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#1376 Re: Help Me ! » Factor Theorem? » 2005-12-30 10:04:26

It has one zero but it isn't rational:

#1381 Re: Help Me ! » Help » 2005-12-30 09:04:44

Number of the form


is nested radical.
Herschfeld (1935) proved that a nested radical of real nonnegative  terms converges iff
is bounded

#1383 Re: Help Me ! » Factor Theorem? » 2005-12-30 08:51:57

Actually the katy's polynomial is unractorizable.

#1384 Re: Help Me ! » Factor Theorem? » 2005-12-30 08:50:55

irspow wrote:

If you divide your equation by x-1 you will get x² - 2x - 15

   This means that (x-1)( x² - 2x - 15) = x³ + 3x² +13x - 15

   Factoring the second part of the product gives (x-5)(x+3)

   So (x-1)(x-5)(x+3) =  x³ + 3x² +13x - 15

   This is completely factored because there are no powers of x above one.

Not.
(x-1)(x^2-2x-15)=x^3-2x^2-15x-x^2+2x+15=x^3-3x^2-13x+15

#1385 Re: Help Me ! » Help » 2005-12-30 08:19:07

Great!
Siva is right!
Well done, Siva!

#1386 Re: Help Me ! » Very interesting problems.. » 2005-12-30 07:49:45

Let look at the neightbours of n in chain with length n amd maxNumber n:
{...,x,n,y,...}
We take that n>x>y. Then n+x and n+y are perfect squares, so:


n+x > n+y =>

Let


Then

so



So when
there does not exist chain with Length n and MaxNumber n.
From the upper equation we get:

=>

=>
for 0 ≤ n < 5.8 there doesn't exist that kind of chain, so

n>5.

#1389 Re: Help Me ! » Very interesting problems.. » 2005-12-30 07:05:34

I got interesting results. But first I have to define a function:
Let


is the smallest perfect square greater than n.

We have that sq↑[x]>x.
Let sq↑[sq↑[sq↑[...sq↑[x]]]]=sq↑n[x]

#1390 Re: Help Me ! » Very interesting problems.. » 2005-12-30 06:39:40

When you have the proof please don't post it immediately. I want to test myself.

#1392 Re: Help Me ! » Very interesting problems.. » 2005-12-30 06:27:22

Let start.
for any square s
s==0,1(mod 4).

So the sum of two conseuense numbers must be ==0,1(mod 4).
We have this cases:
If n == 0 (mod 4) then the neightbours of n must be == 0,1 (mod 4)
If n == 1 (mod 4) then the neightbours of n must be == 3,0 (mod 4)
If n == 2 (mod 4) then the neightbours of n must be == 2,3 (mod 4)
If n == 3 (mod 4) then the neightbours of n must be == 1,2 (mod 4)

That's a begining.

#1393 Re: Help Me ! » Very interesting problems.. » 2005-12-30 06:15:08

So the problem is mathematical, not programmical.

#1394 Re: Help Me ! » Very interesting problems.. » 2005-12-30 05:31:56

Yes, but actually length =0 and length=1 are trivial.

#1395 Re: Help Me ! » monotonic polynomial » 2005-12-30 05:29:59

Hi. I don't know. And don't delete your e-mail and phone number.

#1396 Re: Help Me ! » Very interesting problems.. » 2005-12-30 04:43:57

I thing for computing these we might use come algoritm simular to the labyrinth algoritm.

#1400 Maths Is Fun - Suggestions and Comments » "At least 60 seconds have to pass between posts..." » 2005-12-29 21:52:32

krassi_holmz
Replies: 4

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