Maclaurin Series

Calculus Level 2

A function f f is defined for all real numbers and satisfies

f ( 0 ) = 1 , f ( 1 ) ( 0 ) = 0 , f ( 2 ) ( 0 ) = 0 , f ( 3 ) ( 0 ) = − 1 , f ( 4 ) ( 0 ) = 0 , f(0) = 1,\quad f^{(1)}(0) = 0,\quad f^{(2)}(0) = 0,\quad f^{(3)}(0) = -1,\quad f^{(4)}(0) = 0,

and, in general, f ( k ) ( 0 ) = ( − 1 ) k f^{(k)}(0) = (-1)^{k} if k k is divisible by 3, and f ( k ) ( 0 ) = 0 f^{(k)}(0) = 0 otherwise.

What is the Maclaurin series of f ? f?

Clarification: In the answer choices, ! ! denotes the factorial function. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

∑ k = 1 ∞ ( − 1 ) k k ! x 3 k \sum_{k=1}^\infty \frac{(-1)^k}{k!} x^{3k} ∑ k = 0 ∞ ( − 1 ) k ( 3 k ) ! x 3 k \sum_{k=0}^\infty \frac{(-1)^k}{(3k)!} x^{3k} ∑ k = 0 ∞ ( − 1 ) 3 k x 3 k \sum_{k=0}^\infty (-1)^{3k} x^{3k} ∑ k = 0 ∞ ( − 1 ) k k ! x 3 k \sum_{k=0}^\infty \frac{(-1)^k}{k!} x^{3k} ∑ k = 1 ∞ ( − 1 ) 3 k x k \sum_{k=1}^\infty (-1)^{3k} x^{k}

Henry Maltby by Henry Maltby , 26, USA

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1 solution

Henry Maltby
May 18, 2016

Relevant wiki: Maclaurin Series

Note that the summation should start at k = 0 k = 0 to include the constant term. The coefficient of x n x^n is ( − 1 ) n n ! \frac{(-1)^n}{n!} , and the only terms that need to be included in the summation are those with degree divisible by 3 3 , e.g. n = 3 k n = 3k . Thus,

f ( x ) = ∑ k = 0 ∞ ( − 1 ) 3 k ( 3 k ) ! x 3 k = ∑ k = 0 ∞ ( − 1 ) k ( 3 k ) ! x 3 k . f(x) = \sum_{k=0}^\infty \frac{(-1)^{3k}}{(3k)!} x^{3k} = \sum_{k=0}^\infty \frac{(-1)^{k}}{(3k)!} x^{3k}.

Note further that f ( x ) = cos ⁡ ( x 3 / 2 ) f(x) = \cos(x^{3/2}) .

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