Find a function whose maclaurin expansion is 1 + x3 + x6 /2! + x9 /3! + x12 /4! +...
Solution:
Given Maclaurin expansion is 1 + x3 + x6 /2! + x9 /3! + x12 /4! +... …….(1)
Maclaurin’s series is given as;
“If f(x) can be expanded as an infinite series then,
f(x) = f(0) + x.f'(0) + x²/ 2!. f''(0) + x³/3!.f'''(0) + x⁴/4!.f''''(0) +...…….(2)
We have Maclaurin expansion of ex = 1 + x + x²/2! + x³/3! + x4/4! +.... ………(3)
Differentiation of ex = ex = f'(x) = f''(x) = f'''(x) = f''''(x) …
Also f'(0) = f''(0) = f'''(0) = f''''(0) = e0 = 1.
Comparing equations (1) and (3), let each term in the expansion of ex be raised to e3x,
⇒ e3x = 1 +x3 + (x3)2/2! + (x3)3/3! + (x3)4/4! +....+ (x3)n/n!
e3x = 1 + x3 + x6/2! + x9/3! + x12/4! +...+ (x3)n/n!
Find a function whose maclaurin expansion is 1 + x3 + x6 /2! + x9 /3! + x12 /4! +...
Summary:
The function whose maclaurin expansion is 1 + x3 + x6/2! + x9/3! + x12/4! +...+ (x3)n/n! is e3x.
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