I want to find the Taylor expansion of $y=\frac {V_t^2}{g} \ln(\cosh(\frac{gt}{V_t}))$
I have tried using the fact $\cosh x= \frac {e^x}{2}$ for large t, which works, I just need help on small values for t.
Using maple17 for small t, I was able to see it expands to $\frac{1}{2}gt^2 - \frac{1}{12}\frac{g^3}{V_t^2}t^4 - o(t^6)$ but I cannot see how it got to this.
For generality, we work with the function supplied by the OP, but with arbitrary constants,
$$f(t)=\ln [\cosh(at)]$$
Applying the chain rule, we may compute the first derivative,
$$f'(t)=\frac{1}{2\cosh(at)}\frac{d}{dt}\left( e^{at} +e^{-at}\right) = a \tanh(at)$$
To compute higher order terms in the Taylor expansion, we require higher derivatives analogously computed using repeated application of the chain rule:
$$f''(t)=a^2 \mathrm{sech}^2 (at) \quad \quad f'''(t)=-2a^3\tanh(at)\mathrm{sech}^2(at)$$ $$f^{(4)}(t)=2a^4(\cosh(2at)-2)\mathrm{sech}^4(at)$$
The Taylor series of the original $f(t)$ is given by the usual expression,
$$f(t)=\sum_{n=0}^{\infty}\frac{f^{(n)}(p)}{n!}(t-a)^n$$
If we choose to center our series at $p=0$, after some algebraic manipulations, we obtain,
$$f(t)=\frac{1}{2}a^2 t^2 -\frac{1}{12}a^4 t^4 + \mathcal{O}\left(t^6\right)$$
For small $|t| \leq 1$, as the powers of $t$ increase, the corrections will become smaller and smaller. Hence higher order terms may be truncated, and the approximation,
$$f(t)\approx \frac{1}{2}a^2 t^2$$
would be sufficient. Plot of various Taylor series orders:
Black: to order $\mathcal{O}(t^2)$, Blue: to order $\mathcal{O}(t^4)$, Red: to order $\mathcal{O}(t^6)$. The behavior of a Taylor approximation after a finite region around the center $p$ is common; I believe it is known as Runge's phenomenon.
