Frobenijusov metod predstavlja jedan od metoda rešavanja diferencijalnih jednačina drugoga reda oblika:
z
2
u
″
+
p
(
z
)
z
u
′
+
q
(
z
)
u
=
0
{\displaystyle z^{2}u''+p(z)zu'+q(z)u=0\,}
gde su:
u
′
≡
d
u
d
z
{\displaystyle u'\equiv {{du} \over {dz}}}
u
″
≡
d
2
u
d
z
2
{\displaystyle u''\equiv {{d^{2}u} \over {dz^{2}}}}
z=0 . Podelimo li sa z 2 dobijamo diferencijalnu jednačinu:
u
″
+
p
(
z
)
z
u
′
+
q
(
z
)
z
2
u
=
0
{\displaystyle u''+{p(z) \over z}u'+{q(z) \over z^{2}}u=0}
Metoda je dobila ime po nemačkom matematičaru Ferdinandu Frobenijusu .
Prema Frobenijusovoj metodi tražimo rešenje u obliku reda:
u
(
z
)
=
∑
k
=
0
∞
A
k
z
k
+
r
,
(
A
0
≠
0
)
{\displaystyle u(z)=\sum _{k=0}^{\infty }A_{k}z^{k+r},\qquad (A_{0}\neq 0)}
Diferenciranjem dobijamo:
u
′
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z
)
=
∑
k
=
0
∞
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k
+
r
)
A
k
z
k
+
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−
1
{\displaystyle u'(z)=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}}
u
″
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z
)
=
∑
k
=
0
∞
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k
+
r
−
1
)
(
k
+
r
)
A
k
z
k
+
r
−
2
{\displaystyle u''(z)=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}}
Posle toga gore dobiujene redove supstituiramo u diferencijalnu jednačinu i dobijamo:
z
2
∑
k
=
0
∞
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+
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A
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A
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A
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∞
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A
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∑
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∞
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A
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+
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∑
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∞
A
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z
k
+
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=
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k
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0
∞
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+
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−
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)
(
k
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A
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k
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A
k
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A
k
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k
+
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∑
k
=
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∞
[
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)
(
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)
+
p
(
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(
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+
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]
A
k
z
k
+
r
=
[
r
(
r
−
1
)
+
p
(
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)
r
+
q
(
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)
]
A
0
z
r
+
∑
k
=
1
∞
[
(
k
+
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1
)
(
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+
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+
p
(
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)
(
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+
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+
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(
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)
]
A
k
z
k
+
r
{\displaystyle {\begin{aligned}&{}\quad z^{2}\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}+zp(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)(k+r)A_{k}z^{k+r}+q(z)A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\\&=\left[r(r-1)+p(z)r+q(z)\right]A_{0}z^{r}+\sum _{k=1}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\end{aligned}}}
Inicijalni polinom je sledeći izraz:
r
(
r
−
1
)
+
p
(
0
)
r
+
q
(
0
)
=
I
(
r
)
{\displaystyle r\left(r-1\right)+p\left(0\right)r+q\left(0\right)=I(r)\,}
Prema opštoj definiciji inicijalni polinomi su koeficijenti najnižega stepena po z .
Opšti izraz za koeficijente od z k + r
I
(
k
+
r
)
A
k
+
∑
j
=
0
k
−
1
[
(
j
+
r
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p
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−
j
)
+
q
(
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−
j
)
]
A
j
{\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}}
Ti koeficijenti treba da budu jednaki nuli, jer oni treba da predstavljaju rešenja diferencijalne jednačine, pa sledi:
I
(
k
+
r
)
A
k
+
∑
j
=
0
k
−
1
[
(
j
+
r
)
p
(
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−
j
)
+
q
(
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−
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)
]
A
j
=
0
{\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=0}
∑
j
=
0
k
−
1
[
(
j
+
r
)
p
(
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−
j
)
+
q
(
k
−
j
)
]
A
j
=
−
I
(
k
+
r
)
A
k
{\displaystyle \sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=-I(k+r)A_{k}}
1
−
I
(
k
+
r
)
∑
j
=
0
k
−
1
[
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j
+
r
)
p
(
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−
j
)
+
q
(
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−
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)
]
A
j
=
A
k
{\displaystyle {1 \over -I(k+r)}\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=A_{k}}
Gornje rešenje sa A k
U
r
(
z
)
=
∑
k
=
0
∞
A
k
z
k
+
r
{\displaystyle U_{r}(z)=\sum _{k=0}^{\infty }A_{k}z^{k+r}}
i zadovoljava:
z
2
U
r
(
z
)
″
+
p
(
z
)
z
U
r
(
z
)
′
+
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U
r
(
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)
=
I
(
r
)
z
r
{\displaystyle z^{2}U_{r}(z)''+p(z)zU_{r}(z)'+q(z)U_{r}(z)=I(r)z^{r}\!\;}
Odaberemo li jedan od korena inicijalnoga polinoma, tada dobijamo rešenje diferencijalne jednačine.
Pokušamo li da rešimo sledeći diferencijalnu jednačinu:
z
2
f
″
−
z
f
′
+
(
1
−
z
)
f
=
0
{\displaystyle z^{2}f''-zf'+(1-z)f=0\,}
Podelimo li je sa z 2 dobijamo:
f
″
−
1
z
f
′
+
1
−
z
z
2
f
=
f
″
−
1
z
f
′
+
(
1
z
2
−
1
z
)
f
=
0
{\displaystyle f''-{1 \over z}f'+{1-z \over z^{2}}f=f''-{1 \over z}f'+\left({1 \over z^{2}}-{1 \over z}\right)f=0}
Pretpostavljamo rešenja u obliku reda:
f
=
∑
k
=
0
∞
A
k
z
k
+
r
{\displaystyle f=\sum _{k=0}^{\infty }A_{k}z^{k+r}}
f
′
=
∑
k
=
0
∞
(
k
+
r
)
A
k
z
k
+
r
−
1
{\displaystyle f'=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}}
f
″
=
∑
k
=
0
∞
(
k
+
r
)
(
k
+
r
−
1
)
A
k
z
k
+
r
−
2
{\displaystyle f''=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}}
i ta rešenja supstituiramo u gornju jednačinu:
∑
k
=
0
∞
(
k
+
r
)
(
k
+
r
−
1
)
A
k
z
k
+
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−
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−
1
z
∑
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=
0
∞
(
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+
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A
k
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+
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−
1
+
(
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−
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∑
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∞
A
k
z
k
+
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=
∑
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∞
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A
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∑
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∞
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A
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+
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2
∑
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+
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−
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z
∑
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∞
A
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z
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+
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=
∑
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=
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∞
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+
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(
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+
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)
A
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z
k
+
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−
∑
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=
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∞
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+
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A
k
z
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+
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+
∑
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∞
A
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z
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+
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−
∑
k
=
0
∞
A
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z
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+
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−
1
{\displaystyle {\begin{aligned}&\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-{1 \over z}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+\left({1 \over z^{2}}-{1 \over z}\right)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-{1 \over z}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+{1 \over z^{2}}\sum _{k=0}^{\infty }A_{k}z^{k+r}-{1 \over z}\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }A_{k}z^{k+r-1}\end{aligned}}}
Pomeramo indekse poslednje sume, tako da se dobija:
=
∑
k
=
0
∞
(
k
+
r
)
(
k
+
r
−
1
)
A
k
z
k
+
r
−
2
−
∑
k
=
0
∞
(
k
+
r
)
A
k
z
k
+
r
−
2
+
∑
k
=
0
∞
A
k
z
k
+
r
−
2
−
∑
k
−
1
=
0
∞
A
k
−
1
z
k
+
r
−
2
=
∑
k
=
0
∞
(
k
+
r
)
(
k
+
r
−
1
)
A
k
z
k
+
r
−
2
−
∑
k
=
0
∞
(
k
+
r
)
A
k
z
k
+
r
−
2
+
∑
k
=
0
∞
A
k
z
k
+
r
−
2
−
∑
k
=
1
∞
A
k
−
1
z
k
+
r
−
2
{\displaystyle {\begin{aligned}&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k-1=0}^{\infty }A_{k-1}z^{k+r-2}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\end{aligned}}}
Startni indeks za k =0 se posebno piše, pa se dobija:
=
(
(
r
)
(
r
−
1
)
A
0
z
r
−
2
)
+
∑
k
=
1
∞
(
k
+
r
)
(
k
+
r
−
1
)
A
k
z
k
+
r
−
2
−
(
(
r
)
A
0
z
r
−
2
)
−
∑
k
=
1
∞
(
k
+
r
)
A
k
z
k
+
r
−
2
+
(
A
0
z
r
−
2
)
+
∑
k
=
1
∞
A
k
z
k
+
r
−
2
−
∑
k
=
1
∞
A
k
−
1
z
k
+
r
−
2
=
(
r
(
r
−
1
)
−
r
+
1
)
A
0
z
r
−
2
+
∑
k
=
1
∞
(
(
(
k
+
r
)
(
k
+
r
−
1
)
−
(
k
+
r
)
+
1
)
A
k
−
A
k
−
1
)
z
k
+
r
−
2
{\displaystyle {\begin{aligned}&=((r)(r-1)A_{0}z^{r-2})+\sum _{k=1}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-((r)A_{0}z^{r-2})-\sum _{k=1}^{\infty }(k+r)A_{k}z^{k+r-2}\\&{}+(A_{0}z^{r-2})+\sum _{k=1}^{\infty }A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=(r(r-1)-r+1)A_{0}z^{r-2}+\sum _{k=1}^{\infty }\left(((k+r)(k+r-1)-(k+r)+1)A_{k}-A_{k-1}\right)z^{k+r-2}\end{aligned}}}
Jedno rešenje dobijamo rešavanjem inicijalnoga polinoma r (r − 1) − r + 1 = r 2 − 2r + 1 = 0, odnosno dobijamo da je 1 dvostruki koren. Koristeći taj koren koeficijenti od z k + r − 2
(
(
k
+
1
)
(
k
)
−
(
k
+
1
)
+
1
)
A
k
−
A
k
−
1
=
(
k
2
)
A
k
−
A
k
−
1
=
0
{\displaystyle ((k+1)(k)-(k+1)+1)A_{k}-A_{k-1}=(k^{2})A_{k}-A_{k-1}=0\,}
A
k
=
A
k
−
1
k
2
{\displaystyle A_{k}={A_{k-1} \over k^{2}}}
Pošto je omer
A
k
/
A
k
−
1
{\displaystyle A_{k}/A_{k-1}}
Literatura
uredi
![{\displaystyle {\begin{aligned}&{}\quad z^{2}\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}+zp(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)(k+r)A_{k}z^{k+r}+q(z)A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\\&=\left[r(r-1)+p(z)r+q(z)\right]A_{0}z^{r}+\sum _{k=1}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59021a42b9f1dec0512b28467b1abfc140730760)
![{\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7efec331f2e6e77f78455546fc998cbf1e1e21c)
![{\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a218dde8e50f2d768a034f659ef894b89dc21a40)
![{\displaystyle \sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=-I(k+r)A_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6d7682c2b8141a6304aed64358183ee34817b09)
![{\displaystyle {1 \over -I(k+r)}\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=A_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8ff3cd35b251557d0b766f66a2f326b03825140)
