Revival-Math Q&A

Hoempa

Math Team
Apr 2010
2,780
361
Your strategy for integer terms seems not to work, if I understand you correctly. You'd have 2014 mod 740 = 534 factors of 2?
Or what value of n do you have in mind?
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
For non-integer partitions we want all $a_i = \frac{a}{n}$.

$\left(\frac{2014}{740}\right)^{740} =$
5922872722214280204042951220535754568748058229975192062033840923815272292035394738358455364694736810891532414406767108776194951506270012601558499660738318680911061252684466883547210975319716231165745599434988964631692907377989456352505403024247023897144609460199612426516556662825736950523993459021787690963655417612195602 to 322 significant digits.

$\left(\frac{2014}{741}\right)^{741} =$
5926146105753072515104708783979408196380566594842659587510171850058064755378424268281865513738912510215463783772413755194307761467756648299208993881101427837565501747859767506772812193125024860980353781117084266088048848035314302562152521410943926373304674119330548433253990147211280514102240883367119202019688622231219548 to 322 significant digits

So the 741-partition is best.
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
Your strategy for integer terms seems not to work, if I understand you correctly. You'd have 2014 mod 740 = 534 factors of 2?
Or what value of n do you have in mind?
I think I've edited the post since you looked. I've got a full answer now.
 

Hoempa

Math Team
Apr 2010
2,780
361
Indeed.
To get that result I compared 740 * log(2014/740) with 741 * log(2014/741).
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
That would be the easy way! I'd just found this online BigInt Calculator so I carried on using it.

When is $n \left( \log{a} - \log{n} \right) \lt (n+1) \left( \log{a} - \log{(n+1)} \right)$?
 
Last edited:

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
\begin{align*}
n \left( \log{a} - \log{n} \right) &\lt (n+1) \left( \log{a} - \log{(n+1)} \right) \\
n \log{a} - n\log{n} &\lt n\log{a} - n\log{(n+1)} + \log{a} - \log{(n+1)} \\
n \log \frac{n+1}{n} &\lt \log \frac{a}{n+1} \\
\end{align*}

So, for large $a$ the answer will be "always", since $\frac{a}{n+1} \approx e$ and $\frac{n+1}{n} \approx 1$ we get $0 \lt 1$ for large $a$.
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
Would that mean that \(\displaystyle e^{i\pi} \ne -1\)
No, it means that -1 is not a root of $z^\pi = -1$, but then -1 is not a root of $z^2 = -1$, so we needn't be surprised.
 
Last edited:

Hoempa

Math Team
Apr 2010
2,780
361
I missed a part from you earlier post in which you elaborated your answer for 1.) Sorry, there is a higher number to get.

It's harder indeed with that calculator! First, we have to compute 2014^740, store it in memory since we can't type 2014^740 in the textbox for x.
Then compute 740^740. Put back the results and divide.

I'm sort of surprised that (-1)^pi isn't -1 (If that's to be read from it). But than again, powers with negative bases are fishy.
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
Did you see the point where I got to $2^{206}3^{534}$? I proved that the 741-partition has a smaller product.

If I'm not correct there, I guess the problem is with going from $e^\frac{a}{e}$ to an expression with integers being raised to a power. But since the integers need to be as close as possible to each other, I can't see how I'd be making an error there.
 
Last edited:

Hoempa

Math Team
Apr 2010
2,780
361
Yes, I did see how you got there, but you have a good hint in your own post.
 
Similar Math Discussions Math Forum Date
Calculus
Math Books
Number Theory
Number Theory