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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. | |
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Subject:
Re: Proof that something is greater than or equal to zero
Answered By: rbnn-ga on 25 Sep 2002 16:29 PDT Rated: |
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. | |
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ponderer-ga
rated this answer:
Thanks rbnn-ga Thanks to everyone else also who made comments and contributed. |
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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. |
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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 |
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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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