Let $P(x)$ be a polynomial with real coefficients such that $P(\sin^2x) = P(\cos^2x)$ for all x in interval [0,π/2] Find which of the following statements are true ?

  1. $P(x)$ is an even function
  2. $P(x)$ can be expressed as a polynomial in $(2x - 1)^2 $
  3. $P(x)$ is a polynomial of even degree.

I tried taking a general poynomial and tried equating bpth sides. I am confused whether to replace x by π/2 - x or replace it by 1 - x while equating and how to move further. Or is there any other simpler approach ?


Firstly, we see that setting $t=\sin^2(x)$ for $x \in [0, \pi/2]$ then $t \in [0, 1]$ and the condition above becomes $P(t) = P(1-t)$ for all $t \in [0, 1]$.

  1. False. Let $P(x) = (x - \frac{1}{2})^2$.

  2. If we consider the polynomial $P(x) - P(1-x)$ the above condition tells us that it is $0$ for all $x \in [0,1]$. Since this polynomial has infinitely many roots it must be that $P(x) = P(1-x)$ for all $x$. Now let $Q$ be a polynomial such that $P(x) = Q(2x - 1)$ (you should think about why we always can do this). Our condition then implies that $$ Q(x) = P\left(\frac{x+1}{2}\right) = P\left(\frac{1-x}{2} \right) = Q(-x) $$ and thus Q is even. Since $Q$ is an even polynomial it must only contain even powers and therefore we can write $Q(x) = R(x^2)$ and thus $P(x) = R((2x-1)^2)$.

  3. True. This follows from 2.

  • $\begingroup$ Why we can always take any polynomial P(x) equal to any polynomial in 2x - 1 $\endgroup$ – Matt Nov 25 '16 at 12:03
  • $\begingroup$ If this was true then I can always write any polynomial in terms of another polynomial in (2x-1)^2 ? $\endgroup$ – Matt Nov 25 '16 at 12:06
  • $\begingroup$ Consider a polynomial $P(x) = a_n x^n + a_{n-1}x^{n-1} + \ldots + a_0$. We can write $x = ((2x-1) + 1)/2$ and expand the polynomial leaving the term $(2x - 1)$ in tact to get $P(x) = Q(2x - 1)$. We could not do this for $(2x-1)^2$ because we would have to write $x = (\sqrt{(2x-1)^2} + 1)/2$ and expanding this term will not result in a polynomial generally since it may include square roots. $\endgroup$ – Homotopy Nov 25 '16 at 19:50
  • $\begingroup$ Is that clear? If not I can elaborate further $\endgroup$ – Homotopy Nov 25 '16 at 19:51
  • $\begingroup$ Yes I got it now. Thank you. Your help is much appreciated. Giving you a plus one. $\endgroup$ – Matt Nov 27 '16 at 16:01

Let $$y=x-\frac{1}{2}$$ and $$P(x)=Q(y)$$ Then because $$P(x)=P(1-x)$$ for $0 \le x \le 1$, we have $$Q(y)=Q(-y)$$ for $-\frac{1}{2}\le y \le \frac{1}{2}$. Hence $Q(y)$ has only even powers in $y$. Therefore $$P(x)=Q(y)=R(y^2)=R((x-1/2)^2)=S((2x-1)^2)$$

  • $\begingroup$ Can you please elaborate ? $\endgroup$ – Matt Nov 16 '16 at 14:12
  • $\begingroup$ @RaghavSingal Which step? $\endgroup$ – velut luna Nov 16 '16 at 14:13
  • $\begingroup$ The first step and its conclusion. $\endgroup$ – Matt Nov 16 '16 at 14:14
  • $\begingroup$ What is R and S and which statement would be true. Also is P(x) even or odd ? $\endgroup$ – Matt Nov 16 '16 at 14:53

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