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2021 Paper 2 Q3
D: 1500.0 B: 1500.0
floor function fractional part simultaneous equations integer and fractional decomposition casework

In this question, \(x\), \(y\) and \(z\) are real numbers. Let \(\lfloor x \rfloor\) denote the largest integer that satisfies \(\lfloor x \rfloor \leqslant x\) and let \(\{x\}\) denote the fractional part of~\(x\), so that \(x = \lfloor x \rfloor + \{x\}\) and \(0 \leqslant \{x\} < 1\). For example, if \(x = 4.2\), then \(\lfloor x \rfloor = 4\) and \(\{x\} = 0.2\) and if \(x = -4.2\), then \(\lfloor x \rfloor = -5\) and \(\{x\} = 0.8\).

  1. Solve the simultaneous equations \begin{align*} \lfloor x \rfloor + \{y\} &= 4.9, \\ \{x\} + \lfloor y \rfloor &= -1.4. \end{align*}
  2. Given that \(x\), \(y\) and \(z\) satisfy the simultaneous equations \begin{align*} \lfloor x \rfloor + y + \{z\} &= 3.9, \\ \{x\} + \lfloor y \rfloor + z &= 5.3, \\ x + \{y\} + \lfloor z \rfloor &= 5, \end{align*} show that \(\{y\} + z = 3.2\) and solve the equations.
  3. Solve the simultaneous equations \begin{align*} \lfloor x \rfloor + 2y + \{z\} &= 3.9, \\ \{x\} + 2\lfloor y \rfloor + z &= 5.3, \\ x + 2\{y\} + \lfloor z \rfloor &= 5. \end{align*}

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2018 Paper 3 Q8
D: 1700.0 B: 1516.0
integration Taylor series fractional part substitution partial fractions series manipulation periodic function floor function

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 \[ \int_0^1 \frac {x^2\{x^{-1}\}}{1-x^2} \, \d x \,. \]

Solution:

  1. \begin{align*} && I &= \int_0^1 \frac{f(x^{-1})}{1+x} \d x \\ u = x^{-1}, \d u = -x^{-2} \d x: &&&= \int_{\infty}^1 \frac{f(u)}{1+\frac{1}{u}} \frac{-1}{u^2} \d u \\ &&&= \int_1^{\infty} \frac{f(u)}{u(1+u)} \d u \\ &&&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(u)}{u(u+1)} \d u \\ \\ \text{if} f(x) = f(x+1)\, \forall x && &=\sum_{n=1}^{\infty} \int_{0}^1 \frac{f(x+n)}{(x+n)(x+n+1)} \d x \\ &&&= \sum_{n=1}^\infty \int_0^1 \frac{f(x)}{(x+n)(x+n+1)} \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n+1)}\r \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \l \frac{1}{x+n} - \frac{1}{x+n+1} \r\r \d x \\ &&&= \int_0^1 f(x) \l \frac{1}{x+1} \r \d x \\ &&&= \int_0^1\frac{f(x)}{x+1} \d x \\ \end{align*}
  2. Since the fractional part is periodic with period \(1\), we can say \begin{align*} && \int_0^1 \frac{\{ x^{-1} \} }{1+x} \d x &= \int_0^1 \frac{\{ x\}}{x+1} \d x \\ &&&= \int_0^1 \frac{x}{x+1} \d x \\ &&&= \int_0^1 1-\frac{1}{x+1} \d x \\ &&&= [x - \ln (1+x) ]_0^1 \\ &&&= 1 - \ln 2 \end{align*} \begin{align*} && \int_0^1 \frac{\{ 2x^{-1}\}}{1+x} \d x &= \int_0^1 \frac{\{2x\}}{x+1} \d x \\ &&&= \int_0^{1/2} \frac{2x}{x+1} \d x +\int_{1/2}^{1} \frac{2x-1}{x+1} \d x \\ &&&= 2\int_0^1 \frac{x}{x+1} \d x + \int_{1/2}^1 \frac{-1}{x+1} \d x \\ &&&= 2 - 2\ln 2 - \l \ln 2 - \ln \tfrac32 \r \\ &&&= 2 - 4 \ln 2 + \ln 3 \\ &&&= 2 + \ln \tfrac {3}{16} \end{align*}
  3. \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x\r \end{align*} Consider for \(f\) periodic with period \(1\) \begin{align*} \int_0^1 \frac{f(x^{-1})}{1-x} \d x &= \int_1^{\infty} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{0}^{1} \frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} \sum_{n=1}^{\infty}\frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} f(u) \sum_{n=1}^{\infty}\l\frac{1}{u+n-1} - \frac{1}{u+n} \r\d u \\ &= \int_0^1 \frac{f(u)}{u} \d u \end{align*} So we have \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x \r \\ &&&= \frac12 \int_0^1 \frac{\{ x \}}{x} \d x - \frac12 (1 - \ln 2) \\ &&&= \frac12 - \frac12 + \frac12 \ln 2 \\ &&&= \frac12 \ln 2 \end{align*}

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