Home » Hyperbolic functions

Tag: Hyperbolic functions

Prove some properties of the complex sine and cosine functions

by RoRi

The following definitions extend the sine and cosine functions to take arguments z \in \mathbb{C}:

    \[ \cos z = \frac{e^{iz} + e^{-iz}}{2}, \qquad \sin z = \frac{e^{iz} - e^{-iz}}{2i}. \]

Prove the following formulas, where u,v,z are complex numbers and z = x + iy.

  1. \sin (u+v) = \sin u \cos v + \cos u \sin v.
  2. \cos (u+v) = \cos u \cos v - \sin u \sin v.
  3. \sin^2 z + \cos^2 z = 1.
  4. \cos (iy) = \cosh y, \qquad \sin(iy) = i \sinh y.
  5. \cos z = \cos x \cosh y - i \sin x \sinh y.
  6. \sin z = \sin x \cosh y + i \cos x \sinh y.
  1. Proof. Using the given definition of the sine of complex numbers we have

        \begin{align*} \sin (u+v) &= \frac{e^{i(u+v)} - e^{-i(u+v)}}{2i} \\[9pt] &= \frac{e^{iu}e^{iv} - e^{-iu}e^{-iv}}{2i} \\[9pt] &= \frac{2 e^{iu}e^{iv} - 2 e^{-iu}e^{-iv}}{2(2i)} \\[9pt] &= \frac{2e^{iu}e^{iv} + e^{iu}e^{-iv} - e^{iu}e^{-iv} + e^{-iu}e^{iv} - e^{-iu}e^{iv} - 2e^{-iu}e^{-iv}}{2(2i)} \\[9pt] &= \frac{e^{iu}(e^{iv} + e^{-iv}) - e^{-iu}(e^{iv} + e^{-iv}) + e^{iv}(e^{iu} + e^{-iu}) - e^{-iv}(e^{iu} + e^{-iv})}{2(2i)} \\[9pt] &= \frac{(e^{iu} - e^{-iu})(e^{iv} + e^{-iv})}{2(2i)} + \frac{(e^{iv} - e^{-iv})(e^{iu} + e^{-iu})}{2(2i)} \\[9pt] &= \sin u \cos v + \sin v \cos u. \qquad \blacksquare \end{align*}

  2. Proof. Similar to part (a) we compute

        \begin{align*} \cos (u+v) &= \frac{1}{2} \big( e^{i(u+v)} + e^{-i(u+v)} \big) \\[9pt] &= \frac{1}{2} \big( e^{iu} e^{iv} + e^{-iu} e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( 2 e^{iu} e^{iv} + 2 e^{-iu} e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( 2e^{iu}e^{iv} + e^{-iu}e^{iv} - e^{-iu}e^{iv} + e^{iu}e^{-iv} - e^{iu}e^{-iv} + 2 e^{-iu}e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( e^{iu}(e^{iv} + e^{-iv}) + e^{-iu}(e^{iv} + e^{-iv}) + e^{iu}(e^{iv} - e^{-iv}) - e^{-iu}(e^{iv} - e^{-iv}) \big) \\[9pt] &= \frac{1}{4} \big( (e^{iu} + e^{-iu})(e^{iv} + e^{-iv}) \big) + \frac{1}{4} \big( (e^{iu} - e^{-iu})(e^{iv} - e^{-iv})\big) \\[9pt] &= \cos u \cos v - \frac{1}{4i^2} \big( (e^{iu} - e^{-iu})(e^{iv} - e^{-iv})\big)\\[9pt] &= \cos u \cos v - \sin u \sin v. \qquad \blacksquare \end{align*}

  3. Proof. We compute

        \begin{align*} \sin^2 z + \cos^2 z &= \left( \frac{1}{2i} \right)^2 (e^z - e^{-z})^2 + \left( \frac{1}{2} \right)^2 (e^z + e^{-z})^2 \\[9pt] &= -\frac{1}{4} (e^z - e^{-z})^2 + \frac{1}{4}(e^z + e^{-z})^2 \\[9pt] &= \frac{1}{4} \left( (e^z+e^{-z})^2 - (e^z - e^{-z})^2 \right) \\[9pt] &= \frac{1}{4} \left( (e^z + e^{-z} + e^z - e^{-z})(e^z + e^{-z} - e^z + e^{-z}) \right) \\[9pt] &= \frac{1}{4} \left( (2e^z)(2e^{-z}) \right) \\[9pt] &= 1. \qquad \blacksquare \end{align*}

  4. Proof. The two computations are as follows,

        \begin{align*} \cos(iy) &= \frac{1}{2} \big( e^{i(iy)} + e^{-i(iy)} \big) \\[9pt] &= \frac{1}{2}(e^{-y} + e^y) \\[9pt] &= \frac{1}{2}(e^y + e^{-y}) \\[9pt] &= \cosh y. \\[9pt] \sin (iy) &= \frac{1}{2i} \big( e^{i(iy)} - e^{-i(iy)} \big) \\[9pt] &= \frac{1}{2i} (e^{-y} - e^y) \\[9pt] &= -\frac{1}{2i} (e^y - e^{-y}) \\[9pt] &= i \sinh y. \qquad \blacksquare \end{align*}

  5. Proof. We have,

        \[ \cos z = \cos (x+iy) = \cos x \cos (iy) - \sin x \sin (iy) = \cos x \cosh y - i \sin x \sinh y. \qquad \blacksquare \]

  6. Proof. We have,

        \[ \sin z = \sin(x+iy) = \sin x \cos (iy) + \cos x \sin (iy) = \sin x \cosh y + i \cos x \sinh y. \qquad \blacksquare \]

Determine a differential equation governing a falling body in resisting medium

by RoRi

Modify the equations (Example 2 on page 314 of Apostol) for the velocity of a falling body in a resisting medium if the resistance of the medium is proportional to v^2 instead of to v. Prove that the resulting differential equation can be written in each of the following forms:

    \[ \frac{ds}{dv} = \frac{m}{k} \frac{v}{c^2 - v^2}, \qquad \frac{dt}{dv} = \frac{m}{k} \frac{1}{c^2-v^2}, \]

where c = \sqrt{\frac{mg}{k}}. By integrating these find the following formulas for v:

    \begin{align*} v^2 &= \frac{mg}{k} \left( 1 - e^{\frac{-2ks}{m}} \right) \\[10pt] v &= c \frac{e^{bt} - e^{-bt}}{e^{bt} + e^{-bt}} = c \tanh (bt), \end{align*}

where b = \sqrt{\frac{kg}{m}}. Determine the value of v as t \to +\infty.

Starting with the equation

    \[ ma =mg - kv \]

in example 2 and modifying it so that the resistance is proportional to v^2 we have

    \begin{align*} ma = mg - kv^2 && \implies && mv' &= mg - kv^2 \\ && \implies && v' = g - \frac{k}{m} v^2. \end{align*}

Using the chain rule as we did in the previous exercise we know

    \[ \frac{dv}{dt} = v \cdot \frac{dv}{ds}. \]

Therefore,

    \begin{align*} &&v \cdot \frac{dv}{ds} &= g - \frac{k}{m} v^2 \\[9pt] \implies && \frac{dv}{ds} &= \frac{g}{v} - \frac{k}{m} v \\[9pt] \implies && \frac{dv}{ds} &= \frac{gm-kv^2}{mv} \\[9pt] \implies && \frac{ds}{dv} &= \frac{mv}{gm-kv^2}. \end{align*}

Letting c = \sqrt{\frac{mg}{k}} we then have

    \[ \frac{ds}{dv} = \frac{m}{k} \left( \frac{v}{c^2-v^2} \right). \]

This is the first requested equation.

Alternatively,

    \begin{align*} v' = g - \frac{k}{m} v^2 && \implies && \frac{dv}{dt} &= \frac{mg-kv^2}{m} \\[9pt] && \implies && \frac{dt}{dv} &= \frac{m}{mg-kv^2} \\[9pt] && \implies && \frac{dt}{dv} &= \frac{m}{k} \left( \frac{1}{c^2-v^2} \right). \end{align*}

Integrating the first equation we find,

    \begin{align*} && \frac{ds}{dv} &= \frac{m}{k} \frac{v}{c^2-v^2} \\[9pt] \implies && s &= \frac{m}{k} \int \frac{v}{c^2-v^2} \, dv \\[9pt] \implies && s &= -\frac{m}{2k} \log (c^2 - v^2) \\[9pt] \implies && e^{-\frac{2ks}{m}} &= c^2 - v^2 \\[9pt] \implies && v^2 &= \frac{mg}{k} - e^{-2ks}{m}. \end{align*}

Integrating the second equation,

    \begin{align*} && \frac{dt}{dv} &= \frac{m}{k} \cdot \frac{1}{c^2-v^2} \\[9pt] \implies && t &= \frac{m}{k} \int \frac{1}{c^2-v^2} \, dv \\[9pt] \implies && t &= \frac{1}{2} \frac{m}{k} \frac{1}{c} log \left(\frac{c+v}{c-v}\right) \\[9pt] \implies && 2bt &= \log \left( \frac{c+v}{c-v} \right) &(b = \sqrt{\frac{kg}{m}} )\\[9pt] \implies && e^{2bt} &= \frac{c+v}{c-v} \\[9pt] \implies && v &= c \cdot \frac{e^{2bt} - 1}{e^{2bt} + 1} \\[9pt] \implies && v &= c \cdot \tanh (bt). \end{align*}

This implies

    \[ \lim_{t \to \infty} v = 1. \qquad \blacksquare\]

Find all x satisfying equations given in terms of sinh

by RoRi

Let c be the number such that \sinh c = \frac{3}{4}. Find all x that satisfy the given equations.

  1. \log (e^x + \sqrt{e^{2x} + 1}) = c.
  2. \log (e^x - \sqrt{e^{2x} - 1}) = c.
  1. We are given \sinh c = \frac{3}{4}. From the formula for \sinh this means

        \[ \frac{e^c - e^{-c}}{2} = \frac{3}{4}. \]

    Then, from the given equation we have

        \[ \log (e^x + \sqrt{e^{2x} + 1}) = c \quad \implies \quad e^x + \sqrt{e^{2x} + 1} = e^c. \]

    Thus,

        \[ e^{-c} = \frac{1}{e^x + \sqrt{e^{2x} + 1}}. \]

    So, then we have

        \begin{align*} \frac{3}{4} = \frac{e^c - e^{-c}}{2} &= \frac{1}{2} \left( e^x + \sqrt{e^{2x} + 1} - \frac{1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt] &= \frac{1}{2} \left( \frac{(e^x + \sqrt{e^{2x}+1})^2 - 1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt] &= \frac{e^{2x} + 2e^x \sqrt{e^{2x}+1} + e^{2x}+1 - 1}{2(e^x + \sqrt{e^{2x}+1})} \\[9pt] &= \frac{e^{2x} + e^x \sqrt{e^{2x}+1}}{e^x + \sqrt{e^{2x}+1}} \\[9pt] &= e^x \end{align*}

    Therefore we have

        \[ e^x = \frac{3}{4} \quad \implies \quad x = \log 3 - \log 4 = \log 3 - 2 \log 2.\]

  2. There can be no x which satisfy the given equation. As in part (a), we use the definition of \sinh x to obtain the equation,

        \[ \sinh c = \frac{3}{4} \quad \implies \quad e^c - e^{-c} = \frac{3}{2}. \]

    Next, we use the equation given in the problem to write,

        \begin{align*} &&\log (e^x - \sqrt{e^{2x} - 1}) &= c \\[9pt] \implies && e^x - \sqrt{e^{2x} -1} &= e^c \\[9pt] \implies && \frac{(e^x - \sqrt{e^{2x} - 1})(e^x + \sqrt{e^{2x}-1})}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt] \implies && \frac{e^{2x} - e^{2x} + 1}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt] \implies && \frac{1}{e^x + \sqrt{e^{2x}-1}} &= e^c. \end{align*}

    Furthermore, we can obtain an expression for e^{-c} by considering

        \[ e^x - \sqrt{e^{2x} - 1} &= e^c \quad \implies \quad \frac{1}{e^x - \sqrt{e^{2x} - 1}} &= e^{-c}. \]

    Putting these expressions for e^c and e^{-c} into our original equation we have

        \begin{align*} \frac{3}{2} &= e^c - e^{-c} \\[9pt] &= \left( \frac{1}{e^x + \sqrt{e^{2x}-1}} \right) - \left( \frac{1}{e^x - \sqrt{e^{2x}-1}} \right) \\[9pt] &= \frac{e^x - \sqrt{e^{2x}-1} - e^x - \sqrt{e^{2x}-1}}{(e^x + \sqrt{e^{2x}-1})(e^x - \sqrt{e^{2x}-1})} \\[9pt] &= \frac{ -2 \sqrt{e^{2x}-1}}{e^{2x} - e^{2x} + 1} \\[9pt] &= -2\sqrt{e^{2x}-1} \end{align*}

    But this implies

        \[ \sqrt{e^{2x}-1} = -\frac{3}{4} \]

    which is impossible. Hence, there can be no real x satisfying this equation.