1. Prove that if

\(-1 < x < 1\) then

\[\frac{1}{\sqrt{1 + x^{2}}} = 1 – \frac{1}{2}x^{2} + \frac{1\cdot3}{2\cdot4}x^{4} – \dots,\quad \frac{1}{\sqrt{1 – x^{2}}} = 1 + \frac{1}{2}x^{2} + \frac{1\cdot3}{2\cdot4}x^{4} + \dots.\]
2. Approximation to quadratic and other surds. Let \(\sqrt{M}\) be a quadratic surd whose numerical value is required. Let \(N^{2}\) be the square nearest to \(M\); and let \(M = N^{2} + x\) or \(M = N^{2} – x\), \(x\) being positive. Since \(x\) cannot be greater than \(N\), \(x/N^{2}\) is comparatively small and the surd \(\sqrt{M} = N\sqrt{1 \pm (x/N^{2})}\) can be expressed in a series \[= N\left\{ 1 \pm \frac{1}{2}\left(\frac{x}{N^{2}}\right) – \frac{1\cdot1}{2\cdot4}\left(\frac{x}{N^{2}}\right)^{2} \pm \dots \right\},\] which is at any rate fairly rapidly convergent, and may be very rapidly so. Thus \[\sqrt{67} = \sqrt{64 + 3} = 8\left\{ 1 + \frac{1}{2}\left(\frac{3}{64}\right) – \frac{1\cdot1}{2\cdot4}\left(\frac{3}{64}\right)^{2} + \dots \right\}.\]

Let us consider the error committed in taking \(8\frac{3}{16}\) (the value given by the first two terms) as an approximate value. After the second term the terms alternate in sign and decrease. Hence the error is one of excess, and is less than \(3^{2}/64^{2}\), which is less than \(.003\).

3. If \(x\) is small compared with \(N^{2}\) then \[\sqrt{N^{2} + x} = N + \frac{x}{4N} + \frac{Nx}{2(2N^{2} + x)},\] the error being of the order \(x^{4}/N^{7}\). Apply the process to \(\sqrt{907}\).

[Expanding by the binomial theorem, we have

\[\sqrt{N^{2} + x} = N + \frac{x}{2N} – \frac{x^{2}}{8N^{3}} + \frac{x^{3}}{16N^{5}},\] the error being less than the numerical value of the next term, viz.

\(5x^{4}/128N^{7}\). Also

\[\frac{Nx}{2(2N^{2} + x)} = \frac{x}{4N} \left(1 + \frac{x}{2N^{2}}\right)^{-1} = \frac{x}{4N} – \frac{x^{2}}{8N^{3}} + \frac{x^{3}}{16N^{5}},\] the error being less than

\(x^{4}/32N^{7}\). The result follows. The same method may be applied to surds other than quadratic surds,

*e.g.* to

\(\sqrt[3]{1031}\).]

4. If \(M\) differs from \(N^{3}\) by less than \(1\) per cent. of either then \(\sqrt[3]{M}\) differs from \(\frac{2}{3}N + \frac{1}{3}(M/N^{2})\) by less than \(N/90,000\).

5. If \(M = N^{4} + x\), and \(x\) is small compared with \(N\), then a good approximation for \(\sqrt[4]{M}\) is \[\frac{51}{56} N + \frac{5}{56}\, \frac{M}{N^{3}} + \frac{27Nx}{14(7M + 5N^{4})}.\] Show that when \(N = 10\), \(x = 1\), this approximation is accurate to \(16\) places of decimals.

6. Show how to sum the series \[\sum_{0}^{\infty} P_{r}(n) \binom{m}{n} x^{n},\] where \(P_{r}(n)\) is a polynomial of degree \(r\) in \(n\).

[Express

\(P_{r}(n)\) in the form

\(A_{0} + A_{1}n + A_{2}n(n – 1) + \dots\) as in

Ex. XC. 7.]

7. Sum the series \(\sum\limits_{0}^{\infty} n \dbinom{m}{n} x^{n}\), \(\sum\limits_{0}^{\infty} n^{2} \dbinom{m}{n} x^{n}\) and prove that \[\sum_{0}^{\infty} n^{3} \binom{m}{n} x^{n} = \{m^{3}x^{3} + m(3m – 1)x^{2} + mx\}(1 + x)^{m-3}.\]