1}{2k+3}-\frac{(-1)^{2k+3}}{2k+3})=1$, which reduces to $c(\frac{1}{2k+3}-\frac{1}{2k+3})=1$. This has no solution.

Now consider the case when $n$ is odd (of the form $n=2k+1$) we have $c(\frac{1}{(2k+1)+3}-\frac{(-1)^{(2k+1)+3}}{(2k+1)+3})=1$, which becomes $c(\frac{1}{2k+4}-\frac{(-1)^{2k+4}}{(2k+4})=1$. This reduces to $c(\frac{1}{2k+3}-\frac{-1}{2k+3})=c(\frac{2}{2k+3})=1$.

This always has a solution. For a polynomial of degree $n$, we have $c(\frac{2}{n+3})=1$, which becomes $2c=n+3$, and $c=\frac{n+3}{2}$. This will give you all monic polynomials parallel to $x^2$.

You could tweak this argument for non-monic polynomials, but it would be algebraically cumbersome.

Also, as an aside, if we considered this inner product over all continuous functions, I don't believe it possible to find a characterization for all $g\in C([-1,1])$ which are parallel to $x^2$. The fact that we are working only with polynomials makes this problem doable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.