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

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tthsqe



Joined: 20 May 2009
Posts: 721
OK. It is time for a serious problem. Prove this:
Code:
For all real numbers a>-7/8, there is an integer N(a) such that,
for all integers n>N(a), the coefficients of the polynomial
(x^4 + x^3 + a*x^2 + x + 1)^n
are all non negative.    
Post 21 Jan 2016, 22:40
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tthsqe



Joined: 20 May 2009
Posts: 721
In case that is too difficult, here is a generalization.
Code:
For all pairs of real numbers a, b  with
                         2
            max(0, 4 - b)
b > 0,  a > -------------- - 2
                   8
there is an integer N(a,b) such that, 
for all integers n>N(a,b), the coefficients of the polynomial 
(x^4 + b*x^3 + a*x^2 + b*x + 1)^n 
are all non negative.    
Post 22 Jan 2016, 01:26
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MHajduk



Joined: 30 Mar 2006
Posts: 5994
Location: Poland
tthsqe wrote:
MHajduk,
Why don't you put my problem (on the other thread) on a vase or another nice presentation?
I would also be nice if you write the solution on a vase as well.
I have made a 3D picture for you - although your problem isn't placed on a vase I think you may like the idea of placing it on the pages of a book.

The book model was made using a few dozen of Bezier curves, each for every single page, the liner and the book cover. Despite of the fact of doubles removal, recalculation of normals the entire render time exceeded 2 hours and 47 minutes. I think I have pushed my machine to the limits and I see that for truly good 3D modelling I should buy something more proper than my current laptop.

Image

Detailed view:
Image
Post 28 Jan 2016, 22:48
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MHajduk



Joined: 30 Mar 2006
Posts: 5994
Location: Poland
Referring to the problem itself, I am almost sure that it is absolutely non-trivial thing. Seems that you have been somehow inspired by a paper written by Colleen Ackermann and John d'Angelo from Univeristy of Illinois (http://www.math.illinois.edu/REGS/reports11/ackermann.pdf). Interesting, do you know the proof of the implication from Property 2 to Property 1. It is not explained there, nor they have given a proof there.

I can understand the implication from the Property 1 to Property 2 and even equivalence between Theorem 2 and Property 2 for p(x) = x^4 + b*x^3 + a*x^2 + b*x + 1 but here you ask about implication Property 2 => Property 1 which is "nearly true" for the mentioned authors. Is the answer so straightforward that they even didn't want to sketch a simple proof or maybe such proof doesn't exist (yet)?
Post 28 Jan 2016, 23:44
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tthsqe



Joined: 20 May 2009
Posts: 721
Im glad you found a reference for this! I was let to the generalization through some numerical testing and a heuristic argument - seems that the paper has the same conclusion. Will look into this further.
Post 29 Jan 2016, 03:17
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Picnic



Joined: 05 May 2007
Posts: 1269
Location: countryside
Lovely models MHajduk, i like your Ptolemy's theorem disc. Such drawings could decorate instructional textbooks pages. May i ask what modeling software you are using?
Post 29 Jan 2016, 10:45
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MHajduk



Joined: 30 Mar 2006
Posts: 5994
Location: Poland
Picnic wrote:
May i ask what modeling software you are using?
Blender, a freeware for 3D modelling. Very good, but its user interface looks a bit "exotic" at first glance. Many unusual hot-keys and work modes. Anyway it's a very good software for such a purpose. Smile
Post 29 Jan 2016, 12:57
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Picnic



Joined: 05 May 2007
Posts: 1269
Location: countryside
Yes, it does look somewhat complicated at first glance. I think i have to upgrade my hardware first. Rolling Eyes
Post 30 Jan 2016, 06:58
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
I came across this interesting geometry problem today.

A right triangle has a hypotenuse equal to 10 and an altitude to the hypotenuse equal to 6. What is the area of the triangle?

Wink
Post 30 Aug 2016, 04:46
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 16576
Location: Earth 2.0 beta
YONG wrote:
I came across this interesting geometry problem today.

A right triangle has a hypotenuse equal to 10 and an altitude to the hypotenuse equal to 6. What is the area of the triangle?
Seems too easy, what is the catch?

6 * (10^2 - 6^2)^(1/2) / 2 = 24 units square
Post 30 Aug 2016, 05:46
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
You seem to have misunderstood the meaning of "an altitude to the hypotenuse equal to 6".

It means if we treat the hypotenuse as the base of the triangle, the altitude of the triangle is 6.

Now, just find the area.

Wink
Post 30 Aug 2016, 06:43
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 16576
Location: Earth 2.0 beta
Are you sure that is possible? A circle of diameter 10 can't fit a right triangle of more than 5 units height.

Maybe I misunderstand something again.
Post 30 Aug 2016, 06:55
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YONG



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

Correct answer: Such a right triangle does not exist.

Wink
Post 30 Aug 2016, 07:28
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shoorick



Joined: 25 Feb 2005
Posts: 1605
Location: Ukraine
kinda area of triangle with sides 8, 12 and 4
Post 30 Aug 2016, 17:56
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
shoorick wrote:
kinda area of triangle with sides 8, 12 and 4
Yeah, the triangle inequality dictates that the sum of any two sides must be greater than the remaining side.

Wink
Post 31 Aug 2016, 02:57
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
Here is another tricky question:

Consider two circles P & Q.

The radius of P is 1/5 the radius of Q.

P rolls, without slipping, around Q one trip, and returns to its starting position.

How many cycles does P revolve in total?

Wink
Post 31 Aug 2016, 03:05
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 16576
Location: Earth 2.0 beta
YONG wrote:
Here is another tricky question:

Consider two circles P & Q.

The radius of P is 1/5 the radius of Q.

P rolls, without slipping, around Q one trip, and returns to its starting position.

How many cycles does P revolve in total?
This is similar to the Venusian year. Many people get it wrong the first time but once the error is seen they always get it right afterwards.
Post 31 Aug 2016, 04:19
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sleepsleep



Joined: 05 Oct 2006
Posts: 8184
Location: ˛                              ⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣ Posts: 6699
https://www.quantamagazine.org/mathematicians-discover-the-perfect-way-to-multiply-20190411/

Perfect way to multiply, why it works, kinda amazing if we could see through the logic of this perfect way.
Post 12 Apr 2019, 13:48
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