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Question Number 211029 by BaliramKumar last updated on 26/Aug/24
Answered by Ar Brandon last updated on 26/Aug/24
f(2)=5 therefore f(x) must be a non-zero polynomial. Let f(x)=a_0 +a_1 x+a_2 x^2 +∙∙∙+a_n x^n , a_n ≠0 Suppose f(x)+f((1/x))=f(x)f((1/x)) for all x≠0 Then Σ_(r=0) ^n a_r x^r +Σ_(r=1) ^n (a_r /x^r )=(Σ_(r=0) ^n a_r x^r )(Σ_(r=1) ^n (a_r /x^r )) Multiplying through by x^n , Σ_(r=0) ^n a_r x^(n+r) +Σ_(r=0) ^n a_r x^(n−r) =(Σ_(r=0) ^n a_r x^r )(Σ_(r=0) ^n a_r x^(n−r) ) That is, (a_0 x^n +a_1 x^(n+1) +∙∙∙a_n x^(2n) )+(a_0 x^n +a_1 x^(n−1) +∙∙∙+a_(n−1) x+a_n ) =(a_0 +a_1 x+∙∙∙a_n x^n )(a_0 x^n +a_1 x^(n−1) +∙∙∙+a_(n−1) x+a_n ) Equating the corresponding coefficients of powers of x, we have a_n =a_0 a_n , a_(n−1) =a_0 a_(n−1) +a_1 a_n a_(n−2) =a_2 a_n +a_1 a_(n−1) +a_(n−2) a_0 2a_0 =a_0 ^2 +a_n ^2 a_n =a_0 a_n ⇒a_0 =1 (since a_n ≠0) a_(n−1) =a_0 a_(n−1) +a_1 a_n ⇒a_1 a_n =0⇒a_1 =0 a_(n−2) =a_2 a_n +a_1 a_(n−1) +a_(n−2) a_0 ⇒a_(n−2) =a_2 a_n +a_(n−2) ⇒a_2 =0 Continuing this process, we get a_(n−1) =0 and 2=1+a_n ^2 . Hence a_n =±1. Therefore f(x)=1±x^n But f(2)=5 hence 5=1±2^n Therefore f(x) cannot be 1−x^n . Thus, f(x)=1+x^n and 5=1+2^n ⇒ n=2. So f(x)=1+x^2 and determinant (((f(3)=10))).
$${f}\left(\mathrm{2}\right)=\mathrm{5}\:\mathrm{therefore}\:{f}\left({x}\right)\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{polynomial}. \\ $$$$\mathrm{Let}\:{f}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{a}_{{n}} {x}^{{n}} ,\:{a}_{{n}} \neq\mathrm{0} \\ $$$$\mathrm{Suppose}\:{f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)\:\mathrm{for}\:\mathrm{all}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{Then}\:\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{\mathrm{r}} +\underset{\mathrm{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{\mathrm{r}} }{{x}^{\mathrm{r}} }=\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{\mathrm{r}} \right)\left(\underset{\mathrm{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{\mathrm{r}} }{{x}^{\mathrm{r}} }\right) \\ $$$$\mathrm{Multiplying}\:\mathrm{through}\:\mathrm{by}\:{x}^{{n}} , \\ $$$$\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{{n}+\mathrm{r}} +\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{{n}−\mathrm{r}} =\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{\mathrm{r}} \right)\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{{n}−\mathrm{r}} \right) \\ $$$$\mathrm{That}\:\mathrm{is}, \\ $$$$\left({a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +\centerdot\centerdot\centerdot{a}_{{n}} {x}^{\mathrm{2}{n}} \right)+\left({a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +\centerdot\centerdot\centerdot+{a}_{{n}−\mathrm{1}} {x}+{a}_{{n}} \right) \\ $$$$\:\:\:\:\:\:=\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+\centerdot\centerdot\centerdot{a}_{{n}} {x}^{{n}} \right)\left({a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +\centerdot\centerdot\centerdot+{a}_{{n}−\mathrm{1}} {x}+{a}_{{n}} \right) \\ $$$$\mathrm{Equating}\:\mathrm{the}\:\mathrm{corresponding}\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{powers}\:\mathrm{of}\:{x}, \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\:\:\:\:\:\:\:\:\:{a}_{{n}} ={a}_{\mathrm{0}} {a}_{{n}} ,\:{a}_{{n}−\mathrm{1}} ={a}_{\mathrm{0}} {a}_{{n}−\mathrm{1}} +{a}_{\mathrm{1}} {a}_{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{2}} {a}_{{n}} +{a}_{\mathrm{1}} {a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} {a}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}{a}_{\mathrm{0}} ={a}_{\mathrm{0}} ^{\mathrm{2}} +{a}_{{n}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}_{{n}} ={a}_{\mathrm{0}} {a}_{{n}} \Rightarrow{a}_{\mathrm{0}} =\mathrm{1}\:\left(\mathrm{since}\:{a}_{{n}} \neq\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}_{{n}−\mathrm{1}} ={a}_{\mathrm{0}} {a}_{{n}−\mathrm{1}} +{a}_{\mathrm{1}} {a}_{{n}} \Rightarrow{a}_{\mathrm{1}} {a}_{{n}} =\mathrm{0}\Rightarrow{a}_{\mathrm{1}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{2}} {a}_{{n}} +{a}_{\mathrm{1}} {a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} {a}_{\mathrm{0}} \Rightarrow{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{2}} {a}_{{n}} +{a}_{{n}−\mathrm{2}} \Rightarrow{a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{Continuing}\:\mathrm{this}\:\mathrm{process},\:\mathrm{we}\:\mathrm{get}\:{a}_{{n}−\mathrm{1}} =\mathrm{0}\:\mathrm{and}\:\mathrm{2}=\mathrm{1}+{a}_{{n}} ^{\mathrm{2}} . \\ $$$$\mathrm{Hence}\:{a}_{{n}} =\pm\mathrm{1}.\:\mathrm{Therefore} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{1}\pm{x}^{{n}} \\ $$$$\mathrm{But}\:{f}\left(\mathrm{2}\right)=\mathrm{5}\:\mathrm{hence} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}=\mathrm{1}\pm\mathrm{2}^{\mathrm{n}} \\ $$$$\mathrm{Therefore}\:{f}\left({x}\right)\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{1}−{x}^{{n}} .\:\mathrm{Thus},\:{f}\left({x}\right)=\mathrm{1}+{x}^{{n}} \:\mathrm{and} \\ $$$$\mathrm{5}=\mathrm{1}+\mathrm{2}^{{n}} \:\Rightarrow\:{n}=\mathrm{2}.\:\mathrm{So}\:{f}\left({x}\right)=\mathrm{1}+{x}^{\mathrm{2}} \:\mathrm{and}\:\begin{array}{|c|}{{f}\left(\mathrm{3}\right)=\mathrm{10}}\\\hline\end{array}. \\ $$
Commented by Frix last updated on 26/Aug/24
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Commented by Ar Brandon last updated on 26/Aug/24
Greetings, Sir! it's been quite a long time��
Commented by BaliramKumar last updated on 27/Aug/24
Thanks sir
$${Thanks}\:{sir} \\ $$
Commented by Ar Brandon last updated on 27/Aug/24
You're welcome!
Answered by Frix last updated on 26/Aug/24
We know 3 things: (1) f(x)f((1/x))=f(x)+f((1/x)) for all x∈R (2) f(2)=5 ((1) ⇒ f((1/x))=((f(x))/(f(x)+1))) ∧ (2) ⇒ (3) f((1/2))=(5/4) ⇒ we can have 3 unknowns ⇒ we have a polynomial of degree 2 ⇒ f(x)=ax^2 +bx+c (2) 4a+2b+c=5 (3) (a/4)+(b/2)+c=(5/4) ⇒ b=−((5(a−1))/2)∧a=c Insert into (1) & transform c^2 −((2(x^2 −5x+1))/((x−2)(2x−1)))c−((5x)/((x−2)(2x−1)))=0 ⇒ c=1∨c=−((5x)/((x−2)(2x−1))) Since c is a constant which must be valid for all x∈R ⇒ c=1 ⇒ a=1∧b=0 ⇒ f(x)=x^2 +1 f(3)=10
$$\mathrm{We}\:\mathrm{know}\:\mathrm{3}\:\mathrm{things}: \\ $$$$\:\:\:\:\:\left(\mathrm{1}\right)\:{f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)\:\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R} \\ $$$$\:\:\:\:\:\left(\mathrm{2}\right)\:{f}\left(\mathrm{2}\right)=\mathrm{5} \\ $$$$\left(\left(\mathrm{1}\right)\:\Rightarrow\:{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{{f}\left({x}\right)}{{f}\left({x}\right)+\mathrm{1}}\right)\:\wedge\:\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\:\:\:\:\:\left(\mathrm{3}\right)\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{have}\:\mathrm{3}\:\mathrm{unknowns}\:\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{a} \\ $$$$\mathrm{polynomial}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{2}\:\Rightarrow \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{4}{a}+\mathrm{2}{b}+{c}=\mathrm{5} \\ $$$$\left(\mathrm{3}\right)\:\frac{{a}}{\mathrm{4}}+\frac{{b}}{\mathrm{2}}+{c}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow\:{b}=−\frac{\mathrm{5}\left({a}−\mathrm{1}\right)}{\mathrm{2}}\wedge{a}={c} \\ $$$$ \\ $$$$\mathrm{Insert}\:\mathrm{into}\:\left(\mathrm{1}\right)\:\&\:\mathrm{transform} \\ $$$${c}^{\mathrm{2}} −\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}\right)}{\left({x}−\mathrm{2}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)}{c}−\frac{\mathrm{5}{x}}{\left({x}−\mathrm{2}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)}=\mathrm{0} \\ $$$$\Rightarrow\:{c}=\mathrm{1}\vee{c}=−\frac{\mathrm{5}{x}}{\left({x}−\mathrm{2}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)} \\ $$$$\mathrm{Since}\:{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{which}\:\mathrm{must}\:\mathrm{be}\:\mathrm{valid} \\ $$$$\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R} \\ $$$$\Rightarrow\:{c}=\mathrm{1} \\ $$$$\Rightarrow\:{a}=\mathrm{1}\wedge{b}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{10} \\ $$
Commented by BaliramKumar last updated on 27/Aug/24
Thanks sir
$${Thanks}\:{sir} \\ $$

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