I have 2 questions with regards to solving of hyperbolic functions. I have presented my current solutions to the best of my ability.

Q1: Show that the real solution $x$ of $\tanh(x) = \operatorname{csch}(x)$ can be written in the form $x=\ln(u)+ \sqrt{u},$ where $u$ is to be determined.

My attempt: write $\tanh x =\operatorname{csch}(x)$ as \begin{align*} \dfrac{\sinh(x)}{\cosh (x)}&= \dfrac{1}{\sinh(x)}\iff \\ \sinh^{2}(x)&=\cosh(x)\iff \\ \cosh ^{2}(x)-1&= \cosh(x) \iff\\ \cosh ^{2}(x)- \cosh(x)-1&=0 \iff \\ \cosh(x)&= \dfrac{1\pm\sqrt{5}}{2}. \end{align*}

Writing $\cosh(x) =\dfrac{e^{x}+e^{-x}}{2}$, we have $ \dfrac{1+\sqrt{5}}{2}=\dfrac{e^{x}+e^{-x}}{2} \iff e^{2x}-(1+\sqrt{5})e^{x}+1=0$. Am I on the right track? This is where I am stuck because by applying the quadratic formula to solve for $e^{x}$ yields a double square root.

Q2: Solve $\cosh(4x)+4\cosh(2x)-125=0$.

My attempt: By using the identities $\cosh(4x)= \cosh^{2}(2x)+ \sinh^{2}(2x)$ and $\sinh^{2}(2x)=1+\cosh^{2}(2x)$ and substituting into the original equation and simplifying, we obtain: $$\cosh^{2}(2x)+ 2\cosh(2x)-62=0.$$ Solving, we obtain $\cosh(2x)=-1+3\sqrt{7}$ or $\cosh(2x)=-1-3\sqrt{7}$. Like the above problem, I am stuck but am I on the right track?

  • $\begingroup$ I don't see the form $\ln(u)+\sqrt u$ in your proposal. But in fact, it is very unlikely, there must be a typo. $\endgroup$ – user65203 Jun 11 '18 at 12:57
  • $\begingroup$ Part 1 is trivial. Every real number can be written as $ f(u) = \ln u + \sqrt{u}$ with $ u > 0,$ because $f(u)$ is strictly increasing, $f(u) \rightarrow -\infty$ for $u \rightarrow 0$ and $f(u) \rightarrow \infty$ for $u \rightarrow \infty.$ $\endgroup$ – gammatester Jun 11 '18 at 13:00

You're on the right track: the quadratic formula tells you that $$ e^x=\frac{1+\sqrt{5}\pm\sqrt{(1+\sqrt{5})^2-4}}{2}= \frac{1+\sqrt{5}\pm\sqrt{2(1+\sqrt{5})}}{2} $$ If you set $u=\frac{1+\sqrt{5}}{2}$, then you get either $$ e^x=u+\sqrt{u} $$ or $$ e^x=u-\sqrt{u} $$ On the other hand $$ u-\sqrt{u}=\frac{u^2-u}{u+\sqrt{u}}=\frac{1}{u+\sqrt{u}} $$ So the first solution is $$ x=\ln(u+\sqrt{u}) $$ and the second solution is $$ x=-\ln(u+\sqrt{u}) $$ The positive real solution of your equation is of the stated form. Notice that $$ \frac{\sinh(-x)}{\cosh(-x)}=-\frac{\sinh x}{\cosh x} \qquad \frac{1}{\sinh(-x)}=-\frac{1}{\sinh x} $$ so any positive solution is accompanied by a negative one.

In both problems 1 and 2, the negative solution for $\cosh x$ must be discarded.

  • $\begingroup$ thanks for your help. For the second question, can you Kindly check if my workings are correct as I still can't seem to arrive at an answer. Likewise, I let $\cosh 2x = \dfrac{e^{2x}+e^{-2x}}{2}=-1+3\sqrt{7}$. Simplifying and using the quadratic equation I obtain an ugly expression. $\endgroup$ – Cleytus Jun 11 '18 at 13:50
  • $\begingroup$ You get $e^{4x}-2(3\sqrt{7}-1)e^{2x}+1=0$, which is solved by the quadratic formula in the same way: finally you get $2x=\ln(3\sqrt{7}-1+\sqrt{63-6\sqrt{7}})$ and its negative. $\endgroup$ – egreg Jun 11 '18 at 13:58

If $\cosh(x) = y$, then, as stated, $e^x+e^{-x} = 2y$ or $e^{2x}-2ye^x+1 = 0$.

Solving, $e^x =\dfrac{2y\pm\sqrt{4y^2-4}}{2} =y\pm\sqrt{y^2-1} $ or $x =\ln(y\pm\sqrt{y^2-1}) $.

If $y^2-1 = y$ (i.e., $y=\dfrac{1\pm\sqrt{5}}{2} $), then $x=\ln(y\pm\sqrt{y})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.