52 Difference Equations = 0 in equations (2.56) follow from the relationships obtained by setting k and (2.57); that is, -1 ed 1 + 2! Bn(0). (2.59) - 1 n=0 Comparison of the powers of A gives Bo(0) = 1, B1(0) = –1/½, B2(0) = /, B3(0) = 0, B4(0) = -1/30, B3(0) = 0, B6(0) = /42, .... The corresponding Bernoulli polynomials are Bo(k) = 1, B1 (k) = k – 1/2, B2(k) = k2 – k + 1/6, B3(k) = k³ – 3/2k² + 1/½k, B4(k) = k – 2k3 + k² – 1/30, B3 (k) = k° – 5/2k* + 5/3k³ – 1/sk, B6(k) = k° – 3k³ + 5/½k* – 1/2k² + !/42, B7(k) = k" – 7/2k® + 7/½k5 – 7/sk³ + 1/ok, Bs(k) = k – 4k" + 14/3k® – 7/3k* + ?/3k? – 1/30, %3D |

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Explain the determain 

52
Difference Equations
0 in equations (2.56)
follow from the relationships obtained by setting k
and (2.57); that is,
%3D
:) -E B. (0).
1+
+
(2.59)
2!
3!
n!
n=0
- 1
Comparison of the powers of A gives
Bo(0) = 1, B1(0) = -1/2, B2(0) = /6,
B3 (0) = 0,
B4(0) = -1/30, B5(0) = 0, B6(0) = 1/42, ....
The
rresponding Bernoulli polynomials are
Bo(k) = 1,
B1(k) = k – 1/2,
B2 (k) = k? – k + 6,
B3(k) = k³ – 3/2k² + 1/½k,
B4(k) = k4 – 2k³ + k² – 1/30,
B5 (k) = k³ – 5/2k“ + 5/3k³ – 1/ok,
B6(k) = k® – 3k5 + 5/½k4 – /½k² + /42,
B7(k) = k" – 7/2k° + 7/2k³ – 7/sk³ + 1/ak,
Bs(k) = k® – 4k7 + 14/3k® – 7/3k“ + 2/3k? – 1/30,
-
etc.
It should be clear that the inhomogeneous difference equation
n
Yk+1 – Yk =
2 amkm,
(2.60)
m=0
Transcribed Image Text:52 Difference Equations 0 in equations (2.56) follow from the relationships obtained by setting k and (2.57); that is, %3D :) -E B. (0). 1+ + (2.59) 2! 3! n! n=0 - 1 Comparison of the powers of A gives Bo(0) = 1, B1(0) = -1/2, B2(0) = /6, B3 (0) = 0, B4(0) = -1/30, B5(0) = 0, B6(0) = 1/42, .... The rresponding Bernoulli polynomials are Bo(k) = 1, B1(k) = k – 1/2, B2 (k) = k? – k + 6, B3(k) = k³ – 3/2k² + 1/½k, B4(k) = k4 – 2k³ + k² – 1/30, B5 (k) = k³ – 5/2k“ + 5/3k³ – 1/ok, B6(k) = k® – 3k5 + 5/½k4 – /½k² + /42, B7(k) = k" – 7/2k° + 7/2k³ – 7/sk³ + 1/ak, Bs(k) = k® – 4k7 + 14/3k® – 7/3k“ + 2/3k? – 1/30, - etc. It should be clear that the inhomogeneous difference equation n Yk+1 – Yk = 2 amkm, (2.60) m=0
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