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Now, from the definition of k(m), given by equation (1.135), we find on setting n = 0, 1, 2, ..., and expanding k(0) = 1, %3D k(1) = k, %3D k(2) = k2 – k, | k(3) = k3 – 3k2 + 2k, (1.166) k(4) = k4 – 6k3+ 11k2 – 6k, k(5) = k5 – 10k4 + 35k3 – 50k2 + 24k, etc. These relations can be inverted to give the various powers of k in terms of the factorial functions 1 = k(0), k = k(1), k2 = k(2) + k(1), 3 = k(3) + 3k(2) + k(1), (1.167) k4 = k(4) + 7k(3) + 6k(2) + k(1), k5 = k(5) + 15k(4) + 25k(3) + 10k(2) + k(1), etc.

Now, from the definition of k(m), given by equation (1.135), we find on setting n = 0, 1, 2, ..., and expanding k(0) = 1, %3D k(1) = k, %3D k(2) = k2 – k, | k(3) = k3 – 3k2 + 2k, (1.166) k(4) = k4 – 6k3+ 11k2 – 6k, k(5) = k5 – 10k4 + 35k3 – 50k2 + 24k, etc. These relations can be inverted to give the various powers of k in terms of the factorial functions 1 = k(0), k = k(1), k2 = k(2) + k(1), 3 = k(3) + 3k(2) + k(1), (1.167) k4 = k(4) + 7k(3) + 6k(2) + k(1), k5 = k(5) + 15k(4) + 25k(3) + 10k(2) + k(1), etc.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain this and the definition from eq(1.35)

Now, from the definition of k("), given by equation (1.135), we find on setting n = 0, 1, 2, ..., and expanding k(0) = 1, k(1) = k, k(2) = k2 – k, k(3) = k3 – 3k2 + 2k, (1.166) k(4) = k4 – 6k³ + 11k² – 6k, k(5) = k5 – 10k4 + 35k³ – 50k² + 24k, etc. These relations can be inverted to give the various powers of k in terms of the factorial functions 1 = k(0), k = k(1), k² = k(2) + k(1), 3 = k(3) + 3k(2) + k(1), (1.167) k4 = k(4) + 7k(3) + 6k(2) + k(1). _k = k(5) + 15k(4) + 25k(3) + 10k(2) + k(1), etc.
Transcribed Image Text:Now, from the definition of k("), given by equation (1.135), we find on setting n = 0, 1, 2, ..., and expanding k(0) = 1, k(1) = k, k(2) = k2 – k, k(3) = k3 – 3k2 + 2k, (1.166) k(4) = k4 – 6k³ + 11k² – 6k, k(5) = k5 – 10k4 + 35k³ – 50k² + 24k, etc. These relations can be inverted to give the various powers of k in terms of the factorial functions 1 = k(0), k = k(1), k² = k(2) + k(1), 3 = k(3) + 3k(2) + k(1), (1.167) k4 = k(4) + 7k(3) + 6k(2) + k(1). _k = k(5) + 15k(4) + 25k(3) + 10k(2) + k(1), etc.
1.7 FACTORIAL POLYΝΟΜΙΑL kn leads to We saw in Example 1.6.1 that taking the first difference of yk = the cumbersome expression Ak" = nk"-1+ k"-4 +...+1. (1.134) If we wished to determine the difference of a polynomial function of k, then the expressions obtained would be rather complex. We now show that it is possible to define a particular function of k such that its various differences have a simple structure. This function is called the factorial function. Let n be a positive integer and define k(n) as k(n) = k(k – 1)(k – 2) . . · (k – n – 1). (1.135) This expression is read "k, n factorial" and can be rewritten in the form k! k(n) (1.136) (k – n)!"
Transcribed Image Text:1.7 FACTORIAL POLYΝΟΜΙΑL kn leads to We saw in Example 1.6.1 that taking the first difference of yk = the cumbersome expression Ak" = nk"-1+ k"-4 +...+1. (1.134) If we wished to determine the difference of a polynomial function of k, then the expressions obtained would be rather complex. We now show that it is possible to define a particular function of k such that its various differences have a simple structure. This function is called the factorial function. Let n be a positive integer and define k(n) as k(n) = k(k – 1)(k – 2) . . · (k – n – 1). (1.135) This expression is read "k, n factorial" and can be rewritten in the form k! k(n) (1.136) (k – n)!"
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