18.01 PROBLEM SET 3

Hand in to Room 2-108 by 1:00 pm Monday, March 11.

The due date and time are absolute. Late homeworks are accepted until the solutions are posted, but the score will be multiplied by 1/2.

Everything included here may be used later. You do not need to memorize the proofs, but you need to know the facts.

1. A.8/6, 8, 12. 91/20. 94/18. 101/2, 16. 108/11. 112/8. 122/87.

(10 points)

2. Let f(x) = 2 − \frac{50}{x} − 2 · \frac{50^{2}}{x^{2}} . Easy computation shows that f(100) = 1. Use

linear approximation to estimate the value f(101). (3 points)

3. Prove the following statements.

(1) lim_{θ→0} \frac{tan θ}{θ} = 1.

(2) lim_{x→0} \frac{tan^{−1} x}{x} = 1.

(3) Let f(x) = \{ \begin{matrix}\frac{tan^{−1} x}{x} & for x ≠ 0, \\ 1 & for x = 0. \end{matrix} Then f′(x) = \frac{\frac{1}{1 + x^{2}} − f(x)}{x}.

(4) f′′(x) = −\frac{2}{(1 + x^{2})^{2}} − 2 · \frac{f′(x)}{x}.

(5) f(x) is an even function. In the rest of the exercise assume that f′(x) and f′′(x) are continuous functions.

(6) f′(0) = 0.

(7) f′′(0) = −\frac{2}{3}.

(8) The quadratic approximation for \frac{tan^{−1} x}{x} at a = 0 is

\frac{tan^{−1} x}{x} ≈ 1 − \frac{x^{2}}{3} .

(9) tan \frac{π}{12} = 2 − \sqrt{3}. [Hint: Use the double (or the half) angle formula for tangent.]

(10) Substitute x = 2 − \sqrt{3} into (8) to obtain

\frac{π}{12} ≈ \frac{4}{3}(2 − \sqrt{3})(\sqrt{3} − 1).

(12 points)

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