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A prime producing polynomial.

 

Observations on the trinomial n2 + n + 41.

 

by Matt C. Anderson

 

September 3, 2016

 

The story so far

 

We assume that n is an integer.  We focus our attention on the polynomial n^2 + n + 41.

 

Furthur, we analyze the behavior of the factorization of integers of the form

 

h(n) = n2 + n + 41                                          (expression 1)

 

where n is a non-negative integer.  It was shown by Legendre, in 1798 that if 0 ≤ n < 40 then h(n) is a prime number.

 

Certain patterns become evident when considering points (a,n) where

 

h(n) ≡ 0 mod a.                                             (expression 2)

 

The collection of all such point produces what we are calling a "graph of discrete divisors" due to certain self-similar features.  From experimental data we find that the integer points in this bifurcation graph lie on a collection of parabolic curves indexed by pairs of relatively prime integers.  The expression for the middle parabolas is –

 

p(r,c) = (c*x – r*y)2 – r*(c*x – r*y) – x + 41*r2.           (expression 3)

 

The restrictions are that 0<r<c and gcd(r,c) = 1 and all four of r,c,x, and y are integers.

 

Each such pair (r,c) yields (again determined experimentally and by observation of calculations) an integer polynomial a*z2 + b*z + c, and the quartic h(a*z2 + b*z + c) then factors non-trivially over the integers into two quadratic expressions.  We call this our "parabola conjecture".  Certain symmetries in the bifurcation graph are due to elementary relationships between pairs of co-prime integers.  For instance if m<n are co-prime integers, then there is an observable relationship between the parabola it determines that that formed from (n-m, n).

 

We conjecture that all composite values of h(n) arise by substituting integer values of z into h(a*z2 + b*z + c), where this quartic factors algebraically over Z for a*z2 + b*z + c a quadratic polynomial determined by a pair of relatively prime integers.  We name this our "no stray points conjecture" because all the points in the bifurcation graph appear to lie on a parabola.

 

We further conjecture that the minimum x-values for parabolas corresponding to (r, c) with gcd(r, c) = 1 are equal for fixed n.  Further, these minimum x-values line up at 163*c^2/4 where c = 2, 3, 4, ...  The numerical evidence seems to support this.  This is called our "parabolas line up" conjecture.

 

The notation gcd(r, c) used above is defined here.  The greatest common devisor of two integers is the smallest whole number that divides both of those integers.

 

Theorem 1 - Consider h(n) with n a non negative integer. 

h(n) never has a factor less than 41.

 

We prove Theorem 1 with a modular construction.  We make a residue table with all the prime factors less than 41.  The fundamental theorem of arithmetic states that any integer greater than one is either a prime number, or can be written as a unique product of prime numbers (ignoring the order).  So if h(n) never has a prime factor less than 41, then by extension it never has an integer factor less than 41.

 

For example, to determine that h(n) is never divisible by 2, note the first column of the residue table.  If n is even, then h(n) is odd.  Similarly, if n is odd then h(n) is also odd.  In either case, h(n) does not have factorization by 2.

 

Also, for divisibility by 3, there are 3 cases to check.  They are n = 0, 1, and 2 mod 3. h(0) mod 3 is 2.  h(1) mod 3 is 1. and h(2) mod 3 is 2.  Due to these three cases, h(n) is never divisible by 3.  This is the second column of the residue table.

 

The number 0 is first found in the residue table for the cases h(0) mod 41 and h(40) mod 41.  This means that if n is congruent to 0 mod 41 then h(n) will be divisible by 41.  Similarly, if n is congruent to 40 mod 41 then h(n) is also divisible by 41.

After the residue table, we observe a bifurcation graph which has points when h(y) mod x is divisible by x.  The points (x,y) can be seen on the bifurcation graph.

 

< insert residue table here >

 

Thus we have shown that h(n) never has a factor less than 41.

 

Theorem 2

 

Since h(a) = a^2 + a + 41, we want to show that h(a) = h( -a -1).

 

Proof of Theorem 2

Because h(a) = a*(a+1) + 41,

Now h(-a -1) = (-a -1)*(-a -1 +1) + 41.

So h(-a -1) = (-a -1)*(-a) +41,

And h(-a -1) = h(a).

Which was what we wanted.

End of proof of theorem 2.

 

Corrolary 1

Further, if h(b) mod c ≡ = then h(c –b -1) mod c ≡ 0.

 

We can observe interesting patterns in the “graph of discrete divisors” on a following page.

 

Hello

I have experienced that maple does not save all of the varibles. But some it does.

I calculate with units, could that be the reason?

I have allso been thinking that it has something to do with saving the document online in onenote. But that works like the file is saved on the Pc's harddrive.

Are there anybody else that has experienced this?

I calculate with units, but as the varible does not appear in the calculation with units, I make one varible with the same result, to get the next calculation to work.   

Regards

Heide

 

restart

int(sin(x), x)

-cos(x)

(1)

int(x^2, x)

x^2*_X

(2)

``

 

Download 12.mw

Dears, When I run calculation in Maple I found an error in matrices. See the file

 

 

 

 

 

could you help me about maple
i try to calculating using chevypade rational approximating and the answer for cos(x) xe is(-.221091073962959*T(1, x-1)+.7710737338*T(0, x-1)-0.4212446689e-1*T(2, x-1))/(0.836360586596837e-1*T(1, x-1)+T(0, x-1)+0.3360079945e-1*T(2, x-1)) i can not to convert to rational form as x^^n .maple is not very friendship
Thanks

Hi, I am completely new to Maple, and I need to use it to optimize my equations in order to make my PLC codes more compressed. I am calculating forward kinematics with the Denavit-Hartenberg method and as such I get long expressions. After a lot of google'ing and frustration, I thought I'd ask here in the hope that one of you might be able to assist me.

I have the following equations;

X := L10*cos(q5) - L16*(sin(q10)*(sin(q5)*sin(q8) - cos(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) - cos(q10)*(sin(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + cos(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6)))) - d2*(cos(q10)*(sin(q5)*sin(q8) - cos(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + sin(q10)*(sin(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + cos(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6)))) + L15*(sin(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + cos(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6))) - L11*cos(q5)*sin(q6) + d1*cos(q5)*cos(q6) - L13*sin(q5)*sin(q8) + L14*cos(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + L13*cos(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7)) - L14*sin(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6)) + L12*cos(q5)*cos(q6)*cos(q7) - L12*cos(q5)*sin(q6)*sin(q7);

Y := L10*sin(q5) - L9 + L16*(sin(q10)*(cos(q5)*sin(q8) - cos(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q10)*(sin(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6)))) + d2*(cos(q10)*(cos(q5)*sin(q8) - cos(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) + sin(q10)*(sin(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6)))) - L15*(sin(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6))) + L13*cos(q5)*sin(q8) - L11*sin(q5)*sin(q6) + d1*cos(q6)*sin(q5) - L14*cos(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - L13*cos(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5)) - L14*sin(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6)) + L12*cos(q6)*cos(q7)*sin(q5) - L12*sin(q5)*sin(q6)*sin(q7);

Z := L15*(cos(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - sin(q8)*sin(q9)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - L11*cos(q6) - L8 - d1*sin(q6) + L16*(cos(q10)*(cos(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - sin(q8)*sin(q9)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - cos(q8)*sin(q10)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - d2*(sin(q10)*(cos(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - sin(q8)*sin(q9)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) + cos(q8)*cos(q10)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - L13*cos(q8)*(cos(q6)*sin(q7) + cos(q7)*sin(q6)) - L14*sin(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - L12*cos(q6)*sin(q7) - L12*cos(q7)*sin(q6) - L14*cos(q9)*sin(q8)*(cos(q6)*sin(q7) + cos(q7)*sin(q6));

 

I need to optimize these equations, but still keep them separate. I would like to use mutual expressions for the calculations within, but still as I said keep the outputs of X, Y and Z separate.

This is MATLAB code.

 

Thanks in advance for any help.

https://social.msdn.microsoft.com/Forums/vstudio/en-US/cc2a85ad-30ec-44ed-8c75-636ff71eade2/how-to-convert-integer-or-decimal-number-into-any-base-number?forum=csharpgeneral

1. for example how to convert decimal or integer number into base 3 number, base 5 number etc.

2.how to do logical operation with custom logic table for example,

 

120 special operator 235 

01111000

11101011

 

special operator according to logical table is

1st op 2nd op output
0 0 1
0 1 0
1 0 1
1 1 0

 

  01111000

  11101011

=00010100 = 20

Hello everybody

I am working on the bending of FGM plate, and I just used maple for the calculation functions of my problem of the bending of FGM plates.

Can any from the members of mapleprimes one help me to write Navier's equation include below for the bending of the plate because I am new in using Maple 18

alfa=m*pi/a, beta=n*pi/b with m=1,3,5..100 and n=1,3,5...100

Thank's for all 

Bending

I set a physical model for my reseach,

equ1 := x^4-5*x

equ2 := 1/x+x^2+3

x is a function of time t and it meets diff(x(t), t) = equi-equ2.

I want to plot the curve of x(t) varying with time

When I use the following command

DEplot({x(0) = 0, diff(x(t), t) = equi-equ2},x(t), t = 0 .. 20)   it shows: Error, (in DEtools/DEplot) called with too few arguments

Who can tell me what is wrong with my calculation? Thanks

Hi there,

 

I'd like to solve 7th order implicit simultaneous equation such as below, so I tried to do it by solve command.

However the calculation wasn't over although three hours passed.

 

eq1 := f1(a,b,c,d,e,f,g) = 0;

eq2 := f2(a,b,c,d,e,f,g) = 0;

.

.

.

eq7 := f7(a,b,c,d,e,f,g) = 0;

 

Just for your information, the eq1 and eq6 are written as follows specifically.

eq1 := -a-b-c-d-e-f-g+0.501857 = 0

eq6 := a*b*c*d*e*f+a*b*c*d*e*g+a*b*c*d*f*g+a*b*c*e*f*g+a*b*d*e*f*g+a*c*d*e*f*g+b*c*d*e*f*g+a*b*c*d*e+a*b*c*d*f+a*b*c*d*g+a*b*c*e*f+a*b*c*e*g+a*b*c*f*g+a*b*d*e*f+a*b*d*e*g+a*b*d*f*g+a*b*e*f*g+a*c*d*e*f+a*c*d*e*g+a*c*d*f*g+a*c*e*f*g+a*d*e*f*g+b*c*d*e*f+b*c*d*e*g+b*c*d*f*g+b*c*e*f*g+b*d*e*f*g+c*d*e*f*g-0.5281141885e-3+1.01894577*10^(-12)*I = 0

 

And the program code I used is:

solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7],[a,b,c,d,e,f,g]);

 

Here is the specification of my computer.

OS: Windows 7 Enterprise 64bit

CPU: Intel Core i7-3520M 2.90 GHz

Memory: 4.00 GB

 

How can I handle this problem? Is the specification not enough to solve the equation? Do I need to leave my computer more and more time?

Any help would be appriciated.

Hello everybody,

I'm trying to solve for a challenging problem : a moving inclined plane with a block

I want to solve for the acceleration components for the block and the plane and the normal force acting on the block.

Let O=(0,0) be an external origin.

Let h be the upper left height of the inclined plane.

Let x1 be the x-position of the center of gravity of the inclined plane.

Let x2 be the x-postion of the center of gravity of the block.

Let y be the y-position of the center of gravity of the block.

Let m1 be the mass of the plane. Let m2 be the mass of the block.

Let  mu[1] be the coeffiction of kinetic friction between the bottom of the inclined plane and the level surface.

Let mu[2]  be the coeffiction of kinetic friction between the block and the upper surface of the inclined plane.

Let theta be the angle of the plane with the horizontal.

Let Fp a force applied to the inclined plane.

 

With those defined variables, I make two separable free body diagrams for the block and for the inclined plane, indicating all of the external forces acting on each. It then comes those two vectorial equations :

Block : m2a2=Wweight of block+Fplan acting on block+Ffriction from plan to block+Nnormal from plan to block

Plane : m1a1=Wweight of plane+Fpushing force+Fblock acting on plane+Ffriction from level to plan+Nnormal from level to plane+Ffriction from block to plane+Nnormal from block to plane

I am quite not sure whether I should include the Ffriction from block to plane and the Nnormal from block to plane into the plane's acceleration calculation. Am I right ?

I notice that from the geometry of the figure, I can write down the relation : tan(theta)=(h-y)/(x2-x1)

This implies the relation : -a2y=tan(theta)(a2x-a1x) (equation 1)

Writing down the equations for the x- and y- components of the accelerations of the block and of the plane , this yields :

( equation 2) : m2a2x=m1 sqrt(a1x2+a1y2) cos(theta) -    mu[2]  N1 sin(theta) +Ncos(theta

 

(equation 3) : m2a2y= m2g+m1 sqrt(a1x2+a1y2) sin(theta) +  mu[2]  N1 cos(theta) +Nsin(theta

(equation 4) : m1a1x=Fp - m2 sqrt(a2x2+a2y2)  sin(theta) +  mu[2]  N1 cos(theta) - Ncos(theta

(equation 5) : m1a1y=-m1g  - m2 sqrt(a2x2+a2y2)  cos(theta) - mu[1] N1 + N1 -  mu[2]  N1 sin(theta) -  mu[2]  N1 cos(theta)

Since N1=m1g,  equation 5 becomes : m1a1y= - m2 sqrt(a2x2+a2y2)  cos(theta)   -  mu[2]  N1 sin(theta) -  mu[2]  N1 cos(theta)

I am confused at this stage because a1y=0, that is to say, the plane remains at the ground level surface.

Where am I wrong ? Does this comes from my previous question ?

 

I want to solve this problem with Maple and plot the solutions. Thank you for any answer !

 

Hi Everybody.

 

Doing some calculation in quantum mechanics, I stuble on that integral:

I see immediately that the answer is 1/2.  But Maple 18 cannot give an answer other than a limit that he cannot evaluate.  I even try assumption that p and hbar are realcons.  I get infinity.

Any idea?

Thank you in advance for your help.

 

--------------------------------------
Mario Lemelin
Maple 18 Ubuntu 13.10 - 64 bits
Maple 18 Win 7 - 64 bits messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987

how maple calculate exp(x) with e.g. 100000 decimal numbers

a divsion of the series x^k/k! with e.g. 1/25000!/25001 lasts longer than the exp(1.xx) calculation

 

is there a faster way to calculate exp(x) than with the x^k/k! series

 

thanks

 

 

 

 

 

 

 

 

Hi everybody,

When doing calculations, I often run in the following problem.  I have the final solution wich I simplify symbolically so many terms are cancelling.  But I get this:

sqrt(2)*sqrt(g)/sqrt(b)

While I would like to regroup all the terms into the square root.  But look, even in this forum, Maple get sqrt(2) automatically out of the square root.

sqrt(2*g/b)

I know that it is the simple form.  But in some instances, I need the square root to stay together so I can show a property.  But is there a way to be able, sometimes, to tell Maple to leave all the terms under the square root?

Thank you in advance for your help.

 

--------------------------------------
Mario Lemelin
Maple 17.01 Ubuntu 13.10 - 64 bits
Maple 17 Win 7 - 64 bits messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987
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