# $\lim\limits_{n\to\infty}\lim\limits_{x\to\ 0}f^{(n)}(x)$

Let $f(x)=\exp(\sqrt{x})+\exp(-\sqrt{x})=2\cosh(\sqrt{x})$. How to calculate $\lim\limits_{n\to\infty}\lim\limits_{x\to\ 0}f^{(n)}(x)$

Using power series, we have $$f(x)=2\sum\limits_{k=0}^{\infty}\frac{x^k}{(2k)!}$$ so the $n$th derivative is: $$f^{(n)}(x)=2\sum\limits_{k=n}^{\infty}\frac{k!}{(k-n)!(2k)!}x^{k-n}$$ so $$\lim\limits_{x\to 0}f^{(n)}(x)=\frac{2n!}{(2n)!}$$ and hence $$\lim\limits_{n\to\infty}\lim\limits_{x\to\ 0}f^{(n)}(x)=0$$ Can one do it by finding a closed form expression for $f^{(n)}(x)$?

• Are you allowed to use the Taylor series? – Empy2 Jul 11 '13 at 16:36
• @Michael This isn't a homework, you can use anything. You gave me an idea... – llllllllllllllllllllllllllllll Jul 11 '13 at 16:46
• "Can one do it by finding a closed form expression...?" But that is usually very tortuous! – Pedro Tamaroff Jul 11 '13 at 17:20
• @ThomasAndrews $$f(x^2)\longrightarrow 2xf'(x^2)\longrightarrow 2f'(x^2)+4x^2f''(x^2)$$ and the expression becomes more and more complicated. – llllllllllllllllllllllllllllll Jul 11 '13 at 17:20
• Yeah, I was talking about an inductive method, @metacompactness. – Thomas Andrews Jul 11 '13 at 17:21

Using the fact that $\bigl(\sqrt x\bigr)^{(j)}=\frac{(-1)^{j-1}}{2^j}\,(2j-3)!!x^{-(2j-1)/2}$ for all $j\geq1$ (here $!!$ denotes the double factorial; in particular $(-1)!!=1$), together with Faà di Bruno's formula, we obtain, for all $n\geq1$, a formula for the $n$-th derivative of the function $g(x)=\exp(\sqrt x)$. In the formula below, the tuple $\mathbf m=(m_1,\dots,m_n)$ ranges over the tuples in $\mathbb N^n$ such that $\sum_{j=1} ^njm_j=n$ (partitions of the number $n$), and $|\mathbf m|=\sum_{j=1}^nm_j$:

\begin{align*} g^{(n)}(x)=&\,g(x)\sum_{\mathbf m}\binom n{m_1,\dots,m_n}\prod_{j=1}^n\Biggl[\frac{(-1)^{j-1}(2j-3)!!x^{-(2j-1)/2}}{2^jj!}\Biggr]^{m_j}\\ =&\,g(x)\sum_{\mathbf m}a_{\mathbf m}\,\frac{x^{|\mathbf m|/2}} {x^n}\,. \end{align*}

Since the value of $|\mathbf m|$ varies between $1$ and $n$ as $\mathbf m$ varies over all the partitions on $n$, it follows that for $x>0$ we can write

$$g^{(n)}(x^2)=\frac{P(x)e^x}{x^{2n-1}}\,,$$

where $P(T)=P_n(T)$ is a polynomial in $T$ of degree $n-1$. On the other hand, the $n$-th derivative of the function $h(x)=\exp(-\sqrt x)$ is very similar, because of the factor $-1$ that multiplies the inner square root:

$$h^{(n)}(x)=h(x)\sum_{\mathbf m}(-1)^{|\mathbf m|}\,a_{\mathbf m}\frac{x^{|\mathbf m|/2}}{x^n}\,,$$

which implies, for $x>0$:

\begin{align*} h^{(n)}(x^2)=&\,e^{-x}\sum_{\mathbf m}a_{\mathbf m}\frac{(-x)^{|\mathbf m|}}{(-x)^{2n}}\\ =&\,-\frac{P(-x)e^{-x}}{x^{2n-1}}\,. \end{align*}

Since we are interested at the limit $\lim_{x\to0^+}\bigl[g^{(n)}(x)+h^{(n)}(x)\bigr]$, we can change $x$ by $x^2$ (with $x>0$), and so the desired limit is equal to

$$L=\lim_{x\to0^+}\frac{Q(x)-Q(-x)}{x^{2n-1}}=\lim_{x\to0^+}\frac{H(x)}{x^{2n-1}}\,,$$

where $Q(x)=P(x)e^x$ and $H(x)=Q(x)-Q(-x)$. Since $H(0)=0$ and because of the term $x^{2n-1}$, we are led to use L'Hôpital's rule, hopefully $2n-1$ times (well, not hopefully but certainly instead, because we already know the result). We have

$$H^{(r)}(0)=\bigl[Q^{(r)}(x)-(-1)^rQ^{(r)}(-x)\bigr]\Bigl|_{x=0}=\begin{cases} 0,&\ \style{font-family:inherit;}{\text{if}}\ r\ \style{font-family:inherit;}{\text{is even}};\\ 2Q^{(r)}(0),&\,\ \style{font-family:inherit;}{\text{if}}\ r\ \style{font-family:inherit;}{\text{is odd}}. \end{cases}$$

Finally, if $P(T)=\sum_{k=0}^{n-1}b_kT^k$, then by general Leibniz rule we have

\begin{align*} Q^{(r)}(0)=&\,\biggr[\sum_{k=0}^r\binom rkP^{(k)}(x)\ \frac{d^{r-k}\ e^x}{dx^{r-k}}\biggr]\Biggl|_{x=0}=\sum_{k=0}^r\binom rkP^{(k)}(0)\\ =&\,\sum_{k=0}^r\binom rk\,k!b_k\,.\tag{\boldsymbol\ast} \end{align*}

At this point it is necessary to determine the coefficients of the polynomial $P(T)$. Recall that actually $P(T)$ is a polynomial that depends on $n$, and that for all $n\geq1$ we have

$$g^{(n)}(x^2)=\frac{P_n(x)e^x}{x^{2n-1}}\,.$$

Defining $R_n(T)=2^nP_n(T)$ and using the equality $2xg^{(n)}(x^2)=\bigl[g^{(n-1)}(x^2)\bigr]^\prime$ for $n\geq2$ we obtain (exercise) the recurrence

$$R_n(T)=(T-2n+3)R_{n-1}(T)+TR_{n-1}^\prime(T),\ \text{for all}\ n\geq2\,,$$

and initial value $R_1(T)=1$ . Writing $R_n(T)=\sum_{k=0}^{n-1}r_{n,k}T^k$, the recurrence becomes (verify)

\begin{align*} r_{n,n-1}=r_{n-1,n-2},&\quad\style{font-family:inherit;}{\text{for}}\ n\geq2;\\ r_{n,k}=(3-2n+k)r_{n-1,k}+r_{n-1,k-1},&\quad\style{font-family:inherit;}{\text{for}}\ n\geq2\ \style{font-family:inherit;}{\text{and}}\ k=1,\dots,n-2;\\ r_{n,0}=(3-2n)r_{n-1,0},&\quad\style{font-family:inherit;}{\text{for}}\ n\geq2\,. \end{align*}

From this we see that $r_{n,n-1}=1$ for all $n\geq1$ and $r_{n,0}=(-1)^{n-1}(2n-3)!!$ for all $n\geq1$. Moreover, for all $m\geq1$ and all $t\geq2$ we get

$$r_{m+t,m}=r_{t,0}+\sum_{k=1}^m(r_{k+t,k}-r_{(k-1)+t,k-1})=(-1)^{t-1}(2t-3)!!+\sum_{k=1}^m(3-2t-k)r_{k+(t-1),k}\,.$$

With this we are able to iteratively determine the values $r_{m+t,m}$, starting with $t=2$. I used Mathematica to do this, and after some trials I discovered the following formula:

$$r_{m+t,m}=\frac{(-1)^{t-1}}{(2t-2)!!}\,(m+1)\cdots(m+2t-2)=\frac{(-1)^{t-1}(m+2t-2)!}{2^{t-1}(t-1)!m!}\,,$$

which can be rewritten as

\begin{align*} r_{n,k}=&\,\frac{(-1)^{n-k-1}(2n-k-2)!}{2^{n-k-1}(n-k-1)!k!}\\ =&\,(n-1)!\frac{(-1)^{n-k-1}}{2^{n-k-1}k!}\binom{2n-2-k}{n-1}\,,\ \style{font-family:inherit;}{\text{for}}\ n\geq2\ \style{font-family:inherit;}{\text{and}}\ k=1,\dots,n-2\,. \end{align*}

Actually, the formula above continue to hold at the remaining cases. Therefore we have

$$P_n(T)=(n-1)!\sum_{k=0}^{n-1}\frac{(-1)^{n-k-1}}{2^{2n-k-1}k!}\binom{2n-2-k}{n-1}\,T^k\,,$$

and we would like to show directly (see $(\boldsymbol\ast)$) that for all $r$ odd we have

$$\sum_{k=0}^r\binom rk\,\frac{(-1)^{n-k-1}}{2^{2n-k-1}}\binom{2n-2-k}{n-1}=\,\begin{cases} 0,&\ \text{if}\ r<2n-1;\\ 1/2,&\ \text{if}\ r=2n-1. \end{cases}\tag{\boldsymbol{\ast\ast}}$$

I don't have any idea about how to prove equality $(\boldsymbol{\ast\ast})$ above.

SUMMARY AND MORAL

Your desired, explicit formula (that is, without using power series) for $f^{(n)}(x)$ is as follows:

\begin{align*} f^{(n)}(x)=&\,g^{(n)}(x)+h^{(n)}(x)\\ =&\,\frac{(n-1)!}{x^{n-\frac12}}\sum_{k=0}^{n-1}\frac{(-1)^{n-k-1}}{2^{2n-k-1}k!}\binom{2n-2-k}{n-1}\,x^{k/2}\bigl[e^{\sqrt x}-(-1)^ke^{-\sqrt x}\,\bigr]\,. \end{align*}

The moral of the story is: it is a lot better to use power series!!!!!

Maple does this in terms of a Bessel function $$2\,\sum _{k=n}^{\infty }{\frac {{x}^{k-n}k!}{ \left( k-n \right) !\, \left( 2\,k \right) !}}={\frac {n!\, {{\rm I}_{-1/2+n}\left(\sqrt {x}\right)}\Gamma \left( 1/2+n \right) { 2}^{1/2+n}}{ \left( 2\,n \right) !\,{x}^{-1/4+1/2\,n}}}$$

edit

To do this by hand, recall $${\it I_q} \left( y \right) =\sum _{k=0}^{\infty }{\frac { \left( y/2 \right) ^{2 k+q}}{k!\left( k+q \right) !}}$$ where non-integer factorial is to be expressed in terms of the Gamma function.

• Wow, a product of a Bessel function and a gamma function, that's something you don't see everyday; these CAS gives unexpected results sometimes. I think it's hard to prove it without the power series but can we prove it (without Maple's help) from the power series $2\sum\limits_{k=n}^{\infty}\frac{k!}{(k-n)!(2k)!}x^{k-n}$ ? – llllllllllllllllllllllllllllll Jul 11 '13 at 19:46
• Two factorials in the denominator suggests a Bessel series. Recall the Bessel function series for $I_q(x)$ and try to put our series into this form. – GEdgar Jul 11 '13 at 21:03
• What about the series of the gamma function? – llllllllllllllllllllllllllllll Jul 11 '13 at 21:06
• That Gamma function, together with the factorials, are just to get the gamma functions needed in the Bessel series. – GEdgar Jul 11 '13 at 21:09