# Math question

**URL:** <https://boards.straightdope.com/t/math-question/486689>\
**Category:** Factual Questions\
**Created:** [February 20, 2009, 6:22am UTC](https://boards.straightdope.com/t/math-question/486689 "2009-02-20T06:22:24Z")\
**Posts on this page:** 14\
**Page:** 1

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**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [February 20, 2009, 6:22am UTC](https://boards.straightdope.com/t/math-question/486689/1 "2009-02-20T06:22:24Z")

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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?

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**Author:** ![Oslo\_Ostragoth](https://avatars.discourse-cdn.com/v4/letter/o/a9a28c/32.png) [@Oslo\_Ostragoth](https://boards.straightdope.com/u/Oslo_Ostragoth)\
**Post date:** [February 20, 2009, 6:32am UTC](https://boards.straightdope.com/t/math-question/486689/2 "2009-02-20T06:32:59Z")

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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.

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**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [February 20, 2009, 6:34am UTC](https://boards.straightdope.com/t/math-question/486689/3 "2009-02-20T06:34:12Z")

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> [@Oslo\_Ostragoth](#):
>
> I’m getting:
> 
> .693 x 1.38 x 1.79 x 2.08 …
> 
> Either this doesn’t converge to 1, or I’m misreading your post.

The base of the logarithms is n

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [February 20, 2009, 7:14am UTC](https://boards.straightdope.com/t/math-question/486689/4 "2009-02-20T07:14:11Z")

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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].

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [February 20, 2009, 7:24am UTC](https://boards.straightdope.com/t/math-question/486689/5 "2009-02-20T07:24:44Z")

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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).

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [February 20, 2009, 8:02am UTC](https://boards.straightdope.com/t/math-question/486689/6 "2009-02-20T08:02:20Z")

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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.

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [February 20, 2009, 8:12am UTC](https://boards.straightdope.com/t/math-question/486689/7 "2009-02-20T08:12:10Z")

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> [@Indistinguishable](#):
>
> Break it up as follows:  
> [log(2) \* log(2n - 2)] \* [log(4) \* log(2n - 4)] \* [log(6) \* log(2n - 6)] \* …  
> …  
> Well, all of these pieces are of the form log(n - x) \* log(n + x)

[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.]

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<div class="post-metadata">

**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [February 20, 2009, 8:22am UTC](https://boards.straightdope.com/t/math-question/486689/8 "2009-02-20T08:22:29Z")

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> [@Indistinguishable](#):
>
> 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.

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.

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<div class="post-metadata">

**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [February 20, 2009, 8:29am UTC](https://boards.straightdope.com/t/math-question/486689/9 "2009-02-20T08:29:49Z")

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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

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<div class="post-metadata">

**Author:** ![Oslo\_Ostragoth](https://avatars.discourse-cdn.com/v4/letter/o/a9a28c/32.png) [@Oslo\_Ostragoth](https://boards.straightdope.com/u/Oslo_Ostragoth)\
**Post date:** [February 21, 2009, 7:16am UTC](https://boards.straightdope.com/t/math-question/486689/10 "2009-02-21T07:16:43Z")

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> [@Polerius](#):
>
> The base of the logarithms is n

Um, yeah, is my TI calculator malfunctioning?

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<div class="post-metadata">

**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [February 21, 2009, 7:21am UTC](https://boards.straightdope.com/t/math-question/486689/11 "2009-02-21T07:21:27Z")

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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.)

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<div class="post-metadata">

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [February 21, 2009, 7:22pm UTC](https://boards.straightdope.com/t/math-question/486689/12 "2009-02-21T19:22:42Z")

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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).

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<div class="post-metadata">

**Author:** ![Oslo\_Ostragoth](https://avatars.discourse-cdn.com/v4/letter/o/a9a28c/32.png) [@Oslo\_Ostragoth](https://boards.straightdope.com/u/Oslo_Ostragoth)\
**Post date:** [February 22, 2009, 2:25am UTC](https://boards.straightdope.com/t/math-question/486689/13 "2009-02-22T02:25:41Z")

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Jeez, can you tell that it has been a while for me? I used a slide rule in school… :smack:

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<div class="post-metadata">

**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [February 24, 2009, 8:26pm UTC](https://boards.straightdope.com/t/math-question/486689/14 "2009-02-24T20:26:49Z")

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> [@Polerius](#):
>
> I think I got it:
> 
> Jensen: sum(f(x\_k)) \<= N \* f(sum(x\_k)/N)  
> when f is concave

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.

```auto

(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
           

```