Theorem 1.5. Consider the absolutely convergent series S(x) = ao + a1x + a2x2 + ...+ akx* + ·.., (1.246) where ak is a given function of k. Then S(x) can be expressed in the form xAao + ao S(x) = +.. (1.247) (1 – x)² (1 – x)3 1- x This result is known as Montmort's theorem on infinite summation. Note that if ar is a polynomial in k of degree n, then A"ao will be zero for all m > n and thus a finite number of terms for the series S(x) will occur. 32 Difference Equations Proof. We have = ao + a1x + a2x2 + + akx* + ... .. = (1+xE +x²E² + + x* E* + ·.. · )ao .. = (1 – xE)¯'ao = [1 – x(1+A)]¬'ao %3D -1 1 1 1- x ao (1.248) - 1- x 1 1+ 1 - x + (1 – x)² ao .. 1- x ao xAao +.. (1 – x)² ' (1 – x)3 1- x

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Theorem 1.5. Consider the absolutely convergent series
56
S(x) = ao + a1x+ a2x² +
+ akxk + ..
(1.246)
..
where ak is a given function of k. Then S(x) can be expressed in the form
ao
xAao
S(x)
(1.247)
(1 – x)²
(1 – æ)3
1 - x
This result is known as Montmort's theorem on infinite summation. Note
that if ak is a polynomial in k of degree n, then Amao will be zero for all
m > n and thus a finite number of terms for the series S(x) will occur.
32
Difference Equations
Proof. We have
S(x) = ao + a1x + a2x² + •..
+ akx* + ...
(1+ xE + x² E² + · ..
+ x* Ek + ...·
||
(1 – xE)¯'ao = [1 – x(1+A)]¯'ao
-1
1
1
1 - x
(1.248)
ao
1 - x
1
1+
1- x
ao
1- x
(1 – x)2
xAao
+
ao
1- x
(1 – x)² ' (1 – æ)
Transcribed Image Text:Theorem 1.5. Consider the absolutely convergent series 56 S(x) = ao + a1x+ a2x² + + akxk + .. (1.246) .. where ak is a given function of k. Then S(x) can be expressed in the form ao xAao S(x) (1.247) (1 – x)² (1 – æ)3 1 - x This result is known as Montmort's theorem on infinite summation. Note that if ak is a polynomial in k of degree n, then Amao will be zero for all m > n and thus a finite number of terms for the series S(x) will occur. 32 Difference Equations Proof. We have S(x) = ao + a1x + a2x² + •.. + akx* + ... (1+ xE + x² E² + · .. + x* Ek + ...· || (1 – xE)¯'ao = [1 – x(1+A)]¯'ao -1 1 1 1 - x (1.248) ao 1 - x 1 1+ 1- x ao 1- x (1 – x)2 xAao + ao 1- x (1 – x)² ' (1 – æ)
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