Summation bounds using hyperbolic integrals

A question is this type if and only if it asks to find upper or lower bounds for a summation by considering rectangles under a curve involving hyperbolic or inverse hyperbolic functions.

4 questions · Challenging +1.8

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CAIE Further Paper 2 2023 June Q7
11 marks Challenging +1.8
7
  1. Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\).
    \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-12_778_1548_1007_296} The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
  2. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
  3. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).
CAIE Further Paper 2 2023 June Q7
11 marks Challenging +1.8
7
  1. Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\).
    \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-12_778_1548_1007_296} The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
  2. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
  3. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).
CAIE Further Paper 2 2022 November Q4
12 marks Challenging +1.2
4
  1. Starting from the definitions of cosh and sinh in terms of exponentials, prove that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1 .$$
  2. Show that \(\frac { \mathrm { d } } { \mathrm { dx } } \left( \tan ^ { - 1 } ( \sinh x ) \right) = \operatorname { sech } x\).
  3. Sketch the graph of \(y = \operatorname { sechx }\), stating the equation of the asymptote.
  4. By considering a suitable set of \(n\) rectangles of unit width, use your sketch to show that $$\sum _ { r = 1 } ^ { n } \operatorname { sechr } < \tan ^ { - 1 } ( \operatorname { sinhn } )$$
  5. Hence state an upper bound, in terms of \(\pi\), for \(\sum _ { r = 1 } ^ { \infty }\) sech \(r\).
CAIE Further Paper 2 2023 November Q5
10 marks Hard +2.3
5
\includegraphics[max width=\textwidth, alt={}, center]{1fa404d4-5e14-4356-9b6d-f176d5a9f6db-08_773_1161_278_443} The diagram shows part of the curve \(\mathrm { y } = \mathrm { xsech } ^ { 2 } \mathrm { x }\) and its maximum point \(M\).
  1. Show that, at \(M\), $$2 x \tanh x - 1 = 0$$ and verify that this equation has a root between 0.7 and 0.8 .
  2. By considering a suitable set of rectangles, use the diagram to show that
    \(\sum _ { r = 2 } ^ { n } r \operatorname { sech } ^ { 2 } r < n \tanh n + \operatorname { lnsechn } - \tanh 1 - \operatorname { lnsech } 1\).