I have the following expression
$$\sqrt\frac{\left(1+e^{k(c_2-c_1)}\right)^2}{e^{k(c_2-c_1)}}$$
Why is it that the expression equals the following:
$$\cosh\left(\frac{k}{2}(c_2-c_1)\right)$$
I know that $\cosh(x)= \cosh(-x)=(e^x+e^{-x})/2$, but I cannot relate this the the first expression.
The first expression, think equals $$e^{-k/2(c_2-c_1)}+e^{k/2(c_2-c_1)}$$
On
The formula's not right as is. But I can still explain the "trick" here.
Given a formula of the form
$$ x + \frac{1}{x} $$
one might instead prefer to collect the terms and write it as
$$ \frac{x^2 + 1}{x} $$
That's basically what's going on here; the formula for the hyperbolic cosine can be rearranged to
$$ \cosh(x) = \frac{e^{2x} + 1}{2 e^x} $$
On
The hyperbolic functions can also be written as $$ \cosh x=\frac{\mathrm e^{2x}+1}{2\mathrm e^x},\quad\sinh x=\frac{\mathrm e^{2x}-1}{2\mathrm e^x},\quad\tanh x=\frac{\mathrm e^{2x}-1}{\mathrm e^{2x}+1}.$$
On
Let $x = c_2-c_1$ , then:$$ \cosh\left(\frac{k}{2}x\right)=\frac{e^{\frac{k}{2}x}+e^{-\frac{k}{2}x}}{2}=\frac{e^{-\frac{k}{2}x}\left(1+e^{kx}\right)}{2}=\frac{1+e^{kx}}{2e^{\frac{k}{2}x}}=\frac{1}{2}\sqrt\frac{\left(1+e^{kx}\right)^2}{e^{kx}}$$
On
Divide the denominator into each term of the numerator, noting the square. $$\sqrt\frac{\left(1+e^{k(c_2-c_1)}\right)^2}{e^{k(c_2-c_1)}}=\sqrt{\left(e^{-\frac k2(c_2-c_1)}+e^{\frac k2(c_2-c_1)}\right)^2}\\=e^{-\frac k2(c_2-c_1)}+e^{\frac k2(c_2-c_1)} \\=2\cosh\left(\frac k2(c_2-c_1)\right)$$ The equality is off by the factor $2$
set $a=k(c_2-c_1)$ then I obtain as follow