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?

  • $\begingroup$ 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

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.


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