How to “directly” obtain Maclaurin series of $\exp(x-1+\sqrt{x^2+1})$

Question

Consider the Taylor expansion centered around $x_0 = 0$ for $$f(x) = e^{x-1 + \sqrt{x^2+1}} ~,\qquad x\in\mathbb{R}$$

The goal is to arrive at $\sum c_k x^k$ by hand, and I wonder if there is an efficient way to obtain the coefficients of each power $x^k$ directly?

Below are examples of three approaches I know of, which are NOT efficient because they involve cumbersome combinatorial identities or further rearrangements that are too complicated (to me).

1. Take the whole exponent into the formula $e^z = \sum z^k / k!$ where $z = x-1 + \sqrt{x^2+1}$

2. Split the exponent into two so that $\displaystyle f(x) = \left( \sum \frac{ z_1^k }{ k! }\right) \left( \sum \frac{ z_2^m }{ m! }\right)$ where $z_1 = x-1$ and $z_2 = \sqrt{x^2-1}$.

3. Back to the definition of Taylor expansion $f(x) = \sum (x - x_0)^k f^{(k)}(x_0)\, /\, k!$ and do the derivatives with chain rules, for every order.

Note that wherever $\sqrt{x^2 + 1}$ appears, it has to be expanded into power series $1 + \frac12 x^2 - \frac18 x^4 + \ldots$ sooner or later to yield the final result in the desired form $f(x) = \sum x^k / k!$

I consider these calculations as indirect because the procedure itself by definition doesn't yield $c_k$ "directly", and one has to go through more rearrangements that are error prone.

When one wants only the first few terms, these might be okay, but when it comes to exact expressions (in summations), I personally don't find them to be practical.

There are many posts on this site pertaining to finding series expansion of various functions. However, so far what I've seen are all "indirect" methods like described above. For example, this post.

One can always throw the expression to Wolfram Alpha if one cares only about the result. However, sometimes one needs the exact expression for some further steps of proving etc, that is, doing the analysis based on the series.

Thank you for your time.

Answer 1

I'm trying to provide another way to do it, but I cannot say it is efficient enough.

Let $$f(x)=e^{x-1+\sqrt{x^2+1}}=\sum_{k=0}^\infty a_k x^k=a_0+a_1x+a_2x^2+\cdots$$ as usual. Since $$\dfrac d{dx}e^{x-1+\sqrt{x^2+1}}=\bigg(1+\dfrac x{\sqrt{x^2+1}}\bigg)e^{x-1+\sqrt{x^2+1}}=\sum_{k=0}^\infty(k+1)a_{k+1}x^k$$ We have $$\frac{x}{\sqrt{x^2+1}}e^{x-1+\sqrt{x^2+1}}=\sum_{k=0}^\infty [(k+1)a_{k+1}-a_k]x^k$$ Turning back to Taylor series aspect, since when $x=0$, we have $f(0)=1$ and $f'(0)=1$, hence $a_0=a_1=1$, then we can write $$\dfrac{x}{\sqrt{x^2+1}}f(x)=\sum_{k=1}^\infty [(k+1)a_{k+1}-a_k]x^k$$ By shifting the index we have $$\sum_{k=0}^\infty a_kx^k=\sqrt{x^2+1}\sum_{k=0}^\infty[(k+2)a_{k+2}-a_{k+1}]x^k$$ By using the Binomial expansion of $\sqrt{x^2+1}$ and solving some simultaneous equation system, we can solve up to $a_n$ for any natural number $n$, despite you would required to have a powerful brain or computer.

Answer 2

I think that directly would be very difficult since it is already hard to find the general expression of the coefficients for the expansion of $e^{f(x)}$ $$\frac{e^{f(x)}}{e^{f(0)} }=1+x f'(0)+\frac{1}{2} x^2 \left(f''(0)+f'(0)^2\right)+\frac{1}{6} x^3 \left(f^{(3)}(0)+f'(0)^3+3 f'(0) f''(0)\right)+$$ $$\frac{1}{24} x^4 \left(f^{(4)}(0)+3 f''(0)^2+f'(0)^4+4 f^{(3)}(0) f'(0)+6 f'(0)^2 f''(0)\right)+O\left(x^5\right)$$

On the other side, composing first the Taylor series of $f(x)$ is in general simple. In your specific case, we can write $$\sqrt{x^2+1}=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k} x^{2 k}$$ $$x-1+\sqrt{x^2+1}=x+\sum_{k=1}^\infty \binom{\frac{1}{2}}{k} x^{2 k}$$

$$\exp(x-1+\sqrt{x^2+1})=e^x \prod_{k=1}^\infty \exp\left(\binom{\frac{1}{2}}{k} x^{2 k} \right)$$ Doing the development of each term of the infinite product would again lead to an infinite product of series and finding explicitely the coefficient of $x^n$ could be a real problem.

Edit

Back to the problem five years later, writing $$f(x) = e^{x-1 + \sqrt{x^2+1}}=\sum_{n=0}^\infty a_n x^n$$ the $a_n$ are defined by the recurrence relation $$a_n=\frac{(2 n-3)\, a_{n-1}-\left(n^2-5 n+7\right)\, a_{n-2}+2 (n-3)\, a_{n-3}}{(n-2)\, n}$$ where $a_0=a_1=a_2=1$.

Answer 3

You have $f'(x)=f(x)\left(1+{x\over\sqrt{x^2+1}}\right)$ and therefore $$\left({f'(x)\over f(x)}-1\right)^2={x^2\over x^2+1}\ .$$ It follows that $f$ satisfies the ODE $$(x^2+1)\bigl(f'^2-2ff'\bigr)+f^2=0\ .\tag{1}$$ Now plug $f(x):=\sum_{k=0}^\infty a_k x^k$ into $(1)$. Using $$f^2(x)=\sum_{j\geq0} a_jx^j\cdot \sum_{k\geq0} a_kx^k=\sum_{r\geq0}\left(\sum_{k=0}^r a_{r-k} a_k\right)x^r\ ,$$ and similarly for $f'^2$and $ff'$, you can obtain a recursion scheme for the $a_k$.

Answer 4

Don't you have the definition $c_k = f^{(k)}(0)/k!$ from Taylor's theorem? This gives you a "direct" way to compute $c_k$: just compute $k$ derivatives of $f$, plug in $0$, and divide by $k!$.

(Of course, computing $k$ derivatives of your given $f$ will get hairy very quickly, which is why people try the "indirect" methods as well.)