Taylor series

In this question, you should ignore issues of convergence.

1. Write down the binomial expansion, for \(\vert x \vert<1\,\), of \(\;\dfrac{1}{1+x}\,\) and deduce that %. By considering %$ %\displaystyle \int \frac 1 {1+x} \, \d x %\,, %$ %show that \[ \displaystyle \ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n \, \] for \(\vert x \vert <1 \,\).
2. Write down the series expansion in powers of \(x\) of \(\displaystyle \e^{-ax}\,\). Use this expansion to show that \[ \int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x \,\d x = \ln(1+a) \ \ \ \ \ \ \ (\vert a \vert <1)\,. \]
3. Deduce the value of \[ \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x \ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1) \,. \]

In this question, you should ignore issues of convergence.

1. Let \[ I = \int_0^1 \frac{\f(x^{-1}) } {1+x} \, \d x \,, \] where \(\f(x)\) is a function for which the integral exists. Show that \[ I = \sum_{n=1}^\infty \int_n^{n+1} \frac{\f(y) } {y(1+y)}\, \d y \] and deduce that, if \(\f(x) = \f(x+1)\) for all \(x\), then \[ I= \int_0^1 \frac{\f(x)} {1+x} \, \d x \,. \]
2. The {\em fractional part}, \(\{x\}\), of a real number \(x\) is defined to be \(x-\lfloor x\rfloor\) where \(\lfloor x \rfloor\) is the largest integer less than or equal to \(x\). For example \(\{3.2\} = 0.2\) and \(\{3\}=0\,\). Use the result of part (i) to evaluate \[ \displaystyle \int _0^1 \frac { \{x^{-1}\}}{1+x}\, \d x \text{ and } \displaystyle \int _0^1 \frac { \{2x^{-1}\}}{1+x}\, \d x \,. \]
3. (Bonus) Use the same method to evaluate \[ \int_0^1 \frac {x\{x^{-1}\}}{1-x^2} \, \d x \,. \]
4. (Bonus - harder) Use the same method to evaluate \[ \int_0^1 \frac {x^2\{x^{-1}\}}{1-x^2} \, \d x \,. \]

1. By use of calculus, show that \(x- \ln(1+x)\) is positive for all positive \(x\). Use this result to show that \[ \sum_{k=1}^n \frac 1 k > \ln (n+1) \,. \]
2. By considering \( x+\ln (1-x)\), show that \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2 \,. \]

In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]

In this question, you may assume that the infinite series \[ \ln(1+x) = x-\frac{x^2}2 + \frac{x^3}{3} -\frac {x^4}4 +\cdots + (-1)^{n+1} \frac {x^n}{n} + \cdots \] is valid for \(\vert x \vert <1\).

1. Let \(n\) be an integer greater than 1. Show that, for any positive integer \(k\), \[ \frac1{(k+1)n^{k+1}} < \frac1{kn^{k}}\,. \] Hence show that \(\displaystyle \ln\! \left(1+\frac1n\right) <\frac1n\,\). Deduce that \[ \left(1+\frac1n\right)^{\!n}<\e\,. \]
2. Show, using an expansion in powers of \(\dfrac1y\,\), that $ \displaystyle \ln \! \left(\frac{2y+1}{2y-1}\right) > \frac 1y %= \sum _{r=0}^\infty \frac 1{(2r+1)(2y)^{2r}}\,. \( for \)y>\frac12$. Deduce that, for any positive integer \(n\), \[ \e < \left(1+\frac1n\right)^{\! n+\frac12}\,. \]
3. Use parts (i) and (ii) to show that as \(n\to\infty\) \[ \left(1+\frac1n\right)^{\!n} \to \e\,. \]

1. Show that \[ \sum_{n=1} ^\infty \frac{n+1}{n!} = 2\e - 1 \] and \[ \sum _{n=1}^\infty \frac {(n+1)^2}{n!} = 5\e-1\,. \] Sum the series $\displaystyle \sum _{n=1}^\infty \frac {(2n-1)^3}{n!} \,.$
2. Sum the series $\displaystyle \sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)}\,$, giving your answer in terms of natural logarithms.

The function \(\f(t)\) is defined, for \(t\ne0\), by \[ \f(t) = \frac t {\e^t-1}\,. \] \begin{questionparts} \item By expanding \(\e^t\), show that \(\displaystyle \lim _{t\to0} \f(t) = 1\,\). Find \(\f'(t)\) and evaluate \(\displaystyle \lim _{t\to0} \f'(t)\,\). \item Show that \(\f(t) +\frac12 t\) is an even function. [{\bf Note:} A function \(\g(t)\) is said to be {\em even} if \(\g(t) \equiv \g(-t)\).] \item Show with the aid of a sketch that \( \e^t( 1-t)\le 1\,\) and deduce that \(\f'(t)\ne 0\) for \(t\ne0\). \end{questionpart} Sketch the graph of \(\f(t)\).

Show Solution

1. Claim \(f(t) + \frac12 t\) is an even function. Proof: Consider \(f(-t) - \frac12t\), then \begin{align*} f(-t) - \frac12t &= \frac{-t}{e^{-t}-1} - \frac12t \\ &= \frac{-te^t}{1-e^t} - \frac12 t \\ &= \frac{t(1-e^t) -t}{1-e^t} - \frac12 t \\ &= t - \frac{t}{1-e^t} - \frac12 t \\ &= \frac{t}{e^t-1} + \frac12 t \end{align*} So it is even.
2. 

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   Drawing the tangent to \(y = e^{-x}\) at \((0,1)\) we find that \(e^{-t} \geq (1-t)\) for all \(t\), in particular, \(e^t(1-t) \leq 1\) \(f'(t) = \frac{(e^t(1-t) -1}{(e^t-1)^2} \leq 0\) and \(f'(t) = -\frac12\) when \(t = 0\)

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[Note: This is the exponential generating function for the Bernoulli numbers]

1. Let \[ \tan x = \sum\limits_{n=0}^\infty a_n x^n \text{ and } \cot x = \dfrac 1 x +\sum\limits_{n=0}^\infty b_nx^n \] for \(0< x < \frac12\pi\,\). Explain why \(a_n=0\) for even \(n\). Prove the identity \[ \cot x - \tan x \equiv 2 \cot 2x\, \] and show that \[a_{n} = (1-2^{n+1})b_n\,.\]
2. Let $ \displaystyle {\rm cosec}\, x = \frac1x +\sum\limits _{n=0}^\infty c_n x^n\,$ for \(0< x < \frac12\pi\,\). By considering \(\cot x + \tan x\), or otherwise, show that \[ c_n = (2^{-n} -1)b_n \,. \]
3. Show that \[ \left(1+x{ \sum\limits_{n=0}^\infty} b_n x^n \right)^2 +x^2 = \left(1+x{ \sum\limits_{n=0} ^\infty} c_n x^n \right)^2\,. \] Deduce from this and the previous results that \(a_1=1\), and find \(a_3\).

The function \(f\) satisfies the identity \begin{equation} f(x) +f(y) \equiv f(x+y) \tag{\(*\)} \end{equation} for all \(x\) and \(y\). Show that \(2\f(x)\equiv \f(2x)\) and deduce that \(f''(0)=0\). By considering the Maclaurin series for \(\f(x)\), find the most general function that satisfies \((*)\). [{\it Do not consider issues of existence or convergence of Maclaurin series in this question.}]

1. By considering the function \(\G\), defined by \(\ln\big(\g(x)\big) =\G(x)\), find the most general function that, for all \(x\) and \(y\), satisfies the identity \[ \g(x) \g(y) \equiv \g(x+y)\,. \]
2. By considering the function \(H\), defined by \(\h(\e^u) =H(u)\), find the most general function that satisfies, for all positive \(x\) and \(y\), the identity \[ \h(x) +\h(y) \equiv \h(xy) \,. \]
3. Find the most general function \(t\) that, for all \(x\) and \(y\), satisfies the identity \begin{equation*} t(x) + t(y) \equiv t(z)\,, \end{equation*} where \(z= \dfrac{x+y}{1-xy}\,\).

Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.

The exponential of a square matrix \({\bf A}\) is defined to be $$ \exp ({\bf A}) = \sum_{r=0}^\infty {1\over r!} {\bf A}^r \,, $$ where \({\bf A}^0={\bf I}\) and \(\bf I\) is the identity matrix. Let $$ {\bf M}=\left(\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array} \right) \,. $$ Show that \({\bf M}^2=-{\bf I}\) and hence express \(\exp({\theta {\bf M}})\) as a single \(2\times 2\) matrix, where \(\theta\) is a real number. Explain the geometrical significance of \(\exp({\theta {\bf M}})\). Let $$ {\bf N}=\left(\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right) \,. $$ Express similarly \(\exp({s{\bf N}})\), where \(s\) is a real number, and explain the geometrical significance of \(\exp({s{\bf N}})\). For which values of \(\theta\) does $$ \exp({s{\bf N}})\; \exp({\theta {\bf M}})\, = \, \exp({\theta {\bf M}})\;\exp({s{\bf N}}) $$ for all \(s\)? Interpret this fact geometrically.

Sketch the graph of \({\rm f}(s)={ \e}^s(s-3)+3\) for \(0\le s < \infty\). Taking \({\e\approx 2.7}\), find the smallest positive integer, \(m\), such that \({\rm f}(m) > 0\). Now let $$ {\rm b}(x) = {x^3 \over \e^{x/T} -1} \, $$ where \(T\) is a positive constant. Show that \({\rm b}(x)\) has a single turning point in \(0 < x < \infty\). By considering the behaviour for small \(x\) and for large \(x\), sketch \({\rm b}(x)\) for \(0\le x < \infty\). Let $$ \int_0^\infty {\rm b}(x)\,\d x =B, $$ which may be assumed to be finite. Show that \(B = K T^n\) where \(K\) is a constant, and \(n\) is an integer which you should determine. Given that \(\displaystyle{B \approx 2 \int_0^{Tm} {\rm b}(x) {\,\rm d }x}\), use your graph of \({\rm b}(x)\) to find a rough estimate for \(K\).

1. By considering the series expansion of \((x^2+5x+4){\rm \; e}^x\) show that \[10{\rm\, e}=4+\frac{3^2}{1!}+\frac{4^2}{2!}+\frac{5^2}{3!}+\cdots\;.\]
2. Show that \[5{\rm\, e}=1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\;.\]
3. Evaluate \[1+\frac{2^3}{1!}+\frac{3^3}{2!}+\frac{4^3}{3!}+\cdots\;.\]

The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).

1. Evaluate \[ \sum_{r=1}^{n}\frac{6}{r(r+1)(r+3)}. \]
2. Expand \(\ln(1+x+x^{2}+x^{3})\) as a series in powers of \(x\), where \(\left|x\right|<1\), giving the first five non-zero terms and the general term.
3. Expand \(\mathrm{e}^{x\ln(1+x)}\) as a series in powers of \(x\), where \(-1 < x\leqslant1\), as far as the term in \(x^{4}\).

The points \(P\,(0,a),\) \(Q\,(a,0)\) and \(R\,(a,-a)\) lie on the curve \(C\) with cartesian equation \[ xy^{2}+x^{3}+a^{2}y-a^{3}=0,\qquad\mbox{ where }a>0. \] At each of \(P,Q\) and \(R\), express \(y\) as a Taylor series in \(h\), where \(h\) is a small increment in \(x\), as far as the term in \(h^{2}.\) Hence, or otherwise, sketch the shape of \(C\) near each of these points. Show that, if \((x,y)\) lies on \(C\), then \[ 4x^{4}-4a^{3}x-a^{4}\leqslant0. \] Sketch the graph of \(y=4x^{4}-4a^{3}x-a^{4}.\) Given that the \(y\)-axis is an asymptote to \(C\), sketch the curve \(C\).

Show Solution

\begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ \frac{\d }{\d x} : && 0 &= y^2+2xyy' + 3x^2+a^2y' \\ \Rightarrow && y' &= -\frac{y^2+3x^2}{a^2+2xy} \\ \\ \frac{\d^2 }{\d x^2}: && 0 &= 2yy'+2yy'+2x(y')^2+2xyy''+6x+a^2y'' \\ \Rightarrow && y'' &= -\frac{4yy'+2x(y')^2+6x}{a^2+2xy} \\ \\ P: && y &= a \\ && y' &= -\frac{a^2}{a^2} = -1 \\ && y'' &= -\frac{-4a}{a^2} = \frac{4}{a} \\ \Rightarrow && y &\approx a - h+\frac{2}{a}h^2 \\ \\ Q: && y &= 0 \\ && y' &= -\frac{3a^2}{a^2} = -3 \\ && y'' &= -\frac{18a+6a}{a^2} = -\frac{24}{a} \\ \Rightarrow && y &\approx 0-3h-\frac{12}{a}h \\ \\ R: && y &= -a \\ && y' &= -\frac{a^2+3a^2}{a^2-2a^2} = 4 \\ && y'' &= -\frac{-16a+32a+6a}{a^2-2a^2} = \frac{22}{a} \\ \Rightarrow && y &\approx -a+4h + \frac{11}{a}h^2 \end{align*} Alternatively: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ P(0,a): && y &\approx a + c_1h + c_2h^2 \\ && 0 &= h(a+c_1h)^2 + a^2(a + c_1h + c_2h^2)-a^3 \\ &&&= a^3-a^3 + (a^2+a^2c_1)h+(2ac_1+a^2c_2)h^2 \\ \Rightarrow && c_1 &=-1, c_2 =\frac{2}{a} \\ \Rightarrow && y &\approx a - h + \frac{2}{a}h^2 \\ \\ Q(a,0): && y &\approx c_1h + c_2h^2 \\ && 0 &= (a+h)(c_1h)^2+(a+h)^3+a^2(c_1h + c_2h^2 )-a^3 \\ &&&= a^3-a^3+(3a^2+a^2c_1)h + (ac_1^2+3a+a^2c_2)h^2 + \cdots \\ \Rightarrow && c_1 &=-3, c_2 = -\frac{12}{a} \\ \Rightarrow && y &\approx -3h -\frac{12}{a}h \\ \\ R(a,-a): && y &\approx -a + c_1h + c_2h^2 \\ && 0 &= (a+h)(-a + c_1h+c_2h^2)^2+(a+h)^3+a^2(-a + c_1h + c_2h^2)-a^3 \\ &&&= (a^2-2a^2c_1+3a^2+a^2c_1)h+(-2ac_1+c_1^2+\cdots)h^2 \\ \Rightarrow && c_1 &=4, c_2 = \frac{11}{a} \\ \Rightarrow && y &\approx -a + 4h + \frac{11}{a} \end{align*}

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If \((x,y)\) lies on the curve, then viewing it as a quadratic in \(y\) we must have \(\Delta = (a^2)^2-4\cdot x \cdot (x^3-a^3) \geq 0 \Rightarrow a^4-4x^4+4xa^3 \geq 0 \Rightarrow 4x^4-4a^3x-a^4 \leq 0\)

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Let \begin{alignat*}{2} \tan x & =\ \ \, \quad{\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}} & & \text{ for small }x,\\ x\cot x & =1+\sum_{n=1}^{\infty}b_{n}x^{n}\quad & & \text{ for small }x\text{ and not zero}. \end{alignat*} Using the relation \[ \cot x-\tan x=2\cot2x,\tag{*} \] or otherwise, prove that \(a_{n-1}=(1-2^{n})b_{n}\), for \(n\geqslant1\). Let \[ x\mathrm{cosec}x=1+{\displaystyle \sum_{n=1}^{\infty}c_{n}x^{n}\quad\text{ for small }x\neq0. \qquad \qquad \, } \] Using a relation similar to \((*)\) involving \(2\mathrm{cosec}2x\), or otherwise, prove that \[ c_{n}=\frac{2^{n-1}-1}{2^{n}-1}\frac{1}{2^{n-1}}a_{n-1}\qquad(n\geqslant1). \]