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Question Number 94337 by mathmax by abdo last updated on 18/May/20

developp at integr serie f(x) =(1/((x−1)(x−2)))

$${developp}\:{at}\:{integr}\:{serie}\:{f}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)} \\ $$

Answered by Rio Michael last updated on 18/May/20

if you mean develope a series then: f(x) = (1/((x−1)(x−2))) ≡ (1/(x−2)) −(1/(x−1)) ⇒f(x) = (x−2)^(−1) −(x−1)^(−1) = x^(−1) (1−(2/x))^(−1) −x^(−1) (1−(1/x))^(−1) = x^(−1) [1 + (2/x) + (4/x^2 )+...]−x^(−1) [1 + (1/x) + (1/x^2 ) + ...] = (1/x) + (2/x^2 ) + ... −(1/x)−(1/x^2 ) −... = (1/x^2 ) +... ∣(2/x)∣ < 1

$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{develope}\:\mathrm{a}\:\mathrm{series}\:\mathrm{then}: \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}\:\equiv\:\frac{\mathrm{1}}{{x}−\mathrm{2}}\:−\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$\Rightarrow{f}\left({x}\right)\:=\:\left({x}−\mathrm{2}\right)^{−\mathrm{1}} −\left({x}−\mathrm{1}\right)^{−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}^{−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)^{−\mathrm{1}} −{x}^{−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}^{−\mathrm{1}} \left[\mathrm{1}\:+\:\frac{\mathrm{2}}{{x}}\:+\:\frac{\mathrm{4}}{{x}^{\mathrm{2}} }+...\right]−{x}^{−\mathrm{1}} \left[\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:...\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\:+\:...\:−\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−...\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+...\:\mid\frac{\mathrm{2}}{{x}}\mid\:<\:\mathrm{1} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 18/May/20

sir rio this is not developpement at integr serie ...

$$\mathrm{sir}\:\mathrm{rio}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{developpement}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie}\:... \\ $$

Answered by mathmax by abdo last updated on 18/May/20

we have f(x) =Σ_(n=0) ^∞ ((f^n (0))/(n!)) x^n let determine f^n (0) we have f(x) =(1/(x−2))−(1/(x−1)) ⇒f^((n)) (x) =(((−1)^n n!)/((x−2)^(n+1) ))−(((−1)^n n!)/((x−1)^(n+1) )) ⇒f^((n)) (0) =(−1)^n n!{(1/((−2)^(n+1) ))−(1/((−1)^(n+1) ))} =(((−1)^n n!)/((−1)^(n+1) )){ (1/2^(n+1) )−1} =−n!{(1/2^(n+1) )−1}⇒ f(x) =−Σ_(n=0) ^∞ {(1/2^(n+1) )−1} x^n =−Σ_(n=0) ^∞ (x^n /2^(n+1) ) +Σ_(n=0) ^∞ x^n with ∣x∣<1 .

$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}^{\mathrm{n}} \left(\mathrm{0}\right)}{\mathrm{n}!}\:\mathrm{x}^{\mathrm{n}} \:\:\mathrm{let}\:\mathrm{determine}\:\mathrm{f}^{\mathrm{n}} \left(\mathrm{0}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{n}+\mathrm{1}} }−\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!\left\{\frac{\mathrm{1}}{\left(−\mathrm{2}\right)^{\mathrm{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\left\{\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\mathrm{1}\right\}\:=−\mathrm{n}!\left\{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\mathrm{1}\right\}\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left\{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\mathrm{1}\right\}\:\mathrm{x}^{\mathrm{n}} \:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\:+\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:\:\:\mathrm{with} \\ $$$$\mid\mathrm{x}\mid<\mathrm{1}\:\:. \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 18/May/20

another way we have f(x)=(1/(x−2))−(1/(x−1)) =(1/(1−x))−(1/(2−x)) =(1/(1−x))−(1/(2(1−(x/2)))) so for ∣x∣<1 we get f(x) =Σ_(n=0) ^∞ x^n −(1/2)Σ_(n=0) ^∞ ((x/2))^n =Σ_(n=0) ^∞ x^n −Σ_(n=0) ^∞ (x^n /2^(n+1) ) =Σ_(n=0) ^∞ (1−(1/2^(n+1) ))x^n

$$\mathrm{another}\:\mathrm{way}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{2}−\mathrm{x}}\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{2}}\right)}\:\:\mathrm{so}\:\mathrm{for}\:\mid\mathrm{x}\mid<\mathrm{1}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{n}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{n}} −\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\right)\mathrm{x}^{\mathrm{n}} \\ $$

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