Problems

Solution:

1. \(\,\)
   

   + - ↺
   Notice that \(1+r+r^2\) ranges from \(\frac34\) to \(3\) over \((-1,1)\) therefore \(\frac34 \leq p < 3\) attaining its minimum but not its maximum. If \(p > 1\) we know we must be on the right branch and hence we can determine \(r\) and \(S\) uniquely. If \(\frac34 < p < 1\) then we must have two possible values for \(r\), satisfying \(r_i^2 + r+1 = p\) and so our two possible values for \(S_i = \frac{1}{1-r_i}\) or \(r_i = 1-\frac{1}{S_i}\) and so \begin{align*} && p &= \left ( 1 - \frac{1}{S} \right)^2 + 1 - \frac1S + 1 \\ \Rightarrow && pS^2 &= (S-1)^2 + S^2-S + S^2 \\ \Rightarrow && 0 &= (3-p)S^2 -3S + 1 \end{align*}
2. \(\,\)
   

   + - ↺
   So \(\frac23 \leq q < 6\) and \(r\) and hence \(T\) is uniquely determined if \(2 \leq q < 6\). if \(\frac23

Solution: Let \(X \sim Po(\lambda)\), then

1. \(\,\)
   

   + - ↺
   Suppose \(\mathbb{P}(X \geq 2) = 1-p\) then \(\mathbb{P}(X=0) + \mathbb{P}(X=1) = p\), ie \(e^{-\lambda} +\lambda e^{-\lambda} = p\) If \(f(x) = (1+x)e^{-x}\) we have see it is strictly decreasing on \(x \geq 0\) and takes all values from \(1\) to \(0\), therefore we can find a unique value such that \(f(\lambda) = p\) which is our desired \(\lambda\).
2. Note that \(\mathbb{P}(X = 1) = \lambda e^{-\lambda}\)
   

   + - ↺
   Sketching \(y = xe^{-x}\) and finding it's turning point we can see that there is a unique value of \(\lambda = 1\) when \(q = \frac{1}{e}\), otherwise there is either no solution (\(p > \frac1{e}\)) or two solutions (\(0 < q > \frac1{e}\)).
3. Suppose \(\mathbb{P}(X = 1 | X \leq 2) = r\), ie \begin{align*} && r &= \frac{\lambda e^{-\lambda}}{e^{-\lambda} + \lambda e^{-\lambda} + \frac12 \lambda^2 e^{-\lambda}} \\ &&&= \frac{2\lambda}{2+2\lambda+\lambda^2} \\ \Rightarrow && 0 &= r\lambda^2 + 2(r-1) \lambda + 2r\\ \Rightarrow && \Delta &= 4(r-1)^2 - 4\cdot r \cdot 2 r \\ &&&= 4((r-1)^2-2r^2) \\ &&&= 4(r-1-\sqrt{2}r)(r-1+\sqrt{2}r) \\ &&&= -4((\sqrt{2}-1)r + 1)((1+\sqrt{2})r-1) \end{align*} Therefore our quadratic in \(r\) has a unique solution if \(r = \frac{1}{1+\sqrt{2}}\). If it has a positive solution then note since \(2r > 0\) both solutions are positive, so \(\lambda\) is not unique by excluding other solutions.

Solution: Notice that \(\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}\), so any unit fraction can be expressed as the sum of two distinct unit fractions. \begin{align*} && \frac{1}N &= \frac1a + \frac1b \\ \Leftrightarrow && ab&= Nb+Na \\ \Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\ \Leftrightarrow && N^2 &= (a-N)(b-N) \end{align*} If \(N\) is prime then the only factors of \(N^2\) are \(1,N\) and \(N^2\). if \(a-N = b-N = N\) then \(a=b\) and we don't have distinct fractions. Therefore \(a-N = 1\) and \(b-N = N^2\) and we obtain the decomposition earlier (and it must be the only solution). \begin{align*} && \frac2N &= \frac1a+\frac1b \\ \Leftrightarrow && 2ab &= Nb+Na \\ \Leftrightarrow && 4ab &= 2Na+2Nb \\ \Leftrightarrow && N^2 &= (2a-N)(2b-N) \end{align*} Therefore for \(a,b\) to be distinct we must have \(2a = N+1\) and \(2b = N+N^2\) as the only possible factorisation. Both of the right hand sides are even so we can write \[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \] and this is unique

Solution: Let \(z = x^\alpha, \frac{\d z}{\d x}=\alpha x^{\alpha-1} \), then \begin{align*} \frac{\d y}{\d x} &= \frac{\d y}{\d z} \frac{\d z}{\d x} \\ &= \alpha x^{\alpha-1}\frac{\d y}{\d z} \\ \\ \frac{\d^2 y}{\d x^2} &= \frac{\d }{\d x} \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) \\ &= \alpha (\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha x^{\alpha-1} \frac{\d ^2 y}{\d z^2} \frac{\d z}{\d x} \\ &= \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2} \end{align*} \begin{align*} && 0 &=x\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-b\frac{\mathrm{d}y}{\mathrm{d}x}+x^{2b+1}y \\ &&&= x \left ( \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2}\right) - b \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) + x^{2b+1}y \\ &&&= \alpha^2 x^{2\alpha-1} \frac{\d^2 y}{\d z^2} +\left (\alpha(\alpha-1)x^{\alpha-1}-b\alpha x^{\alpha-1} \right) \frac{\d y}{\d z} + x^{2b+1} y \\ \end{align*} If we set \(\alpha = b +1\) the middle term disappears, so we get \begin{align*} && 0 &= (b+1)^2 x^{2b+1} \frac{\d^2 y}{\d z^2} + x^{2b+1} y \\ \Rightarrow && 0 &= (b+1)^2 \frac{\d^2 y}{\d z^2} + y \\ \Rightarrow && y &= A \sin \left (\frac{z}{b+1} \right) + B \cos \left (\frac{z}{b+1} \right) \\ &&&= \boxed{A \sin \left (\frac{x^{b+1}}{b+1} \right) + B \cos \left (\frac{x^{b+1}}{b+1} \right)} \\ \\ \lim_{x \to 0}: && y &\to B \\ && \frac{\d y}{\d x} &= A x^b \cos\left (\frac{x^{b+1}}{b+1} \right) - B x^b \sin\left (\frac{x^{b+1}}{b+1} \right) \\ b>0: && \frac{\d y}{\d x} &\to 0 \\ \end{align*} So there are infinitely many different solutions with \(B = 1\) and \(A\) is anything it wants to be. If \(b = 0\) \(y' \to A\) so \(A =0 \) and unique. If \(b < 0\) \(x^b \to \infty\) so we need \(A = 0\), unique. However, we also need \(y' \to 0\), so we need to check \(y' = -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \to 0\), \begin{align*} y' &= -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \\ &\approx -x^b \left ( \frac{x^{b+1}}{b+1}\right) \\ &= - \frac{x^{2b+1}}{b+1} \end{align*} so we need \(2b+1>0 \Rightarrow b > -\frac12\). Therefore the solution is unique on \((-\frac12,0]\)

Solution: First notice that \(u_n = \alpha^n\) and \(u_n = \beta^n\) both satisfy the recurrence, since: \begin{align*} && a \alpha^2 + b \alpha + c &= 0 \\ \Rightarrow && a \alpha^{n+2} + b \alpha^{n+1} + c \alpha^n &= 0 \\ \Rightarrow && a u_{n+2} + bu_{n+1} + cu_n &=0 \end{align*} Notice also that if \(u_n\) and \(v_n\) both satisfy the recurrence, then any linear combination of them will satisfy the recurrence: \begin{align*} && \begin{cases} au_{n+2} + bu_{n+1} + cu_n &= 0 \\ av_{n+2} + bv_{n+1} + cv_n &= 0 \\ \end{cases} \\ \Rightarrow && a (\lambda u_{n+2}+ \mu v_{n+2}) + b (\lambda u_{n+1}+ \mu v_{n+1}) + c (\lambda u_{n}+ \mu v_{n}) &= 0 \end{align*} by adding a linear combination of the top two equations. Therefore it suffices to check that the constants \(A\) and \(B\) are such that we match \(u_0\) and \(u_1\). \(\frac{u_1 - \beta u_0}{\alpha - \beta} + \frac{u_1 - \alpha u_0}{\beta - \alpha} = u_0\) and \(\frac{u_1 - \beta u_0}{\alpha - \beta}\alpha + \frac{u_1 - \alpha u_0}{\beta - \alpha}\beta = u_1\). So we are done. Suppose we have another sequence, then we first notice that the first and second terms must be identical to each other. Suppose the first \(k\) terms are identical, then since the \(k+1\)th term depends only on the \(k\) and \(k-1\)th terms (both of which are equal) the \(k+1\)th term is the same. Therefore, by the principle of mathematical induction, all terms are the same. First notice that if you put \(v_n = (n+1)w_n\) we have \begin{align*} && 8(n+3)(n+2)(n+1)w_{n+2} - 2(n+3)(n+2)(n+1)w_{n+1} - (n+3)(n+2)(n+1)w_n &= 0 \\ \Rightarrow && 8w_{n+2}-2w_{n+1}-w_n &= 0 \end{align*} This has characteristic equation \(8\lambda^2 - 2\lambda - 1 = 0 \Rightarrow \lambda = \frac12, -\frac14\). Therefore the general solution is \(w_n = A \l \frac12 \r^n + B \l -\frac14\r^n\) and \(v_n = (n+1)\l A \l \frac12 \r^n + B \l -\frac14\r^n \r\). When \(n = 0\) we have \(A+B = 0 \Rightarrow B =-A\). When \(n=1\) we have \(1 = 2 \l \frac{A}{2} + \frac{A}{4} \r \Rightarrow A = \frac{4}{3}\), therefore \[ v_n = \frac{4}{3}(n+1) \l \frac{1}{2^n} + \l -\frac14\r^n \r\]