- A Function F X Satisfies 1 X2 F X Xf X F X 0 A If N Is A Positive Integer What Is The Value Of K In T 1 (272.53 KiB) Viewed 25 times
A function f(x) satisfies (1 – x2)f"(x) – xf'(x) + f(x) = 0. (a) If n is a positive integer, what is the value of K in t
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A function f(x) satisfies (1 – x2)f"(x) – xf'(x) + f(x) = 0. (a) If n is a positive integer, what is the value of K in t
A function f(x) satisfies (1 – x2)f"(x) – xf'(x) + f(x) = 0. (a) If n is a positive integer, what is the value of K in the following (6 marks) 2 (1 – x)f(n+2)(x) – X (2n + 1)f(n+1)(x) + K f(n)(x) = 0. (+ = n n n Hint: (89)" (x) = Ï (3) s«>7) g'a-y(s), where () ) = = n! k!(n – k)! k k k=0 (b) Given that f(0) 1 and f'(0) = 0 find the values of f"(0), f(3)0), f(4)(0) and f(5)(0). Hence, determine the Maclaurin series of f(x) as far as the terms in 25. (4 marks)
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