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# Polynomial Relation - Problem 2

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

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

Note by Cody Johnson
3 years, 3 months ago

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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 $$p(x)$$. · 3 years, 3 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. Staff · 3 years, 3 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 $$p(x)=k\cdot x$$. · 3 years, 3 months ago

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." Staff · 3 years, 3 months ago

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. · 3 years, 3 months ago

The calculus argument is more or less the same, by deriving $$f(2x)=2f(x)$$ we get $$f'(2x)=f'(x)$$. · 3 years, 3 months ago