If $a(x)a(-x)=b(x)$, who is $a(x)$?

63 Views Asked by At

If we have $a(x)a(-x)=b(x)$, how can we find $a(x)$?

Or, if we have $c(x)+c(-x)=d(x)$, how can we find $c(x)$?

Please note that these functions should be continuous ones.


For example, if

$a(x)a(-x)=-x^2$, then $a(x)$ could be $x$ itself.

Also, if $c(x)+c(-x)=2x^2$, then $c(x)$ could be $x^2$.

How can we generically solve this problem?


Especially, if $f(x)f(-x)=\sin(x)/x$, who is $f(x)$?

What is the polynomial expansion of this function?


As we can see, $b$ and $d$ are even functions, and using logs or exponentials we could transform one problem to the other one.

Edit

As pointed by @AHusain, we could always create a $c'(x)=c(x)+e(x)$ where $e(x)$ is an odd function, creating infinite solutions. So instead, I'd like to ask, how can we find at least a base function $c(x)$ that works and from this one we could derive other $c'$, $c''$ solutions as we wish?

2

There are 2 best solutions below

1
On BEST ANSWER

$a(x)a(-x)=b(x) \rightarrow a(-x)a(x)=b(-x)$

So, $b(x)=b(-x)$, $b(x)$ is an even function.

  1. For $b(x)$ a positive function, $a(x)=\sqrt{b(x)}$.

But for $b(x)$ not always positive, it's harder to define $a(x)$.

For, $$ b(x)=\frac{\text{sin}(x)}{x}=\prod_{n=1}^{\infty} \left(1-\frac{x}{n \pi}\right)\left(1+\frac{x}{n \pi}\right)$$ we can define $a(x)$ like this

$$ f(x)= a(x)=\prod_{n=1}^{\infty} \left(1-\frac{x}{n \pi}\right) $$.....(1)

But unfortunately this is useless.

It should be checked whether there are other representation other than this

Suppose we write $$f(x)=P(x)=\sum_{k=0}^{\Lambda} a_kx^k=\prod_{i=0}^{\lambda} \left(1-\frac{x}{\alpha_i}\right)^{m_i}=$$

then

$$ b(x)=f(x)f(-x) =$$

$$ {a_0}\prod_{i=1}^{\lambda} \left(1-\left(\frac{x}{\alpha_i}\right)^2\right)^{m_i}$$.

($\lambda$ is the measure of number of roots)

For the roots $\alpha_i, i=1,2,3....$

This means that we have to use all roots of $b(x)$, $\alpha_i$ in $a(x)$. And that's the only way of writing.

So, $\frac{\text{sin}(x)}{x}$ can be expressed only in that way. So (1) is the only way to write $\frac{\text{sin}(x)}{x}$ and is useless.

0
On

Because you can add odd functions to $c(x)$ arbitrarily, we might as well assume $c(x)$ is even. Then we have

$$ c(x) + c(-x) = d(x)\\ c(x) + c(x) = d(x)\\ c(x) = \frac{1}{2} d(x)\\ $$

Reduce to this case for multiplicative version with $\log$