flat assembler
Message board for the users of flat assembler. Index > Heap > A simple mathemagical trick for interested ones. Goto page Previous  1, 2, 3, 4, 5  Next
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 Thread  MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk Yeah, this is the second analytic proof I have seen today (I have posted the same exercise also in other places and one person solved this by calculating the first derivative of the function f(x) = (a^x - b^x)/(a^x + b^x) and proved that f'(x) > 0 for any x, basically it's the same approach as yours). There are also proofs that don't involve differential calculus, only basic arithmetic operations. 19 Nov 2015, 01:05
 YONG Joined: 16 Mar 2005 Posts: 8000 Location: 22° 15' N | 114° 10' E YONG tthsqe wrote:(x-1)/(x+1) is increasing for x>1 (a/b)^p > (a/b)^q > 1 by hypothesis "(a/b)^q > 1" Is this really true? Suppose q = 0. (a/b)^q = 1. Suppose q = -1. (a/b)^q = b/a < 1.   19 Nov 2015, 04:59
tthsqe

Joined: 20 May 2009
Posts: 724
tthsqe
oops - please change my statement to
(x-1)/(x+1) is increasing for x>0
(a/b)^p > (a/b)^q > 0 by hypothesis

it is also obvious that (x-1)/(x+1) = 1-2/(x+1) is increasing for x>0 without doing any analysis. 19 Nov 2015, 05:23  MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk Below I present my solution of the aforementioned math problem. We prove the given inequality showing step-by-step its equivalence with the condition a > b. The number of the condition (placed near the "if and only if" connective) used in the particular transition signalizes the importance of the condition for assuring the current step is correct.  19 Nov 2015, 19:40
 revolution When all else fails, read the source Joined: 24 Aug 2004 Posts: 17247 Location: In your JS exploiting you and your system revolution MHajduk wrote:Below I present my solution of the aforementioned math problem. We prove the given inequality showing step-by-step its equivalence with the condition a > b. The number of the condition (placed near the "if and only if" connective) used in the particular transition signalizes the importance of the condition for assuring the current step is correct. Perhaps I misunderstood something but how do you prove that you can eliminate the negative roots? 20 Nov 2015, 01:23
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk revolution wrote:Perhaps I misunderstood something but how do you prove that you can eliminate the negative roots? Very good question, you're the first person that noticed this fact. Of course, we have that from the first condition a > b > 0 and this assures as that we won't calculate roots from the negative numbers. We should all the time bear in mind that we are in the "world" of the positive numbers it's a consequence of the first condition. Or maybe you meant something else? 20 Nov 2015, 16:13
 revolution When all else fails, read the source Joined: 24 Aug 2004 Posts: 17247 Location: In your JS exploiting you and your system revolution I was thinking that for example sqrt(4) = +/- 2. You have (p-q)root(some positive value). But for integral (p-q) there can be both negative and positive solutions. 20 Nov 2015, 20:43
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk  20 Nov 2015, 21:17
 revolution When all else fails, read the source Joined: 24 Aug 2004 Posts: 17247 Location: In your JS exploiting you and your system revolution Let's make a=2 and (p-q)=2. (p-q)root(a^(p-q)) = sqrt(4) = +/- 2. I think you can't run away from the negative result by some algebraic manipulations. Otherwise, from your description above, there would never be any negative roots, ever. 20 Nov 2015, 21:55
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk  20 Nov 2015, 23:54
 revolution When all else fails, read the source Joined: 24 Aug 2004 Posts: 17247 Location: In your JS exploiting you and your system revolution I was taught: When taking roots we have to account for both negative and positive roots. The mathematics doesn't know that we only want positive roots, it still gives the valid result of negative roots. We can choose to ignore negative roots but then that makes any proofs based upon it invalid or incomplete. I still claim -2 is a perfectly valid solution to the square root of 4. Last edited by revolution on 21 Nov 2015, 05:50; edited 1 time in total 21 Nov 2015, 01:47
TmX

Joined: 02 Mar 2006
Posts: 821
Location: Jakarta, Indonesia
TmX
I guess this is a sort of convenience/convention.

When asked what are the solutions of x^2 = 25, you will be expected to answer both 5 and -5.
But when asked what is the square root of 25, just answer 5.
Maybe most people prefer injective function  21 Nov 2015, 05:47  YONG Joined: 16 Mar 2005 Posts: 8000 Location: 22° 15' N | 114° 10' E YONG Let's try another route. TRUE <=> a > b <=> ln a > ln b [since a & b are +ve and ln is monotonic increasing] <=> (p-q) ln a > (p-q) ln b [since (p-q) > 0] <=> ln [a^(p-q)] > ln [b^(p-q)] [by properties of logarithm] <=> a^(p-q) > b^(p-q) [since ln is monotonic increasing]  21 Nov 2015, 06:04
 revolution When all else fails, read the source Joined: 24 Aug 2004 Posts: 17247 Location: In your JS exploiting you and your system revolution I think the trick is simply to realise that (p-q) > 0. And simplify to an equivalent a^c > b^c where c=(p-q) and a>b>0. YONG's solution also works for me. 21 Nov 2015, 07:25
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk revolution wrote:I was taught: When taking roots we have to account for both negative and positive roots. The mathematics doesn't know that we only want positive roots, it still gives the valid result of negative roots. We can choose to ignore negative roots but then that makes any proofs based upon it invalid or incomplete. I didn't ignore the negative roots, only have said that they have no influence on the logical value of the equivalence / biconditional a^(p-q) > b^(p-q) <=> a > b From the assumption a^(p-q) > b^(p-q) , a > b> 0 , p > q we can prove many facts, among them is the inequality a > b. The implication a > b => a^(p-q) > b^(p-q) has been proved above by you.The points 1 and 2 give us a proof of the mentioned biconditional. revolution wrote:I still claim -2 is a perfectly valid solution to the square root of 4. It depends on what you understand by the sign of the "square root". If we are talking about sqrt(x) as a function from R to R we can't accept negative values. If we are talking about solutions of the relation y^2 = x in R^2 then we have to accept both pairs (x, -sqrt(x)) and (x, sqrt(x)). The widely accepted practice is to treat the sign of the square root as a symbol of a function (as long as we don't move to the field of the complex numbers where we have to accept all n roots of the n-th root of the given numbers, both real and complex). 21 Nov 2015, 16:55
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk A stained glass effect test. This was something truly laborious from the technical point of view (a lot of work with lighting).  27 Nov 2015, 00:14
tthsqe

Joined: 20 May 2009
Posts: 724
tthsqe
I would say taylor's inequality for sqrt(x) about x=1. This works for both sides of the inequality if you dehomogenize it both ways. 28 Nov 2015, 03:41  Xorpd! Joined: 21 Dec 2006 Posts: 161 Xorpd! I've never heard of Taylor's inequality, but here is an approach which seems to work: Let f(a,b) = 4a(a+b)-8√(a³b)-(a-b)² = 3a²+6ab-b²-8√(a³b) Then on the left boundary of the domain, a = b, f(a,a) = 0. ∂f/∂a = 6a+6b-12√(ab) = 6(√a-√b)² > 0 for a > b, so the function increases from left to right and so is nonnegative over its domain. It's easy to get from there to the left inequality. Let g(a,b) = (a-b)²-4b(a+b)+8√(ab³) = a²-6ab-3b²+8√(ab³) On the upper boundary of the domain, a = b, g(b,b) = 0. ∂g/∂b = -6a-6b+12√(ab) = -6(√a-√b)² < 0 for a > b, so the function increases from top to bottom and so is nonnegative over its domain. This yields the right inequality. Edit: corrected 2 typos. 28 Nov 2015, 05:15
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk tthsqe wrote:I would say taylor's inequality for sqrt(x) about x=1. This works for both sides of the inequality if you dehomogenize it both ways. That's interesting. Could you explain it in details? 29 Nov 2015, 17:49
 MHajduk Joined: 30 Mar 2006 Posts: 6034 Location: Poland MHajduk The approach presented by Xorpd! gives me a new perspective of solving inequalities. I dared to formulate his solution in a more descriptive manner (as I understood it):   29 Nov 2015, 17:52
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