Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

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

 

 

 

 

 

 

 

 

I've been poking around with convolutions on Maple, and some weird behavior came up---if I let it compute the convolution of a piecewise function, then take the convolution of that, it comes out differently than if I enter a function from scratch as the middle step---file attached (PiecewiseProblem.mw).  I'm not really a Maple pro, so am I'm doing something crazy here?

Thanks!

Dear collegues

I wrote the following code

 


restart:
Digits := 15;
a[k]:=0;
b[k]:=7.47;
a[mu]:=39.11;
b[mu]:=533.9;
mu[bf]:=9.93/10000;
k[bf]:=0.597;
ro[p]:=3880 ;
ro[bf]:= 998.2;
c[p]:= 773;
c[bf]:= 4182;
#mu[bf]:=1;
Gr[phi]:=0; Gr[T]:=0;
#dp:=0.1;
Ree:=1;
Pr:=1;
Nbt:=cc*NBTT+(1-cc^2)*6;

#######################
slip:=0.1;         ####
NBTT:=2;           ####
lambda:=0.1;       ####
phi_avg:=0.02;    ####
#######################


eq1:=diff( (1+a[mu]*phi(eta)+b[mu]*phi(eta)^2)*diff(u(eta),eta),eta)+dp/mu[bf]+Gr[T]*T(eta)-Gr[phi]*phi(eta);
eq2:=diff((1+a[k]*phi(eta)+b[k]*phi(eta)^2)*diff(T(eta),eta),eta)+lambda*T(eta)/k[bf];
eq3:=diff(phi(eta),eta)+1/Nbt*diff(T(eta),eta);
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({subs(dp=pp2,eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=fi0}, numeric,output=listprocedure,continuation=cc);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int(F0(eta),eta=0..1));
INT10:=evalf(Int(F0(eta)*F1(eta),eta=0..1));
a[1]:=evalf(Int(F0(eta),eta=0..1))-Ree*Pr;;
a[2]:=INT10/INT0-phi_avg;
Q1(_passed):=a[1];
Q2(_passed):=a[2];
if type(procname,indexed) then a[op(procname)] else a[1],a[2] end if
end proc;
Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[0.3,0.0007]);




se:=%[2];
res2 := dsolve({subs(dp=se[1],eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=se[2]}, numeric,output=listprocedure,continuation=cc);
G0,G1,G2:=op(subs(res2,[u(eta),phi(eta),T(eta)])):
TTb:=evalf(Int(G0(eta)*G2(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1))/evalf(Int(G0(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1));
with(plots):
odeplot(res2,[[eta,phi(eta)/phi_avg]],0..1);
odeplot(res2,[[eta,T(eta)/TTb]],0..1);
odeplot(res2,[[eta,u(eta)/(Ree*Pr)]],0..1);

res2(1);
Nuu:=(1/TTb);
1/((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2));
(1/TTb)*(((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2)));
>

I want to run the code for the value of NBTT in the range of 0.2 to 10. this code gave the results in the range of 4-10 easily. So, I used the continuation which improve the range of the results between 2-10. However, I coudnt gave the results when 0.2<NBTT<2. Would you please help me in this situation.

Also, It is to be said that the values of phi should be positive. in some ranges, I can see that phi(1) is negative. Can I place a condition in which the values phi restricted to be positive.

Thanks for your attentions in advance

Amir

Dear all;

Than you for help.

how  many steps are required to achieve a error of 1.e-3 in the numerical value of y(1).

Here The 3 -step procedure  Range Kutta Method.

## Exact  solution  

### We will modifty N ( number of steps to get error =10^(-3). )

 

## Procedure Range Kutta

> RK3 := proc (f, a, b, y0, N)

local x, y, n, h, k, vectRK3;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf(b-a)/N;

x[0] := a; y[0] := 1;

for n from 0 to N-1 do

x[n+1] := a+(n+1)*h;

k[1] := f(x[n], y[n]);

k[2] := f(x[n]+(1/2)*h, y[n]+(1/2)*h*k[1]);

k[3] := f(x[n]+h, y[n]+h*(-k[1]+2*k[2]));

y[n+1] := y[n]+(1/6)*h*(k[1]+4*k[2]+k[3])

end do;

[seq([x[n], y[n]], n = 0 .. N)]; y[1];

end proc;

## Now  we compute the error between y(1) and exact  solution for different value of  N

### I have a problem in this part


 errorRk3 := array(1 .. 29);
 for N from  2 to 30 do

errorrRk3[N] := abs(eval(rhs(res), x = 1)-RK3((x,y)->-y,0,1,N));

if errorrRk3[N] =10^{-3} end ;
end  do ;

 

 

write a maple package for quaternion polynomials.it must include the following procedures:

1) for quaternion polynomials f, g:  find degree of f , compute f +g ,f-g, fg.

2)for matrices over quarternion polynomials A,B: compute A+B,A-B, AB.

hint using records to represent quaternions.

Greetings folks,

 

I need to perform Gauss Jordan Elimination on a 7x10 Matrix, returning the row reduced echolon form. The matrix entries itself are sums of several variable products, where the variables itself are sometimes exponential. The link provides an excerpt of said matrix for illustration.

https://www.dropbox.com/s/p9lyg8tmk0hpbbj/polyGLS7rows.png

I tried solving this with MatLab Mupad, but to no avail, at some point the calculation runs out of memory. Simplification of the expressions didn't help either.

Tried the same with Maple. It doesn't run out of memory but looses connection with the kernel at some point.

I'd be glad about some suggestion about how to solve this problem. Basically I want to reproduce the process of the following paper: http://www.inf.ethz.ch/personal/pomarc/pubs/FraundorferECCV10.pdf

The paper itself cites using the Gröbner basis package of Maple for reaching a solution so maybe I am missing something out.

Any help is greatly appreciated

Regards,

JCR

 

Edit: Exponential functions reside in the lower part of the matrix. Path to Maple Worksheet https://www.dropbox.com/s/xb99xlddba57cs7/GLSPoly7rows.mw

Maple crashed while saving and now there is nothing in my file. Lost a bunch of work, any way to fix it?

Thanks
EquationSheet357.mw

(x+y)(x2+y2) = 5500

(x-y)(x2-y2) = 352

hi,i want to take differential with respect to another differential using physics package,but using D instead of diff,could anyone help me do that ? for example :

restart; with(Physics):
A1 := -(1/24)*1*rho*((diff(phi[1](x, t), t))^2)*(h^3)-(1/2)*1*rho*((diff(u[ref](x, t), t))^2)*h-(1/2)*rho*((diff(w(x, t), t))^2)*h+(1/24)*1*1*((diff(phi[1](x, t), x))^2)*(h^3)+(1/2)*1*(1*((diff(u[ref](x, t), x)+(1/2)*(diff(w(x, t), x))^2)^2)+K*1*((diff(w(x, t), x)+phi[1](x, t))^2))*h-1*q*w(x, t):

A2:=-diff(diff(A1,diff(u[ref](x,t),x)),x);

here i want to compute A2 using D command,not diff and i do not want use convert command ! i just need to calculate A2 directly using D command. tnx for your help.

 

how to transform

Matrix([[1,0,0],[1,0,0],[0,0,0]])

to

Matrix([[0,0,0],[0,0,1],[0,0,1]])

(a) Design your own 3-stage explicit Runge-Kutta method with one-step error O(h4).

(b) Test your method by solving y= −y. Confirm that the global error in your numerical solution

is O(h3).

Write a Maple procedure that solves for y(1) in the initial value problem

                     y= f(y),     y(0) = 1,

 

 

using a numerical stencil based on the nth order Taylor series expansion of y. The procedure’s arguments should include an arbitrary function f, an integer n representing the accuracy of the Taylor series expansion, another integer N representing the number of steps between x = 0 and x = 1. Pick a test problem and compare your results with the output of dsolve/numeric.

restart:

Eq1:=S*diff(f(x,t),x,t)+diff(f(x,t),x)^2-f(x,t)*diff(f(x,t),x$2)=diff(f(x,t),x$3);

BCs := {D[1](f)(0,t)=cos(t), f(0,t)=0,D[1](f)(L,t)=0};

ICs := {f(x,0)=0};

S:=10:L:=5:
smod3:= pdsolve(Eq1,ICs union BCs,numeric,range=0..L);

smod3:-plot(t=0,  color=red):

it seems to me that the problem is due to the mixed bcs. Any way around?

Cheers!

1.which rules or theorems can guide to generate relations for words in group theory?

2.Is topological method such as complexes the direction to answer Question 1?

I am running a set of diffrential equation , but i receive below error

"Error, (in LL) cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up"

I can't understand the problem

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