Problems

Solution: \begin{align*} && I_n &= \int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x\\ && I_0 &= \int_0^a x^{\frac12}(a-x)^{\frac12} \d x \\ x = a \sin^2 \theta, \d x = 2a \sin \theta \cos \theta \d \theta &&&= \int_{\theta =0}^{\theta = \pi/2} \sqrt{a}\sin \theta\sqrt{a} \cos \theta 2a \sin \theta \cos \theta \d \theta \\ &&&= \frac{a^2}{2} \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac{a^2}{4} \int_0^{\pi/2}(1- \underbrace{\cos 4\theta}_{\text{runs round the whole unit circle}}) \d \theta \\ &&&= \frac{\pi a^2}{8} \\ \\ && I_n &= \int_0^a x^{n+\frac12}(a-x)^{\frac12} \d x \\ &&&=\underbrace{\left [-\frac23x^{n+\frac12}(a-x)^\frac32 \right]_0^a}_{=0} + \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)^\frac32 \d x \\ &&&= \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)(a-x)^\frac12 \d x \\ &&&= \frac23 \left(n+\frac12\right)aI_{n-1}-\frac23 \left(n+\frac12\right)I_{n} \\ \Rightarrow && \left(n+\frac12+\frac32\right)I_{n} &= \left(n+\frac12\right)aI_{n-1}\\ \Rightarrow && (2n+4)I_n &= (2n+1)aI_{n-1} \\ \\ \Rightarrow && I_n &= \frac{2n+1}{2n+4}a I_{n-1} \\ &&&=\frac{2n+1}{2n+4}\frac{2n-1}{2n+2}a^2 I_{n-2} \\ &&&= \frac{(2n+1)!!}{(2n+4)!!} \pi a^{n+2} \end{align*}

Solution: If $\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\(, them \)\mathbf{J}^2=\begin{pmatrix}2 & 2\\ 2 & 2 \end{pmatrix} = 2\mathbf{J}\(. Therefore \)\mathbf{J}^n = 2\mathbf{J}^{n-1} = 2^{n-1}\mathbf{J}$ Let \(\mathbf{A}=\mathbf{I}+a\mathbf{J}\) then \begin{align*} \mathbf{A}^2 &=\l \mathbf{I}+a\mathbf{J}\r^2 \\ &= \mathbf{I}+2a\mathbf{J} + a^2\mathbf{J}^2 \\ &= \mathbf{I}+2a\mathbf{J} + 2a^2\mathbf{J} \\ &= \mathbf{I}+(2a+ 2a^2)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4a+ 4a^2-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^2-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^3 &=\l \mathbf{I}+a\mathbf{J}\r^3 \\ &= \mathbf{I}+3a\mathbf{J} + a^2\mathbf{J} + a^3\mathbf{J}^3 \\ &= \mathbf{I}+3a\mathbf{J} + 6a^2\mathbf{J} + 4a^3\mathbf{J} \\ &= \mathbf{I}+(3a+ 6a^3+4a^3)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+3\cdot2a+3\dot4a^2+ 8a^3-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^3-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^4 &=\l \mathbf{I}+a\mathbf{J}\r^4 \\ &= \mathbf{I}+4a\mathbf{J} + 6a^2\mathbf{J}^2 + 4a^3\mathbf{J}^3+a^4\mathbf{J}^4 \\ &= \mathbf{I}+4a\mathbf{J} + 12a^2\mathbf{J} + 16a^3\mathbf{J}+8a^4\mathbf{J}\\ &= \mathbf{I}+(4a+ 12a^3+16a^3+8a^4)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4\cdot2a+6\cdot4a^2+ 4\cdot8a^3+16a^4-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^4-1)\mathbf{J} \\ \end{align*} Claim: \(\mathbf{A}^k = \mathbf{I} + \frac12 ((1+2a)^{k}-1)\mathbf{J}\) Proof: Firstly, note that \(\mathbf{I}\) commutes with everything, so we can just apply the binomial theorem as if we were using real numbers: \begin{align*} \mathbf{A}^k &=\l \mathbf{I}+a\mathbf{J}\r^k \\ &= \sum_{i=0}^k \binom{k}{i}a^i\mathbf{J}^i \\ &= \mathbf{I} + \sum_{i=1}^k \binom{k}{i}a^i2^{i-1}\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=1}^k \binom{k}{i}a^i2^{i}\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=0}^k \binom{k}{i}a^i2^{i} - 1\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l(1+2a)^k - 1\r\mathbf{J} \end{align*} as required

Solution: Claim: \(S_p = q^p\) for \(1 \leq p \leq n-1\) Proof: When \(p = 1\), \(S_p = \Sigma_1 = q\) as expected. Note that \(\Sigma_i = 0\) for \(i = 2, \cdots, n-1\). Using \(S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}\), we can see that \(S_p = qS_{p-q}\) when \(1 \leq p \leq n-1\), ie \(S_p = q^p\). Note that \begin{align*} S_n &= \sum \alpha_i^n \\ &= q\sum \alpha_i^{n-1} - \sum r \\ &= qS_{n-1} - nr \\ &= q^n - nr \\ \\ S_{n+1} &= \sum \alpha_i^{n+1} \\ &= q \sum \alpha_i^{n} - r \sum \alpha_i \\ &= q^{n+1} - rq \\ \\ S_{n+2} &= \sum \alpha_i^{n+2} \\ &= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\ &= q^{n+2} - rq^2 \\ \end{align*} Claim: \(S_{n+m} = q^{n+m} - rq^{m}\) Proof: The obvious

Solution: \begin{align*} \binom{n}{r} + \binom{n}{r-1} &= \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\ &= \frac{n!}{(r-1)!(n-r)!} \left ( \frac{1}{r} + \frac{1}{n-r+1} \right) \\ &= \frac{n!}{(r-1)!(n-r)!} \frac{(n-r+1)+r}{r(n-r+1)} \\ &= \frac{n! (n+1)}{r! (n-r+1)!} \\ &= \frac{(n+1)!}{r!(n+1-r)!} \\ &= \binom{n+1}{r} \end{align*} Claim: \(\displaystyle (uv)^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\) Proof: (By induction on \(n\)). Base case: \(n = 0\) is clear. Inductive step: Suppose it is true for \(n = k\), then consider \begin{align*} (uv)^{(k+1)} &= \left ( (uv)^{(k)} \right)' \\ &= \left ( \sum_{r=0}^k \binom{k}{r} u^{(k-r)} v^{(r)} \right)' \tag{by assumption} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r)} v^{(r)}\right)' \tag{linearity} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r+1)} v^{(r)} + u^{(k-r)}v^{(r+1)}\right) \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=0}^{k} \binom{k}{r} u^{(k-r)}v^{(r+1)} \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=1}^{k+1} \binom{k}{r-1} u^{(k-r+1)}v^{(r)} \\ &= u^{(k+1)}v + \sum_{r=1}^k \left (\binom{k}{r} + \binom{k}{r-1} \right)u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= u^{(k+1)}v + \sum_{r=1}^k \binom{k+1}{r} u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= \sum_{r=0}^{k+1} \binom{k+1}{r} u^{(k-r+1)}v^{(r)}\\ \end{align*} Therefore if our statement is true for \(n = k\) it is true for \(n = k+1\). Since it is true for \(n = 0\) by the principle of mathematical induction it is true for all integer \(n \geq 0\) Suppose \( y = \sin^{-1} x\), then \(y' = \frac{1}{\sqrt{1-x^2}}\), \(y'' = \frac{x}{(1-x^2)^{3/2}}\). Not that this means that \((1-x^2)y'' - xy' = 0\) (which is our formula when \(n = 0\)). Now apply Leibniz's formula to this. \begin{align*} 0 &= \left ( (1-x^2)y'' - xy' \right)^{(n)} \\ &= \left ( (1-x^2)y'' \right)^{(n)} -\left ( xy' \right)^{(n)} \\ &= \left ( (1-x^2)y^{(n+2)} - n\cdot 2x \cdot y^{(n+1)}-\binom{n}{2} \cdot 2 \cdot y^{(n)} \right )- \left (xy^{(n+1)}+ny^{(n)} \right) \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - \left ( n(n-1)+n \right)y^{(n)} \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - n^2y^{(n)} \\ \end{align*} as required

Solution: \begin{align*} && \cot x - \tan x &= 2 \cot 2x \\ \Rightarrow && x\cot x - x\tan x &= 2x\cot 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1} &= 1 + \sum_{n=1}^{\infty} b_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty}(1-2^n)b_nx^n &= \sum_{n=1}^{\infty} a_{n-1}x^n \\ \Rightarrow && a_{n-1} &= (1-2^n)b_n \quad \text{if }n \geq 1 \end{align*} \begin{align*} \cot x + \tan x &= 2 \cosec 2x \end{align*} So \begin{align*} && \cot x + \tan x &= 2 \cosec 2x \\ \Rightarrow && x \cot x + x\tan x &= 2x \cosec 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} &= 1+\sum_{n=1}^\infty c_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty} \frac{1}{1-2^n}a_{n-1} +\sum_{n=1}^{\infty}a_{n-1}x^n &= \sum_{n=1}^{\infty} 2^nc_n x^n \\ \Rightarrow && c_n &= \frac{1}{2^n} \left ( 1 + \frac{1}{1-2^n} \right)a_{n-1} \\ &&&= \frac1{2^n} \frac{2^n-2}{2^n-1} a_{n-1}\\ &&&= \frac1{2^{n-1}}\frac{2^{n-1}-1}{2^n-1} a_{n-1} \end{align*}

Solution: \(\mathbf{M}^m = \mathbf{I}\), we also have \(\mathrm{det}(\mathbf{M - I}) = \cos^2(2\pi/m) - 2\cos(2\pi/m) + 1 + \sin^2(2\pi/m) = 2(1-\cos(2\pi/m))\) therefore \(\mathbf{M-I}\) is invertible. Therefore since \(\mathbf{(M-I)(M^{m-1} + M^{m-1} + \cdots + M^2 + M + I)= M^m-I = 0}\) we can cancel the \(\mathbf{M-I}\) to obtain the desired result. \(\mathbf{X_0 = X_0}\) \(\mathbf{X_1 = PX_0+Q}\) \(\mathbf{X_2 = P(PX_0+Q)+Q = P^2X_0 + PQ + Q}\) Claim: \(\mathbf{X_k = P^k X_0 + (P^{k-1} + P^{k-2} + \cdots + I)Q}\) Proof: (By induction on \(k\)). Base case \(k = 0\) is true. Assume it's true for some \(k = l\), then consider \(k = l+1\) \(\mathbf{X_{l+1} = PX_l + Q = P( P^l X_0 + (P^{l-1} + P^{l-2} + \cdots + I)Q) + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P)Q + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P + I)Q}\) Suppose \(\mathbf{P} = \mathbf{M}\), then consider the set \(\{\mathbf{X_1, X_2}, \ldots\}\) with the operation \(*\) as defined. \(\mathbf{X_i * X_j} = M^{i}(X_j) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i}(M^jX_0 + (M^{j-1} + M^{j-2} + \cdots + M + I)Q) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i+j}X_0 + (M^{i+j-1}+\cdots + M + I)Q = X_{i+j}\) Since \(X_m = X_0\) we can check all the requirements of the group, but this is going to be isomorphic to the cyclic group with \(m\) elements.

Solution: If the population in a given year is \(\mathbf{v} = \begin{pmatrix}A \\ B \end{pmatrix}\) then the population the next year is \(\mathbf{Mv}\) where \(\mathbf{M} = \begin{pmatrix} \frac14 & \frac34 \\ -\frac12 &\frac32 \end{pmatrix}\) The ratio is the same if \(\mathbf{Mv} = \lambda \mathbf{v}\) ie if \(\mathbf{v}\) is an eigenvector of \(\mathbf{M}\). The eigenvalues will be \(1\) and \(\frac38\) (by inspection) so we should be able to solve for the eigenvectors: \(\lambda = 1\) we have \(\frac14A + \frac34B = A \Rightarrow A = B\) a ratio of \(1\). \(\lambda = \frac38\) we have \(\frac14A + \frac34B = \frac38A \Rightarrow \frac34B = \frac18A \Rightarrow A = 6B\) a ratio of \(6\). If we write \(\begin{pmatrix} a \\ b \end{pmatrix}\) as \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) we find that after \(n\) years, we have: \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \l \frac38 \r^n x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) for the populations. Therefore if \(x_1\) is \(< 0\) then in finite time we will end up with one population being 0. If \(x_1 > 0\) are positive we tend to a finite population and if \(x_1 = 0\) then over time the population will tend to \(0\) at infinity. In our diagram these areas correspond to (red) - die out in finite time, (green) population stable and the thick black line where the population goes extinct as \(t \to \infty\)

+ - ↺

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 \[ v_n = \frac{4}{3}(n+1) \l \frac{1}{2^n} + \l -\frac14\r^n \r\]

Solution: After \(n\) days they will have borrowed \(c\) for \(n-1\) days, \(c\) for \(n-2\) days, etc until \(c\) for no days. Therefore the outstanding balance will be: \begin{align*} c + (1+k)\cdot c+ (1+k)^2 \cdot c + \cdots + (1+k)^{n-1} \cdot c &= c\frac{(1+k)^n-1}{(1+k)-1} \\ &= \frac{c[(1+k)^n-1]}{k} \end{align*} At the end of \(T\) days the outstanding balance will be \(S_T = \frac{c[(1+k)^T-1]}{k}\). We can think of each payment of \(d\) during the subsequent period as being equivalent of a payment of \(d (1+k)^{m-1}\) \(m\) days later (as otherwise they would have accrued the equivalent amount in interest. Therefore after \(m\) days the amount paid back (equivalent) is: \begin{align*} (1+k)^{m-1} \cdot d + (1+k)^{m-2} \cdot d + \cdots + d &= \frac{d[(1+k)^m-1]}{k} \end{align*} Therefore the net position, \(S_{T+m}\) will be: \begin{align*} S_{T+m} &= \frac{c[(1+k)^T-1](1+k)^m-d[(1+k)^m-1]}{k} \\ &= \frac{(1+k)^m [c ((1+k)^T-1)-d]+d}{k} \end{align*} Therefore they will eventually pay back their debts if \( [c ((1+k)^T-1)-d]\) is negative. ie \(d > c((1+k)^T-1) \Rightarrow d/c > (1+k)^T-1\)

Solution:

1. \(M(0) = 0\) since if there's no space on the shelf, we wont be able to put any books on the shelf.
2. If the shelf has length \(1\) it can only fit a thin book. For a thin book to be placed on the shelf, the very first book which comes to be placed must be thin. But this happens with probability \(q\). Therefore \(M(1) = q\).
3. Suppose no books have been placed on the shelf, then with probability \(p\) a large book gets placed on the shelf, and the expected number of books to be placed on the shelf is equivalent to how many books will be placed on the shelf if the shelf only had \(n-2\) spaces. This is \(M(n-2)\). Similar if the book which arrives first is thin (with probability \(q\)) then there will be \(M(n-1)\) more books placed on the shelf in expectation. We've just added \(1\) more book, therefore \(M(n) = 1+pM(n-2) + qM(n-1)\) or rearranging \(M(n) - qM(n-1) - pM(n-2) = 1\).