Surdaki yontem kullanilarak farkli bir cozum yapilabilir.
1=(1−x2)∞∑n=0an(x−3)n olsun. n=0 icin
1=(1−x2)a0 ve x=3 icin a0=−18 olur.
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1=(1−x2)∞∑n=0an(x−3)n her iki tarafin turevini alalim.
0=−2x∞∑n=0an(x−3)n+(1−x2)∞∑n=0ann(x−3)n−1 ve n=0,1 icin
0=−2xa0(x−3)0−2xa1(x−3)1+(1−x2)a00(x−3)−1+(1−x2)a11(x−3)0
0=−2xa0−2xa1(x−3)1+(1−x2)a1 ve x=3 icin
0=−6a0−8a1⟹0=−6(−18)−8a1⟹a1=332
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1=(1−x2)∞∑n=0an(x−3)n esitligini asagidaki gibi yazarsak
1=[−8−6(x−3)−(x−3)2]∞∑n=0an(x−3)n
katsayilarin indirgeme formulu soyle olur
a0=−18a1=332 olmak uzere,
∀n≥2−8an−6an−1−an−2=0⟹−8an+2−6an+1−an=0.
−8an+2−6an+1−an=0⟹−8r2−6r−1=0
⟹r1=−12r2=−14
an=(−12)nc1+(−14)nc2
n=0:a0=−18=c1+c2
n=1:a1=332=(−12)c1+(−14)c2
⟹c1=−14 ve c2=18
⟹an=(−12)n(−14)+(−14)n18
11+x2=∞∑n=0an(x−3)n=∞∑n=0[−14(−12)n+18(−14)n](x−3)n
=−18+332(x−3)−7128(x−3)2+15512(x−3)3−312048(x−3)4+638192(x−3)5+⋯
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∞∑n=0[−14(−12)n+18(−14)n](x−3)n
=12[∞∑n=0[−12(−12)n−(−14)(−14)n](x−3)n]
=12[∞∑n=0[(−12)n+1−(−14)n+1](x−3)n]
yazarak bir onceki cozume denk oldugu gosterileilir.