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If f(x)f(1÷x)=f(x)+f(1÷x).Then Why f(x)=1+x^n?Here f(x) is a polynomial function.

mycroft holmes
272 Points
4 years ago
Multiply the equation by xn to obtain.
$x^n f(x) f \left(\frac{1}{x} \right) = x^n f(x)+ x^n f \left(\frac{1}{x} \right )$

Notice that if is a root of f(x), it is also a root of the ‘reciprocal polynomial’ xn f(1/x) and vice-versa. This means that

$f(x) = c x^n f \left(\frac{1}{x} \right)$

for some complex number c.

Using this above relation in the original equation gives
$f^2(x) = f(x) (1+cx^n)$

So either f(x) is the zero polynomial or $f(x)= (1+cx^n)$

Again plugging back in the original equation and solving for c, we get c2 =1 so c=1, or -1.

So that the only solutions in polynomials is $f(x)= 1 \pm x^n$