In the Maclaurin expansion for ln(1+x) around 0, what is the coefficient of x^2?

Boost your skills in A Level Further Mathematics Core Pure. Study with flashcards and multiple choice questions, each question is followed by hints and explanations. Get prepared for your exam with confidence!

Multiple Choice

In the Maclaurin expansion for ln(1+x) around 0, what is the coefficient of x^2?

The coefficient of x^2 is found from how the function behaves up to second order around 0. In a Maclaurin series, the x^2 term comes from the second derivative at 0: it is f''(0)/2!. For ln(1+x), f'(x) = 1/(1+x) and f''(x) = -1/(1+x)^2. Evaluating at 0 gives f''(0) = -1, so the x^2 term is (-1)/2 = -1/2. This matches the standard expansion ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ..., valid for |x| < 1. So the coefficient of x^2 is -1/2.

The coefficient of x^2 is found from how the function behaves up to second order around 0. In a Maclaurin series, the x^2 term comes from the second derivative at 0: it is f''(0)/2!.

For ln(1+x), f'(x) = 1/(1+x) and f''(x) = -1/(1+x)^2. Evaluating at 0 gives f''(0) = -1, so the x^2 term is (-1)/2 = -1/2.

This matches the standard expansion ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ..., valid for |x| < 1. So the coefficient of x^2 is -1/2.

More practice questions