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I want to show that $P(x)$ is a third degree polymomial: $P(x) = P(x+2) -x^2-2$ for every real $x$

It is not so difficult, but I am not seeing the proof straight away.

Blue
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Alexandre Tourinho
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  • check what happens to the degree when in any polynomial $P(x)$ you let $x=x+2$ then consider what @GitGud said above to look at $P(x+2)-P(x)$ which is a rearrangement of what you were given – Jepsilon Jun 09 '18 at 12:27
  • @Jepsilon I've removed my hint because I realised I can't deal with a detail that comes up. – Git Gud Jun 09 '18 at 12:29
  • @GitGud Was your detail something regarding the differentiability of $P$ itself? – String Jun 09 '18 at 12:33
  • @String Precisely! – Git Gud Jun 09 '18 at 12:33
  • Wait why use differentiability? Just try out a couple of polynomials of different degrees with the substitution $x=x+2$ (or $x=x+a$ to be a bit more general) and compare the degrees from before and after the transformation. – Jepsilon Jun 09 '18 at 12:36
  • @Jepsilon again assuming $P$ is a polynomial. Which is already assuming differentiability and some more. – String Jun 09 '18 at 12:38
  • Wait so you were given no info about $P$? I thought it was "here's a polynomial show that its degree is 3 given...". I guess I misread the question apologies – Jepsilon Jun 09 '18 at 12:53

4 Answers4

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For any polynomial with degree $n$ and for each constant $k\neq0$, $P(x+k)-P(x)$ is a polynomial whose degree is $n-1$; to see why, just expand $P(x+k)$.

Since, in your case, $P(x+2)-P(x)$ has degree $2$, $P(x)$ is a cubic polynomial.

Note that this assumes that $P(x)$ is a polynomial function. Otherwise, the statement is false. A counter-example is this:$$P(x)=\begin{cases}\displaystyle\frac{x^3}{6}-\frac{x^2}{2}+\frac{4 x}{3}&\text{ if $x$ is algebraic}\\[3mm]\displaystyle\frac{x^3}{6}-\frac{x^2}{2}+\frac{4 x}{3}+1&\text{ otherwise.}\end{cases}$$

José Carlos Santos
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Assuming $P$ is $C^3$, you have $P'''(x) = P'''(x+2)$ for all $x$. Since $P'''$ is continuous, we conclude that it's constant. When we integrate 3 times, we get a cubic polynomial.

B. Goddard
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  • Can you make this argument work with the third differences rather than the third derivative, so that you can avoid assuming differentiability? You may need to assume continuity. – Ethan Bolker Jun 09 '18 at 12:34
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The statement, as written, is false.

Take any function on $[0,2]$ with $p(2)=p(0)+2$. The linear function $p(x) = x$ between 0 and 2 will do. Again, it linear on this domain. Now use the relations $p(x+2) = p(x)+x^2+2$ to determine the function of $[2,4]$ then on $[4,6]$, etc. and $p(x-2)=p(x)-x^2-2$ on $[-2,0]$, $[-4,-2]$, etc.

For example, between $[2,4]$, $$p(x)=p(x-2)+x^2-4x+6=(x-2)+x^2-4x+6$$ So you have a piece-wise continuous function that satisfies the criterion but is linear on $[0,2]$ and second degree on other intervals.

Obviously, you can take any nasty function you want on $[0,2]$ that satisfies $p(2)=p(0)+2$ and get a non-polynomial solution of the recurrence relation.

Now, if one assumes that $p(x)$ is a polynomial, I don't think you can do it in any way other than simply assuming $$p(x)=a_nx^n+\cdots+a_1x+a_0.$$ Then use the binomial theorem to expand $p(x+2)$ then subtract $p(x)$ and set it equal to $x^2+2$. You will get $a_n =a_{n-1} =\cdots = a_4 = 0$ and $6a_3=1$, $12a_3+4a_2=0$ and $8a_3+4a_2+2a_1=2$ and $a_0$ arbitrary.

Final answer: $p(x)=\frac16x^3-\frac12x^2+\frac{13}3x+C$

Robert McLean MD PhD
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Let: $$\begin{align}P(x+2)&=a_0(x+2)^n+a_1(x+2)^{n-1}+a_2(x+2)^{n-2}+\cdots+a_n\\ P(x)&=\color{red}{a_0}x^n+\color{blue}{a_1}x^{n-1}+a_2x^{n-2}+\cdots+a_n \\ x^2+2=P(x+2)-P(x)&=2na_0x^{n-1}+(2n(n-1)a_0+2(n-1)a_1)x^{n-2}+\cdots \Rightarrow \\ x^2&=2na_0x^{n-1} \Rightarrow \\ 2&=n-1 \quad \text{and} \quad 2na_0=1\Rightarrow \\ n&=3 \quad \text{and} \quad \color{red}{a_0}=\frac16. \end{align}$$ Thus $n=3$. You can continue to find the next coefficient: $$2n(n-1)a_0+2(n-1)a_1=0 \Rightarrow \color{blue}{a_1}=\frac{-2}{4}=-\frac12.$$

farruhota
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