(no subject)

[jack]

One of Gerald Duck's question was: Which is greater, pi^e or e^pi? I happened to be doing something related yesterday, and was sufficiently amused to continue here.

Spoilers

SNAPE KILLS DUMBLEDORE! Just kidding.

e^pi is greater for any value of pi.

The other question was, "Given a set of n positive integers which sum to 100, what is their maximum product?"

If you relax the constraint on being integers, and there being an integer number of them, and assume they are all the same, you get "Observe that (100/n)^n is low at n=1 and n=100. Where is the maximum?" and differentiating by n (with product and chain rule and letting n=e^logn), you find, unsurprisingly, the answer is always e.

Of course, for the problem with integers, it suggests a sum of 3s is correct, but it's easier to prove it by showing any change makes it smaller.

But it incidentally solves Gerald Duck's problem -- e^x is inherently a maximum :)

Comments

[megamole.livejournal.com]

But pi^e is tastier.

[rochvelleth.livejournal.com]

Mmm, mathmo pi^e :)

[drswirly.livejournal.com]

for the problem with integers, it suggests a sum of 3s is correct

Well, not always just 3s. The solution for 100 isn't 1*3³³, after all.

[cartesiandaemon.livejournal.com]

It's ok, I had a proper proof for integers before thinking about reals (none =1 or >4; at most two 2s; a most one 4; can't have both 2 and 4). But I forgot to explain there.

[drswirly.livejournal.com]

(none =1 or >4; at most two 2s; a most one 4; can't have both 2 and 4)

More concisely: since 2+2 = 2*2, you can ignore 4 as well, so you're left with just 2s and 3s, and at most two 2s. (We're only after the biggest total, not the number of ways of writing it!)

[cartesiandaemon.livejournal.com]

Mmm, yes, ok. That would eliminate one more step.

[cartesiandaemon.livejournal.com]

OK, two steps.

(i) Can't have 1s.
(ii) WLOG can't have 4s or bigger.
(iii) Can't have three 2s.

Doesn't give you all both possibilities, but gives you the max.

[drswirly.livejournal.com]

Indeed.

(i) n > 1*(n-1), so 1s are always useless

(ii) pulling out a 2 is at least as good if and only if 2(n-2) >= n, which is true iff n >= 4, therefore no point in using 4 or more

(iii) 2+2+2 = 3+3 but 2*2*2 < 3*3 (which is something I love writing down in a degree-level supervision!), so at most two 2s.

[mair_in_grenderich]

what?

[simont]

A couple of months ago, gerald_duck posted a list of mathematical problems, the tenth of which is what Jack is referring to here without a link.

I'll just eat pie instead.

[mair_in_grenderich]

I realised that, but I'm utterly baffled by the content.

[simont]

I think the derivation goes like this:

Raise both e^pi and pi^e to the power 1/(pi*e). This is monotonic in the positive reals, so it doesn't affect which of them is greater. But now the question becomes, is e^(1/e) greater or less than pi^(1/pi)? And the nice thing about the question in that form is that we've turned it into a function of one variable which we can investigate using the usual calculus tools: differentiating x^(1/x) gives (if I haven't made a mistake) x^(1/x-2)*(1-log(x)), which is positive for x < e, negative for x > e, and zero at x = e itself. Thus, it has a unique global maximum at x=e, and hence e^(1/e) > pi^(1/pi), and therefore we can raise that to the power e*pi and find the answer to the original question is that e^pi > pi^e. And, as Jack says, this proof demonstrates that the key thing is that one of the numbers is e; pi is arbitrary and could have been any other number you like.

[cartesiandaemon.livejournal.com]

Thank you for filling in! Yes, that's a good summary.

Mair -- sorry for filling my journal with incomprehensible wittering :)