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Question Number 198850 by mustafazaheen last updated on 25/Oct/23

$$$$$$\mathrm{p}(\mathrm{x}+1)+\mathrm{p}(\mathrm{x}-1)={4\mathrm{x}}^2-2\mathrm{x}+10$$$$\mathrm{p}\left(\mathrm{x}\right)=?$$$$$$$$$$

Answered by Rasheed.Sindhi last updated on 25/Oct/23

$$LetP\left(x\right)={ax}^2+bx+c$$$$P(x+1)+P(x-1)$$$$=(a{(x+1)}^2+b(x+1)+c)-(a{(x-1)}^2+b(x-1)+c)$$$$=a({(x+1)}^2+{(x-1)}^2)+b((x+1)+(x-1))+2c$$$$=2a(x^2+1)+2b\left(x\right)+2c$$$$=2{ax}^2+2a+2bx+2c={4\mathrm{x}}^2-2\mathrm{x}+10$$$$2a=4,2b=-2,2a+2c=10$$$$a=2,b=-1,c=5-a=5-2=3$$$$P\left(x\right)=2x^2-x+3$$

Commented by mr W last updated on 25/Oct/23

$$arethereotherpossibilities?$$$$forexample$$$$p\left(x\right)=2x^2-x+3+f\left(x\right)$$$$withf\left(x\right)=-f(x-2)$$$$but{i}^{\prime }mnotsureifsuchaf\left(x\right)exists.$$

Commented by mr W last updated on 25/Oct/23

$$great!$$$$thatmeans$$$$p\left(x\right)=2x^2-x+3+k\sin \frac{(2n+1)\pi x}{2}$$$$isalsoasolution.$$

Commented by Rasheed.Sindhi last updated on 25/Oct/23

$$I^{\prime }malsonotsure\mathit{s}\mathit{i}\mathit{r}.$$

Commented by Frix last updated on 25/Oct/23

$$f\left(x\right)=\sin \frac{(2n+1)\pi x}{2}$$

Commented by witcher3 last updated on 25/Oct/23

$$\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{f}(\mathrm{x}-2)$$$$\mathrm{f}(\mathrm{x}-2)=-\mathrm{f}(\mathrm{x}-4)=-\mathrm{f}\left(\mathrm{x}\right)$$$$\Rightarrow \mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}(\mathrm{y}+4),\mathrm{y}\mathrm{periodic}4$$$$\sin \left(\frac{\pi }{2}\mathrm{x}\right)$$$$\sin \left(\frac{\pi }{2}(\mathrm{x}-2)\right)=\sin (\frac{\pi \mathrm{x}}{2}-\pi )=-\sin \left(\frac{\pi \mathrm{x}}{2}\right)..\mathrm{worck}$$

Commented by mr W last updated on 26/Oct/23

$$great!$$

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