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Question Number 179302 by mnjuly1970 last updated on 28/Oct/22
Commented by Frix last updated on 28/Oct/22
p(x)p((1/x))=p(x)+p((1/x)) ⇒ p(x)=((p((1/x)))/(p((1/x))−1)) p(x) is a polynomial ⇒ p(x)=c_0 +c_1 x+c_2 x^2 +...+c_n x^n ⇒ ((p((1/x)))/(p((1/x))−1))=((c_0 x^n +c_1 x^(n−1) +...+c_n )/((c_0 −1)x^n +c_1 x^(n+1) +...+c_n )) this seems only possible if c_0 =1∧c_n =±1∧c_j =0∀j∣0<j<n p(x)=±x^n +1 p((1/x))=1±(1/x^n ) p(x)p((1/x))=p(x)+p((1/x))=2±(x^n +(1/x^n )) p(2)=33 ⇔ 2^n +1=33 ⇒ n=5 p(x)=x^5 +1
$${p}\left({x}\right){p}\left(\frac{\mathrm{1}}{{x}}\right)={p}\left({x}\right)+{p}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\Rightarrow \\ $$$${p}\left({x}\right)=\frac{{p}\left(\frac{\mathrm{1}}{{x}}\right)}{{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}} \\ $$$${p}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\Rightarrow \\ $$$${p}\left({x}\right)={c}_{\mathrm{0}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{c}_{{n}} {x}^{{n}} \\ $$$$\Rightarrow \\ $$$$\frac{{p}\left(\frac{\mathrm{1}}{{x}}\right)}{{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}}=\frac{{c}_{\mathrm{0}} {x}^{{n}} +{c}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +…+{c}_{{n}} }{\left({c}_{\mathrm{0}} −\mathrm{1}\right){x}^{{n}} +{c}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +…+{c}_{{n}} } \\ $$$$\mathrm{this}\:\mathrm{seems}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{if} \\ $$$${c}_{\mathrm{0}} =\mathrm{1}\wedge{c}_{{n}} =\pm\mathrm{1}\wedge{c}_{{j}} =\mathrm{0}\forall{j}\mid\mathrm{0}<{j}<{n} \\ $$$${p}\left({x}\right)=\pm{x}^{{n}} +\mathrm{1} \\ $$$${p}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}\pm\frac{\mathrm{1}}{{x}^{{n}} } \\ $$$${p}\left({x}\right){p}\left(\frac{\mathrm{1}}{{x}}\right)={p}\left({x}\right)+{p}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}\pm\left({x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }\right) \\ $$$${p}\left(\mathrm{2}\right)=\mathrm{33}\:\Leftrightarrow\:\mathrm{2}^{{n}} +\mathrm{1}=\mathrm{33}\:\Rightarrow\:{n}=\mathrm{5} \\ $$$${p}\left({x}\right)={x}^{\mathrm{5}} +\mathrm{1} \\ $$
Commented by mnjuly1970 last updated on 28/Oct/22
thanks alot sir ....great
$$\:\:\:{thanks}\:{alot}\:{sir}\:….{great} \\ $$
Answered by a.lgnaoui last updated on 28/Oct/22
p(x)[p((1/x))−1]=p((1/x)) p(x)=((p((1/x)))/(p((1/x))−1)) p(2)=((p((1/2)))/(p((1/2))−1))=33 33p((1/2))−33=p((1/2)) p((1/2))=((33)/(32)) p(3)=p((1/(1/3)))=((p(3))/(p(3)−1))⇒p(3)^2 −p(3)=p(3) p(3)=2 p(4)=((p(4))/(p(4)−1))⇒p(4)={0,2} p(x)^2 −2p(x)=p(x) pour x=1 p(1)=2 pour x=2 p(x)=33 pour x>2 p(x)=2
$${p}\left({x}\right)\left[{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}\right]={p}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${p}\left({x}\right)=\frac{{p}\left(\frac{\mathrm{1}}{{x}}\right)}{{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}} \\ $$$${p}\left(\mathrm{2}\right)=\frac{{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}}=\mathrm{33}\:\:\:\mathrm{33}{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{33}={p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{33}}{\mathrm{32}} \\ $$$$ \\ $$$${p}\left(\mathrm{3}\right)=\mathrm{p}\left(\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}}}\right)=\frac{\mathrm{p}\left(\mathrm{3}\right)}{\mathrm{p}\left(\mathrm{3}\right)−\mathrm{1}}\Rightarrow{p}\left(\mathrm{3}\right)^{\mathrm{2}} −{p}\left(\mathrm{3}\right)={p}\left(\mathrm{3}\right) \\ $$$${p}\left(\mathrm{3}\right)=\mathrm{2}\:\:\:\:\:\:\:{p}\left(\mathrm{4}\right)=\frac{{p}\left(\mathrm{4}\right)}{{p}\left(\mathrm{4}\right)−\mathrm{1}}\Rightarrow{p}\left(\mathrm{4}\right)=\left\{\mathrm{0},\mathrm{2}\right\} \\ $$$${p}\left({x}\right)^{\mathrm{2}} −\mathrm{2}{p}\left({x}\right)={p}\left({x}\right) \\ $$$$ \\ $$$${pour}\:{x}=\mathrm{1}\:\:\:\:\:\:\:\:{p}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$${pour}\:\:{x}=\mathrm{2}\:\:\:\:\:\:\:{p}\left({x}\right)=\mathrm{33} \\ $$$${pour}\:{x}>\mathrm{2}\:\:\:\:\:\:\:{p}\left({x}\right)=\mathrm{2} \\ $$$$ \\ $$
Answered by mr W last updated on 28/Oct/22
say p(x)=a_0 +a_1 x+a_2 x^2 +...+a_n x^n with a_n ≠0 p(x)p((1/x))=p(x)+p((1/x)) p(x)=((p((1/x)))/(p((1/x))−1))=((a_0 x^n +a_1 x^(n−1) +...+a_n )/((a_0 −1)x^n +a_1 x^(n+1) +...+a_n )) p(0)=a_0 ⇒(a_n /a_n )=a_0 ⇒a_0 =1 p(x)=((p((1/x)))/(p((1/x))−1))=1+(x^n /(a_1 x^(n+1) +a_2 x^n +...+a_n )) 1+a_1 x+a_2 x^2 +...+a_n x^n =1+(x^n /(a_1 x^(n+1) +a_2 x^n +...+a_n )) (a_1 x+a_2 x^2 +...+a_n x^n )(a_1 x^(n+1) +a_2 x^n +...+a_n )=x^n a_1 a_n x+a_2 a_n x^2 +...+a_(n−1) a_n x^(n−1) +a_n ^2 x^n +(...)x^(n+1) +(...)x^(n+2) +...+(...)x^(2n+1) =x^n comparing coef. of all terms a_1 =a_2 =...=a_(n−1) =0 a_n ^2 =1 ⇒a_n =±1 ⇒p(x)=1±x^n p(2)=1±2^n =33 ±2^n =32=2^5 only solution is 2^n =2^5 ⇒n=5 ⇒f(x)=1+x^5
$${say}\:{p}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{{n}} {x}^{{n}} \\ $$$${with}\:{a}_{{n}} \neq\mathrm{0} \\ $$$${p}\left({x}\right){p}\left(\frac{\mathrm{1}}{{x}}\right)={p}\left({x}\right)+{p}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${p}\left({x}\right)=\frac{{p}\left(\frac{\mathrm{1}}{{x}}\right)}{{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}}=\frac{{a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +…+{a}_{{n}} }{\left({a}_{\mathrm{0}} −\mathrm{1}\right){x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +…+{a}_{{n}} } \\ $$$${p}\left(\mathrm{0}\right)={a}_{\mathrm{0}} \:\Rightarrow\frac{{a}_{{n}} }{{a}_{{n}} }={a}_{\mathrm{0}} \:\Rightarrow{a}_{\mathrm{0}} =\mathrm{1} \\ $$$${p}\left({x}\right)=\frac{{p}\left(\frac{\mathrm{1}}{{x}}\right)}{{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}}=\mathrm{1}+\frac{{x}^{{n}} }{{a}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +{a}_{\mathrm{2}} {x}^{{n}} +…+{a}_{{n}} } \\ $$$$\mathrm{1}+{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{{n}} {x}^{{n}} =\mathrm{1}+\frac{{x}^{{n}} }{{a}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +{a}_{\mathrm{2}} {x}^{{n}} +…+{a}_{{n}} } \\ $$$$\left({a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{{n}} {x}^{{n}} \right)\left({a}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +{a}_{\mathrm{2}} {x}^{{n}} +…+{a}_{{n}} \right)={x}^{{n}} \\ $$$$\cancel{{a}_{\mathrm{1}} {a}_{{n}} }{x}+\cancel{{a}_{\mathrm{2}} {a}_{{n}} }{x}^{\mathrm{2}} +…+\cancel{{a}_{{n}−\mathrm{1}} {a}_{{n}} }{x}^{{n}−\mathrm{1}} +{a}_{{n}} ^{\mathrm{2}} {x}^{{n}} +\left(\cancel{…}\right){x}^{{n}+\mathrm{1}} +\left(\cancel{…}\right){x}^{{n}+\mathrm{2}} +…+\left(\cancel{…}\right){x}^{\mathrm{2}{n}+\mathrm{1}} ={x}^{{n}} \\ $$$${comparing}\:{coef}.\:{of}\:{all}\:{terms} \\ $$$${a}_{\mathrm{1}} ={a}_{\mathrm{2}} =…={a}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${a}_{{n}} ^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{a}_{{n}} =\pm\mathrm{1} \\ $$$$\Rightarrow{p}\left({x}\right)=\mathrm{1}\pm{x}^{{n}} \\ $$$${p}\left(\mathrm{2}\right)=\mathrm{1}\pm\mathrm{2}^{{n}} =\mathrm{33} \\ $$$$\pm\mathrm{2}^{{n}} =\mathrm{32}=\mathrm{2}^{\mathrm{5}} \: \\ $$$${only}\:{solution}\:{is}\:\mathrm{2}^{{n}} =\mathrm{2}^{\mathrm{5}} \:\Rightarrow{n}=\mathrm{5} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{1}+{x}^{\mathrm{5}} \\ $$
Commented by mnjuly1970 last updated on 28/Oct/22
grateful sir W....excellent solution...
$$\:\:\:{grateful}\:\:{sir}\:\:{W}….{excellent} \\ $$$${solution}… \\ $$
Commented by Tawa11 last updated on 28/Oct/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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