If $f(x) = e^{x^a}$, what is $f^{(n)}(x)$, the $n$-th derivative of $f$ and what is $\lim_{x \to 0^+} f^{(n)}(x)/x^{a-n}$.

Question

This is inspired by Finding the $18th$ Derivative of a Particular Product at $x = 0$

If $f(x) = e^{x^a}$, what is $f^{(n)}(x)$, the $n$-th derivative of $f$ and what is $\lim_{x \to 0^+} f^{(n)}(x)/x^{a-n}$.

I'm sure that this is a duplicate, but I haven't been able to find it.

My conjecture is that $f^{(n)}(x) =f(x)ax^{a-n}g_n(x, a) $ where $g_n(x, a) =a^{n-1}x^{a(n-1)}+\sum_{k=0}^{n-1}c(a, n, k)x^{ka} $, the $c(a, n, k) $ are polynomials in $a$ of degree $n-1$, and $g_n(0,a) =c(a, n, 0) =\prod_{k=1}^{n-1} (a-k) =\dfrac{(a-1)!}{(a-n)!} $.

Note: I just added my derivation of the recurrence for the $c(a, n, k)$. It's messy, so there is a fair chance of error(s).

Here is what I've done.

The following was done with Wolfy and https://www.derivative-calculator.net

$\begin{array}\\ f'(x) &=ax^{a-1}f(x)\\ f''(x) &=a x^{a - 2} (a x^a + a - 1)f(x)\\ f'''(x) &=f(x) (a^3 x^{3 a - 3} + (a - 1) a^2 x^{2 a - 3} + a^2 (2 a - 2) x^{2 a - 3} + (a - 2) (a - 1) a x^{a - 3})\\ &=f(x)x^{a-3} (a^3 x^{2 a} + (a - 1) a^2 x^{ a} + a^2 (2 a - 2) x^{ a} + (a - 2) (a - 1)a)\\ &=f(x)ax^{a-3} (a^2 x^{2 a} + ((a - 1) a+a (2 a - 2) ) x^{ a} + (a - 2) (a - 1))\\ &=f(x)ax^{a-3} (a^2 x^{2 a} + 3(a - 1) a x^{ a} + (a - 2) (a - 1))\\ f''''(x) &=ax^{a-4}f(x)\left(a^3x^{3a}+\left(6a^3-6a^2\right)x^{2a}+\left(7a^3-18a^2+11a\right)x^a+a^3-6a^2+11a-6\right)\\ &=ax^{a-4}f(x)\left(a^3x^{3a}+6a^2(a-1)x^{2a}+a (a - 1) (7 a - 11)x^a+(a - 1) (a - 2) (a - 3)\right)\\ ...\\ f^{(n)}(x) &=f(x)ax^{a-n}g_n(x, a)\\ g_1(x, a) &= 1\\ g_2(x, a) &= ax^a+a-1\\ g_3(x, a) &= a^2 x^{2 a} + 3(a - 1) a x^{ a} + (a - 2) (a - 1)\\ g_4(x) &=a^3x^{3a}+6a^2(a-1)x^{2a}+a (a - 1) (7 a - 11)x^a+(a - 1) (a - 2) (a - 3)\\ ...\\ g_n(x, a) &=a^{n-1}x^{a(n-1)}+\sum_{k=0}^{n-1}c(a, n, k)x^{ka}\\ \end{array} $

It looks like $g_n(0, a) =c(a, n, 0) =\prod_{k=1}^{n-1} (a-k) =\dfrac{(a-1)!}{(a-n)!} $ (the last for integer $a$).

I'm quite sure that the existence of the $g_n(x, a)$ can be confirmed by induction, but finding the form of the recurrence, though probably straightforward, would take more work than I am willing to do right now.

Maybe later.

And here it is.

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If $f(x) =\prod_{k=1}^m f_k(x) $, then, removing the "$(x)$", $\ln f =\sum_{k=1}^m f_k $ so, differentiating, $\dfrac{f'}{f} =\sum_{k=1}^m \dfrac{f_k'}{f_k} $ so that $f' =\sum_{k=1}^m f_k'\prod_{j=1, j\ne k}^m f_j $.

Since

$\begin{array}\\ f^{(n)}(x) &=f(x)ax^{a-n}g_n(x, a),\\ f^{(n+1)}(x) &=(f(x)ax^{a-n}g_n(x, a))'\\ &=f'(x)ax^{a-n}g_n(x, a)+f(x)a(x^{a-n})'g_n(x, a)+f(x)ax^{a-n}g_n'(x, a)\\ &=ax^{a-1}f(x)ax^{a-n}g_n(x, a)+f(x)a(a-n)x^{a-n-1}g_n(x, a)+f(x)ax^{a-n}g_n'(x, a)\\ &=a^2x^{2a-n-1}f(x)g_n(x, a)+f(x)a(a-n)x^{a-n-1}g_n(x, a)+f(x)ax^{a-n}g_n'(x, a)\\ &=f(x)ax^{a-n-1}(ax^{a}g_n(x, a)+(a-n)g_n(x, a)+xg_n'(x, a))\\ &=f(x)ax^{a-n-1}((ax^{a}+a-n)g_n(x, a)+xg_n'(x, a))\\ \text{so}\\ g_{n+1}(x, a) &=(ax^{a}+a-n)g_n(x, a)+xg_n'(x, a)\\ \text{Since}\\ g_n(x, a) &=a^{n-1}x^{a(n-1)}+\sum_{k=0}^{n-2}c(a, n, k)x^{ka}\\ g_n'(x, a) &=a^{n-1}a(n-1)x^{a(n-1)-1}+\sum_{k=0}^{n-2}kac(a, n, k)x^{ka-1}\\ &=a^{n}(n-1)x^{a(n-1)-1}+\sum_{k=0}^{n-2}kac(a, n, k)x^{ka-1}\\ \text{so}\\ g_{n+1}(x, a) &=(ax^{a}+a-n)(a^{n-1}x^{a(n-1)}+\sum_{k=0}^{n-2}c(a, n, k)x^{ka})\\ &+x(a^{n}(n-1)x^{a(n-1)-1}+\sum_{k=0}^{n-2}kac(a, n, k)x^{ka-1})\\ &=(ax^{a}+a-n)a^{n-1}x^{a(n-1)}+(ax^{a}+a-n)\sum_{k=0}^{n-2}c(a, n, k)x^{ka})\\ &+a^{n}(n-1)x^{a(n-1)}+\sum_{k=0}^{n-2}kac(a, n, k)x^{ka}\\ &=ax^{a}a^{n-1}x^{a(n-1)}+(a-n)a^{n-1}x^{a(n-2)}+ax^{a}\sum_{k=0}^{n-2}c(a, n, k)x^{ka}\\ &+(a-n)\sum_{k=0}^{n-2}c(a, n, k)x^{ka}\\ &+a^{n}(n-1)x^{a(n-2)}+\sum_{k=1}^{n-2}kac(a, n, k)x^{ka}\\ &=ax^{a}a^{n-1}x^{a(n-1)}+(a-n)a^{n-1}x^{a(n-2)} +a\sum_{k=0}^{n-2}c(a, n, k)x^{(k+1)a}\\ &+(a-n)\sum_{k=0}^{n-2}c(a, n, k)x^{ka}\\ &+a^{n}(n-1)x^{a(n-2)}+\sum_{k=1}^{n-2}kac(a, n, k)x^{ka}\\ &=ax^{a}a^{n-1}x^{a(n-1)}+((a-n)a^{n-1}+a^{n}(n-1))x^{a(n-2)}\\ &+\sum_{k=1}^{n-1}ac(a, n, k-1)x^{ka}\\ &+(a-n)\sum_{k=0}^{n-2}c(a, n, k)x^{ka}\\ &+\sum_{k=1}^{n-2}kac(a, n, k)x^{ka}\\ &=ax^{a}a^{n-1}x^{a(n-1)}+(a^n-na^{n-1}+na^{n}-a^n)x^{a(n-2)}\\ &+\sum_{k=1}^{n-2}ac(a, n, k-1)x^{ka} +ac(a, n, n-2)x^{(n-1)a}\\ &+(a-n)c(a, n, 0)+(a-n)\sum_{k=0}^{n-2}c(a, n, k)x^{ka}\\ &+\sum_{k=1}^{n-2}kac(a, n, k)x^{ka}\\ &=a^{n}x^{an}+ac(a, n, n-2)x^{(n-1)a}+na^{n-1}(a-1)x^{a(n-2)}\\ &+(a-n)c(a, n, 0)\\ &+\sum_{k=1}^{n-2}(ac(a, n, k-1)+(a-n)+kac(a, n, k))x^{ka}\\ \text{Matching}\\ g_{n+1}(x, a) &=a^{n}x^{an}+\sum_{k=0}^{n-1}c(a, n+1, k)x^{ka}\\ c(a, n+1, n-1) &=ac(a, n, n-2)\\ c(a, n+1, n-2) &=na^{n-1}(a-1)+ac(a, n, n-3)+(a-n)+(n-2)ac(a, n, n-2)\\ c(a, n+1, 0) &=(a-n)c(a, n, 0)\\ c(a, n+1, k) &=ac(a, n, k-1)+(a-n)+kac(a, n, k) \quad\text{for }k=1..n-3\\ \end{array} $

Answer 1

If that's all you want, then it is simple enough to verify this inductively:

$$\frac{\mathrm d}{\mathrm dx}f(x)\big((a)_nx^{a-n}+\mathcal O\big(x^{2a-n}\big)\big)=f(x)\big((a)_{n+1}x^{a-(n+1)}+\mathcal O\big(x^{2a-(n+1)}\big)\big)$$

where $(a)_n$ is the falling factorial.

Answer 2

Let \begin{equation*}%\label{Fall-Factorial-Dfn-Eq} \langle\alpha\rangle_n= \prod_{k=0}^{n-1}(\alpha-k)= \begin{cases} \alpha(\alpha-1)\dotsm(\alpha-n+1), & n\ge1\\ 1,& n=0 \end{cases} \end{equation*} denotes the falling factorial of $\alpha\in\mathbb{C}$ respectively.

In Remark 3.1 of the paper [1] below, the formula \begin{equation}\label{Bell-1-lambda}\tag{1} B_{n,k}\Biggl(1, 1-\alpha, (1-\alpha)(1-2\alpha),\dotsc, \prod_{\ell=0}^{n-k}(1-\ell\alpha)\Biggr) =\frac{(-1)^k}{k!} \sum_{\ell=0}^k (-1)^{\ell} \binom{k}{\ell} \prod_{q=0}^{n-1}(\ell-q\alpha) \end{equation} for $\alpha\in\mathbb{C}$ and $n\ge k\ge0$ was concluded.

In Theorem 2.1 of the paper [2] below, the formula \begin{equation}\label{Bell-fall-Eq}\tag{2} B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-k+1}) =\frac{(-1)^k}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\langle\alpha\ell\rangle_n \end{equation} for $n\ge k\ge0$ and $\alpha\in\mathbb{R}$ was established.

The formulas \eqref{Bell-1-lambda} and \eqref{Bell-fall-Eq} were reviewed in Section 1.3 of the article [3] below.

The formulas \eqref{Bell-1-lambda} and \eqref{Bell-fall-Eq} are equivalent to \begin{equation}\tag{6} B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-k+1}) =\sum _{j=k}^n s(n,j)\alpha^jS(j,k) \end{equation} for $n\ge k\ge0$ and $\alpha\in\mathbb{R}$ at https://mathoverflow.net/a/88071/147732, where $s(n,j)$ and $S(j,k)$ stand for the Stirling numbers of the first and second kinds respectively.

The Faa di Bruno formula (see Theorem 11.4 in the book [4] and Theorem C on page 139 in the monograph [5] below) can be described in terms of the Bell polynomials of the second kind $B_{n,k}\bigl(x_1,x_2,\dotsc,x_{n-k+1}\bigr)$ by \begin{equation}\label{Bruno-Bell-Polynomial}\tag{1} \frac{\textrm{d}^n}{\textrm{d}x^n}f\circ h(x)=\sum_{k=0}^nf^{(k)}(h(x)) B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr), \quad n\ge0. \end{equation} Then we have \begin{align*} \bigl(\textrm{e}^{x^\alpha}\bigr)^{(n)} &=\textrm{e}^{x^\alpha}\sum_{k=0}^nB_{n,k}\bigl(\langle\alpha\rangle_1x^{\alpha-1}, \langle\alpha\rangle_2x^{\alpha-2}, \dotsc, \langle\alpha\rangle_{n-k+1}x^{\alpha-(n-k+1)}\bigr)\\ &=\textrm{e}^{x^\alpha}\sum_{k=0}^n x^{k\alpha-n} B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc, \langle\alpha\rangle_{n-k+1})\\ &=\textrm{e}^{x^\alpha}\sum_{k=0}^n x^{k\alpha-n} \sum _{j=k}^n s(n,j)\alpha^jS(j,k) \end{align*} for $n\ge0$, where we used the identity \begin{equation}\label{Bell(n-k)} B_{n,k}\bigl(abx_1,ab^2x_2,\dotsc,ab^{n-k+1}x_{n-k+1}\bigr) =a^kb^nB_{n,k}(x_1,x_2,\dotsc,x_{n-k+1}) \end{equation} for $n\ge k\ge0$ on page 135 in the monograph [5]. Accordingly, we obtain \begin{align} \lim_{x\to0^+}\frac{\bigl(\textrm{e}^{x^\alpha}\bigr)^{(n)}}{x^{\alpha-n}} &=\lim_{x\to0^+}\Biggl[\textrm{e}^{x^\alpha}\sum_{k=1}^n x^{(k-1)\alpha} \sum _{j=k}^n s(n,j)\alpha^jS(j,k)\Biggr]\\ &=\begin{cases}\displaystyle \sum _{j=1}^n s(n,j)\alpha^jS(j,k), & \alpha>0;\\ 0, & \alpha=0;\\ (-1)^n\infty, & \alpha<0. \end{cases} \end{align}

References

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2. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Bai-Ni Guo, Closed formulas and identities for the Bell polynomials and falling factorials, Contributions to Discrete Mathematics 15 (2020), no. 1, 163--174; available online at https://doi.org/10.11575/cdm.v15i1.68111.
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