Mathematics

Problem

Which statement best explains why the Maclaurin polynomial for

\cos x

contains only even powers of

x

? a) Because

\cos x

is a periodic function, its Maclaurin series repeats every

2\pi

, which forces all odd powers of

x

to cancel. b) Only even–order derivatives are used when building a Maclaurin polynomial, so odd–power terms never appear. c) All odd–order derivatives of

\cos x

evaluated at

x=0

are equal to

0

. d) Because

\cos x

is even, it cannot have non-zero derivatives at

x=0

, so all coefficients of odd powers vanish.

Intuitive explanation

The key is recognizing that $\cos x$ is an even function—it satisfies $f(-x) = f(x)$ . When you build a Maclaurin polynomial by taking derivatives and evaluating at $x=0$ , all odd derivatives of $\cos x$ equal zero at that point, so their coefficients vanish. This forces the polynomial to contain only even powers, perfectly mirroring the symmetry of the original function.

Intuitive explanation

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