1

Let c>0, find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ P(x+c)+P(x-c)=2P(x)$$

Thank you, dxiv.

Let $Q(x) = P(x+c)-P(x)$,

since $P(x+c)-P(x)=P(x)-P(x-c)$ so $Q(x) = Q(x-c)$ i.e., $Q(x+c) = Q(x)$

Then $Q(x)=Q(x+c)=Q(x+2c)=\ldots$

Assume $Q(x_0)=a$, where $a\in \mathbb{R}$, for some $x_0$.

$Q(x)-a$ has infinitely many roots $ x_0, x_0+c, x_0+2c, \ldots$

so $Q(x)-a $ is zero polynomial $\to Q(x)=a $, so $Q(x)$ is constant polynomial.

$\deg(P(x+c)-P(x))=0 \to \deg(P(x)) \leq1$

Therefore $P(x)=ax+b$, where $a,b \in\mathbb{R}$

user403160
  • 3,286
  • 2
    so x,x+c,x+2c,… are roots No, that's not given, and it doesn't otherwise follow that $,Q(x)=0,$ for that $x$. What does follow, however, is that for a given $,a = Q(x_0),$ the polynomial $,Q(x)-a,$ has infinitely many roots $x_0, x_0+c, x_0+2c, \dots,$ and is therefore the zero polynomial, which implies $Q(x)=a,$ i.e. constant. – dxiv Aug 27 '17 at 05:45
  • You could also say that $Q(x+c) = Q(x)$ means that $Q$ is periodic (completely determined by its values on the half-open interval $[0, c)$.), so it is bounded... but the only bounded polynomials are constant. – user357980 Aug 27 '17 at 05:54
  • You could also note that the identity for $P$ is unchanged by differentiating, so if $P$ is not linear, then you either have $P$ beign a quadratic polynomial or you can keep differentiating until you get some derivative of $P$ being a quadratic. Then show that there is no quadratic polynomial that satisfies that condition by simple algebra. – user357980 Aug 27 '17 at 05:56
  • 3
    @carat Q(x) has infinitely many roots so Q(x) is constant polynomial No, the only polynomial with infinitely many roots is the zero polynomial, not just any constant polynomial. To cover that "gap", you need to go back one step to if r is root and add "otherwise if $Q$ has no roots then it must be a constant non-zero polynomial". As a side note, and this may sound like nitpicking, but often times it's not enough to just have the right idea, you must also write it down it in the right, clearest terms. – dxiv Aug 27 '17 at 06:18
  • @dxiv, Edited. May I write in this way ? is there any error ? Thank you. – user403160 Aug 27 '17 at 07:18
  • 1
    @carat If deg(Q(x))≠0, let r be root The edit does still not cover the zero polynomial. Conventions vary, but a polynomial of degree $0$ is commonly taken to be a constant non-zero polynomial, while the zero polynomial has its degree defined to be $-\infty$ (so that $\deg P \cdot Q = \deg P + \deg Q$ in all cases), or sometimes $-1$ (see here and here). I still think that looking at $,Q(x)-Q(x_0),$ (per my very 1st comment) is the more direct way. – dxiv Aug 27 '17 at 07:24
  • I think I got it now. Please recheck my edited work :) – user403160 Aug 27 '17 at 07:58
  • 1
    @carat Looks good now, only I'd write the last step as $\deg P \color{red}{\le} 1,$ to cover the constant solutions, too. – dxiv Aug 27 '17 at 16:11
  • @dxiv. Edited. Thank you very much for your kind help. – user403160 Aug 27 '17 at 16:20

1 Answers1

2

I think dxiv's comment is enough to answer your main question.

Here is a less_typing solution:

By Jensen inequality, for $x\in\mathbb{R}$,

$\dfrac{1}{2}P(x+c)+\dfrac{1}{2}P(x-c)=P\Big(\dfrac{1}{2}(x+c)+\dfrac{1}{2}(x-c)\Big)=P(x)$, iff $P''(x)=0$.

Since this is true for all $x\in\mathbb{R}$, Then $P(x)=ax+b$, for some $a,b\in \mathbb{R}$.

MAN-MADE
  • 5,381
  • Thank you for the alternative solution. Please do not delete. Many readers understand your answer :) – user403160 Aug 27 '17 at 08:02
  • This is a nice solution, however Jensen's cannot be applied for all $x\in \mathbb{R}$, since you need convexity. However, this is not a problem. – clark Aug 27 '17 at 13:05
  • @clark I change my argument a bit. Can you check now! – MAN-MADE Aug 27 '17 at 13:12