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Question Number 139532 by mathdanisur last updated on 28/Apr/21

(1/3) − ((x−3)/9) + (((x−3)^2 )/(27)) + ... =?

$$\frac{\mathrm{1}}{\mathrm{3}}\:−\:\frac{{x}−\mathrm{3}}{\mathrm{9}}\:+\:\frac{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{27}}\:+\:...\:=? \\ $$

Commented by mr W last updated on 28/Apr/21

=(1/x)

$$=\frac{\mathrm{1}}{{x}} \\ $$

Commented by mathdanisur last updated on 28/Apr/21

how sir?

$${how}\:{sir}? \\ $$

Commented by mr W last updated on 28/Apr/21

a_0 +a_0 q+a_0 q^2 +...=(a_0 /(1−q)) with ∣q∣<1 now a_0 =(1/3),q=−((x−3)/3) (1/3) − ((x−3)/9) + (((x−3)^2 )/(27)) + ... =(1/3)×(1/(1−(−((x−3)/3))))=(1/x)

$${a}_{\mathrm{0}} +{a}_{\mathrm{0}} {q}+{a}_{\mathrm{0}} {q}^{\mathrm{2}} +...=\frac{{a}_{\mathrm{0}} }{\mathrm{1}−{q}}\:{with}\:\mid{q}\mid<\mathrm{1} \\ $$$${now}\:{a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{3}},{q}=−\frac{{x}−\mathrm{3}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\:−\:\frac{{x}−\mathrm{3}}{\mathrm{9}}\:+\:\frac{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{27}}\:+\:...\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{1}−\left(−\frac{{x}−\mathrm{3}}{\mathrm{3}}\right)}=\frac{\mathrm{1}}{{x}} \\ $$

Commented by mathdanisur last updated on 30/Apr/21

thanks sir

$${thanks}\:{sir} \\ $$

Answered by mathmax by abdo last updated on 28/Apr/21

S(x)=Σ_(n=0) ^∞ (((−1)^n (x−3)^n )/3^(n+1) ) =w(x−3) withw(u)=Σ_(n=0) ^∞ (((−1)^n u^n )/3^(n+1) ) we have w(u)=(1/3)Σ_(n=0) ^∞ (−1)^n ((u/3))^n =(1/3)Σ_(n=0) ^∞ (−(u/3))^n =(1/3)(1/(1+(u/3))) =(1/(u+3)) ⇒S(x)=w(x−3)=(1/(x−3+3)) =(1/x) ⇒ S(x)=(1/x) and condition of convergence ∣u∣<3 ⇒∣x−3∣<3

$$\mathrm{S}\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}+\mathrm{1}} }\:=\mathrm{w}\left(\mathrm{x}−\mathrm{3}\right)\:\mathrm{withw}\left(\mathrm{u}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}+\mathrm{1}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{w}\left(\mathrm{u}\right)=\frac{\mathrm{1}}{\mathrm{3}}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\frac{\mathrm{u}}{\mathrm{3}}\right)^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{3}}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{u}}{\mathrm{3}}\right)^{\mathrm{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{u}}{\mathrm{3}}}\:=\frac{\mathrm{1}}{\mathrm{u}+\mathrm{3}}\:\Rightarrow\mathrm{S}\left(\mathrm{x}\right)=\mathrm{w}\left(\mathrm{x}−\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{3}+\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow \\ $$$$\mathrm{S}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}\:\:\:\:\mathrm{and}\:\mathrm{condition}\:\mathrm{of}\:\mathrm{convergence}\:\mid\mathrm{u}\mid<\mathrm{3}\:\Rightarrow\mid\mathrm{x}−\mathrm{3}\mid<\mathrm{3} \\ $$

Commented by mathdanisur last updated on 30/Apr/21

thanks sir

$${thanks}\:{sir} \\ $$

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