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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. 

22 Jan 2016, 01:26 

tthsqe wrote: MHajduk, 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. Detailed view: 

28 Jan 2016, 22:48 

Referring to the problem itself, I am almost sure that it is absolutely nontrivial 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)? 

28 Jan 2016, 23:44 

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.


29 Jan 2016, 03:17 

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?


29 Jan 2016, 10:45 

Picnic wrote: May i ask what modeling software you are using? 

29 Jan 2016, 12:57 

Yes, it does look somewhat complicated at first glance. I think i have to upgrade my hardware first.


30 Jan 2016, 06:58 

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? 

30 Aug 2016, 04:46 

YONG wrote: I came across this interesting geometry problem today. 6 * (10^2  6^2)^(1/2) / 2 = 24 units square 

30 Aug 2016, 05:46 

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. 

30 Aug 2016, 06:43 

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. 

30 Aug 2016, 06:55 

You are smart!
Correct answer: Such a right triangle does not exist. 

30 Aug 2016, 07:28 

kinda area of triangle with sides 8, 12 and 4


30 Aug 2016, 17:56 

shoorick wrote: kinda area of triangle with sides 8, 12 and 4 

31 Aug 2016, 02:57 

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? 

31 Aug 2016, 03:05 

YONG wrote: Here is another tricky question: 

31 Aug 2016, 04:19 

https://www.quantamagazine.org/mathematiciansdiscovertheperfectwaytomultiply20190411/
Perfect way to multiply, why it works, kinda amazing if we could see through the logic of this perfect way. 

12 Apr 2019, 13:48 

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