UFM Additional Further Pure

Let \(a\) and \(b\) be positive integers such that \(b<2a-1\). For any given positive integer \(n\), the integers \(N\) and \(M\) are defined by \[ [a+\sqrt{a^{2}-b}]^{n}=N-r, \] \[ [a-\sqrt{a^{2}-b}]^{n}=M+s, \] where \(0\leqslant r<1\) and \(0\leqslant s<1\). Prove that \begin{questionparts} \item \(M=0\), \item \(r=s\), \item \(r^{2}-Nr+b^{n}=0.\) \end{questionpart} Show that for large \(n\), \(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by about \(2^{-4n}\).

1. The coefficients in the series \[ S= \tfrac13 x + \tfrac 16 x^2 + \tfrac1{12} x^3 + \cdots + a_rx^r + \cdots \] satisfy a recurrence relation of the form \(a_{r+1} + p a_r =0\). Write down the value of \(p\). By considering \((1+px)S\), find an expression for the sum to infinity of \(S\) (assuming that it exists). Find also an expression for the sum of the first \(n+1\) terms of \(S\).
2. The coefficients in the series \[ T=2 + 8x +18x^2+37 x^3 +\cdots + a_rx^r + \cdots \] satisfy a recurrence relation of the form \(a_{r+2}+pa_{r+1} +qa_r=0\). Find an expression for the sum to infinity of \(T\) (assuming that it exists). By expressing \(T\) in partial fractions, or otherwise, find an expression for the sum of the first \(n+1\) terms of \(T\).

The sequence \(a_n\) is defined by \(a_0 = 1\) , \(a_1 = 1\) , and $$ a_n = {1 + a_{n - 1}^2 \over a_{n - 2} } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( n \ge 2 ) . $$ Prove by induction that $$ a_n = 3 a_{n - 1} - a_{n - 2} \ \ \ \ \ \ \ \ \ \ \ ( n \ge2 ) . $$ Hence show that $$ a_n = {\alpha^{2 n - 1} + \alpha^{- ( 2 n - 1 )} \over \sqrt 5} \ \ \ \ \ \ (n\ge1), $$ where \(\displaystyle{\alpha = {1 + \sqrt 5 \over 2}}\).

The value \(V_N\) of a bond after \(N\) days is determined by the equation $$ V_{N+1} = (1+c) V_{N} -d \qquad (c>0, \ d>0), $$ where \(c\) and \(d\) are given constants. By looking for solutions of the form \(V_T= A k^T + B\) for some constants \(A,B\) and \(k\), or otherwise, find \(V_N\) in terms of \(V_0\). What is the solution for \(c=0\)? Show that this is the limit (for fixed \(N\)) as \(c\rightarrow 0\) of your solution for \(c>0\).

Each day, books returned to a library are placed on a shelf in order of arrival, and left there. When a book arrives for which there is no room on the shelf, that book and all books subsequently returned are put on a trolley. At the end of each day, the shelf and trolley are cleared. There are just two-sizes of book: thick, requiring two units of shelf space; and thin, requiring one unit. The probability that a returned book is thick is \(p\), and the probability that it is thin is \(q=1-p.\) Let \(M(n)\) be the expected number of books that will be put on the shelf, when the length of the shelf is \(n\) units and \(n\) is an integer, on the assumption that more books will be returned each day than can be placed on the shelf. Show, giving reasoning, that

1. \(M(0)=0;\)
2. \(M(1)=q;\)
3. \(M(n)-qM(n-1)-pM(n-2)=1,\) for \(n\geqslant2.\)

The real numbers \(u_{0},u_{1},u_{2},\ldots\) satisfy the difference equation \[ au_{n+2}+bu_{n+1}+cu_{n}=0\qquad(n=0,1,2,\ldots), \] where \(a,b\) and \(c\) are real numbers such that the quadratic equation \[ ax^{2}+bx+c=0 \] has two distinct real roots \(\alpha\) and \(\beta.\) Show that the above difference equation is satisfied by the numbers \(u_{n}\) defined by \[ u_{n}=A\alpha^{n}+B\beta^{n}, \] where \[ A=\frac{u_{1}-\beta u_{0}}{\alpha-\beta}\qquad\mbox{ and }\qquad B=\frac{u_{1}-\alpha u_{0}}{\beta-\alpha}. \] Show also, by induction, that these numbers provide the only solution. Find the numbers \(v_{n}\) \((n=0,1,2,\ldots)\) which satisfy \[ 8(n+2)(n+1)v_{n+2}-2(n+3)(n+1)v_{n+1}-(n+3)(n+2)v_{n}=0 \] with \(v_{0}=0\) and \(v_{1}=1.\)

A secret message consists of the numbers \(1,3,7,23,24,37,39,43,43,43,45,47\) arranged in some order as \(a_{1},a_{2},\ldots,a_{12}.\) The message is encoded as \(b_{1},b_{2},\ldots,b_{12}\) with \(0\leqslant b_{j}\leqslant49\) and \begin{alignat*}{1} b_{2j} & \equiv a_{2j}+n_{0}+j\pmod{50},\\ b_{2j+1} & \equiv a_{2j+1}+n_{1}+j\pmod{50}, \end{alignat*} for some integers \(n_{0}\) and \(n_{1}.\) If the coded message is \(35,27,2,36,15,35,8,40,40,37,24,48,\) find the original message, explaining your method carefully.

The integers \(a,b\) and \(c\) satisfy \[ 2a^{2}+b^{2}=5c^{2}. \] By considering the possible values of \(a\pmod5\) and \(b\pmod5\), show that \(a\) and \(b\) must both be divisible by \(5\). By considering how many times \(a,b\) and \(c\) can be divided by \(5\), show that the only solution is \(a=b=c=0.\)

Find the ratio, over one revolution, of the distance moved by a wheel rolling on a flat surface to the distance traced out by a point on its circumference.

The curve \(P\) has the parametric equations $$ x= \sin\theta, \quad y=\cos2\theta \qquad\hbox{ for }-\pi/2 \le \theta \le \pi/2. $$ Show that \(P\) is part of the parabola \(y=1-2x^2\) and sketch \(P\). Show that the length of \(P\) is \(\surd (17) + {1\over 4} \sinh^{-1}4\). Obtain the volume of the solid enclosed when \(P\) is rotated through \(2\pi\) radians about the line \(y=-1\).

Show Solution

First notice that \(y = \cos 2 \theta = 1 - 2\sin^2 \theta = 1- 2x^2\), therefore \(P\) is lies on that parabola.

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The arc length is \begin{align*} && s &= \int_{-\pi/2}^{\pi/2} \sqrt{\left ( \frac{\d x}{\d \theta} \right)^2+\left ( \frac{\d y}{\d \theta} \right)^2} \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \sqrt{\cos^2 \theta+16 \sin^2 \theta \cos^2 \theta } \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \cos \theta\sqrt{1+16 \sin^2 \theta} \d \theta\\ u = \sin \theta, \d u = \cos \theta \d \theta && &= \int_{u=-1}^{u=1} \sqrt{1+16 u^2} \d u\\ 4u = \sinh v, 4\d u = \cosh v: && &= \int_{v=-\sinh^{-1} 4}^{v=\sinh^{-1} 4} \sqrt{1+\sinh^2 v} \tfrac14\cosh v \d v\\ && &= \frac14 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} \cosh^2 v \d v\\ && &= \frac18 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} (1 + \cosh 2v) \d v\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \frac12\sinh 2v \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \sinh v \sqrt{1 + \sinh^2 v} \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \left (\frac18 \cdot 4 \sqrt{17} \right) - \left (\frac18 \cdot (-4) \sqrt{17} \right)\\ && &= \frac14 \sinh^{-1} 4 + \sqrt{17}\\ \end{align*} The volume of revolution is \begin{align*} && V &=\pi \int_{-1}^1 (2-2x^2)^2 \d x \\ &&&= \pi \left [4x-\frac83x^3+\frac45x^5 \right]_{-1}^1 \\ &&&= \pi \left ( 8-\frac{16}3+\frac85 \right) \\ &&&= \frac{64}{15}\pi \end{align*}

A kingdom consists of a vast plane with a central parabolic hill. In a vertical cross-section through the centre of the hill, with the \(x\)-axis horizontal and the \(z\)-axis vertical, the surface of the plane and hill is given by \[ z=\begin{cases} \dfrac{1}{2a}(a^{2}-x^{2}) & \mbox{ for }\left|x\right|\leqslant a,\\ 0 & \mbox{ for }\left|x\right|>a. \end{cases} \] The whole surface is formed by rotating this cross-section about the \(z\)-axis. In the \((x,z)\) plane through the centre of the hill, the king has a summer residence at \((-R,0)\) and a winter residence at \((R,0)\), where \(R>a.\) He wishes to connect them by a road, consisting of the following segments: \begin{itemize} \item a path in the \((x,z)\) plane joining \((-R,0)\) to \((-b,(a^{2}-b^{2})/2a),\) where \(0\leqslant b\leqslant a.\) \item a horizontal semicircular path joining the two points \((\pm b,(a^{2}-b^{2})/2a),\) if \(b\neq0;\) \item a path in the \((x,z)\) plane joining \((b,(a^{2}-b^{2})/2a)\) to \((R,0).\) \end{itemiz} The king wants the road to be as short as possible. Advise him on his choice of \(b.\)

Let \((G,*)\) and \((H,\circ)\) be two groups and \(G\times H\) be the set of ordered pairs \((g,h)\) with \(g\in G\) and \(h\in H.\) A multiplication on \(G\times H\) is defined by \[ (g_{1},h_{1})(g_{2},h_{2})=(g_{1}*g_{2},h_{1}\circ h_{2}) \] for all \(g_{1},g_{2}\in G\) and \(h_{1},h_{2}\in H\). Show that, with this multiplication, \(G\times H\) is a group. State whether the following are true or false and prove your answers.

1. \(G\times H\) is abelian if and only if both \(G\) and \(H\) are abelian.
2. \(G\times H\) contains a subgroup isomorphic to \(G\).
3. \(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\) is isomorphic to \(\mathbb{Z}_{4}.\)
4. \(S_{2}\times S_{3}\) is isomorphic to \(S_{6}.\)

The set \(S\) consists of \(N(>2)\) elements \(a_{1},a_{2},\ldots,a_{N}.\) \(S\) is acted upon by a binary operation \(\circ,\) defined by \[ a_{j}\circ a_{k}=a_{m}, \] where \(m\) is equal to the greater of \(j\) and \(k\). Determine, giving reasons, which of the four group axioms hold for \(S\) under \(\circ,\) and which do not. Determine also, giving reasons, which of the group axioms hold for \(S\) under \(*\), where \(*\) is defined by \[ a_{j}*a_{k}=a_{n}, \] where \(n=\left|j-k\right|+1\).

Let \(a\) be a non-zero real number and define a binary operation on the set of real numbers by $$ x*y = x+y+axy \,. $$ Show that the operation \(*\) is associative. Show that \((G,*)\) is a group, where \(G\) is the set of all real numbers except for one number which you should identify. Find a subgroup of \((G,*)\) which has exactly 2 elements.

1. Let \(S\) be the set of matrices of the form \[ \begin{pmatrix}a & a\\ a & a \end{pmatrix}, \] where \(a\) is any real non-zero number. Show that \(S\) is closed under matrix multiplication and, further, that \(S\) is a group under matrix multiplication.
2. Let \(G\) be a set of \(n\times n\) matrices which is a group under matrix multiplication, with identity element \(\mathbf{E}.\) By considering equations of the form \(\mathbf{BC=D}\) for suitable elements \(\mathbf{B},\) \(\mathbf{C}\) and \(\mathbf{D}\) of \(G\), show that if a given element \(\mathbf{A}\) of \(G\) is a singular matrix (i.e. \(\det\mathbf{A}=0\)), then all elements of \(G\) are singular. Give, with justification, an example of such a group of singular matrices in the case \(n=3.\)

Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication is associative. In each case state, giving adequate reasons, whether or not the set is a group.

1. the complex numbers of unit modulus;
2. the integers modulo 4;
3. the matrices \[ \mathrm{M}(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}, \] where \(0\leqslant\theta<2\pi\);
4. the integers \(1,3,5,7\) modulo 8;
5. the \(2\times2\) matrices all of whose entries are integers;
6. the integers \(1,2,3,4\) modulo 5.

Let \(S_{3}\) be the group of permutations of three objects and \(Z_{6}\) be the group of integers under addition modulo 6. List all the elements of each group, stating the order of each element. State, with reasons, whether \(S_{3}\) is isomorphic with \(Z_{6}.\) Let \(C_{6}\) be the group of 6th roots of unity. That is, \(C_{6}=\{1,\alpha,\alpha^{2},\alpha^{3},\alpha^{4},\alpha^{5}\}\) where \(\alpha=\mathrm{e}^{\mathrm{i}\pi/3}\) and the group operation is complex multiplication. Prove that \(C_{6}\) is isomorphic with \(Z_{6}.\) Is there any (multiplicative or additive) subgroup of the complex numbers which is isomorphic with \(S_{3}\)? Give a reason for your answer.

The set \(S\) consists of ordered pairs of complex numbers \((z_1,z_2)\) and a binary operation \(\circ\) on \(S\) is defined by $$ (z_1,z_2)\circ(w_1,w_2)= (z_1w_1-z_2w^*_2, \; z_1w_2+z_2w^*_1). $$ Show that the operation \(\circ\) is associative and determine whether it is commutative. Evaluate \((z,0)\circ(w,0)\), \((z,0)\circ(0,w)\), \((0,z)\circ(w,0)\) and \((0,z)\circ(0,w)\). The set \(S_1\) is the subset of \(S\) consisting of \(A\), \(B\), \(\ldots\,\), \(H\), where \(A=(1,0)\), \(B=(0,1)\), \(C=(i,0)\), \(D=(0,i)\), \(E=(-1,0)\), \(F=(0,-1)\), \(G=(-i,0)\) and \(H=(0,-i)\). Show that \(S_1\) is closed under \(\circ\) and that it has an identity element. Determine the inverse and order of each element of \(S_1\). Show that \(S_1\) is a group under \(\circ\). \hfil\break [You are not required to compute the multiplication table in full.] Show that \(\{A,B,E,F\}\) is a subgroup of \(S_1\) and determine whether it is isomorphic to the group generated by the \(2\times2\) matrix $\begin{pmatrix}0 & 1\\ -1 & 0 \end{pmatrix}$ under matrix multiplication.

Explain what is meant by the order of an element \(g\) of a group \(G\). The set \(S\) consists of all \(2\times2\) matrices whose determinant is \(1\). Find the inverse of the element \(\mathbf{A}\) of \(S\), where \[ \mathbf{A}=\begin{pmatrix}w & x\\ y & z \end{pmatrix}. \] Show that \(S\) is a group under matrix multiplication (you may assume that matrix multiplication is associative). For which elements \(\mathbf{A}\) is \(\mathbf{A}^{-1}=\mathbf{A}\)? Which element or elements have order 2? Show that the element \(\mathbf{A}\) of \(S\) has order 3 if, and only if, \(w+z+1=0.\) Write down one such element.

Let \(G\) be the set of all matrices of the form \[ \begin{pmatrix}a & b\\ 0 & c \end{pmatrix}, \] where \(a,b\) and \(c\) are integers modulo 5, and \(a\neq0\neq c\). Show that \(G\) forms a group under matrix multiplication (which may be assumed to be associative). What is the order of \(G\)? Determine whether or not \(G\) is commutative. Determine whether or not the set consisting of all elements in \(G\) of order \(1\) or \(2\) is a subgroup of \(G\).

The elements \(a,b,c,d\) belong to the group \(G\) with binary operation \(*.\) Show that

1. if \(a,b\) and \(a*b\) are of order 2, then \(a\) and \(b\) commute;
2. \(c*d\) and \(d*c\) have the same order;
3. if \(c^{-1}*b*c=b^{r},\) then \(c^{-1}*b^{s}*c=b^{sr}\) and \(c^{-n}*b^{s}*c^{n}=b^{sr^{n}}.\)

The matrix \(\mathbf{M}\) is given by \[ \mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\ \sin(2\pi/m) & \cos(2\pi/m) \end{pmatrix}, \] where \(m\) is an integer greater than \(1.\) Prove that \[ \mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O}, \] where $\mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( and \)\mathbf{O}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}.$ The sequence \(\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots\) is defined by \[ \mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q}, \] where \(\mathbf{P,Q}\) and \(\mathbf{X}_{0}\) are given \(2\times2\) matrices. Suggest a suitable expression for \(\mathbf{X}_{k}\) in terms of \(\mathbf{P},\) \(\mathbf{Q}\) and \(\mathbf{X}_{0},\) and justify it by induction. The binary operation \(*\) is defined as follows: \[ \mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. } \] Show that if \(\mathbf{P=M},\) the set \(\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}\) forms a finite group under the operation \(*\).

Let \(\Omega=\exp(\mathrm{i}\pi/3).\) Prove that \(\Omega^{2}-\Omega+1=0.\) Two transformations, \(R\) and \(T\), of the complex plane are defined by \[ R:z\longmapsto\Omega^{2}z\qquad\mbox{ and }\qquad T:z\longmapsto\dfrac{\Omega z+\Omega^{2}}{2\Omega^{2}z+1}. \] Verify that each of \(R\) and \(T\) permute the four point \(z_{0}=0,\) \(z_{1}=1,\) \(z_{2}=\Omega^{2}\) and \(z_{3}=-\Omega.\) Explain, without explicitly producing a group multiplication table, why the smallest group of transformations which contains elements \(R\) and \(T\) has order at least 12. Are there any permutations of these points which cannot be produced by repeated combinations of \(R\) and \(T\)?

Let \(G\) be a finite group with identity \(e.\) For each element \(g\in G,\) the order of \(g\), \(o(g),\) is defined to be the smallest positive integer \(n\) for which \(g^{n}=e.\)

1. Show that, if \(o(g)=n\) and \(g^{N}=e,\) then \(n\) divides \(N.\)
2. Let \(g\) and \(h\) be elements of \(G\). Prove that, for any integer \(m,\) \[ gh^{m}g^{-1}=(ghg^{-1})^{m}. \]
3. Let \(g\) and \(h\) be elements of \(G\), such that \(g^{5}=e,h\neq e\) and \(ghg^{-1}=h^{2}.\) Prove that \(g^{2}hg^{-2}=h^{4}\) and find \(o(h).\)

Give a careful argument to show that, if \(G_{1}\) and \(G_{2}\) are subgroups of a finite group \(G\) such that every element of \(G\) is either in \(G_{1}\) or in \(G_{2},\) then either \(G_{1}=G\) or \(G_{2}=G\). Give an example of a group \(H\) which has three subgroups \(H_{1},H_{2}\) and \(H_{3}\) such that every element of \(H\) is either in \(H_{1},H_{2}\) or \(H_{3}\) and \(H_{1}\neq H,H_{2}\neq H,H_{3}\neq H\).

Let \({\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}\) where \(m\) and \(n\) are non-negative integers. Prove that for \(m,n\geqslant1,\) \[ (m+n)I_{m,n}=-\cos^{m}x\cos nx+mI_{m-1,n-1}. \]

1. Show that \({\displaystyle \int_{0}^{\pi}\cos^{m}x\sin nx\,\mathrm{d}x=0}\) whenever \(m,n\) are both even or both odd.
2. Evaluate \({\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x.}\)

Let \[ \displaystyle I_n= \int_{-\infty}^\infty \frac 1 {(x^2+2ax+b)^n} \, \d x \] where \(a\) and \(b\) are constants with \(b > a^2\), and \(n\) is a positive integer.

1. By using the substitution \(x + a = \sqrt{b- a^2} \, \tan u\,\), or otherwise, show that \[ I_1 = \dfrac \pi {\sqrt{b-a^2}}\, . \]
2. Show that \(2n(b - a^2)\, I_{n+1} =(2n - 1) \, I_n\,\).
3. Hence prove by induction that \[ I_n =\frac{\pi}{2^{2n-2}( b - a^2)^{n-\frac12}} \, \binom {2n-2}{n-1} \]

1. Let \[ I_n= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u \,, \] where \(n\) is a positive integer. Show that \[ I_n - I_{n+1} = \frac 1 {2n} I_n \] and deduce that \[ I_{n+1} = \frac{(2n)!\, \pi}{2^{2n+1}(n!)^2} \,. \]
2. Let \[ J = \int_0^\infty \f\big( (x- x^{-1})^2\big ) \, \d x \,, \] where \(\f\) is any function for which the integral exists. Show that \[ J = \int_0^\infty x^{-2} \f\big( (x- x^{-1})^2\big) \, \d x \, = \frac12 \int_0^\infty (1 + x^{-2}) \f\big( (x- x^{-1})^2\big ) \, \d x \, = \int_0^\infty \f\big(u^2\big) \,\d u \,. \]
3. Hence evaluate \[ \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x \,, \] where \(n\) is a positive integer.

For \(n\ge 0\), let \[ I_n = \int_0^1 x^n(1-x)^n\d x\,. \]

1. For \(n\ge 1\), show by means of a substitution that \[ \int_0^1 x^{n-1}(1-x)^n\d x = \int_0^1 x^n(1-x)^{n-1}\d x\, \] and deduce that \[ 2 \int_0^1 x^{n-1}(1-x)^n\d x = I_{n-1}\,. \] Show also, for \(n\ge1\), that \[ I_n = \frac n {n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \] and hence that \(I_n = \dfrac{n}{2(2n+1)} I_{n-1}\,.\)
2. When \(n\) is a positive integer, show that \[ I_n = \frac{(n!)^2}{(2n+1)!}\,. \]
3. Use the substitution \(x= \sin^2 \theta\) to show that \(I_{\frac12}= \frac \pi 8\), and evaluate \(I_{\frac32}\).

Let \(m\) be a positive integer and let \(n\) be a non-negative integer.

1. Use the result \(\displaystyle \lim_{t\to\infty}\e^{-mt} t^n=0\) to show that \[ \lim_{x\to0} x^m (\ln x)^n =0\,. \] By writing \(x^x\) as \(\e^{x\ln x}\) show that \[ \lim _{x\to0} x^x=1\,. \]
2. Let \(\displaystyle I_{n} = \int_0^1 x^m (\ln x)^n \d x\,\). Show that \[ I_{n+1} = - \frac {n+1}{m+1} I_{n} \] and hence evaluate \(I_{n}\).
3. Show that \[ \int_0^1 x^x \d x = 1 -\left(\tfrac12\right)^2 +\left(\tfrac13\right)^3 -\left(\tfrac14\right)^4 + \cdots \,. \]

For \(n=1\), \(2\), \(3\), \(\ldots\,\), let \[ I_n = \int_0^1 {t^{n-1} \over \l t+1 \r^n} \, \mathrm{d} t \, . \] By considering the greatest value taken by \(\displaystyle {t \over t+1}\) for \(0 \le t \le 1\) show that \(I_{n+1} < {1 \over 2} I_{n}\,\). Show also that \(\; \displaystyle I_{n+1}= - \frac 1{\; n\, 2^n} + I_{n}\,\). Deduce that \(\; \displaystyle I_n < \frac1 {\; n \, 2^{n-1}}\,\). Prove that \[ \ln 2 = \sum_{r=1}^n {1 \over \; r\, 2^r} + I_{n+1} \] and hence show that \({2 \over 3} < \ln 2 < {17 \over 24}\,\).

If \[ \mathrm{I}_{n}=\int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x, \] show that \(\mathrm{I}_{0}=\pi a^{2}/8.\) Show that \((2n+4)\mathrm{I}_{n}=(2n+1)a\mathrm{I}_{n-1}\) and hence evaluate \(\mathrm{I}_{n}\).

Let \[ u_{n}=\int_{0}^{\frac{1}{2}\pi}\sin^{n}t\,\mathrm{d}t \] for each integer \(n\geqslant0\). By integrating \[ \int_{0}^{\frac{1}{2}\pi}\sin t\sin^{n-1}t\,\mathrm{d}t \] by parts, or otherwise, obtain a formula connecting \(u_{n}\) and \(u_{n-2}\) when \(n\geqslant2\) and deduce that \[ nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2} \] for all \(n\geqslant2\). Deduce that \[ nu_{n}u_{n-1}=\tfrac{1}{2}\pi. \] Sketch graphs of \(\sin^{n}t\) and \(\sin^{n-1}t\), for \(0\leqslant t\leqslant\frac{1}{2}\pi,\) on the same diagram and explain why \(0 < u_{n} < u_{n-1}.\) By using the result of the previous paragraph show that \[ nu_{n}^{2} < \tfrac{1}{2}\pi < nu_{n-1}^{2} \] for all \(n\geqslant1\). Hence show that \[ \left(\frac{n}{n+1}\right)\tfrac{1}{2}\pi < nu_{n}^{2} < \tfrac{1}{2}\pi \] and deduce that \(nu_{n}^{2}\rightarrow\tfrac{1}{2}\pi\) as \(n\rightarrow\infty\).

Show Solution

\begin{align*} && u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\ && &= \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\ && &= \left [ -\cos t \sin^{n-1} t \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\ && &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\ && &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\ && &= (n-1)u_{n-2} - (n-1)u_n \\ \Rightarrow && n u_n &= (n-1)u_{n-2} \\ \end{align*} Mutplying both sides by \(u_{n-1}\) we obtain \(nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}\). Therefore \(nu_nu_{n-1}\) is constant, ie is equal to \(\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}\)

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Since \(0 < \sin t < 1\) for \(t \in (0, \tfrac{\pi}{2})\) we must have \(0 < \sin^n t < \sin^{n-1} t\), in particular \(0 < u_n < u_{n-1}\) Therefore \begin{align*} && nu_{n}u_{n-1} &= \tfrac{1}{2}\pi \\ \Rightarrow && nu_n u_n &< \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_{n-1} u_{n-1} &> \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_n^2 &< \tfrac12 \pi < n u_{n-1}^2 \end{align*} However we also have \(\tfrac12 \pi < (n+1)u_n^2\) (by considering the next inequality), so \(\left ( \frac{n}{n+1}\right) \tfrac12 \pi < n u_n^2 < \tfrac12 \pi\) but since as \(n \to \infty\) the right hand bound is constant and the left hand bound tends to \(\tfrac12 \pi\) therefore \(n u_n^2 \to \tfrac12 \pi\)

Calculate \[ \int_{0}^{x}\mathrm{sech}\, t\,\mathrm{d}t. \] Find the reduction formula involving \(I_{n}\) and \(I_{n-2}\), where \[ I_{n}=\int_{0}^{x}\mathrm{sech}^{n}t\,\mathrm{d}t \] and, hence or otherwise, find \(I_{5}\) and \(I_{6}.\)

Given that \({\displaystyle I_{n}=\int_{0}^{\pi}\frac{x\sin^{2}(nx)}{\sin^{2}x}\,\mathrm{d}x,}\) where \(n\) is a positive integer, show that \(I_{n}-I_{n-1}=J_{n},\) where \[ J_{n}=\int_{0}^{\pi}\frac{x\sin(2n-1)x}{\sin x}\,\mathrm{d}x. \] Obtain also a reduction formula for \(J_{n}.\) The curve \(C\) is given by the cartesian equation \[ y=\dfrac{x\sin^{2}(nx)}{\sin^{2}x}, \] where \(n\) is a positive integer and \(0\leqslant x\leqslant\pi.\) Show that the area under the curve \(C\) is \(\frac{1}{2}n\pi^{2}.\)

1. The integral \(I_{k}\) is defined by \[ I_{k}=\int_{0}^{\theta}\cos^{k}x\,\cos kx\,\mathrm{d}x. \] Prove that \(2kI_{k}=kI_{k-1}+\cos^{k}\theta\sin k\theta.\)
2. Prove that \[ 1+m\cos2\theta+\binom{m}{2}\cos4\theta+\cdots+\binom{m}{r}\cos2r\theta+\cdots+\cos2m\theta=2^{m}\cos^{m}\theta\cos m\theta. \]
3. Using the results of (i) and (ii), show that \[ m\frac{\sin2\theta}{2}+\binom{m}{2}\frac{\sin4\theta}{4}+\cdots+\binom{m}{r}\frac{\sin2r\theta}{2r}+\cdots+\frac{\sin2m\theta}{2m} \] is equal to \[ \cos\theta\sin\theta+\cos^{2}\theta\sin2\theta+\cdots+\frac{1}{r}2^{r-1}\cos^{r}\theta\sin r\theta+\cdots+\frac{1}{m}2^{m-1}\cos^{m}\theta\sin m\theta. \]

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

Show Solution

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Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.