I found this in my papers from way back when:
Show that
log[sub]n[/sub]2 x log[sub]n[/sub]4 x log[sub]n[/sub]6 x … x log[sub]n/sub <= 1
for all n >= 2
I can’t seem to come up with a nice elegant solution to this. Any ideas?
I’m getting:
.693 x 1.38 x 1.79 x 2.08 …
Either your problem doesn’t converge to 1, or I’m misreading your post.
The base of the logarithms is n
Break it up as follows:
[log(2) * log(2n - 2)] * [log(4) * log(2n - 4)] * [log(6) * log(2n - 6)] * …
Each of these logarithms is positive, so to get that this product is <= 1, we just need to show that each of these bracketed pieces is <= 1.
Well, all of these pieces are of the form log(n - x) * log(n + x) (where x is in (-n, n)).
Since log(n) * log(n) = 1, it will suffice to demonstrate that the function which sends x to log(n - x) * log(n + x) is maximized when x = 0.
And this is a matter of simple calculus [take the derivative, observe that it always has the same sign as x].
Sorry, the last line from the above should be “[take the derivative, observe that it always has the opposite sign from x]” (thus establishing a maximum, not a minimum, at x = 0).
Actually, we can skip the calculus:
We want to show that log(n - x) * log(n + x) is maximized when x = 0. Note that log(n - x) * log(n + x) = log(n(1 - x/n)) * log(n(1 + x/n)) = [log(1 - x/n) + log(n)][log(1 + x/n) + log(n)] = [log(1 - x/n) + 1][log(1 + x/n) + 1] = log(1 - x/n) * log(1 + x/n) + log(1 - x/n) + log(1 + x/n) + 1 = log(1 - x/n) * log(1 + x/n) + log(1 - (x/n)^2) + 1. As far as maximization is concerned, the last constant term doesn’t matter; thus, let us just look at log(1 - x/n) * log(1 + x/n) + log(1 - x^2/n^2). If x is non-zero, then this is the product of a positive and a negative (in some order) + a negative = a negative. However, if x is zero, then this is zero. Thus, this is maximized when x = 0. Q.E.D.
[Except if n is even, in which case, the middle piece is just log(n), rather than [log(n) * log(n)]. But since these are both equal to 1, it doesn’t matter.]
Great!
I had gotten it to the point where I had to show that log(n - x) * log(n + x) <= 1, and I wasn’t coming up with any quick way of showing it.
I was hoping to use some sort of Jensen-type inequality to get that last step finished, but your calculus and non-calculus solutions both work.
I think I got it:
Jensen: sum(f(x_k)) <= N * f(sum(x_k)/N)
when f is concave
let f(x) = log log(x)
log is concave, so f(x) is concave
So, we have
log log (n+x) + log log(n-x) <= 2 log log((n+x + n-x)/2) = 2 log log n = 2 log 1 = 0
==> log log (n+x) + log log(n-x) <= 0
==> log (log (n+x) * log(n-x)) <= 0
==> log (n+x) * log(n-x) <= 1
Um, yeah, is my TI calculator malfunctioning?
You’re taking logarithms with a base of e (the usual default). The OP is about logarithms with base n, where n is some arbitrary integer. You couldn’t possibly calculate their actual value until you knew what n was, which you don’t, because it’s a variable…
It may also help to realize that the multiplication in the OP is of a finite series, not an infinite one. (How long is this finite series? It has n-1 many terms. What’s n? It’s the variable we were just talking about.)
You can verify this with a calculator for any specific n by applying the change of base formula and multiplying both sides by log(n).
Jeez, can you tell that it has been a while for me? I used a slide rule in school… :smack:
I can’t believe I missed it: I applied Jensen’s inequality to the sub-problem of proving log (n+x) * log(n-x) <= 1, when I could have removed that step and applied it to the original problem.
(All logs below are base n)
Let
f(x) = log log x
g(n) = log 2 * log 4 * log 6 * ... * log(2(n-1))
Then
log(g(n)) = sum f(2k) k=1..(n-1)
<= (n-1) f(sum(2k)/(n-1)) k=1..(n-1)
= (n-1) f(n)
= (n-1) log log n
= (n-1) * 0
= 0
=> log(g(n)) <= 0
=> g(n) <= 1
i.e log 2 * log 4 * log 6 * ... * log(2(n-1)) <= 1