Finding a exponential generating function

Question

I'm trying to find a generating function of a series expression, any suggestions would be appreciated.

$S=1+\alpha(x-\alpha t)a+[\alpha^2(x-\alpha t)^2+2\alpha\nu(x-2\alpha t)]a^2/2!+[\alpha^3(x-\alpha t)^3+4\alpha\nu^2(x-4\alpha t)+6\alpha^2\nu(x-\alpha t)(x-2\alpha t)]a^3/3!+ [\alpha^4 (x - \alpha t)^4 + 12 \alpha^3 (x - 2 \alpha t) (x - \alpha t)^2 \nu + 4 \alpha^2 (7 x^2 - 32 t x \alpha + 28 t^2 \alpha^2) \nu^2 + 8 \alpha (x - 8 \alpha t) \nu^3] a^4/4! +\ldots$

Clearly, it's an exponential generating function of the form $\sum_{n\geq 0}^\infty B(n) a_n/n!$ where $B(n)$ is some function of $x,\alpha,t$ etc... The first thing that struck me was the $\alpha(x-\alpha t)$ term so I tried taking $\exp(\alpha(x-\alpha t))$ as a common factor by expanding it in a series form and dividing it from the original expression, however it doesn't simplify the expression.

EDIT: To add some context I recently became interested in Lie symmetry groups, I have a symmetry group given by $X=2 \nu x \partial_x + 4 \nu t \partial_t + \alpha (x-\alpha t) u\partial_ u$ where $x,t$ are the independent variables and $u$ is the dependent variable, i.e. $u(x,t)$. Here $\alpha$ and $\nu$ are non-zero constants.

From the above symmetry group I can obtain a Lie-Symmetry point transformation given by $\bar{u}=\exp(aX)u=u[1+\alpha(x-\alpha t)a+[\alpha^2(x-\alpha t)^2+2\alpha\nu(x-2\alpha t)]a^2/2!+\ldots]$

Answer 1

To those interested I think I figured it out. I will lay my thought process out here. Although I believe I solved the closed-form expression I need, I'll leave this open because I'm interested to see how others may arrive at the solution. Part of the process for me was a fluke, to be honest. I feel like there should be a more systematic way.

note: my background is in engineering and not math, so those who want rigor might shudder at my approach.

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1. Let's fully expand $S$ and group the $x$-terms alone, the $t$-terms alone and the $xt$-terms alone:

• $S= 1 + (\alpha x)a + (\alpha^2 x^2) a^2/2! + (2\alpha \nu x)a^2/2! + (\alpha^3 x^3) a^3/3! + (4 \alpha \nu^2 x) a^3/3! + (6 \alpha^2 \nu x^2) a^3/3! \\ - (\alpha^2 t)a + (\alpha^4 t^2) a^2/2! - (4 \alpha^2 \nu t) a^2/2! - (\alpha^4 t^3)a^3/3! - (16 \alpha^2 \nu^2 t) a^3/3! + (12 \alpha^3 \nu t^2) a^3/3! \\ -(2\alpha^3 x t) a^2 /2! - (3 \alpha^4 x^2 t) a^3/3! + (3 \alpha^4 x t^2)a^3/3! - (18 \alpha^3 \nu x t^3) a^3/3! + ... $

Next, let's focus on the first row that includes the $x$-terms, and for the sake of clarity let's put them in a separate series called $S_1$:

• $S_1 = 1 + (\alpha x) a + [(2\nu) \alpha x + \alpha^2 x^2]a^2/2! + [(2 \nu)^2 \alpha x + (2 \nu) 3 \alpha^2 x^2 + \alpha^3 x^3]a^3/3! + ...$

which can be rewritten as

• $S_1 = 1 + [\alpha x/(2 \nu)] (2\nu a) + [\alpha x/(2 \nu) + \alpha^2 x^2/(2 \nu)^2] (2 \nu a)^2/2! + [\alpha x/(2 \nu) + 3 \alpha^2 x^2/(2 \nu)^2 + \alpha^3 x^3/(2 \nu)^3] (2 \nu a)^3/3! + ...$

Let $A=\alpha x /(2\nu)$ then

• $S_1 = 1 + A (2\nu a) + (A + A^2) (2 \nu a)^2/2! + (A + 3 A^2 + A^3) (2 \nu a)^3/3! + ...$

The expression has the flavor of an exponential generating function.

Ok, so here's where luck/chance and some guessing took place. I went for the textbook "generatingfunctionology" by Herbert S. Wilf from the University of Pennsylvania and searched for exponential generating functions and stumbled on pages 20-22.

To simplify my life, I set $A=1$ and disregarded the $2 \nu$ term in $S_1$ which gave:

$1 + a + 2 a^2/2! + 5 a^3/3! + ...$ which have the following coefficients {1,1,2,5} etc.. the Bell numbers of the second kind given by:

$B(a)=\sum_{n \ge 0} b(n)/n! \, a^n$

which has a function of the following form: $e^{e^a-1}$. So with some minor algebraic manipulations, and generalizing back to our original expression we get $e^{A(e^{2\nu a}-1)}$, i.e. :

$S_1 = e^{\frac{\alpha x}{ 2\nu}(e^{2\nu a}-1)}$

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2. Now that we found $S_1$, let's take it out as a common factor from $S$ to find $S/S_1$. Recall that if we have two Taylor series expansions given by $f$ and $g$ respectively then $f/g$ is such that the:

• $0$th-order term is: $\frac{f(0)}{g(0)}$
• $1$st-order term is: $\frac{f'(0)}{g(0)} - \frac{f(0)g'(0)}{g(0)^2}$
• $2$nd-order term is: $-\frac{f'(0)g'(0)}{g(0)^2} + \frac{f'''(0)}{2g(0)}+\frac{f(0)}{g(0)} \bigg [\frac{f'(0)^2}{g(0)^2}- \frac{g''(0)}{2g(0)} \bigg]$
• $3$rd-order term is: $-\frac{f''(0)g'(0)}{2g(0)}+\frac{f'(0)}{g(0)} \bigg[ \frac{g'(0)^2}{g(0)^2}-\frac{g''(0)}{2 g(0)} \bigg] + \frac{f'''(0)}{6g(0)} + \frac{f(0)}{g(0)} \bigg[-\frac{g'(0)^3}{g(0)}+\frac{g'(0)g''(0)}{g(0)^2} - \frac{g'''(0)}{6 g(0)} \bigg] $

Let $f$ and $g$ correspond to $S$ and $S_1$ respectively. We get:

• $f(0) = 1$
• $f'(0) = \alpha(x-\alpha t)$
• $f''(0) = \alpha^2 (x-\alpha t)^2 + 2 \alpha \nu (x-2\alpha t)$
• $f'''(0) = \alpha^3 (x-\alpha t)^3 + 4 \alpha \nu^2 (x-4 \alpha t) + 6 \alpha^2 \nu (x-\alpha t)(x- 2\alpha t)$

and

• $g(0) = 1$
• $g'(0) = \alpha x$
• $g''(0) = \alpha^2 x^2 + 2 \alpha \nu x$
• $g'''(0) = \alpha^3 x^3 + 4 \alpha \nu^2 x + 6 \alpha^2 \nu x^2$

such that

• $0$th-order term is: $1$
• $1$st-order term is: $-\alpha^2 t$
• $2$nd-order term is: $\frac{1}{2} \alpha^2 t (\alpha^2 t - 4 \nu)$
• $3$rd-order term is: $-\frac{\alpha^2 t}{6}(\alpha^4 t^2 - 12 \alpha^2 \nu t + 16 \nu^2)$

e.g. $S_2 =S/S_1 = 1 - (\alpha^2 t) a + \big[\frac{1}{2} \alpha^2 t (\alpha^2 t - 4 \nu) \big] a^2/2! - \big[\frac{\alpha^2 t}{6}(\alpha^4 t^2 - 12 \alpha^2 \nu t + 16 \nu^2) \big] a^3/3! + ... $

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3. Now that we have $S_2$ rewrite the expression as:

• $S_2 = 1 - \big( \alpha^2 t/(4 \nu) \big) (4 \nu a) + \big[ \big( \alpha^2 t/(4 \nu) \big)^2 - \alpha^2 t/(4 \nu) \big] (4 \nu a)^2/2! - \big[ \big(\alpha^2 t / (4 \nu) \big)^3 - 3 \big( \alpha^2 t/(4\nu) \big)^2 + \alpha^2 t/ (4\nu) \big] (4 \nu a)^3/3! + ... $

Let $C=\alpha^2 t/(4 \nu)$ we get:

• $S_2 = 1 - C (4 \nu a) + (C^2 - C) (4 \nu a)^2/2! - (C^3 - 3C + C) (4 \nu a)^3/3! + ... $

which is a similar exponential generating function to $S_1$ that we showed above with a sign difference in the exponent. Without going into the details you can follow suit and get the following expression $e^{-C(e^{4\nu a}-1)}$, i.e.:

$S_2 = e^{-\frac{\alpha^2 t}{ 4\nu}(e^{4\nu a}-1)}$

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4. Therefore, the final expression of $S$ is the product of the two closed-form expressions of the generating functions $S_1$ and $S_2$:

$S = e^{\frac{\alpha x}{ 2\nu}(e^{2\nu a}-1)} \, e^{-\frac{\alpha^2 t}{ 4\nu}(e^{4\nu a}-1)} $

or simplified a little more:

$S = e^{-\frac{\alpha}{2\nu}[x(1-e^{2\nu a})-\frac{\alpha t}{2}(1-e^{4\nu a})]} $

Like I said, this is the answer I'm looking for. I'll keep this open to see how others might arrive to the answer more systematically.