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Question Number 7752 by Tawakalitu. last updated on 14/Sep/16

If f(x) = x + ax^2 + bx^3 + ... .obtain (√(f(x^3 ))) up to x^3

$${If}\:{f}\left({x}\right)\:=\:{x}\:+\:{ax}^{\mathrm{2}} \:+\:{bx}^{\mathrm{3}} \:+\:...\:\:\:\:\:\:.{obtain}\:\sqrt{{f}\left({x}^{\mathrm{3}} \right)} \\ $$$${up}\:{to}\:{x}^{\mathrm{3}} \\ $$

Commented by FilupSmith last updated on 14/Sep/16

f(x)=x^1 a_0 +x^2 a_1 +x^3 a_2 +... ∴f(x^t )=Σ_(n=0) ^k x^(t+n) a_n =x^(t−1) Σ_(n=0) ^k x^(n+1) a_n ∴f(x^3 )=x^3 (a_0 +xa_1 +...)=x^2 f(x) f(x^3 )=x^2 f(x) ∴(√(f(x^3 )))=x(√(f(x)))

$${f}\left({x}\right)={x}^{\mathrm{1}} {a}_{\mathrm{0}} +{x}^{\mathrm{2}} {a}_{\mathrm{1}} +{x}^{\mathrm{3}} {a}_{\mathrm{2}} +... \\ $$$$\therefore{f}\left({x}^{{t}} \right)=\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}{x}^{{t}+{n}} {a}_{{n}} ={x}^{{t}−\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}{x}^{{n}+\mathrm{1}} {a}_{{n}} \\ $$$$ \\ $$$$\therefore{f}\left({x}^{\mathrm{3}} \right)={x}^{\mathrm{3}} \left({a}_{\mathrm{0}} +{xa}_{\mathrm{1}} +...\right)={x}^{\mathrm{2}} {f}\left({x}\right) \\ $$$${f}\left({x}^{\mathrm{3}} \right)={x}^{\mathrm{2}} {f}\left({x}\right) \\ $$$$\therefore\sqrt{{f}\left({x}^{\mathrm{3}} \right)}={x}\sqrt{{f}\left({x}\right)} \\ $$

Commented by Tawakalitu. last updated on 14/Sep/16

I really appreciate sir. thank you.

$${I}\:{really}\:{appreciate}\:{sir}.\:{thank}\:{you}. \\ $$

Commented by Rasheed Soomro last updated on 14/Sep/16

f(x)=x^1 a_0 +x^2 a_1 +x^3 a_2 +... f(x^t )=(x^t )^1 a_0 +(x^t )^2 a_1 +(x^t )^3 a_2 +.... f(x^t )=x^(t×1) a_0 +x^(t×2) a_1 +x^(t×3) a_2 +.... f(x^t )=Σ_(n=0) ^k x^(t×(n+1)) a_n or f(x^t )=Σ_(n=1) ^k x^(t×n) a_(n−1) I may be wrong.I′m not sure.

$${f}\left({x}\right)={x}^{\mathrm{1}} {a}_{\mathrm{0}} +{x}^{\mathrm{2}} {a}_{\mathrm{1}} +{x}^{\mathrm{3}} {a}_{\mathrm{2}} +... \\ $$$${f}\left({x}^{{t}} \right)=\left({x}^{{t}} \right)^{\mathrm{1}} {a}_{\mathrm{0}} +\left({x}^{{t}} \right)^{\mathrm{2}} {a}_{\mathrm{1}} +\left({x}^{{t}} \right)^{\mathrm{3}} {a}_{\mathrm{2}} +.... \\ $$$${f}\left({x}^{{t}} \right)={x}^{{t}×\mathrm{1}} {a}_{\mathrm{0}} +{x}^{{t}×\mathrm{2}} {a}_{\mathrm{1}} +{x}^{{t}×\mathrm{3}} {a}_{\mathrm{2}} +.... \\ $$$${f}\left({x}^{{t}} \right)=\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}{x}^{{t}×\left({n}+\mathrm{1}\right)} {a}_{{n}} \:\:\:{or}\:\:\:\:{f}\left({x}^{{t}} \right)=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}{x}^{{t}×{n}} {a}_{{n}−\mathrm{1}} \\ $$$$\:{I}\:{may}\:\:{be}\:{wrong}.{I}'{m}\:{not}\:{sure}. \\ $$

Commented by ajaenuri86@gmail.com last updated on 14/Sep/16

true

$${true} \\ $$

Commented by Tawakalitu. last updated on 14/Sep/16

Thanks for your contribution sir.

$${Thanks}\:{for}\:{your}\:{contribution}\:{sir}. \\ $$

Commented by Rasheed Soomro last updated on 15/Sep/16

f(x)=x^1 a_0 +x^2 a_1 +x^3 a_2 +... f(x^t )=Σ_(n=0) ^k x^(t+n) a_n ??? −−−−−−−−− f(x)=x^1 a_0 +x^2 a_1 +x^3 a_2 +... f(x^t )=(x^t )^1 a_0 +(x^t )^2 a_1 +(x^t )^3 a_2 +... =x^(t×1) a_0 +x^(t×2) a_1 +x^(t×3) a_2 +...... =^(?) Σ_(n=0) ^k x^(t+n) a_n ???

$${f}\left({x}\right)={x}^{\mathrm{1}} {a}_{\mathrm{0}} +{x}^{\mathrm{2}} {a}_{\mathrm{1}} +{x}^{\mathrm{3}} {a}_{\mathrm{2}} +... \\ $$$${f}\left({x}^{{t}} \right)=\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}{x}^{{t}+{n}} {a}_{{n}} ??? \\ $$$$−−−−−−−−− \\ $$$${f}\left({x}\right)={x}^{\mathrm{1}} {a}_{\mathrm{0}} +{x}^{\mathrm{2}} {a}_{\mathrm{1}} +{x}^{\mathrm{3}} {a}_{\mathrm{2}} +... \\ $$$${f}\left({x}^{{t}} \right)=\left({x}^{{t}} \right)^{\mathrm{1}} {a}_{\mathrm{0}} +\left({x}^{{t}} \right)^{\mathrm{2}} {a}_{\mathrm{1}} +\left({x}^{{t}} \right)^{\mathrm{3}} {a}_{\mathrm{2}} +... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={x}^{{t}×\mathrm{1}} {a}_{\mathrm{0}} +{x}^{{t}×\mathrm{2}} {a}_{\mathrm{1}} +{x}^{{t}×\mathrm{3}} {a}_{\mathrm{2}} +...... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\overset{?} {=}\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}{x}^{{t}+{n}} {a}_{{n}} ??? \\ $$

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