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Question Number 1307 by Rasheed Ahmad last updated on 21/Jul/15
f(2x)−2[ f(x) ]^2 +1=0 f(x)=?
$${f}\left(\mathrm{2}{x}\right)−\mathrm{2}\left[\:{f}\left({x}\right)\:\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$
Commented by Rasheed Soomro last updated on 22/Jul/15
(Rasheed Ahmad) Actually this equation is disguised form of the trigonometric identity given below: cos(2x)=2cos^2 (x)−1 So f(x)=cos x is expected solution for me. It is a matter of interest for me that there exist other functoins also which accept such relation! Anyway THANKS and APPRECIATIONS for the participants.
$$\left({Rasheed}\:{Ahmad}\right) \\ $$$${Actually}\:{this}\:{equation}\:{is}\:{disguised}\:{form}\:{of}\:{the}\:{trigonometric} \\ $$$${identity}\:{given}\:{below}: \\ $$$${cos}\left(\mathrm{2}{x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1} \\ $$$${So}\:{f}\left({x}\right)={cos}\:{x}\:{is}\:{expected}\:{solution}\:{for}\:{me}. \\ $$$${It}\:{is}\:{a}\:{matter}\:{of}\:{interest}\:{for}\:{me}\:{that}\:{there}\:{exist}\:{other}\:\: \\ $$$${functoins}\:{also}\:{which}\:{accept}\:{such}\:{relation}! \\ $$$${Anyway}\:{THANKS}\:{and}\:{APPRECIATIONS}\:{for}\:{the}\: \\ $$$${participants}. \\ $$$$ \\ $$
Commented by 123456 last updated on 21/Jul/15
f(0)−2[f(0)]^2 +1=0 f(0)=x −2x^2 +x+1=0 Δ=(1)^2 −4(−2)(1)=1+8=9 x=((−(1)±(√9))/(2(−2)))=((−1±3)/(−4))=((1∓3)/4) x_1 =((1−3)/4)=−(2/4)=−(1/2) x_2 =((1+3)/4)=(4/4)=1 f(0)=−(1/2)∨f(0)=1
$${f}\left(\mathrm{0}\right)−\mathrm{2}\left[{f}\left(\mathrm{0}\right)\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)={x} \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{2}\right)\left(\mathrm{1}\right)=\mathrm{1}+\mathrm{8}=\mathrm{9} \\ $$$${x}=\frac{−\left(\mathrm{1}\right)\pm\sqrt{\mathrm{9}}}{\mathrm{2}\left(−\mathrm{2}\right)}=\frac{−\mathrm{1}\pm\mathrm{3}}{−\mathrm{4}}=\frac{\mathrm{1}\mp\mathrm{3}}{\mathrm{4}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}−\mathrm{3}}{\mathrm{4}}=−\frac{\mathrm{2}}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\vee{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 22/Jul/15
f(x)=(1/2)(b^(ax) +(1/b^(ax) ))..........(i) f(x)=((b^(ax) +b^(−ax) )/2)................(ii) I think these two answes are same. Answer f(x)=cosh x and answer (ii) has similarity . Although they are not same: f(x)=cosh x=((e^x +e^(−x) )/2) f(x) =((b^(ax) +b^(−ax) )/2) I think latter is general form and former is special case. taking b=e and a=1 in latter (as a and b are arbitrary constants) former can be achieved. one can achieve former. Also note that we may take x ∈ C in these two solutions.
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({b}^{{ax}} +\frac{\mathrm{1}}{{b}^{{ax}} }\right)……….\left({i}\right) \\ $$$${f}\left({x}\right)=\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}}…………….\left({ii}\right) \\ $$$${I}\:{think}\:{these}\:{two}\:{answes}\:\:{are}\:{same}. \\ $$$${Answer}\:{f}\left({x}\right)=\mathrm{cosh}\:{x}\:{and}\:{answer}\:\left({ii}\right)\:{has}\:{similarity}\:.\:{Although} \\ $$$${they}\:{are}\:{not}\:{same}: \\ $$$${f}\left({x}\right)=\mathrm{cosh}\:{x}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$${f}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}} \\ $$$${I}\:{think}\:{latter}\:{is}\:{general}\:{form}\:{and}\:{former}\:{is}\:{special}\:{case}. \\ $$$${taking}\:{b}={e}\:{and}\:\:{a}=\mathrm{1}\:{in}\:{latter}\:\left({as}\:{a}\:{and}\:{b}\:{are}\:{arbitrary}\:{constants}\right)\:{former}\:{can}\:{be}\:{achieved}. \\ $$$${one}\:{can}\:{achieve}\:{former}. \\ $$$${Also}\:{note}\:{that}\:{we}\:{may}\:{take}\:{x}\:\in\:\mathbb{C}\:{in}\:{these}\:{two}\:{solutions}. \\ $$
Commented by 123456 last updated on 23/Jul/15
cos x=((e^(ix) +e^(−ix) )/2) a^(bx) =(a^b )^x ≡c^x ,c=a^b cosh x=((e^x +e^(−x) )/2) also cosh (xi)=cos x cos (xi)=cosh x f(x)=((b^(ax) +b^(−ax) )/2) =(((e^(ln b) )^(ax) +(e^(ln b) )^(−ax) )/2) =((e^(axln b) +e^(−axln b) )/2) =cosh (axln b) aln b=k f(x)=cosh (kx)
$$\mathrm{cos}\:{x}=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}} \\ $$$${a}^{{bx}} =\left({a}^{{b}} \right)^{{x}} \equiv{c}^{{x}} ,{c}={a}^{{b}} \\ $$$$\mathrm{cosh}\:{x}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{also} \\ $$$$\mathrm{cosh}\:\left({xi}\right)=\mathrm{cos}\:{x} \\ $$$$\mathrm{cos}\:\left({xi}\right)=\mathrm{cosh}\:{x} \\ $$$${f}\left({x}\right)=\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left({e}^{\mathrm{ln}\:{b}} \right)^{{ax}} +\left({e}^{\mathrm{ln}\:{b}} \right)^{−{ax}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{{e}^{{ax}\mathrm{ln}\:{b}} +{e}^{−{ax}\mathrm{ln}\:{b}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{cosh}\:\left({ax}\mathrm{ln}\:{b}\right) \\ $$$${a}\mathrm{ln}\:{b}={k} \\ $$$${f}\left({x}\right)=\mathrm{cosh}\:\left({kx}\right) \\ $$
Commented by prakash jain last updated on 23/Jul/15
f(2x)=2f(x)^2 −1 In general when addition in domain results in multiplication in range then function is exponential. The extra 1 on the right side suggests that exponent and its reciprocal are multiplied together. cosh(kx) is general solution. Other form of solution((b^(ax) +b^(−ax) )/2) is also reducible to cosh(kx).
$${f}\left(\mathrm{2}{x}\right)=\mathrm{2}{f}\left({x}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{In}\:\mathrm{general}\:\mathrm{when}\:\mathrm{addition}\:\mathrm{in}\:\mathrm{domain}\:\mathrm{results} \\ $$$$\mathrm{in}\:\mathrm{multiplication}\:\mathrm{in}\:\mathrm{range}\:\mathrm{then}\:\mathrm{function} \\ $$$$\mathrm{is}\:\mathrm{exponential}.\:\mathrm{The}\:\mathrm{extra}\:\mathrm{1}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right} \\ $$$$\mathrm{side}\:\mathrm{suggests}\:\mathrm{that}\:\mathrm{exponent}\:\mathrm{and}\:\mathrm{its}\:\mathrm{reciprocal} \\ $$$$\mathrm{are}\:\mathrm{multiplied}\:\mathrm{together}. \\ $$$$\mathrm{cosh}\left({kx}\right)\:\mathrm{is}\:\mathrm{general}\:\mathrm{solution}. \\ $$$$\mathrm{Other}\:\mathrm{form}\:\mathrm{of}\:\mathrm{solution}\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}}\:\mathrm{is}\:\mathrm{also}\:\mathrm{reducible} \\ $$$$\mathrm{to}\:\mathrm{cosh}\left({kx}\right). \\ $$
Answered by prakash jain last updated on 21/Jul/15
f(x)=(1/( (√2)))((2^(ax) /( (√2)))+(1/( (√2)∙2^(ax) ))) 2(f(x))^2 =(2^(2ax) /2)+(1/(2∙2^(2ax) ))+1 f(2x)=(1/( (√2)))((2^(2ax) /( (√2)))+(1/( (√2)∙2^(2ax) )))=(2^(2ax) /2)+(1/(2∙2^(2ax) )) f(2x)−2(f(x))^2 +1=0 f(x)=(1/( (√2)))((b^(ax) /( (√2)))+(1/( (√2) ∙b^(ax) ))) will also satisfy.
$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{2}^{{ax}} }{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\centerdot\mathrm{2}^{{ax}} }\right) \\ $$$$\mathrm{2}\left({f}\left({x}\right)\right)^{\mathrm{2}} =\frac{\mathrm{2}^{\mathrm{2}{ax}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2}^{\mathrm{2}{ax}} }+\mathrm{1} \\ $$$${f}\left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{2}^{\mathrm{2}{ax}} }{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\centerdot\mathrm{2}^{\mathrm{2}{ax}} }\right)=\frac{\mathrm{2}^{\mathrm{2}{ax}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2}^{\mathrm{2}{ax}} } \\ $$$${f}\left(\mathrm{2}{x}\right)−\mathrm{2}\left({f}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{{b}^{{ax}} }{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:\centerdot{b}^{{ax}} }\right)\:\mathrm{will}\:\mathrm{also}\:\mathrm{satisfy}. \\ $$
Commented by prakash jain last updated on 21/Jul/15
f(x)=(1/2)(b^(ax) +(1/b^(ax) ))
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({b}^{{ax}} +\frac{\mathrm{1}}{{b}^{{ax}} }\right) \\ $$
Commented by 112358 last updated on 21/Jul/15
f(x)=cosx or f(x)=coshx also satistfy the equation.
$${f}\left({x}\right)={cosx}\:\:\:{or}\:\:{f}\left({x}\right)={coshx}\: \\ $$$${also}\:{satistfy}\:{the}\:{equation}. \\ $$
Commented by prakash jain last updated on 21/Jul/15
Thanks. After working thru one solution with base 2, I see any base b can be used. f(x)=((b^(ax) +b^(−ax) )/2)
$$\mathrm{Thanks}.\:\mathrm{After}\:\mathrm{working}\:\mathrm{thru}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\mathrm{with}\:\mathrm{base}\:\mathrm{2},\:\mathrm{I}\:\mathrm{see}\:\mathrm{any}\:\mathrm{base}\:{b}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}. \\ $$$${f}\left({x}\right)=\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}} \\ $$
Commented by 112358 last updated on 21/Jul/15
Nice.
$${Nice}. \\ $$
Commented by Rasheed Soomro last updated on 22/Jul/15
(1/( (√2)))((b^(ax) /( (√2)))+(1/( (√2) ∙b^(ax) ))) can be reduced into ((b^(ax) +b^(−ax) )/2) . Overall two answers have come. f(x)=cos x and f(x)=((b^(ax) +b^(−ax) )/2) Other answers include in former or its special cases including f(x)=cosh x.
$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{{b}^{{ax}} }{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:\centerdot{b}^{{ax}} }\right)\:{can}\:{be}\:{reduced}\:{into}\:\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}}\:\:\:. \\ $$$${Overall}\:{two}\:{answers}\:{have}\:{come}. \\ $$$${f}\left({x}\right)={cos}\:{x}\:{and}\:{f}\left({x}\right)=\frac{{b}^{{ax}} +{b}^{−{ax}} }{\mathrm{2}} \\ $$$${Other}\:{answers}\:{include}\:{in}\:{former}\:{or}\:{its}\:{special}\:{cases}\:{including} \\ $$$${f}\left({x}\right)=\mathrm{cosh}\:{x}. \\ $$
Commented by Rasheed Soomro last updated on 22/Jul/15
Note:Some comments on answers are mistakenly included in Question′s comments. please see there.
$${Note}:{Some}\:{comments}\:{on}\:{answers}\:{are}\:{mistakenly}\:{included}\:{in} \\ $$$${Question}'{s}\:{comments}.\:{please}\:{see}\:{there}. \\ $$

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