Why does the distance between approach $\ln(2)$ as y increases?

Question

I was playing with some online graphing tools and I ended up with this:

I was wondering why the x value for the blue curve subtracted by the green/black curve approached $\ln2$ as y increased. Is there an explanation for this? Or did I just miss something really obvious?
Thanks in advance.

Answer 1

You can use the following formulas:

• $\cosh(x) = \frac{e^x + e^{-x}}{2}$
• $\sinh(x) = \frac{e^x - e^{-x}}{2}$
• $\cosh^{-1}(x) = \ln(x + \sqrt{x+1}\sqrt{x-1})$

Then what you ask for is a limit of $\cosh^{-1}(\cosh(x) + \sinh(x)) - x$. By inserting the above formulas and taking the limit for $x\to\infty$ (i.e. $e^x\pm1\approx e^x$) you exactly obtain $\ln(2)$.

Answer 2

Considering the positive quadrant, you have

• $\cosh(x)+\sinh(x)= \frac{e^x+e^{-x}}{2} + \frac{e^x-e^{-x}}{2} =e^x$ with inverse function $\log_e \left(y \right)$

• $\operatorname {arcosh} (y) -\log_e(y)= \log_e \left(y + \sqrt{y^{2} - 1}\right) -\log_e(y) = \log_e\left(1+ \sqrt{1 - \frac{1}{y^2}}\right) \to \log_e \left(2\right)$ as $y\to \infty$

• so the difference between the two tends to $\log_e \left(2\right)$ as $y$ increases

The negative $x$ case is similar, with the first bullet using $\cosh(x)-\sinh(x)=e^{-x}$ and so its inverse being $-\log_e(y)$ so the difference becomes $-\log_e\left(1+ \sqrt{1 - \frac{1}{y^2}}\right) \to -\log_e \left(2\right)$ as $y$ increases

Answer 3

Starting from @NDewolf's answer

$$\Delta(x)=\cosh^{-1}(\cosh(x) + \sinh(x)) - x=\cosh^{-1}(e^x) - x$$ Using expansions for infinitely large values of $x$, you would get $$\Delta(x)=\log(2)-\frac{1}{4} e^{-2 x}-\frac{3}{32} e^{-4 x}-\frac{5}{96} e^{-6 x}-\cdots$$

Just use it for $x=1$ (quite far away from $\infty$). The exact value is $0.657454$ while the above truncated expansion gives $0.657467$.

More rigorously, $$\Delta(x)=\log(2)-\sum_{n=1}^\infty \frac{ (2 n-1)!}{2^{2 n}\,(n!)^2}e^{-2nx}$$

Answer 4

Presumably you mean the horizontal distance between the curves, which are respectively

$$y_B=e^x,y_g=e^{-x}\text{ and }y_b=\cosh(x)=\frac{e^x+e^{-x}}2.$$

It is clear that for large $x$, $y_B\approx2y_b$, and we have $2e^x=e^{x\color{red}{+\log 2}}$. Similarly, for negative $x$, $y_B\approx2y_g$.