Computing Series Product: a-i+1^a-i+1

[oscaralive]

Hi all,
anyone knows how to compute the following serie?

\prod_{i=1}^{a}(a-i+1)^(a-i+1)

Many thanks in advance!

 

[tiny-tim]

hi oscaralive!

(try using the X² icon just above the Reply box )

isn't that just 1¹2²3³4⁴…a^a …

why not leave it as it is? ​

(or you could write it as (a!)^a over something)

 

[oscaralive]

Because I'm trying to find an upper bound to define the complexity of an algorithm...and I cannot put it that way...it would be great to find an appropriate upper bound to this product

Thanks

 

[tiny-tim]

well, the log would be ∑nlogn … does that help?

 

[oscaralive]

In fact, I come from the log serie...

$$\sum_{i=1}^{a}(a-i+1)log(a-i+1)$$

$$\sum_{i=1}^{a}log((a-i+1)^{(a-i+1)})$$

$$log(\prod_{i=1}^{a}{(a-i+1)}^{(a-i+1)})$$
which now has been transformed to the product...

thanks,

 

[tiny-tim]

i was wondering whether it would be close to ∫ xlogx dx

 

[oscaralive]

I'm stuck here :(