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.


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.

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    $\begingroup$ I really like your creative approach.Thanks. There's a later answer with essentially the same observation, but yours is closer to my level. $\endgroup$ – Charlie Mosby Feb 16 '19 at 5:27

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.


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 $a_0=a_1=a_2=1$.

  • $\begingroup$ Regarding your opening line ...it is already hard to find the general expression of... Would you mind take a look at my following question post? $\endgroup$ – Charlie Mosby Mar 11 '19 at 7:47
  • $\begingroup$ @CharlieMosby. Thanks to your comment, I went back to the problem and found the recurrence relation. I shall now have a look at your post. Thanks and cheers $\endgroup$ – Claude Leibovici Mar 11 '19 at 8:26
  • $\begingroup$ Thank you very much for the recurrence relation. It will be a good trip for me trying to understand how it works and how it came to be. $\endgroup$ – Charlie Mosby Mar 11 '19 at 8:47

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 $a_k$.


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.)

  • $\begingroup$ Thanks for taking an interest in my question. I mentioned (in passing) in the post that: there are terms involving $\sqrt{x^2 + 1}$ and its derivatives which have to be expanded, producing sub-series for each order. This requires rearrangement or recognizing some obscure combinatorial identity (to arrive at the final "clean" power series). $\endgroup$ – Charlie Mosby Feb 14 '19 at 4:32
  • $\begingroup$ My apology....I just realized I mistakenly mixed things up in the comment above (and in the post). Indeed as you said (the very basic thing) that calculating $f^{(k)}(x)$ is merely taking repeated derivatives (albeit hairy). Evaluated at $x_0 = 0$ it is just a constant, and there's no extra expansion needed like I suggested. $\endgroup$ – Charlie Mosby Feb 14 '19 at 5:02

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