1

Let $P(x+2)+P(x-2) = 8x+6$. Find all of the solutions for $P(x)$.

If we put $P(x) = ax + b $ the answer is obvious but how to determine all of the solutions?

Mohsen Shahriari
  • 6,718
S.H.W
  • 4,379

3 Answers3

5

First get rid of the RHS by using $Q(x):=P(x)-4x-3$ so that $$Q(x+2)+Q(x-2)=P(x+2)-4x-8-3+P(x-2)-4x+8-3=0.$$

Then the solution of

$$Q(x+2)=-Q(x-2)$$ is of the form

$$Q(x)=(-1)^{\lfloor x/4\rfloor}Q(x\bmod 4),$$

where $Q(x)$ is absolutely arbitrary in $[0,4)$.

Hence,

$$\color{green}{P(x)=4x+3+(-1)^{\lfloor x/4\rfloor}Q(x\bmod 4)}.$$

Below, an example of a smooth solution:

enter image description here

and another, piecewise linear and discontinuous:

enter image description here

  • This is my way : https://math.stackexchange.com/questions/3091713/formula-for-the-sequence-0-3-8-15-24/3091719#3091719 OR https://math.stackexchange.com/questions/3080437/how-to-solve-vn-2-cdot-vn-1-2-cdot-n – lab bhattacharjee Feb 04 '19 at 12:22
  • @labbhattacharjee: sorry but you are pointing at discrete recurrences. The key point in this answer is to find the general solution on $\mathbb R$. –  Feb 04 '19 at 12:39
  • @YvesDaoust You are assuming that $P$ is defined on $\mathbb{R}$ . Is that right ? – S.H.W Feb 04 '19 at 12:51
  • @S.H.W: right. But the answer remains perfectly valid in $\mathbb Z$ and you can choose $Q_0,Q_1,Q_2,Q_3$ arbitrarily, and the rest follows. –  Feb 04 '19 at 12:54
  • @YvesDaoust And how you get $Q(x)=(-1)^{\lfloor x/4\rfloor}Q(x\bmod 4)$ from $Q(x+2)=-Q(x-2)$ ? – S.H.W Feb 04 '19 at 12:56
  • @S.H.W: the sign alternates every four units. In other words, the homogeneous solution is "antiperiodic". –  Feb 04 '19 at 12:58
  • @YvesDaoust My problem is the meaning of "$x\bmod4$" . I know how it's defined for integers but what about real numbers ? – S.H.W Feb 04 '19 at 13:02
  • @S.H.W: same definition. :) –  Feb 04 '19 at 13:05
  • @YvesDaoust For example what is the value of $\sqrt{2} \bmod 4$ ? – S.H.W Feb 04 '19 at 13:09
  • @S.H.W: $\sqrt2$. What is your definition of the modulo ? –  Feb 04 '19 at 13:10
  • @YvesDaoust $a\equiv b\pmod n \to a - b = kn$ for some integer $k$ . – S.H.W Feb 04 '19 at 13:15
  • @S.H.W: doesn't it work with reals ? –  Feb 04 '19 at 13:16
  • @YvesDaoust I've got it , thanks a lot . – S.H.W Feb 04 '19 at 13:23
0

If $P$ is a polynomial of degree $n$, then $P(x+2)+P(x-2)$ has degree $n$, hence $n=1.$.

Fred
  • 77,394
-1

This difference equation is linear so it's solution can be reprasented as

$$ P(x) = P_h(x) + P_p(x) $$

with

$$ P_h(x+2)+P_h(x-2) = 0\\ P_p(x+2)+P_p(x-2) = 8x+6 $$

for the homogeneous we propose $P_h(x) = \gamma^x$ and after substituting

$$ \gamma^x\left(\gamma^2+\frac{1}{\gamma^2}\right)=0 $$

and the solutions are

$$ \gamma_h = \left\{-\frac{1+i}{\sqrt{2}},\frac{1+i}{\sqrt{2}},\frac{1-i}{\sqrt{2}},-\frac{1-i}{\sqrt{2}}\right\} $$

so the homogeneous solution is

$$ P_h(x) = C_1\sin\left(\frac{\pi x}{4}\right)+C_2\cos\left(\frac{\pi x}{4}\right)+C_3\sin\left(\frac{3\pi x}{4}\right)+C_4\cos\left(\frac{3\pi x}{4}\right) $$

now for the particular making $P_p(x) = 4x+3$ we conclude

$$ P(x) = 4x+3+C_1\sin\left(\frac{\pi x}{4}\right)+C_2\cos\left(\frac{\pi x}{4}\right)+C_3\sin\left(\frac{3\pi x}{4}\right)+C_4\cos\left(\frac{3\pi x}{4}\right) $$

Cesareo
  • 33,252