By Peter R. Turner (auth.)

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8, 0]. 8]. 3 Show that the equation ex = x + 2 has exactly 2 solutions and find intervals of length 1 containing them. Let f(x) = ex - x - 2. Now f(x) -+ oo as x -+ ± oo and f(O) < 0. Since fis continuous it follows from the IVT that there are at least two solutions. Now f (x) = ex - 1 which is zero only for x = 0. It follows that there is exactly one turning point. Hence the equation has at most, and therefore exactly, two solutions. Furthermore we know that one is positive and one is negative.

10. 841405662 which show convergence to at least 8 decimal places using just 7 iterates. en. It follows immediately that corresponding second differences are also the same. Hence from which it follows that e';;+ 21en+ 2 = 1 - (11en+ 1) 21en+ 2 11 2 en The linear convergence of the original sequence implies that there is some non-zero lei < 1 such that en+ 11en ~c. 15 ~ 1- (c - 1}2 2c + 1 = 0 C2 - Write a program to implement Aitken's method for the solution of an equation using a rearrangement x = g(x).

3 Derive the series expansion ln(1 - x) = - I/: 1 xi/i for jxj < 1. Use this to derive the expansion ln2 = I/: 1 T;li. ) We begin with the geometric series and integrate term-by-term as follows: ln(1 - x) =- J x 0 - 1 1- t dt =_ r 1 + t + t 2 + ... dt 0 = - [t + t 212 + t 3/3 + .. ·] ~ 00 as required. To obtain the desired expansion of ln 2, we observe that In 2 24 = -ln(l/2) and hence L L 00 ln2 = - ln(l/2) = (112Yii = T'li i=l i=l which is the required series. 4). Use the fact that arctan 1/"\13" = n/6 to deduce that n = 2"\13" "' ( 1)n L n-o (2n + 1)3n Estimate rt using the first five terms of this series.