$f(x)=\dfrac{13}{x^2-5x-36}$ fonksiyonunu kuvvet serisi olarak, kısmi kesir açılım yöntemiyle, ifade ediniz.

Question

$f(x)=\dfrac{13}{x^2-5x-36} = \dfrac{1}{x-9}-\dfrac{1}{x+4}$ şeklinde ayırıyorum. Sonra hepsini $\dfrac{1}{1-r}$ formatına getiriyorum.

$\dfrac{1}{1-(10-x)}-\dfrac{1}{1-(-x-3)}$. En son $\displaystyle\sum_{n=0}^\infty=(10-x)^n-(-x-3)^n$ buluyorum.

Comments

$\dfrac{1}{x-9}=-\dfrac{1}{9(1-\frac{x}{9})}$ olarak yazmayi dene..

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Hangi nokta merkezli?

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x=0 merkezli

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Onu da denedim ama olmadı: $(\dfrac{1}{x-9}=-\dfrac{1}{9(1-\frac{x}{9})})$

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$\dfrac{1}{1-x}=\displaystyle\sum_{n=0}^\infty x^n\implies\dfrac{1}{x-9}=-\dfrac{1}{9}\dfrac{1}{1-\dfrac x9}\implies\dfrac{1}{x-9}=-\dfrac{1}{9}\sum_{n=0}^\infty \left(\dfrac x9\right)^n=-\dfrac{1}{9}\sum_{n=0}^\infty9^{-n}x^n=-\sum_{n=0}^\infty9^{-n-1}x^n$

 

Benzer seklide digerini yapip birlestirmeyi dene bakalim.

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Denedim. İşe yaradı, teşekkür ederim.

Answer 1

$f(x)=\dfrac{13}{x^2-5x-36} = \dfrac{1}{x-9}-\dfrac{1}{x+4}$

 

$\dfrac{1}{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

 

$\dfrac{1}{x-9}=-\dfrac{1}{9}\dfrac{1}{1-\dfrac x9}\implies$

 

$\dfrac{1}{x-9}=-\dfrac{1}{9}\displaystyle\sum_{n=0}^\infty \left(\dfrac x9\right)^n=-\dfrac{1}{9}\sum_{n=0}^\infty9^{-n}x^n=-\sum_{n=0}^\infty9^{-n-1}x^n$

 

$\dfrac{1}{x+4}=\dfrac{1}{4}\dfrac{1}{1-\left(-\dfrac{x}{4}\right)}\implies$

 

$-\dfrac{1}{x+4}=-\dfrac{1}{4}\displaystyle\sum_{n=0}^\infty \left(-\dfrac{x}{4}\right)^n=-\dfrac{1}{4}\sum_{n=0}^\infty(-4)^{-n}x^n=\sum_{n=0}^\infty(-4)^{-n-1}x^n$

 

$\displaystyle f(x)=\dfrac{13}{x^2-5x-36} = \dfrac{1}{x-9}-\dfrac{1}{x+4}$

 

$=\displaystyle-\sum_{n=0}^\infty9^{-n-1}x^n+\sum_{n=0}^\infty(-4)^{-n-1}x^n=\sum_{n=0}^\infty\left[-9^{-n-1}+(-4)^{-n-1}\right]x^n$