$c=3$ için, $f(x) = \dfrac{1}{1-x^{2}}$ fonksiyonunu Taylor Serisi ile ifade ediniz.

Question

Bu soruya nasıl bir yaklaşım sergilemeliyim?

Comments

$\dfrac{1}{1-x}$ fonksiyonun seri aciliminindan yararlanmalisin..

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İstersen paydayı çarpanlarına ayrıp iki kesir için de deneyebilirsin. Her türlü Ökkeş'in dediği yoldan yap derim.

Answer 1

$f(x)=\dfrac{1}{1-x^2}=\dfrac12\left[\dfrac{1}{1-x}+\dfrac{1}{1+x}\right]$

 

 

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

$c=3:\quad \displaystyle\dfrac{1}{1-x}=\dfrac{1}{-2-(x-3)}=-\dfrac{1}{2}\left(\dfrac{1}{1-\left(\dfrac{x-3}{-2}\right)}\right)=-\dfrac{1}{2}\sum_{n=0}^\infty \left(\dfrac{x-3}{-2}\right)^n=-\dfrac{1}{2}\sum_{n=0}^\infty \left(-\dfrac{1}{2}\right)^n(x-3)^n=\sum_{n=0}^\infty \left(-\dfrac{1}{2}\right)^{n+1}(x-3)^n$

$c=3:\quad \displaystyle\dfrac{1}{1+x}=\dfrac{1}{4+(x-3)}=\dfrac{1}{4}\left(\dfrac{1}{1+\left(\dfrac{x-3}{4}\right)}\right)=\dfrac{1}{4}\left(\dfrac{1}{1-\left(\dfrac{x-3}{-4}\right)}\right)=\dfrac{1}{4}\sum_{n=0}^\infty \left(\dfrac{x-3}{-4}\right)^n=\dfrac{1}{4}\sum_{n=0}^\infty \left(-\dfrac{1}{4}\right)^n(x-3)^n=-\sum_{n=0}^\infty \left(-\dfrac{1}{4}\right)^{n+1}(x-3)^n$

 

$\displaystyle f(x)=\dfrac{1}{1-x^2}=\dfrac12\left[\sum_{n=0}^\infty \left(-\dfrac{1}{2}\right)^{n+1}(x-3)^n-\sum_{n=0}^\infty \left(-\dfrac{1}{4}\right)^{n+1}(x-3)^n\right]=\dfrac12\left[\sum_{n=0}^\infty\left[\left(-\dfrac{1}{2}\right)^{n+1}-\left(-\dfrac{1}{4}\right)^{n+1}\right] (x-3)^n\right]$

Answer 2

Surdaki   yontem kullanilarak farkli bir cozum yapilabilir.

$1=(1-x^2)\displaystyle\sum_{n=0}^\infty a_n(x-3)^n$ olsun.  $n=0$ icin

 

$1=(1-x^2)a_0$  ve $x=3$ icin $a_0=-\dfrac{1}{8}$ olur.

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$1=(1-x^2)\displaystyle\sum_{n=0}^\infty a_n(x-3)^n$ her iki tarafin turevini alalim.

 

$0=-2x\displaystyle\sum_{n=0}^\infty a_n(x-3)^n+(1-x^2)\displaystyle\sum_{n=0}^\infty a_nn(x-3)^{n-1}$ ve   $n=0,1$ icin

 

$0=-2xa_0(x-3)^0-2xa_1(x-3)^1+(1-x^2)a_00(x-3)^{-1}+(1-x^2)a_11(x-3)^{0}$

 

$0=-2xa_0-2xa_1(x-3)^1+(1-x^2)a_1$ ve $x=3$ icin

 

$0=-6a_0-8a_1\implies 0=-6\left(-\dfrac{1}{8}\right)-8a_1\implies a_1=\dfrac{3}{32}$

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$1=(1-x^2)\displaystyle\sum_{n=0}^\infty a_n(x-3)^n$ esitligini asagidaki gibi yazarsak

$1=[-8-6(x-3)-(x-3)^2]\displaystyle\sum_{n=0}^\infty a_n(x-3)^n$

 

katsayilarin indirgeme formulu soyle olur

 

$a_0=-\dfrac{1}{8}\quad a_1=\dfrac{3}{32}$ olmak uzere,

 

$\forall n\geq2 \quad -8a_n-6a_{n-1}-a_{n-2}=0\implies -8a_{n+2}-6a_{n+1}-a_n=0$.

$-8a_{n+2}-6a_{n+1}-a_n=0\implies -8r^2-6r-1=0$

 

$\implies r_1 =-\dfrac{1}{2}\, r_2 =-\dfrac{1}{4}$

 

$a_n=\left(-\dfrac{1}{2}\right)^nc_1+\left(-\dfrac{1}{4}\right)^nc_2$

 

$n=0: \quad a_0=-\dfrac{1}{8}=c_1+c_2$

$n=1:\quad a_1=\dfrac{3}{32}=\left(-\dfrac{1}{2}\right)c_1+\left(-\dfrac{1}{4}\right)c_2$

 

$\implies c_1=-\dfrac{1}{4}$  ve $c_2=\dfrac{1}{8}$

 

$\implies a_n=\left(-\dfrac{1}{2}\right)^n\left(-\dfrac{1}{4}\right)+\left(-\dfrac{1}{4}\right)^n\dfrac{1}{8}$

 

$\dfrac1{1+x^2}=\displaystyle\sum_{n=0}^\infty a_n(x-3)^n=\sum_{n=0}^\infty \left[-\dfrac{1}{4}\left(-\dfrac{1}{2}\right)^n+\dfrac{1}{8}\left(-\dfrac{1}{4}\right)^n\right](x-3)^n$

 

$=-\dfrac{1}{8}+\dfrac{3 }{32}(x-3)-\dfrac{7}{128}    (x-3)^2+\dfrac{15}{512} (x-3)^3-\dfrac{31    }{2048}(x-3)^4+\dfrac{63}{8192}(x-3)^5+\cdots$

 

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$\displaystyle\sum_{n=0}^\infty \left[-\dfrac{1}{4}\left(-\dfrac{1}{2}\right)^n+\dfrac{1}{8}\left(-\dfrac{1}{4}\right)^n\right](x-3)^n$

 

$=\displaystyle\dfrac12\left[\sum_{n=0}^\infty \left[-\dfrac{1}{2}\left(-\dfrac{1}{2}\right)^n-\left(-\dfrac{1}{4}\right)\left(-\dfrac{1}{4}\right)^n\right](x-3)^n\right]$

 

$=\displaystyle\dfrac12\left[\sum_{n=0}^\infty \left[\left(-\dfrac{1}{2}\right)^{n+1}-\left(-\dfrac{1}{4}\right)^{n+1}\right](x-3)^n\right]$

 

yazarak bir onceki cozume denk oldugu gosterileilir.