# Taylor expansion using $a^m$

There is a problem in my textbook that has baffled me for long. The problem says that the Taylor series expansion of $a^m$ is: $$a^m= 1+m(\log_ea)+\frac{m^2}{2!} (\log_ea)^2+.....$$ From this we are required to deduce the Taylor expansion of $$f(x)=\log_e (x+ \sqrt{x^2+1})$$ Though it is pretty easy to calculate the taylor series directly for $f(x)$ but I am unable to think over how can I deduce $f(x)$ from the expansion of $a^m$. I tried to use logarithmic properties but they are just making it much more complex. Kindly guide me to the right direction to reach the suitable answer.

P.S. I wrote my function as $$e^{log_e(log_e (x+ \sqrt{x^2+1})))}$$. But that just seems too peculiar!

• I certainly wouldn't do it the way that textbook suggests. Actually $f$ is a well-known function. If you compute $f'(x)$ you might recognise it. – Lord Shark the Unknown Apr 30 '17 at 5:43
• Yes, it is a well-known function because it cancels out some terms after differentiation and we get $f'(x)=\frac{1}{\sqrt{x^2+1}}$ but how can I convert it suitable in $a^m$ form? – Harsh Sharma Apr 30 '17 at 5:47
• I reiterate my advice: don't do it the way the textbook suggests! – Lord Shark the Unknown Apr 30 '17 at 5:49
• I understand your suggestion truly because there is no need to assault our own intuition in order to follow the textbook but still it would be nice to know some more ways to solve the problem. – Harsh Sharma Apr 30 '17 at 5:51
• @LordSharktheUnknown That can be a dangerous piece of advice. It is certainly true, imo, that sometimes this kind of advices just make things messier, but I think the OP should at least know how in general things work out, if not the whole development. BTW, the function $\;f\;$ in the question is $\;\text{arcsinh}\,x\;$ = the inverse hyperbolic sine. – DonAntonio Apr 30 '17 at 7:16

You are told to use the taylor expansion for $a^x = e^{ln(a)x} = \sum \frac{ln(a)^kx^k}{k!}$
Suppose $f(x) = \sum a_kx^k$
Then $$e^{f(x)} = e^{\sum a_kx^k} = \prod (e^{a_k})^{x^k} = \prod (\text{taylor polynomials}) = x + (x^2 + 1)^{1/2}$$