Explain this and the definition from eq(1.35) Now, from the definition of k("), given by equation (1.135), we find on settingn = 0, 1, 2, ..., and expandingk(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 thefactorial functions1 = 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 POLYΝΟΜΙΑLkn leads toWe saw in Example 1.6.1 that taking the first difference of yk =the cumbersome expressionAk" = nk"-1+k"-4 +...+1.(1.134)If we wished to determine the difference of a polynomial function of k, thenthe expressions obtained would be rather complex. We now show that it ispossible to define a particular function of k such that its various differenceshave a simple structure. This function is called the factorial function.Let n be a positive integer and define k(n) ask(n) = k(k – 1)(k – 2) . . · (k – n – 1).(1.135)This expression is read "k, n factorial" and can be rewritten in the formk!k(n)(1.136)(k – n)!"

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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

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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