Question-83543

Question Number 83543 by Power last updated on 03/Mar/20

Commented by john santu last updated on 03/Mar/20

let 1+2x+3x^2 +4x^3 +5x^4 +... = f(x) ∫ f(x)dx= ∫(1+2x+3x^2 +4x^3 +...)dx ∫f(x) dx = x+x^2 +x^3 +x^4 +... ∫ f(x)dx = (x/(1−x))= (x/(−x+1)) ⇒ f(x) =(1/((1−x)^2 )) (1/((1−x)^2 )) = ((49)/(36 )) ⇒ (1−x)^2 = ((36)/(49)) (1−x−(6/7))(1−x+(6/7))=0 ((1/7)−x)(((13)/7)−x)=0 x = { ((1/7)),(((13)/7)) :}

$$\mathrm{let}\:\mathrm{1}+\mathrm{2x}+\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{3}} +\mathrm{5x}^{\mathrm{4}} +…\:=\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\int\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\:\int\left(\mathrm{1}+\mathrm{2x}+\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{3}} +…\right)\mathrm{dx} \\ $$$$\int\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} +… \\ $$$$\int\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=\:\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}}=\:\frac{\mathrm{x}}{−\mathrm{x}+\mathrm{1}}\:\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{49}}{\mathrm{36}\:}\:\Rightarrow\:\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{36}}{\mathrm{49}} \\ $$$$\left(\mathrm{1}−\mathrm{x}−\frac{\mathrm{6}}{\mathrm{7}}\right)\left(\mathrm{1}−\mathrm{x}+\frac{\mathrm{6}}{\mathrm{7}}\right)=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{x}\right)\left(\frac{\mathrm{13}}{\mathrm{7}}−\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{7}}}\\{\frac{\mathrm{13}}{\mathrm{7}}}\end{cases} \\ $$

Commented by Power last updated on 03/Mar/20

$$\mathrm{thanks} \\ $$

Answered by mr W last updated on 03/Mar/20

f(x)=1+2x+3x^2 +4x^3 +5x^4 +... xf(x)= x+2x^2 +3x^3 +4x^4 +... (1−x)f(x)=1+x+x^2 +x^3 +x^4 +... (1−x)f(x)=(1/(1−x)), if ∣x∣<1 (1−x)f(x)=∞, if ∣x∣≥1 with ∣x∣<1: ⇒f(x)=(1/((1−x)^2 ))=((49)/(36)) ⇒(1/(1−x))=±(7/6) ⇒1−x=±(6/7) ⇒x=1±(6/7) since ∣x∣<1, there is only one solution: ⇒x=(1/7)

$${f}\left({x}\right)=\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{4}} +… \\ $$$${xf}\left({x}\right)=\:\:\:\:\:{x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +… \\ $$$$\left(\mathrm{1}−{x}\right){f}\left({x}\right)=\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +… \\ $$$$\left(\mathrm{1}−{x}\right){f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}},\:{if}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left(\mathrm{1}−{x}\right){f}\left({x}\right)=\infty,\:{if}\:\mid{x}\mid\geqslant\mathrm{1} \\ $$$$ \\ $$$${with}\:\mid{x}\mid<\mathrm{1}: \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\frac{\mathrm{49}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}−{x}}=\pm\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{1}−{x}=\pm\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\Rightarrow{x}=\mathrm{1}\pm\frac{\mathrm{6}}{\mathrm{7}} \\ $$$${since}\:\mid{x}\mid<\mathrm{1},\:{there}\:{is}\:{only}\:{one}\:{solution}: \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$

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