 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-1/4e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

f^{'} (x) = \frac{1}{2} (1 + x)^{- 1/2}

, which is the stated formula at

r = 1

. If the formula holds at

r

, differentiating multiplies its coefficient by

\frac{1}{2} - r = - (2 r - 1) /2

\frac{( 2 r - 2 )!}{2 ^{2 r - 1} ( r - 1 )!} \frac{2 r - 1}{2} = \frac{( 2 r )!}{2 ^{2 r + 1} r !} .

The sign changes, proving by mathematical induction that

f^{(r)} (x) = (- 1)^{r - 1} \frac{( 2 r - 2 )!}{2 ^{2 r - 1} ( r - 1 )!} (1 + x)^{1/2 - r} (r \geq 1) .

x = 1/2

. Its polynomial part is

1 + 2 S_{n}

, where

S_{n}

is the first

n

terms of the requested series, since

\frac{f ^{(r)} ( 0 )}{r !} 2^{- r} = 2 (- 1)^{r - 1} \frac{( 2 r - 2 )!}{8 ^{r} r ! ( r - 1 )!} .

To justify passage to the infinite series, for

0 \leq t \leq 1/2

∣ f^{(n + 1)} (t) ∣ \leq \frac{( 2 n )!}{2 ^{2 n + 1} n !} .

Therefore the integral remainder obeys

∣ R_{n} (1/2) ∣ \leq \frac{( 2 n )!}{2 ^{2 n + 1} ( n ! ) ^{2}} \frac{2 ^{- (n + 1)}}{n + 1} = \frac{( n 2 n )}{2 ^{3 n + 2} ( n + 1 )} \leq \frac{2 ^{- n - 2}}{n + 1} ⟶ 0.

The last inequality follows from

(n 2 n) \leq \sum_{j = 0}^{2 n} (j 2 n) = 4^{n}

, a direct binomial theorem bound. Hence

3/2 = 1 + 2 lim S_{n}

, and

r = 1 \sum \infty (- 1)^{r - 1} \frac{( 2 r - 2 )!}{8 ^{r} r ! ( r - 1 )!} = \frac{6 - 2}{4} .

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