http://img409.imageshack.us/img409/3350/digitproblemdi8.jpg

A bit hard at first, but I came up with a rather elegant solution. Anyone want to give it a try?

It is currently the problem of the week at Purdue University.

\frac{9}{10}*(e^{10}-1)

Even easier than that :)

Even easier than that :)

How so?

Oh. Nevermind. It might not be that easy :)

Where do you get 9/10 from?

I get the same thing, up to that factor of ten. Show me where I screwed up...

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9\cdot\frac{1}{1!} + 90 \cdot \frac{1}{2!} + \cdots

Then this is is just

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9 \Bigg{\sum_{n=0}^{\infty} \frac{10^n}{n!} - 1\Bigg} = 9 \cdot \Big{ e^{10} -1 \Big}

e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + ...

When doing the summation for the above question, it's:

9 + \frac{90}{2!} + \frac{900}{3!} + ...

9*(1 + \frac{10}{2!} + \frac{{100}}{3!} + ...)

9*(1 + \frac{10^{1}}{2!} + \frac{{10^{2}}}{3!} + ...)

Multiply by \frac{10}{10} and you should see it (I'm tired of typing tex).

10^0 = 1?

10^0 = 1?

yes? Why does this matter?

Ugh.. I don't feel like typing out the explanation. Basically by multiplying by ten, you get e^10 -1.

IAh yes, I screwed up.

Where do you get 9/10 from?

I get the same thing, up to that factor of ten. Show me where I screwed up...

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9\cdot\frac{1}{1!} + 90 \cdot \frac{1}{2!} + \cdots

Then this is is just

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9 \Bigg{\sum_{n=0}^{\infty} \frac{10^n}{n!} - 1\Bigg} = 9 \cdot \Big{ e^{10} -1 \Big}

This should be:

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9 \Bigg{\sum_{n=0}^{\infty} \frac{10^n}{n!}\Bigg}-1,

which gives the same answer as yours.

Apologies!

Your original "this" is correct, only that "-1" is not under the sum.

Now we knoe why camilus was asking about d(n). :-)

Now we knoe why camilus was asking about d(n). :-)

Yea I noticed that, too. However, I don't know why he was worrying about formula for it. I suppose he was hoping for an easy "plug and chug" so that the answer would present itself.

When I wrote out the summation formula, I saw pretty quick the taylor expansion for e^x mixed in there... I just needed to do some simple manpulation.

Also, didn't the OP originally post as Arcane?



As a side note... I am so excited! I may be able to write some sort of paper with my graph theory professor. This means that my first paper will be with a guy whose Erdos number is 1. Mine may be 2!

And sorry for last night Ben.. for some reason, I was a bit testy. I really did not feel like explaining the "error." :)

Wow, Erdos number 2! You don't see them everyday.

And sorry for last night Ben.. for some reason, I was a bit testy. I really did not feel like explaining the "error."

No worries man. My arithmetic is pretty dreadful anyway.

How do you calculate Erdos number?

How do you calculate Erdos number?

Erdos himself has the number zero. Anyone that has coauthored a paper with Erdos has Erdos number 1. Anyone that coauthors/coauthored a paper with someone that coauthored a paper with Erdos has Erdos number 2. Etc...

http://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Erdosnumber.png

Congrats! So, are you going to try for a finite Erdős-Bacon number?

wow absane I think you're the only one who got it.

it is indeed {9 \over 10}(e^{10}-1)

it comes directly from

\sum_{n = 0}^{\infty} {x^n \over (n+1)!} = {e^x \over x} - {1 \over x} = {e^x - 1 \over x}

http://i159.photobucket.com/albums/t121/camilus23/etothex.jpg

Bah I was off by a factor of ten.

wow absane I think you're the only one who got it.


I don't mean to be strange, but it would be fair to say Absane was the first to get it.

http://i159.photobucket.com/albums/t121/camilus23/etothex.jpg

I see you that you do mathematics in pen.