Problems

Solution:

1. Note that, \begin{align*} && \sum_{r = 0}^\infty x^r &= \frac{1}{1-x} && |x| < 1\\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \sum_{r = 0}^\infty rx^{r-1} &= \frac{1}{(1-x)^2} && |x| < 1\\ && \sum_{r = 1}^\infty rx^{r-1} &= \frac{1}{(1-x)^2} && |x| < 1\\ \end{align*}
2. 1. \begin{align*} && \mathbb{P}(\text{Ali wins in }s\text{ rounds}) &= \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\ \Rightarrow && \mathbb{P}(\text{Ali wins}) &= \sum_{s=1}^\infty \mathbb{P}(\text{Ali wins in }s\text{ rounds}) \\ &&&=\sum_{s=1}^\infty \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\ &&&= \frac{1}{2k} \sum_{s=0}^\infty \left ( \frac{k-1}{2k} \right)^{s} \\ &&&= \frac{1}{2k} \frac{1}{1 - \frac{k-1}{2k}} \\ &&&= \frac{1}{2k - (k-1)} \\ &&&= \frac{1}{k+1} \end{align*}
   2. \begin{align*} \mathbb{E}(\text{Ali score}) &= \sum_{s=1}^{\infty} s \mathbb{P}(\text{Ali wins in }s\text{ rounds}) \\ &= \sum_{s=1}^{\infty} s \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\ &= \frac{1}{2k} \frac{1}{\left (1 - \frac{k-1}{2k} \right)^2} \\ &= \frac{2k}{(k+1)^2} \end{align*}
3. \begin{align*} && (1+x)^{n} &= \sum_{k=0}^n \binom{n}{k} x^k \\ \Rightarrow && x(1+x)^n &= \sum_{k=0}^n \binom{n}{k} x^{k+1} \\ \Rightarrow && (1+x)^n + nx(1+x)^{n-1} &= \sum_{k=0}^n (k+1)\binom{n}{k} x^k \\ \Rightarrow && (1+x)^{n-1}(1+(n+1)x) &= 1 + 2\binom{n}{1}x + 3\binom{n}{2}x^2 + \ldots + (n+1)x^n \end{align*}
4. \begin{align*} \mathbb{E}(\text{Zen score}) &= \sum_{s=1}^{2k} s \mathbb{P} \left ( \text{Zen gets }s\text{ numbers in increasing order ending with }2k \right) \\ &= \sum_{s=1}^{2k} s \binom{2k-1}{s-1} \frac{1}{(2k)^s} \\ &= \frac{1}{2k}\sum_{s=0}^{2k-1} (s+1) \binom{2k-1}{s} \frac{1}{(2k)^s} \\ &= \frac{1}{2k} \left ( 1 + \frac{1}{2k} \right)^{2k-2} \left ( 1 + (2k-1+1) \frac{1}{2k} \right) \\ &= \frac{1}{k}\left ( 1 + \frac{1}{2k} \right)^{2k-2} \end{align*}
5. Therefore as \(k \to \infty\) \begin{align*} \frac{\mathbb{E}(\text{Zen score})}{\mathbb{E}(\text{Ali score}) } &= \frac{1}{k}\left ( 1 + \frac{1}{2k} \right)^{2k-2} \big / \frac{2k}{(k+1)^2} \\ &= \frac{(k+1)^2}{2k^2} \cdot \left ( 1 + \frac{1}{2k} \right)^{2k} \cdot \left ( 1 + \frac{1}{2k} \right)^{-2} \\ &\to \frac12 e \approx 2.7/2 = 1.35 \end{align*} ie Zen's expected earnings are \(\approx 35\%\) more.

Solution:

1. \(\,\)
   

   + - ↺
   \begin{align*} && y & = x^{1/x} = \exp \left ( \tfrac1x \ln x \right ) \\ \Rightarrow && y' &= \exp \left ( \tfrac1x \ln x \right ) \cdot \left ( \frac{1}{x^2} - \frac{\ln x}{x^2} \right ) \\ &&&= \frac{\exp \left ( \tfrac1x \ln x \right ) }{x^2}(1 - \ln x) \\ y' =0: && x &= e \end{align*} Therefore the largest integer values will be \(2\) or \(3\). Comparing \((2^{\frac12})^6 = 8 < 9 = (3^{1/3})^6\) we see the maximum value of \(n^{1/n}\) where \(n\) is an integer is \(\sqrt[3]{3}\)
2. The number of tests is \(r\) plus however many groups fail times \(k\). The probability of a group failing is \(g = 1-(1-p)^k\) and the number of failing groups is \(\sim B(r, g)\) so the expected number of additional groups is \(rg\) and the expected total number of tests is \[ \frac{N}{k} + N(1-(1-p)^k) \]
3. \(\,\) \begin{align*} && N &\geq E = \frac{N}{k} + N(1-(1-p)^k) \\ \Rightarrow && 1 &\geq \frac{1}{k} + 1-(1-p)^k \\ \Rightarrow && (1-p)^k &\geq \frac1k \\ \Rightarrow && k \ln(1-p) &\geq - \ln k \\ \Rightarrow && \ln(1 - p) &\geq -\frac{1}{k} \ln k \geq -\frac13 \ln 3 \\ \Rightarrow && 1-p &\geq \frac{1}{\sqrt[3]{3}} \\ \Rightarrow && p &\leq 1-\frac{1}{\sqrt[3]{3}} \end{align*} (taking \(k=3\)) Claim: \(1 - \frac{1}{\sqrt[3]{3}} > \frac14\) Proof: This is equivalent to \(\sqrt[3]{3} > \frac43\) or \(81 > 4^3 = 64\) which is clearly true.
4. If \(pk\) is small then \((1-p)^k \approx 1 - pk\) and so we obtain \(N \left ( \frac1k +pk \right)\) as required. If \(p = 0.01\) and \(k = 10\) then \(\frac{1}{10} + 0.01 \cdot 10 = 0.2\) so the expected number of tests is \(\sim 20\%\) of \(N\)

Solution:

1. The probability no-one picks the winning number is \(\left ( 1 - \frac{1}{N}\right)^N \approx \frac1e\). \begin{align*} && \mathbb{E}(\text{profit}) &= Nc - (1-e^{-1})J \\ &&& < Nc -(1- \tfrac12 )J \\ &&& < Nc - \frac12 J \\ &&&= \frac{2Nc-J}{2} \end{align*} Therefore if \(J = 2Nc\) the expected profit is negative.
2. \(\,\) \begin{align*} && 1 &= \sum_{\text{all numbers}} \mathbb{P}(\text{pick }i) \\ &&&= \sum_{\text{popular numbers}} \mathbb{P}(\text{pick }i)+\sum_{\text{unpopular numbers}} \mathbb{P}(\text{pick }i) \\ &&&=\gamma N \frac{a}{N} + (1-\gamma)N \frac{b}{N} \\ &&&= \gamma a + (1-\gamma)b \end{align*} \begin{align*} && \mathbb{P}(\text{no-one picks winning number}) &= \mathbb{P}(\text{no-one picks winning number} | \text{winning number is popular})\mathbb{P})(\text{winning number is popular}) + \\ &&&\quad + \mathbb{P}(\text{no-one picks} | \text{unpopular})\mathbb{P}(\text{unpopular}) \\ &&&= \left (1 - \frac{a}{N} \right)^N \gamma + \left (1 - \frac{b}{N} \right)^N (1-\gamma) \\ &&&\approx \gamma e^{-a} + (1-\gamma)e^{-b} \\ \\ && \mathbb{E}(\text{profit}) &= Nc - (1-\gamma e^{-a} - (1-\gamma)e^{-b})J \\ &&&= Nc-J+J\gamma e^{-a} +J(1-\gamma)e^{-b} \end{align*} If \(\gamma = \frac18\) and \(a=9b\), then \(1=\frac18 a + \frac78b = 2b \Rightarrow b = \frac12, a = \frac92\) and \begin{align*} && \mathbb{E}(\text{profit}) &= Nc-J +J\tfrac18e^{-9/2}+J\tfrac78e^{-1/2} \\ &&&= Nc-J+\tfrac18Je^{-1/2}(e^{-4}+7) \end{align*} If we can show \(e^{-1/2}\frac{e^{-4}+7}{8} > \frac12\) we'd be done, so \begin{align*} && e^{-1/2}\frac{e^{-4}+7}{8} &> \frac12 \\ \Leftrightarrow && e^{-4}+7 &>4e^{1/2} \\ \Leftrightarrow && 49+14e^{-4}+e^{-8} &>16e \\ \end{align*} But clearly the LHS \(>49\) and the RHS \(<48\) so we're done

Solution: \begin{align*} && \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\ &&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\ &&&= \frac{1}{m-\frac12} \end{align*}

+ - ↺

1. \(\,\) \begin{align*} E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\ E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\ E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\ &= \frac54+\frac{2}{5} = \frac{33}{20} \end{align*}
2. The error is \begin{align*} && \epsilon &= \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\ &&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\ &&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\ &&&= \frac{\frac14}{r^2(r^2-\frac14)} \\ &&&\approx \frac{1}{4r^4} \end{align*} Therefore \begin{align*} && \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2} \right) + 1 + \frac{1}{2^4}\\ &&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 1.0825 \approx 1.08 \end{align*}

Solution: \begin{align*} && s &= ut + \frac12at^2 \\ && D &= \frac12gt_D^2 \\ \Rightarrow && t_D &= \sqrt{\frac{2D}{g}} \\ \Rightarrow && t_{D-\delta} &= \sqrt{\frac{2(D-\delta}{g}} \end{align*} Therefore the difference in times of what I hear will be: \begin{align*} t &= \underbrace{\sqrt{\frac{2D}{g}}}_{\text{time for first stone to hit water}} + \underbrace{\frac{D}{u}}_{\text{time to hear about it}} - \left (\underbrace{\sqrt{\frac{2(D-\delta)}{g}}}_{\text{time for second stone to hit water}} + \underbrace{\frac{D-\delta}{u}}_{\text{time to hear about it}} \right) \\ &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \end{align*} \begin{align*} && t &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \\ \Rightarrow && \beta &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} \\ && \beta^2 &= \frac{2D}{g} + \frac{2(D-\delta)}{g} - \frac{4}{g}\sqrt{D(D-g)} \\ &&&= \frac{4D}{g} - \frac{2\delta}{g} - \frac{4}{g} \sqrt{D(D-\delta)}\\ \Rightarrow && g\beta^2 &= 4D-2\delta -4\sqrt{D(D-\delta)}\\ \Rightarrow && (g \beta^2-4D+2\delta)^2 &= 16D(D-\delta) \\ \Rightarrow && g^2\beta^4 + 16D^2 + 4\delta^2 -8g\beta^2D+4g\beta^2 \delta -16D\delta &= 16D^2-16D\delta \\ \Rightarrow && 8g\beta^2D &= g\beta^4 +4\delta^2 +4g\beta^2 \delta \\ \Rightarrow && D &= \frac1{8g\beta^2} \left ( g^2\beta^4 +4\delta^2 +4g\beta^2 \delta\right) \\ &&&= \frac1{8g\beta^2} \left ( g\beta^2 +2\delta \right)^2 \\ &&&= \frac12g \left ( \frac{\beta}{2} + \frac{\delta}{g\beta} \right)^2 \end{align*} If \(u = 300, g = 10, t = \frac15, \delta = 10\), then \begin{align*} && \beta &= \frac15-\frac{10}{300}\\ &&&= \frac15 - \frac1{30} \\ &&&= \frac{1}{6}\\ && D &= \frac12 \cdot 10 \left ( \frac1{12} + 6 \right)^2 \\ &&&= 5\cdot (36 + 1 + \frac{1}{12^2}) \\ &&&\approx 37 \cdot 5 = 185 \end{align*}

Solution: \begin{align*} \P(X = r) &= \P(X_1 = r, X_2 \leq r) + \P(X_2 = r, X_1 < r) \\ &= \P(X_1 = r) \P(X_2 \leq r) + \P(X_2 = r)\P( X_1 < r) \\ &= \frac{1}{N} \frac{r}{N} + \frac{1}{N} \frac{r-1}{N} \\ &= \frac{2r-1}{N^2} \end{align*} \begin{align*} \E(X) &= \sum_{r=1}^N r \P(X = r) \\ &= \sum_{r=1}^N \frac{2r^2 - r}{N^2} \\ &= \frac{1}{N^2} \l \frac{N(N+1)(2N+1)}{3} - \frac{N(N+1)}{2} \r \\ &= \frac{N+1}{N} \l \frac{4N-1}{6} \r \end{align*} When \(N = 100\), this is equal to \(\frac{101 \cdot 399}{6 \cdot 100} = \frac{101 \cdot 133}{200} = 67.165\) \begin{align*} &&\frac12 &\leq \P(X \leq m) \\ &&&=\sum_{r=1}^m \P(X=r) \\ &&&=\sum_{r=1}^m \frac{2r-1}{N^2} \\ &&&= \frac{1}{N^2} \l m(m+1) - m \r \\ &&&= \frac{m^2}{N^2} \\ \Rightarrow && m^2 &\geq \frac{N^2}{2} \\ \Rightarrow && m &\geq \frac{N}{\sqrt{2}} \\ \Rightarrow && m &= \left \lceil \frac{N}{\sqrt{2}} \right \rceil \end{align*} When \(N = 100\), \(100/\sqrt{2} = \sqrt{2}50\). \(\sqrt{2} > 1.4 \Rightarrow 50\sqrt{2} > 70\) \(\sqrt{2} < 1.42 \Rightarrow 50 \sqrt{2} < 71\), therefore \(\displaystyle \left \lceil \frac{100}{\sqrt{2}} \right \rceil = 71\) \begin{align*} \lim_{N \to \infty} \frac{\frac{(N+1)(4N-1)}{6N}}{ \left \lceil\frac{N}{\sqrt{2}} \right \rceil} &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l \frac{4N^2 +3N - 1}{2N^2} \r \tag{since the floor will be irrelevant}\\ &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l 2 + \frac{3}{2N} - \frac{1}{N^2} \r \\ &= \lim_{N \to \infty} \frac{2\sqrt{2}}{3} \end{align*}

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(\text{Arthur took wicket and exactly one wicket}) &= \binom{30}{1} \frac{1}{36} \left ( \frac{35}{36} \right)^{29} \binom{30}{0} \left ( \frac{24}{25} \right)^{30} \binom{30}{0} \left ( \frac{40}{41} \right)^{30}\\ &&&= \frac{30 \cdot 35^{29} \cdot 24^{30} \cdot 40^{30}}{36^{30} \cdot 25^{30} \cdot {41}^{30}}\\ &&&= \frac{1}{35} N\\ && \mathbb{P}(\text{B took wicket and exactly one wicket}) &= \binom{30}{0}\left ( \frac{35}{36} \right)^{30} \binom{30}{1} \frac{1}{25} \left ( \frac{24}{25} \right)^{29} \binom{30}{0} \left ( \frac{40}{41} \right)^{30}\\ &&&= \frac{1}{24} N \\ && \mathbb{P}(\text{C took wicket and exactly one wicket}) &= \binom{30}{0}\left ( \frac{35}{36} \right)^{30} \binom{30}{0}\left ( \frac{24}{25} \right)^{30} \binom{30}{1} \frac{1}{41} \left ( \frac{40}{41} \right)^{29}\\ &&&= \frac{1}{40} N \\ && \mathbb{P}(\text{Arthur took wicket} | \text{exactly one wicket}) &= \frac{ \mathbb{P}(\text{Arthur took wicket and exactly one wicket}) }{ \mathbb{P}(\text{exactly one wicket}) } \\ &&&= \frac{ \frac{1}{35} N}{\frac1{35} N + \frac{1}{24}N + \frac{1}{40} N} \\ &&&= \frac{3}{10} \end{align*} Alternatively, we could look at: \begin{align*} && \mathbb{P}(X_A = 1 | X_A + X_B + X_C =1) &= \frac{\mathbb{P}(X_A = 1, X_B = 0,X_C = 0)}{\mathbb{P}(X_A = 1, X_B = 0,X_C = 0)+\mathbb{P}(X_A = 0, X_B = 1,X_C = 0)+\mathbb{P}(X_A = 0, X_B = 0,X_C = 1)} \\ &&&= \frac{\frac{\mathbb{P}(X_A = 1)}{\mathbb{P}(X_A=0)}}{\frac{\mathbb{P}(X_A = 1)}{\mathbb{P}(X_A=0)}+\frac{\mathbb{P}(X_B = 1)}{\mathbb{P}(X_B=0)}+\frac{\mathbb{P}(X_C = 1)}{\mathbb{P}(X_C=0)}} \end{align*} and we can calculate these relatively likelihoods in a similar way to above.
2. \(\,\) \begin{align*} && \mathbb{E}(\text{number of wickets}) &= \mathbb{E} \left ( \sum_{i=1}^{90} \mathbb{1}_{i\text{th ball is a wicket}} \right) \\ &&&= \sum_{i=1}^{90} \mathbb{E} \left (\mathbb{1}_{i\text{th ball is a wicket}} \right) \\ &&&= 30 \cdot \frac{1}{36} + 30 \cdot \frac{1}{25} + 30 \cdot \frac{1}{41} \\ &&&\approx 1 + 1 + 1 = 3 \end{align*} We might model the number of wickets taken as \(Po(\lambda)\), where \(\lambda\) is the average number of wickets taken. We can think of this roughly as the Poisson approximation to the binomial where \(N\) is large and \(Np\) is small. Assuming we use \(Po(3)\) we have \begin{align*} && \mathbb{P}(\text{at least 5 wickets}) &= 1-\mathbb{P}(\text{4 or fewer wickets}) \\ &&&= 1- e^{-3} \left (1 + \frac{3}{1} + \frac{3^2}{2} + \frac{3^3}{6} + \frac{3^4}{24} \right) \\ &&&= 1 - \frac{1}{20} \left ( 1 + 3 + \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) \\ &&&= 1 - \frac{1}{20} \left (13 + 3\tfrac38 \right) \\ &&&\approx 1 - \frac{16}{20} = \frac15 \end{align*}

Solution:

1. Case \(k \neq 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \left [\frac{1}{k}(x+1)^k \right]_0^1 \\ &&&= \frac{2^k-1}{k} \\ \end{align*} Case \(k = 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \int_0^1 (x+1)^{-1} \d x \\ &&&= \left [\ln(x+1) \right]_0^1 \\ &&&= \ln 2 \end{align*} Therefore for \(k \approx 0\), we must have both integrals being close to each other, since the function is nice on this interval, ie \(\frac{2^k-1}{k} \approx \ln 2\)
2. Case \(m = 0\). \(I = \frac12\) Case \(m \neq 0, -1, -2\) \begin{align*} u = x+1, \d u = \d x && \int_0^1 x(x+1)^m \d x &= \int_{u=1}^{u=2} (u-1)u^m \d u \\ &&&=\left[ \frac{u^{m+2}}{m+2} - \frac{u^{m+1}}{m+1} \right]_1^2 \\ &&&= 2^{m+1}\left ( \frac{2}{m+2} - \frac1{m+1} \right) - \frac{1}{m+2} + \frac{1}{m+1} \\ &&&= 2^{m+1} \frac{m}{(m+1)(m+2)} + \frac{1}{(m+1)(m+2)} \\ &&&= \frac{m2^{m+1}+1}{(m+1)(m+2)} \\ \end{align*} Case \(m = -1\). \begin{align*} && \int_0^1 \frac{x}{x+1} \d x &= \int_0^1 1 - \frac{1}{x+1} \d x \\ &&&= 1 - \ln2 \\ \end{align*} Case \(m = -2\): \begin{align*} && \int_0^1 \frac{x}{(x+1)^2} \d x &= \int_0^1\frac{x+1-1}{(x+1)^2} \d x \\ &&&= \left [ \ln (x+1) +(1+x)^{-1} \right]_0^1 \\ &&&= \ln 2 + \frac12 - 1 \\ &&&= \ln 2 - \frac12 \end{align*}

Solution: First notice that the the probability a page contains fewer than two errors is \(\mathbb{P}(X < 2)\) where \(X \sim Po(\lambda)\), ie \(\mathbb{P}(X<2) = e^{-\lambda} + \lambda e^{-\lambda} = (1+\lambda)e^{-\lambda}\). Therefore the number of pages \(Y\) with fewer than two errors out of our sample of \(n\) is \(Bin(n, p)\) where \(p\) is as before. ie \(\mathbb{P}(Y = k) = \binom{n}{k} p^kq^{n-k}\).

1. \(\,\) \begin{align*} && q &= 1- p = 1-(1+\lambda)e^{-\lambda} \\ &&&= 1 - (1+ \lambda)(1 - \lambda + \tfrac12 \lambda^2 + o(\lambda^3)) \\ &&&= 1 - 1+ \lambda - \lambda+\lambda^2 - \tfrac12 \lambda^2 + o(\lambda^3) \\ &&&= \tfrac12 \lambda^2 + o(\lambda^3) \end{align*}
2. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= p^n \\ &&&= (1+\lambda)^ne^{-\lambda n} \\ &&&= (1 + n \lambda + \frac{n(n-1)}{2} \lambda^2 + \cdots)(1 - \lambda n + \frac{\lambda^2 n^2}{2} + \cdots) \\ &&&= 1 + 0 \lambda + \left ( \frac{n(n-1)}{2} + \frac{n^2}{2} - n^2 \right) \lambda^2 + o(\lambda^3) \\ &&&= 1 - \frac{n}{2} \lambda^2 + o(\lambda^3) \end{align*} So if \(\frac{n}{2} \lambda \leq 1\) or \(n \leq \frac{2}{\lambda}\) \(\mathbb{P}(Y = n) \leq 1- \lambda\).
3. \(\,\) \begin{align*} && \mathbb{P}(Y > 1 | Y > 0) &= \frac{1-(q^n + npq^{n-1})}{1-q^n} \\ &&&= 1 - \frac{npq^{n-1}}{1-q^n} \\ &&&= 1 -n \frac{(1+ \lambda)e^{-\lambda} (\tfrac12 \lambda^2 + o(\lambda^3))^{n-1}}{1-(\tfrac12 \lambda^2 + o(\lambda^3))^n} \\ &&&= 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} \frac{(1+ \lambda)(1-\lambda + \lambda^2/2 - \cdots)(1+o(\lambda)^{n-1}}{1-(\tfrac12 \lambda^2 + o(\lambda^3))^n} \\ &&&= 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} (1 + o(\lambda)) \\ &&&\approx 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} \end{align*}

Solution: \begin{align*} && (1-1/50)^{1/2} &= 1 + \frac12 \cdot \left ( -\frac1{50} \right) + \frac1{2!} \frac12 \cdot \left ( -\frac12 \right)\cdot \left ( -\frac1{50} \right)^2 + \frac1{3!} \frac12 \cdot \left ( -\frac12 \right) \cdot \left ( -\frac32 \right)\cdot \left ( -\frac1{50} \right)^3 + \cdots \\ &&&=1-\frac{1}{100} - \frac12 \frac1{10000} -\frac12 \frac1{1000000} +\cdots \\ &&&= 0.9899495 + \cdots \\ \Rightarrow && \frac{7\sqrt{2}}{10} &\approx 0.9899495 \\ \Rightarrow && \sqrt{2} &\approx \frac{9.899495}{7} \\ &&&\approx 1.414214 \end{align*} \begin{align*} && (1 + 3/125)^{1/3} &= \frac{\sqrt[3]{125+3}}{5} \\ &&& = \frac{8\sqrt[3]{2}}{10} \\ && (1 + 3/125)^{1/3} &= 1 + \frac13 \left ( \frac{3}{125} \right) + \frac1{2!} \cdot \frac{1}{3} \cdot \left ( -\frac23\right) \left ( \frac{3}{125}\right)^2 +\cdots \\ &&&= 1+ \frac{8}{1000} - \frac{64}{1000000} \\ &&&= 1.007936 \\ \Rightarrow && \sqrt[3]{2} &= \frac{10.07936}{8} \\ &&&= 1.259920 \end{align*}