Polynomial Relation - Problem 2

Find all polynomials $p(x)$ such that for all real numbers $a,b,c\neq0$ satisfying $\frac1a+\frac1b=\frac1c$ ,

$\frac1{p(a)}+\frac1{p(b)}=\frac1{p(c)}$

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Only polynomials of the form $p(x)=k\cdot x$ do the job, since by taking $b=a$ and $c=a/2$ we have $p(a)=2\cdot p(a/2)$ for any non-zero real number $a$ such that neither $a$ or $a/2$ belong to the zeroes of $p(x)$ .

Jack D'Aurizio - 7 years, 2 months ago

All that you have shown is that no real non-zero number can be the root.

You have not shown that no complex number can be the root of the polynomial, nor that 0 can be a repeated root.

Calvin Lin Staff - 7 years, 2 months ago

@Calvin Lin : I do not agree. A polynomial is a $C^1$ function, and there are only a finite number of $x$ such that $x=0,p(x)=0$ or $p(x/2)=0$ . The functional equation $f(2x)=2f(x)$ gives that the derivative of $f$ is constant, and the only value that $p$ can take in zero is zero, hence $p(x)=k\cdot x$ .

Jack D'Aurizio - 7 years, 2 months ago

@Jack D'Aurizio – This is now a more complete argument. You did not bring in the calculus aspect (that the derivative is constant) in your initial proof. Can you explain why " $f(2x) = 2f(x)$ gives that the derivative of $f$ is constant"?

Note that this approach doesn't require that "there are only a finite number of $x$ such that $x = 0$ , which was the main gist of your initial statement. All that you had initially, was "if $a$ is a non-zero real number which is the root of the polynomial, then there are infinitely many roots, which means $p(a)$ is a constant, which gives a contradiction. Hence there are no non-zero real roots."

Calvin Lin Staff - 7 years, 2 months ago

@Calvin Lin – That's right, $f(x)$ can be (and, indeed, is) zero only for $x=0$ . In order to avoid calculus arguments, we can just notice that under this assumptions $q(x)=f(x)/x$ is a polynomial satisfying $q(2x) = q(x)$ for any $x\neq 0$ , hence $q(x)-q(1)$ has an infinite number of roots, and $q(x)$ is a constant polynomial.

Jack D'Aurizio - 7 years, 2 months ago

@Jack D'Aurizio – The calculus argument is more or less the same, by deriving $f(2x)=2f(x)$ we get $f'(2x)=f'(x)$ .

Jack D'Aurizio - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$