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Question Number 129489 by mr W last updated on 16/Jan/21

if p(x+2)−2p(x)=x^2 −5x−3 find p(x)

$${if}\:{p}\left({x}+\mathrm{2}\right)−\mathrm{2}{p}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3} \\ $$$${find}\:{p}\left({x}\right) \\ $$

Answered by bramlexs22 last updated on 16/Jan/21

let p(x)=ax^2 +bx+c p(x+2)=a(x+2)^2 +b(x+2)+c = ax^2 +(4a+b)x+(4a+2b+c) 2p(x)= 2ax^2 +2bx+2c p(x+2)−2p(x)= −ax^2 +(4a−b)x+(4a+2b−c) −ax^2 +(4a−b)x+(4a+2b−c)=x^2 −5x−3 { ((a=−1)),((−4−b=−5 ; b=1 )),((−4+2−c=−3 ; c = 1)) :} p(x)=−x^2 +x+1

$$\mathrm{let}\:{p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${p}\left({x}+\mathrm{2}\right)={a}\left({x}+\mathrm{2}\right)^{\mathrm{2}} +{b}\left({x}+\mathrm{2}\right)+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{ax}^{\mathrm{2}} +\left(\mathrm{4}{a}+{b}\right){x}+\left(\mathrm{4}{a}+\mathrm{2}{b}+{c}\right) \\ $$$$\mathrm{2}{p}\left({x}\right)=\:\mathrm{2}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+\mathrm{2}{c}\: \\ $$$${p}\left({x}+\mathrm{2}\right)−\mathrm{2}{p}\left({x}\right)=\:−{ax}^{\mathrm{2}} +\left(\mathrm{4}{a}−{b}\right){x}+\left(\mathrm{4}{a}+\mathrm{2}{b}−{c}\right) \\ $$$$−{ax}^{\mathrm{2}} +\left(\mathrm{4}{a}−{b}\right){x}+\left(\mathrm{4}{a}+\mathrm{2}{b}−{c}\right)={x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3} \\ $$$$\:\begin{cases}{{a}=−\mathrm{1}}\\{−\mathrm{4}−{b}=−\mathrm{5}\:;\:{b}=\mathrm{1}\:}\\{−\mathrm{4}+\mathrm{2}−{c}=−\mathrm{3}\:;\:{c}\:=\:\mathrm{1}}\end{cases} \\ $$$${p}\left({x}\right)=−{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$$ \\ $$

Commented by mr W last updated on 16/Jan/21

thanks sir! but is this solution unique? are there any other possiblities?

$${thanks}\:{sir}! \\ $$$${but}\:{is}\:{this}\:{solution}\:{unique}?\:{are}\:{there} \\ $$$${any}\:{other}\:{possiblities}? \\ $$

Answered by liberty last updated on 16/Jan/21

let p(x+2)=u_(n+2) and p(x)=u_n homogenous equation λ^2 −2=0 ; λ=±(√2) u_n =C_1 ((√2))^n +C_2 (−(√2))^n =λ((√2))^n particular solution : ax^2 +bx+c → { ((u_(n+1) =2ax+b)),((u_(n+2) =2a)) :} comparing : 2a−(ax^2 +bx+c)=x^2 −5x−3 −ax^2 −bx+2a−c=x^2 −5x−3 { ((a=−1 ; b=5)),((−2−c=−3 ; c=1)) :} ∴ p(x)=λ((√2) )^x −x^2 +5x+1

$$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{u}_{\mathrm{n}+\mathrm{2}} \:\mathrm{and}\:\mathrm{p}\left(\mathrm{x}\right)=\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{homogenous}\:\mathrm{equation}\:\lambda^{\mathrm{2}} −\mathrm{2}=\mathrm{0}\:;\:\lambda=\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\mathrm{C}_{\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} +\mathrm{C}_{\mathrm{2}} \left(−\sqrt{\mathrm{2}}\right)^{\mathrm{n}} =\lambda\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \\ $$$$\mathrm{particular}\:\mathrm{solution}\::\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\:\rightarrow\begin{cases}{\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{2ax}+\mathrm{b}}\\{\mathrm{u}_{\mathrm{n}+\mathrm{2}} =\mathrm{2a}}\end{cases} \\ $$$$\mathrm{comparing}\::\:\mathrm{2a}−\left(\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{3} \\ $$$$\:−\mathrm{ax}^{\mathrm{2}} −\mathrm{bx}+\mathrm{2a}−\mathrm{c}=\mathrm{x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{3} \\ $$$$\:\begin{cases}{\mathrm{a}=−\mathrm{1}\:;\:\mathrm{b}=\mathrm{5}}\\{−\mathrm{2}−\mathrm{c}=−\mathrm{3}\:;\:\mathrm{c}=\mathrm{1}}\end{cases} \\ $$$$\therefore\:\mathrm{p}\left(\mathrm{x}\right)=\lambda\left(\sqrt{\mathrm{2}}\:\right)^{{x}} −{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1} \\ $$

Commented by mr W last updated on 16/Jan/21

thanks!

$${thanks}! \\ $$

Commented by liberty last updated on 16/Jan/21

yes...i meant like that.

$$\mathrm{yes}...\mathrm{i}\:\mathrm{meant}\:\mathrm{like}\:\mathrm{that}.\: \\ $$

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