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.