# Finding value of the limit: $\lim_{t \to 0} \frac{\varphi(1+t) - \varphi(t)}{t}$

Let $$\varphi(x) = e^{-x} + \frac{x}{1!} \cdot e^{-2x} + \frac{3x^{2}}{2!}\cdot e^{-3x} + \frac{4^{2} \cdot x^{3}}{3!} \cdot e^{-4x} + \cdots$$

Then what is the value of: $$\displaystyle\lim_{t \to 0} \frac{\varphi(1+t) - \varphi(t)}{t}$$

I am not getting any idea as to how to proceed for this problem. I tried summing up the expression but no avail. I also tried differentiating $\varphi(x)$ since the limit quantity which we require seems to involve derivative but again i couldn't find any pattern. Any ideas on how to solve this problem.

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@chandru1: are you sure the second term is $e^{-3x}$, not $e^{-2x}$? – Qiang Li Feb 2 '11 at 0:21
What is the source of the problem? – Jonas Meyer Feb 2 '11 at 0:40
What is the coefficient in the term with $n!$? $(n+1)^{n-1}$? The question is not very clear. – Aryabhata Feb 2 '11 at 0:54
I presume the individual terms of $\varphi(x)$ are ${n^{n-1}\cdot x^{n-1}\over n!}\cdot e^{-nx}$ - those terms are virtually identical to the Taylor series for the Lambert function W and given the form (it's essentially $e^{-x}$ times a power series in $xe^{-x}$) I strongly suspect that $\varphi$ has some conveniently explicit form... – Steven Stadnicki Feb 2 '11 at 1:02
@All: Question edited now. – anonymous Feb 2 '11 at 2:25

## 2 Answers

The function $\varphi(x)$ can be written as $$\begin{eqnarray} \varphi(x) &=& \frac{1}{x} \sum_{n=1}^{\infty} \frac{n^{n-1}}{n!}\left(xe^{-x}\right)^{n} \\ &=& -\frac{1}{x} \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!}\left(-xe^{-x}\right)^{n} \\ &=& -\frac{1}{x} W_0\left(-xe^{-x}\right), \end{eqnarray}$$ where $W_0$ is the Lambert W-function's main branch. For sufficiently small $y$, the definition of the W-function ensures that $W_0(ye^y)=y$, so for small $t$, $$\begin{eqnarray} \varphi(t) &=& -\frac{1}{t}W_0\left(-te^{-t}\right) \\ &=& -\frac{1}{t}(-t) \\ &=& 1. \end{eqnarray}$$ In fact this equality holds for all $0 \le t < 1$. For values slightly larger than $1$, we use $$-(1+t)e^{-(1+t)} = \frac{-(1+t)}{1+t+\frac{1}{2}t^2+O(t^3)} = -1 + \frac{1}{2}t^2+O(t^3) = -(1-t)e^{-(1-t)},$$ so $$\begin{eqnarray} \varphi(1+t) &=& -\frac{1}{1+t}W_0(-(1+t)e^{-(1+t)}) \\ &\approx& -\frac{1}{1+t}W_0(-(1-t)e^{-(1-t)}) \\ &=& \frac{1-t}{1+t} \\ &=& 1 - 2t + O(t^2). \end{eqnarray}$$

The desired limit, then, is $$\lim_{t\rightarrow 0}\frac{\varphi(1+t) - \varphi(t)}{t} = \lim_{t\rightarrow 0}\frac{(1-2t) - 1}{t} = -2.$$

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Hmm, using Pari/GP, I just tried to find the coefficients of $\varphi (t)$

polcoeffs(exp(-x)*sumalt(k=0,exp(-k*x)*x^k/k!*(k+1)^(k-1)*1.0))


(where the polcoeffs-function just retrieves the whole vector of coefficients up to the defined series-precision and I used x instead of t )

The result of this is an (arbitrary near) approximation to the formal powerseries $\varphi(t) = 1.0$ So it may be useful to analyze the approximation of that powerseries to the constant in order to evaluate your limit-formula.

[update] If one expands the powerseries for $e^{- k x}$ and collect coefficients at like powers of x then indeed all coefficients at a certain power of x sum to zero, thus vanish (at least heuristically up to $x^{64}$, I did not do final analysis) . Then all derivatives are also zero and even the L'Hospital-rule is of no use here. However, I don't know whether there could be some special reason that this expansion into powerseries might not be feasible/allowed because the result is an infinite sum of formal powerseries. [end update]

[update 2]
$\varphi(x)=e^{-x}+xe^{-2x}\frac1{1!} +x^2e^{-3x}\frac3{2!} +x^3e^{-4x}\frac{4^2}{3!} + \ldots$

$\begin{array} {rrrr} \varphi(x)= ( & 1 & -x & + \frac{x^2}{2!}&+ \frac{x^3}{3!} & + \ldots )\\ + \frac{x}{1!}(& 1 & -2x & + \frac{2^2x^2}{2!}&+ \frac{2^3x^3}{3!} & + \ldots ) \\ + \frac{3x^2}{2!}(& 1 & -3x & + \frac{3^2x^2}{2!}&+ \frac{3^3x^3}{3!} & + \ldots ) \\ + \ldots \\ \end{array}$

Collecting equal powers of x means to add coefficients along the antidiagonal so

$\begin{array} {rrrr} \varphi(x)= 1 \\ - x (&\frac{1}{1!}& - \frac{1}{1!} ) \\ + x^2 (&\frac{1}{2!}& - \frac{2}{1!}& + \frac{3^1}{2!}) \\ - x^3 (&\frac{1}{3!}& - \frac{2^2}{2!}& + \frac{3^2}{2!}& - \frac{4^2}{3!}) \\ + \ldots \\ \end{array}$

All the parentheses seem to form binomially weighted sums of like powers with alternating signs which are known to evaluate to zero, so the coefficients at each power of x should be zero and if this is so, the taylorseries is then $\varphi(x)=1$

[end update 2]

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