# Find the function for this Taylor series

1. Jul 23, 2012

### arithmetic

$\sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!}$

Interesting result...

Last edited: Jul 23, 2012
2. Jul 23, 2012

### TheFool

I doubt there is a closed form for it. Plus, it only converges if |x|<1/e. After looking at the graph, it's similar to the Lambert W function when |x|<1/e: $$-W(-x)=\sum_{m=1}^{\infty}{\frac{m^{m-1}x^{m}}{m!}}$$ Subtracting 1 from both sides will make it approximately equal to your sum. However, there is no way to manipulate my series to put yours in terms of the W function.

Last edited: Jul 23, 2012
3. Jul 23, 2012

### TheFool

Another interesting thing of note, if $$f(x)=\sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!}$$ then $$f'(x)=\sum_{n=0}^\infty \frac{n^{n}x^{n}}{n!}$$ as long as |x|<1/e.

Last edited: Jul 24, 2012
4. Jul 24, 2012

### arithmetic

so...
:-)

5. Jul 24, 2012

### arithmetic

Lagrange inversion...

1/(1- ...) = ...

voilá

6. Jul 25, 2012

### TheFool

The Lagrange Inversion Theorem is used to find the inverse of a function. You didn't ask for the inverse in your original post :P

7. Jul 25, 2012

### haruspex

$$\sum_{m=1}^{\infty}{\frac{m^{m-1}x^{m+1}}{(m+1)!}} = \sum_{m=2}^{\infty}{\frac{(m-1)^{m-2}x^{m}}{m!}} = x\sum_{m=2}^{\infty}{\frac{(m-1)^{m-2}x^{m-1}}{m!}}$$
Now divide by x and differentiate:
$$\sum_{m=2}^{\infty}{\frac{(m-1)^{m-1}x^{m-2}}{m!}}$$
.. and finally multiply by x2

 ... finally finally, add in the m = 1 term

Last edited: Jul 25, 2012
8. Jul 25, 2012

### TheFool

Well, it would seem I don't belong posting in this forum. I shouldn't have missed that.

9. Jul 27, 2012

### arithmetic

No, that`s wrong.

Yours is shorter and better. From what you stated, you just need one step further ...
and voilá, 1/ 1-...

10. Jul 27, 2012

### micromass

Did you actually know the answer to this problem??

11. Jul 27, 2012

### haruspex

Is this right?
$f(x) = \sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!} = xG(x) - ∫G(x)$
where G(x) = -W(-x).
Hence $f(x) = ∫x.dG(x)$
But I don't know where we're trying to get to.

12. Jul 27, 2012

### arithmetic

yes, I know

p.3, 14

Last edited: Jul 27, 2012
13. Jul 28, 2012

### TheFool

Even though I showed I really am a fool for missing something so obvious, I decided to finish this.

$$y=\sum_{m=0}^{\infty}\frac{(m-1)^{m-1}x^{m}}{m!}$$
$$y'=\sum_{m=0}^{\infty}\frac{m^{m}x^{m}}{m!}$$
Move on to the W function.
$$-W(-x)=\sum_{m=1}^{\infty}\frac{m^{m-1}x^{m}}{m!}$$
Differentiate.
$$\frac{-W(-x)}{x(W(-x)+1)}=\sum_{m=1}^{\infty}\frac{m^{m}x^{m-1}}{m!}$$
$$\frac{-W(-x)}{W(-x)+1}=\sum_{m=1}^{\infty}\frac{m^{m}x^{m}}{m!}$$
Take note the subscript in that sum is m=1 instead of m=0.
$$y'=1-\frac{W(-x)}{W(-x)+1}$$
$$y=\frac{x}{W(-x)}+C$$
In another form.
$$y=-e^{W(-x)}+C$$
Both are true when |x|<1/e.

Last edited: Jul 28, 2012
14. Jul 28, 2012

### arithmetic

Congratulations FOOL.
There are other ways to get the function, but yours is great and different to others I have seen in the literature.

Also congratulations to haruspex who did a great job, indeed, I wonder how you could posibly have found the function from your first post where you handled the sums, their indexes and the x's and x^2's...

Just another note:
FOOL, you say |x|<1/e

however W(-1/e)=-1 and
y={1/e}/{W(-1/e)}+C= -1/e +C

What do you say about that?

Last edited: Jul 28, 2012
15. Jul 28, 2012

### TheFool

I say that |x|<1/e since that is the radius of convergence of $$y=\sum_{m=0}^{\infty}\frac{(m-1)^{m-1}x^{m}}{m!}$$