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flat assembler > Heap > A simple mathemagical trick for interested ones.

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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
And here you have my solution, a purely arithmetic one:

Image
Post 29 Nov 2015, 18:00
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tthsqe



Joined: 20 May 2009
Posts: 721
MHajduk, If you don't like calculus, maybe you should post and inequality that can't be tackled by an application of it.
Post 29 Nov 2015, 18:38
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
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.
Post 29 Nov 2015, 20:51
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tthsqe



Joined: 20 May 2009
Posts: 721
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                     
Post 30 Nov 2015, 02:01
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
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.

Wink
Post 30 Nov 2015, 04:14
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Xorpd!



Joined: 21 Dec 2006
Posts: 161
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.
Post 30 Nov 2015, 15:25
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tthsqe



Joined: 20 May 2009
Posts: 721
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.    
Post 01 Dec 2015, 01:22
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
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.    
Wink
Post 01 Dec 2015, 03:41
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
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.

Wink
Post 01 Dec 2015, 03:47
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
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.

Wink
Thank you for your words, I appreciate them. Smile
Post 04 Dec 2015, 19:23
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Image

Image
Post 13 Dec 2015, 21:08
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
MHajduk, those visualizations are nice. What did you use to make them?
Post 14 Dec 2015, 15:28
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Tyler wrote:
MHajduk, those visualizations are nice. What did you use to make them?
Blender + GeoGebra + IrfanView + heavily customized LaTeX Smile
Post 14 Dec 2015, 19:09
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Another Archimedean figure:

Image

Image
Post 14 Dec 2015, 22:25
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Image

Image
Post 29 Dec 2015, 00:10
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Image
Post 29 Dec 2015, 11:13
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Image
Post 03 Jan 2016, 02:52
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
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!

Wink
Post 03 Jan 2016, 11:52
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
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!

Wink
OK, I'll try. Wink

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):

Image
Post 03 Jan 2016, 20:16
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MHajduk



Joined: 30 Mar 2006
Posts: 6023
Location: Poland
Image

Image

Image

Image

Image
Post 05 Jan 2016, 01:00
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