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
Author
 Thread  MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk And here you have my solution, a purely arithmetic one:  29 Nov 2015, 18:00
tthsqe

Joined: 20 May 2009
Posts: 721
tthsqe
MHajduk, If you don't like calculus, maybe you should post and inequality that can't be tackled by an application of it. 29 Nov 2015, 18:38  MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk tthsqe wrote:MHajduk, If you don't like calculus, maybe you should post and inequality that can't be tackled by an application of it. I didn't say that I don't like calculus. I have just only shown that this problem may be solved arithmetically, just like that. BTW, you have answered to my question with some supposition. It's not the thing I have expected. 29 Nov 2015, 20:51
tthsqe

Joined: 20 May 2009
Posts: 721
tthsqe
OK, so how would you prove

Code:
```a >= b > 0 implies

2
3*(a-b)      6*a + b     6/7   1/7
--------- <= --------- - a   * b
49*a           7                     ``` 30 Nov 2015, 02:01  YONG Joined: 16 Mar 2005 Posts: 8000 Location: 22° 15' N | 114° 10' E YONG Code:```Define f(x) = 3x^2 - 13x + 49x^(1/7) - 39. f(0) = -39 < 0 f(1) = 0 f'(x) = 6x - 13 + 7/x^(6/7) = 6 (x - 1) + 7 [1/x^(6/7) - 1] Thus, f'(x) > 0 on the interval (0, 1). Since f(x) is increasing on (0, 1), we have f(x) =< f(1) = 0. Now, put x = b/a. Note that 0 < x =< 1. 3(b/a)^2 - 13(b/a) + 49(b/a)^(1/7) - 39 =< 0 Multiply both sides by a^2 (>0): 3b^2 - 13ab + 49 a^(13/7) b^(1/7) - 39a^2 =< 0 Re-arrange the terms: 3a^2 - 6ab + 3b^2 =< 42a^2 + 7ab - 49 a^(13/7) b^(1/7) 3 (a^2 - 2ab + b^2) =< 7a (6a + b) - 49a [a^(6/7) b^(1/7)] 3 (a - b)^2 =< 7a (6a + b) - 49a [a^(6/7) b^(1/7)] 3 (a - b)^2 / 49a =< (6a + b)/7 - [a^(6/7) b^(1/7)] ```We don't really need partial differentiation for questions like this.  30 Nov 2015, 04:14
 Xorpd! Joined: 21 Dec 2006 Posts: 161 Xorpd! The next day after I posted that proof I felt bad because if a = b > 0, it follows that (a-b)²/(8a) ≤ (a+b)/2-√(ab) ≤ (a-b)²/(8b) because all of them are zero. Else if a > b > 0, the way is clear if we multiply through by 2/(√a-√b)². But tthsqe graciously resurrected my technique of proof by offering a theorem which is much more awkward to solve purely algebraically! YONG's method of proof is good except that I can't see the insight that allows us to conclude without further discussion that 6(x-1)+7(1/x^(6/7)-1) > 0 because here 6(x-1) < 0. It is true, however because given that f'(x) = 6x-13+7/x^(6/7), we see that f'(1) = 0 and f"(x) = 6-6/x^(13/7) = 6(1-1/x^(13/7)) < 0 for x ∈ (0,1), so f'(x) > 0 for x ∈ (0,1). The f'(x) that YONG gets is the same as what would be obtained if 49× the original expression were written as f(a,b) ≥ 0 and then ∂f/∂b were taken. An alternative might be to multiply by 49a and then we get f(a,a) = 0 and ∂f/∂a = 78a+13b-91a^(6/7)b^(1/7) > 0 for a > b > 0, which we could prove by analyzing ∂²f/∂a² as before or by noting that ∂f/∂a = 91((6a+b)/7-(a⁶b)^(1/7) > 0 for a > b > 0 because the arithmetic mean of a set of positive numbers is greater than or equal to the geometric mean of the set with equality only if all numbers in the set are equal. 30 Nov 2015, 15:25
tthsqe

Joined: 20 May 2009
Posts: 721
tthsqe
Code:
```a > b > 0 implies
2                                2
3*(a-b)    6*a + b    6/7  1/7   3*(a-b)
-------- < ------- - a   *b    < --------
49*a        7                    49*b

Proof: With f(x) = x^(1/7), Taylor's Theorem says
that there is a t between 1 and x such that
f''(t)      2
f(x) = f(1) + f'(1)*(x-1) + ------*(x-1)  ,
2
or
6 + x    1/7   3   -13/7      2
----- - x    = --*t     *(x-1)  .
7            49
If x < 1, then this r.h.s is > 3*(x-1)^2/49.
Setting x = b/a gives the first inequality after homogenizing.

The second inequality follows from
using f(x) = x^(6/7) and setting x = a/b.    ``` 01 Dec 2015, 01:22  YONG Joined: 16 Mar 2005 Posts: 8000 Location: 22° 15' N | 114° 10' E YONG Xorpd! wrote:YONG's method of proof is good except that I can't see the insight that allows us to conclude without further discussion that 6(x-1)+7(1/x^(6/7)-1) > 0 because here 6(x-1) < 0. I am so sorry because I thought it was an obvious result of YONG's Theorem, which states: Code:```Given a positive integer n. The function g(x) > 0 for all x in (0, 1), where g(x) = n (x - 1) + (n + 1) { 1 / x^[n/(n+1)] - 1 }. Proof: g is continuous on (0, 1). (As x tends to 0+, g(x) tends to positive infinity.) g(1) = 0. g'(x) = n { 1 - 1 / x^[(2n+1)/(n+1)] } < 0 for 0 < x < 1. So, g is decreasing on (0, 1). Thus, g(x) > g(1) = 0 for 0 < x < 1. ```  01 Dec 2015, 03:41
 YONG Joined: 16 Mar 2005 Posts: 8000 Location: 22° 15' N | 114° 10' E YONG A side note: I am happy for the thread-starter, MHajduk, that this topic is gaining more and more interest in the forum. It seems that my suggestion to keep everything in one thread is correct after all.  01 Dec 2015, 03:47
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk YONG wrote:A side note: I am happy for the thread-starter, MHajduk, that this topic is gaining more and more interest in the forum. It seems that my suggestion to keep everything in one thread is correct after all. Thank you for your words, I appreciate them.  04 Dec 2015, 19:23
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk   13 Dec 2015, 21:08
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
MHajduk, those visualizations are nice. What did you use to make them? 14 Dec 2015, 15:28  MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk Tyler wrote:MHajduk, those visualizations are nice. What did you use to make them? Blender + GeoGebra + IrfanView + heavily customized LaTeX  14 Dec 2015, 19:09
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk Another Archimedean figure:   14 Dec 2015, 22:25
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk   29 Dec 2015, 00:10
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk  29 Dec 2015, 11:13
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk  03 Jan 2016, 02:52
 YONG Joined: 16 Mar 2005 Posts: 8000 Location: 22° 15' N | 114° 10' E YONG Please show us the making of the Klein bottle. Also explain, with the aid of 3D illustrations, why it is an example of a non-orientable surface. Thanks!  03 Jan 2016, 11:52
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk YONG wrote:Please show us the making of the Klein bottle. Also explain, with the aid of 3D illustrations, why it is an example of a non-orientable surface. Thanks! OK, I'll try. Below you have a variation on the theme of sphericon - a hexagonal sphericon (the plane going through two apexes has a shape of the regular hexagon):  03 Jan 2016, 20:16
 MHajduk Joined: 30 Mar 2006 Posts: 6032 Location: Poland MHajduk      05 Jan 2016, 01:00
 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First  Jump to: Select a forum Official----------------AssemblyPeripheria General----------------MainDOSWindowsLinuxUnixMenuetOS Specific----------------MacroinstructionsCompiler InternalsIDE DevelopmentOS ConstructionNon-x86 architecturesHigh Level LanguagesProgramming Language DesignProjects and IdeasExamples and Tutorials Other----------------FeedbackHeapTest Area
Goto page Previous  1, 2, 3, 4, 5  Next

Forum Rules:
 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot vote in polls in this forumYou can attach files in this forumYou can download files in this forum