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As a continuation of what I posted in the other thread. (I know someone just posted another one up but this is typed in LaTeX! Not written!
)
DISCLAIMER: I do not guarantee the correctness of these answers. If there is any error, let me know or post up the correction yourself.
Question 8a)
![](https://latex.codecogs.com/png.latex?\bg_white I_m=\displaystyle\int_0^1x^m(x^2-1)^5\,dx\\\\u=x^{m-1}\Rightarrow du=(m-1)x^{m-2}\,dx\\\\dv=x(x^2-1)^5\,dx\Rightarrow v=\dfrac{(x^2-1)^6}{12}\\\\I_m=\left [ \dfrac{x^{m-1}(x^2-1)^6}{12} \right ]_0^1-\dfrac{m-1}{12}\displaystyle\int_0^1x^{m-2}(x^2-1)(x^2-1)^5\,dx\\\\I_m=-\dfrac{m-1}{12}\displaystyle\int_0^1[x^m(x^2-1)^5-x^{m-2}(x^2-1)^5]\,dx\\\\12I_m=(m-1)(I_{m-2}-I_m)\\\\\therefore I_m=\dfrac{m-1}{m+11}I_{m-2})
Question 8b)
i)
![](https://latex.codecogs.com/png.latex?\bg_white $First of all the sample space contains $ 7^7$ possibilities as there are 7 outcomes in each of the 7 trials. In order for each ball to be selected exactly once, the first trial has 7 possibilities. Given the selection of the first trial, the second trial is required to one of either the remaining 6 possibilities and so on. Alternatively, you can think of it as number of ways to arrange the set of numbers ${1, 2, 3, 4, 5, 6, 7}$. Thus, the probability is $\dfrac{7!}{7^7})
ii)
![](https://latex.codecogs.com/png.latex?\bg_white $This is the complementary event of a) hence probability is $ 1 - \dfrac{7!}{7^7})
iii)
![](https://latex.codecogs.com/png.latex?\bg_white $For this to happen, there needs to be exactly one repetition of one the numbers already selected. For example take ${7, 6, 5, 4, 3, 2, 2}$ where 1 is not selected. There are $\dfrac{7!}{2}$ ways of arranging this around 7 trials. In this example, the other possibilities are repetition of 3 or 4 or 5...etc instead (whilst still excluding 1) hence we multiply by 6. Now consider possibilities where numbers other than 1 are not selected which leads to multiplication by 7. \\\\So probability is $\dfrac{\frac{7!}{2}\times 6\times 7}{7^7})
Question 8c)
i)
![](https://latex.codecogs.com/png.latex?\bg_white $If $M = \max(|a_{n-1}|,|a_{n-2}|,...,|a_0| )\\\\M \geq |a_{n-1}|\Rightarrow M|\beta|^{n-1}\geq |a_{n-1}\beta^{n-1}|\\\\M \geq |a_{n-2}|\Rightarrow M|\beta|^{n-2}\geq |a_{n-2}\beta^{n-2}|\\\\M \geq |a_{n-3}|\Rightarrow M|\beta|^{n-3}\geq |a_{n-3}\beta^{n-3}|\\\\...etc\\\\M \geq |a_0|\\\\$Addition of these inequalities $\\\\M(|\beta|^{n-1}+|\beta|^{n-2}+...+1)\geq |a_{n-1}\beta^{n-1}| + |a_{n-2}\beta^{n-2}| + ... + |a_0|\\\\$But from triangle inequality $\\\\|a_{n-1}\beta^{n-1}| + |a_{n-2}\beta^{n-2}| + ... + |a_0| \geq |a_{n-1}\beta^{n-1} + a_{n-2}\beta^{n-2} + ... + a_0|\\\\\Rightarrow M(|\beta|^{n-1}+|\beta|^{n-2}+...+1)\geq |a_{n-1}\beta^{n-1} + a_{n-2}\beta^{n-2} + ... + a_0|\\\\ $But $\beta$ is a root of $P(z)$ \\\\\Rightarrow \beta^n+a_{n-1}\beta^{n-1} + a_{n-2}\beta^{n-2} + ... + a_0 = 0\\\\\Rightarrow a_{n-1}\beta^{n-1} + a_{n-2}\beta^{n-2} + ... + a_0 = -\beta^n\\\\\therefore M(|\beta|^{n-1}+|\beta|^{n-2}+...+1)\geq |-\beta^n| = |\beta|^n)
ii) Courtesy of mma03 after realising my own solution had an error
![](https://latex.codecogs.com/png.latex?\bg_white $Consider $|\beta|^n - 1=(|\beta| - 1)(|\beta|^{n-1}+|\beta|^{n-2}+....+ |\beta| + 1)\\\\\Rightarrow M(|\beta|^{n-1}+|\beta|^{n-2}+...+1) \geq (|\beta| - 1)(|\beta|^{n-1}+|\beta|^{n-2}+....+ |\beta| + 1) + 1 \\\\> (|\beta| - 1)(|\beta|^{n-1}+|\beta|^{n-2}+....+ |\beta| + 1) \\\\\Rightarrow M > |\beta| - 1\\\\\therefore |\beta| < 1 + M)
An alternative, courtesy of largarithmic
![](https://latex.codecogs.com/png.latex?\bg_white M(|\beta|^{n-1}+|\beta|^{n-2}+...+1) \geq |\beta|^n\\\\\Rightarrow M(|\beta|^{n-1}+|\beta|^{n-2}+...+1) + (|\beta|^{n-1}+|\beta|^{n-2}+...+1) \\\\\geq |\beta|^n+|\beta|^{n-1}+|\beta|^{n-2}+...+1\\\\\Rightarrow (M + 1)(|\beta|^{n-1}+|\beta|^{n-2}+...+1) \geq |\beta|^n+|\beta|^{n-1}+|\beta|^{n-2}+...+1 \\\\>|\beta|^n+|\beta|^{n-1}+|\beta|^{n-2}+...+|\beta|\\\\\Rightarrow (M + 1)(|\beta|^{n-1}+|\beta|^{n-2}+...+1) > |\beta| (|\beta|^{n-1}+|\beta|^{n-2}+...+1)\\\\\therefore |\beta| < M + 1)
Question 8d)
![](https://latex.codecogs.com/png.latex?\bg_white S(x)=\displaystyle\sum_{k=0}^nc_k\left ( x+\dfrac{1}{x}\right )^k\\\\\\$Let $z=x+\dfrac{1}{x}$ where $x\in\mathbb{R}\\\\\Rightarrow S(x)\equiv s(z)=\displaystyle\sum_{k=0}^nc_kz^k=c_0+c_1z+c_2z^2+...+c_nz^n\\\\$Suppose that there exists a root $z=\beta$ such that $s(z)=0$ which is equivalent to solving $S(x)=0\\\\c_0+c_1\beta+c_2\beta^2+...+c_n\beta^n=0\\\\\Rightarrow \dfrac{c_0}{c_n}+\dfrac{c_1}{c_n}\beta+\dfrac{c_2}{c_n}\beta^2+...+\beta^n=0\\\\\Rightarrow P(\beta)=0 $ where $a_0=\dfrac{c_0}{c_n}, a_1=\dfrac{c_1}{c_n},..., a_{n-1}=\dfrac{c_{n-1}}{c_n}\\\\$Note that $|c_k|\leq |c_n| \Rightarrow \dfrac{|c_k|}{|c_n|}=|a_k|\leq 1 $ for all $k < n\\\\\Rightarrow M=\max(|a_{n-1}|,|a_{n-2}|,...,|a_0| )\leq 1\\\\\Rightarrow 1 + M \leq 2)
![](https://latex.codecogs.com/png.latex?\bg_white $But $\beta = x + \dfrac{1}{x}$ where $x$ solves $S(x)=0\\\\\Rightarrow x^2-\beta x + 1 = 0\\\\$Since we require $x\in\mathbb{R}$ then $\Delta\geq 0\\\\\Rightarrow \beta^2-4\geq 0\\\\\Rightarrow \beta \geq 2 $ or $ \beta \leq -2 \\\\\Rightarrow |\beta|\geq 2\\\\$This means that $\\\\|\beta|\geq 1 + M $ which contradicts the inequality in (c)(ii) which holds true for any $\beta$ which satisfies that polynomial, hence our assumption that a real root exists is false, thus $S(x)$ has no real solutions.$)
DISCLAIMER: I do not guarantee the correctness of these answers. If there is any error, let me know or post up the correction yourself.
Question 8a)
Question 8b)
i)
ii)
iii)
Question 8c)
i)
ii) Courtesy of mma03 after realising my own solution had an error
An alternative, courtesy of largarithmic
Question 8d)
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