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Q: Proof that something is greater than or equal to zero ( Answered 5 out of 5 stars,   14 Comments )
Question  
Subject: Proof that something is greater than or equal to zero
Category: Science > Math
Asked by: ponderer-ga
List Price: $5.00
Posted: 24 Sep 2002 17:34 PDT
Expires: 24 Oct 2002 17:34 PDT
Question ID: 68672
math_man thought it should be too hard to prove that:
Y = 2 + d + ((2+d)/d)^d is greater than or equal to zero for all
delta.  [Note that ^ means to the power of].

I can't blame him, because it couldn't be too hard either when I first
saw it.

See: https://answers.google.com/answers/main?cmd=threadview&id=68347

I don't think this problem is easily solved or worth trying to "figure
out" for a mere $5 but thought I would post it anyway incase anyone
has seen it before.

Methods already tried:
Have already tried to show it 
* directly as is by splitting and pairing, as well as after taking
logs
* by showing the derivative is strictly negative and calculating value
at d=1, but can't sign the derivative either.

I suspect the solution might involve some kind of power series
expansion which I am not familar with.  Myself (graduate student of
econometrics) and my colleague (PhD in maths) can't get it anyway.

Clarification of Question by ponderer-ga on 24 Sep 2002 17:37 PDT
Sorry, forgot to mention when I specified the question this time, for
all delta BETWEEN 0 and 1.  You can see this is true by plotting the
graph, but of course regretfully thats not a mathematical proof.

Clarification of Question by ponderer-ga on 24 Sep 2002 19:20 PDT
OK I'm really sorry about this, I have done something very stupid.  I
stayed up all night trying to solve this problem 12 hours working on
this problem, but that was a few days ago, and I have misentered it
here.

I meant to say Y = 2 + d - ((2+d)/d)^d.  I put a plus where I meant to
put a minus.  So if I get a stupid answer back to this question, then
its my own fault, sincerest apologies, thanks to the comment from
posted on my other thread which tipped me off to my error.  Obviously
if the minus were a plus it would be elementary - all the terms would
be positive...

https://answers.google.com/answers/main?cmd=threadview&id=68347

Banging head against wall.
Answer  
Subject: Re: Proof that something is greater than or equal to zero
Answered By: rbnn-ga on 25 Sep 2002 16:29 PDT
Rated:5 out of 5 stars
 
We want to show

 2+d-((2+d)/d)^d >=0

for all 0<d<=1

[Note: the original problem stated, for all d "between" 0 and 1, but
the formula is
undefined for d=0, so I interpret between as above.]

Now for d=1, the inequality
 is 3 - 3 >=0 which is true.

So it suffices to prove the formula for 0<d<1, which we assume in the
sequel.


The given formula is true if and only if:

2+d - exp(d log ((2+d)/d)) >= 0

where "log" shall mean natural logarithms here. This formula is true
iff

   -exp( d log ((2+d)/d)) >= -2 -d

which holds iff

   exp( d log ((2+d)/d)) <= 2 + d

which holds iff

   exp (d log (2+d) - d log d) <= 2+d

which holds iff

   d log (2+d) - d log d <= log (2 + d)

which holds iff

   (d-1) log (2+d) <= d log d

which holds iff

   d log d - (d-1)log (d+2) >=0 .

Let f(d)= d log d - (d-1)log (d+2) .

Then it suffices to show:

1. f is continuous at 0 and 1

2. f(0)>f(1)>=0

3. f'(d)<0


Conditions 1,2, and 3 together imply f(d)>=0.

Condition 1 depends on the fact that 

  lim   x log x = 0

This can be verified by L'hopital's rule, see e.g. the computation at:
http://www.mapleapps.com/powertools/calcI/html/L21-LHospital.html .

For condition 2, observe that

  f(0)= log(2)

and f(1) = 0

so condition 2 is true.

Thus, it remains to show condition 3, when we will be done.

We have f'(d) =

  log d + 1 - log (d+2) - (d-1)/(d+2) .

Condition 3 will be implied by the truth of conditions 3(a), 3(b), and
3(c) where:

Condition 3(a):  

  f'(1)<=0

Condition 3(b): 

  lim   f'(d) < f'(1)
 (d->0+)


Condition 3(c)

 f''(d) > 0


We now demonstrate each of these three conditions, which will conclude
the proof.

For condition 3(a), we see f'(1) = 1-log(3) < 0 since e<3 .

For condition 3(b), we see by inspection that f(d)-> -infinity as d->0
from above.

For condition 3(c), we observe:

But f''(d) =

  1/d - 1/(d+2) - ((d+2)-(d-1))/(d+2)^2

 =

  1/d - 1/(d+2) - 3/(d+2)^2

=

  2/(d(d+2)) - 3/(d+2)^2

Thus,

f''(d)>0 iff

  2/(d(d+2)) - 3/(d+2)^2 >0

iff

   2/d(d+2) > 3/(d+2)^2

iff

   2 (d+2)^2 > 3 (d^2+2d)

iff

   2d^2 + 8d + 8 > 3d^2 + 6d

iff

   d^2 - 2d -8 < 0

iff 

  (d-4)(d+2) < 0

which is clearly true for 0<=d<=1 .

Hence, f''(d) > 0, and condition 3(c) holds, so condition 3 holds, and
we are done.

Clarification of Answer by rbnn-ga on 25 Sep 2002 19:40 PDT
f(0) is defined so that f is right-continuous at 0; that is, f(0) may
be defined to be log(2). It is true that f is not defined for negative
arguments however, and "f continuous at 0" should read then "f
right-continuous at 0".

The main point here, is that we wish to show f is non-negative on the
interval (0,1] . Since f(1) is nonnegative, and f(0) > f(1) , all we
have to do is show that f' is negative.
ponderer-ga rated this answer:5 out of 5 stars
Thanks rbnn-ga
Thanks to everyone else also who made comments and contributed.

Comments  
Subject: Re: Proof that something is greater than or equal to zero
From: just4look-ga on 25 Sep 2002 12:56 PDT
 
I solved the solution of this question yesterday. It is a nice question.
Subject: Re: Proof that something is greater than or equal to zero
From: sublime1-ga on 25 Sep 2002 14:51 PDT
 
ponderer...

> Banging head against wall

Why not take a break, and you and your friend can
play with this one for awhile:

Given a = b

multiply both sides by a:

a^2 = ab

subtract b^2 from both sides:

a^2-b^2 = ab-b^2

factor:

(a+b)(a-b) = b(a-b)

divide both sides by (a-b):

(a+b)(a-b)/(a-b) = b(a-b)/(a-b)

a+b = b

substitute b for a, since a = b

2b = b

divide by b:

2 = 1

subtract 1 from each side:

1 = 0

Just a little diversion...
If you know the secret, don't 
post it right away...
It will be fun to see the comments...
 :)
Subject: Re: Proof that something is greater than or equal to zero
From: math_zip_at_yahoo-ga on 25 Sep 2002 18:03 PDT
 
A comment for the answer:

It is not correct that f is continuous at 0 since it is not defined there.
Subject: Re: Proof that something is greater than or equal to zero
From: just4look-ga on 25 Sep 2002 18:33 PDT
 
Here is outline of my answer:

Let f(x)=2+x-((2+x)/x)^x,

f'(x)=1-((2+x)/x)^x(x/(2+x)-1-ln(x/(2+x))).

lim_{x\to 0^+}f'(x)=3(1-ln3)<0, (sorry to use Latex symbol here)

f''(x)=((2+x)/x)^x((1-x/(2+x)+ln(x/(2+x)))^2+4/(x(2+x)^2))>0.

So f'(x) is monotonely increasing based Lagrange mean value theorem, that is 

f'(x)<0, x\in (0, 1].

Hence f(x) is decreasing in (0, 1]. Therefore
         f(x)>f(1)=0, x\in (0, 1)

That's all.
Subject: Re: Proof that something is greater than or equal to zero
From: just4look-ga on 25 Sep 2002 18:36 PDT
 
Sorry for two mistyped:

lim_{x\to 1^-}f'(x)=3(1-ln3)<0

f'(x)<lim_{x\to 1^-}f'(x)=3(1-ln3)<0, x\in (0, 1).
Subject: Re: Proof that something is greater than or equal to zero
From: math_zip_at_yahoo-ga on 25 Sep 2002 18:44 PDT
 
Hi just4look-ga,

Nice answer!
Subject: Re: Proof that something is greater than or equal to zero
From: rbnn-ga on 25 Sep 2002 19:55 PDT
 
just4look-ga : although I have not checked your work, I believe the
method I use is the same as yours. To prove the original formula is
positive, we just look at the endpoints and show the derivative is
positive; we then iterate with the derivative. The sign of the second
derivative is by inspection. In my derivation I simplified the terms
first to make differentiation easier.
Subject: Re: Proof that something is greater than or equal to zero
From: rbnn-ga on 25 Sep 2002 20:06 PDT
 
In point of fact my last comment should perhaps read "derivative is
negative" depending on just how define things. Anyway, the key is, if
the endpoints of a function have the same sign, and if the derivative
is nonzero, then we can conclude in these cases that the function
itself has the same sign as its endpoints. The rest is only a matter
of checking differentiability and performing the differentiation,
twice in these cases.
Subject: Re: Proof that something is greater than or equal to zero
From: math_zip_at_yahoo-ga on 26 Sep 2002 09:23 PDT
 
just4look-ga claimed to have the answer on 24 Sep 2002 22:27 PDT

You can find at

https://answers.google.com/answers/main?cmd=threadview&id=68347
Subject: Re: Proof that something is greater than or equal to zero
From: tne-ga on 28 Sep 2002 11:56 PDT
 
hi sublime1
(a+b)(a-b) = b(a-b) 
a-b = 0 you can't divide

and any number * 0 = any other number * 0 = 0 = a - b

I think you waited long enough for comments
Subject: Re: Proof that something is greater than or equal to zero
From: math_zip_at_yahoo-ga on 02 Oct 2002 20:10 PDT
 
tne-ga, just4look told  sublime1 just after  sublime1 posted the
question. However, just4look put his e-mail address in the message. So
site manager deleted the comment.
Subject: Re: Proof that something is greater than or equal to zero
From: sublime1-ga on 04 Oct 2002 00:01 PDT
 
tne... and just4look (sorry I missed your deleted response)...

Right your are!
Subject: Re: Proof that something is greater than or equal to zero
From: techjed-ga on 30 Oct 2002 09:20 PST
 
ponderer,

I realise you were mainly interested in the web as a tool for finding
proofs in general, but I couldn't resist your problem. The proof does
not require calculus.

Prove 2+d-((2+d)/d)^d >=0 for d in (0,1].

Since d is not zero, we can write 2+d as d((2+d)/d) and factor out
(2+d)/d to get:

    ((2+d)/d)*(d-((2+d)/d)^(d-1))>=0

Again, since (2+d)/d is not zero (or negative) we can divide through
to get:

    d >= (2+d)/d)^(d-1)

Now, 2+d > d is obvious and since d is in (0,1] it follows that

   (2+d)/d > 2+d > d

or

   (2+d)/d > (2+d)/d)^(d-1)

Next multiply both sides by (2+d)/d to get:

   ((2+d)/d)^2 > (2+d)/d)^d

This is true since (2+d)/d > 1 and 2 > d for all d in (0,1].

QED

Why cut cake with a chainsaw?

Concerning your 'real' dilemma - you'd probably be better off with
Schaum's Outlines rather than the web.
Subject: Re: Proof that something is greater than or equal to zero
From: rbnn-ga on 30 Oct 2002 11:43 PST
 
techjed-ga: Thank you for your comment. 

Either I do not understand your comment, or your comment is wrong.

Specifically, you demonstrate that if you can prove the inequality I
will call (*):

d>=((2+d)/d)^(d-1)

then the original inequality will follow.

You then show that (2+d)/d>d and you replace the left hand side of the
inequality (*) with (2+d)/d and show that the resulting inequality is
true. However, the fact that replacing A in an equality A>B with a
term A', and showing that A'>B, does not show that A>B.

Can you clarify your comment or confirm it's incorrect?

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