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mehdi jafari

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use modular if...

mehdi jafari 754
February 08 2024
0 0

if you just want to botain Pi/2 whenever x equals to zero, then you can use if conditional :

> 

restart
> 

f:=(x,y)->`if`(x=0,Pi/2,arctan(y/x))

proc (x, y) options operator, arrow; `if`(x = 0, (1/2)*Pi, arctan(y/x)) end proc

 (1)
> 

f(0,1)

(1/2)*Pi

 (2)
> 

f(1,1)

(1/4)*Pi

 (3)
> 

f(1,0)

0

 (4)

NULL

Download arctan.mw

using the formula...

mehdi jafari 754
August 02 2022
1 0

as it is obvious from the diagrams, decreasing h leads to more accurate results. 

 

> 

restart:with(Student[Calculus1]):
> 

ans:=dsolve([diff(y(t),t)=-cos(y(t))^2, y(0)=0]);

y(t) = -arctan(t)

 (1)
> 

f:=-cos(y(t))^2;

-cos(y(t))^2

 (2)
> 

trap:=proc(h,a,b,y0) local i, n,Y,A,p1,p2;
n:=(b-a)/h;
Y[1]:=y0:
for i to n do
Y[i+1]:=fsolve(MM=Y[i]+h/2*(eval(f,y(t)=Y[i])+eval(f,[y(t)=MM])),MM);
od;
A:=[seq](Y[i],i=1..n):
p1:=plot([seq]([i/n,A[i]],i=1..n),axes=boxed,color=blue):
p2:=plot(-arctan(t),t=0..1,color=green):
plots:-display(p1,p2);
end proc;

proc (h, a, b, y0) local i, n, Y, A, p1, p2; n := (b-a)/h; Y[1] := y0; for i to n do Y[i+1] := fsolve(MM = Y[i]+(1/2)*h*(eval(f, y(t) = Y[i])+eval(f, [y(t) = MM])), MM) end do; A := ([seq])(Y[i], i = 1 .. n); p1 := plot(([seq])([i/n, A[i]], i = 1 .. n), axes = boxed, color = blue); p2 := plot(-arctan(t), t = 0 .. 1, color = green); plots:-display(p1, p2) end proc

 (3)
> 

for i in [0.1,0.01,0.001] do
 trap(i,0,1,0)
od;

 

 

 
> 

 

 

 
> 

 


 

Download trap.mw

with a simple proc...

mehdi jafari 754
June 10 2022
4 1


 

> 

restart:
> 

tangent:=proc(f,a) local Df,ans,b1; global L; # function in x #(a) the point in which the tangent line is plotted
Df:=diff(f,x):
b1:=solve((eval(f,x=a)=eval(diff(f,x)*x+b2,x=a)),b2);
L:=eval(Df,x=a)*x+b1:
print('L'=L);
plot([f,L],x=-5+a..a+5,axes=boxed,gridlines):
end proc:
> 

tangent(x^2+3*x+1+20*sin(x)-log(x),2);

L = (13/2+20*cos(2))*x-2-40*cos(2)+20*sin(2)-ln(2)

  

 
> 

 


 

Download tangent.mw

t*sin(t)...

mehdi jafari 754
January 18 2020
4 1

The growth rate in every period is the difference of the blue points in the diagram.
 

> 

restart
> 

p1:=plot(x*sin(x),x=0..50):p2:=plot([seq]([Pi/2+2*i*Pi,Pi/2+2*i*Pi],i=0..7),style=point,color=blue):
> 

plots:-display(p1,p2)

 
> 

plots[animate](plot, [t*sin(t), t = 0 .. x], x = 0 .. 50)

 
> 

NULL


 

Download MaplePrimes_Question.mw

use RealDomain...

mehdi jafari 754
December 06 2019
1 1


 

> 

restart
> 

f:=(3*x^2+2*x-I)*(x+1)

(3*x^2+2*x-I)*(x+1)

 (1)
> 

solve(f)

-1, -1/3-(1/3)*(1+3*I)^(1/2), -1/3+(1/3)*(1+3*I)^(1/2)

 (2)
> 

RealDomain:-simplify(solve(f))

-1

 (3)

``


 

Download solve.mw

some suggestions...

mehdi jafari 754
November 09 2019
1 0

I have some suggestions for you
for the first question, i think there is NOT as far as i know and only help documentation is availabe. and i think it is enough. since after having the initial knowlegde, rest of it is just using ordinary maple in Matlab interface.
for the second question, do NOT use symbolic computations with numeric ones in Matlab unless there is NOT another way. I have expericne on it and it is not a good way to do computations.
Good luck.

Edited.

use equivalent form...

mehdi jafari 754
June 30 2019
0 2

There is an equivalent form of variation of a functional in form of differentiation. 
You can use this form to find variation of a functional with the corresponding boundary conditions.
but in this form, you should use physics package "diff" since you should differentiate with respect to another function.
see https://en.wikipedia.org/wiki/Euler–Lagrange_equation
 

a procedure with random inputs...

mehdi jafari 754
September 01 2018
0 0


 

> 

restart:
> 

P:=proc(f,n,a,b) local i,V;
V:=LinearAlgebra:-RandomVector[row](n,generator=a..b,outputoptions=[datatype=float[8]]):
seq(eval(f,x=(V[i])),i=1..n);
end proc

proc (f, n, a, b) local i, V; V := LinearAlgebra:-RandomVector[row](n, generator = a .. b, outputoptions = [datatype = float[8]]); seq(eval(f, x = V[i]), i = 1 .. n) end proc

 (1)
> 

P(x^2+sin(x),5,1,10)

HFloat(49.65698659871879), HFloat(49.65698659871879), HFloat(9.141120008059866), HFloat(35.72058450180107), HFloat(24.04107572533686)

 (2)
> 

 


 

Download proce.mw

you should have 2 variabels...

mehdi jafari 754
July 31 2018
0 0

for using plot3d you shouldn't assign x1, and it should remain varibale to get a range. when u assing x1, you can just use plot

 

> 

restart
> 

f2:=3.579167+3.537500*x1-2.645833*x2-0.250000*x1*x1-0.012500*x2*x1+0.175000*x2*x2;

3.579167+3.537500*x1-2.645833*x2-.250000*x1^2-0.12500e-1*x2*x1+.175000*x2^2

 (1)
> 

 plot3d (f2,x1=3..7 ,x2=3..7, axes=boxed);

 
> 

x1:=5;
f1:=3.579167+3.537500*x1-2.645833*x2-0.250000*x1*x1-0.012500*x2*x1+0.175000*x2*x2;

5

  

15.016667-2.708333*x2+.175000*x2^2

 (2)
> 

 plot (f1, x2=3..7, labels=[grape, preference], axes=boxed);

 
> 

 


 

Download plot.mw

solve it using some assumptions...

mehdi jafari 754
July 16 2018
0 2

i think the problem is due to maple can not determine the varibales sign in the "abs" function. so use assumptions to solve it :


 

> 

restart
> 

f:=sqrt(x)*ln(y);evalc(Im(f))

x^(1/2)*ln(y)

  

(1/2)*abs(x)^(1/2)*(1-signum(x))*ln(abs(y))+(1/2)*abs(x)^(1/2)*(1+signum(x))*(1/2-(1/2)*signum(y))*Pi

 (1)
> 

solve(evalc(Im(f)), {x,y}) assuming x>0

{x = 0, y = y}, {x = x, 0 <= y}

 (2)
> 

solve(evalc(Im(f)), {x,y}) assuming y>0

{x = x, y = 1}, {y = y, 0 <= x}, {x = 0, 0 < y}

 (3)
> 

g:=x*ln(y);evalc(Im(g))

x*ln(y)

  

x*(1/2-(1/2)*signum(y))*Pi

 (4)
> 

solve(evalc(Im(g)), {x,y})

{x = 0, y = y}, {x = x, 0 <= y}

 (5)
> 

 


 

Download assum.mw

use select remove...

mehdi jafari 754
July 14 2018
1 0


 

> 

restart
> 

A:=[
y[a0]-y[d0],
k[d1]*y[a1]-k[d2]*y[d2],
k[d1]*y[a1]*x[1]-k[d2]*y[d2]*x[2]];

[y[a0]-y[d0], k[d1]*y[a1]-k[d2]*y[d2], k[d1]*x[1]*y[a1]-k[d2]*x[2]*y[d2]]

 (1)
> 

select(has,A,x);

[k[d1]*x[1]*y[a1]-k[d2]*x[2]*y[d2]]

 (2)
> 

remove(has,A,x);

[y[a0]-y[d0], k[d1]*y[a1]-k[d2]*y[d2]]

 (3)
> 

#or
select(has,A,[seq](x[i],i=1..2));

[k[d1]*x[1]*y[a1]-k[d2]*x[2]*y[d2]]

 (4)
> 

 


 

Download select.mw

it was better to upload the worksheet he...

mehdi jafari 754
July 13 2018
0 0

it was better to upload the worksheet here, i think you should change " := " to " = " or bring the right hand side to the left with minus sign and without any " := " . i think it will work.

another way using if condition ( just fo...

mehdi jafari 754
April 28 2018
0 1


 

> 

restart:printlevel:=2:
> 

M:=5:
> 

for i to M do
if i<=M then
N[i,0](u):=1;N[0,i](u):=0;
fi;od;  

1

  

0

  

1

  

0

  

1

  

0

  

1

  

0

  

1

  

0

 (1)
> 

 


 

Download another_way.mw

numerical solution...

mehdi jafari 754
February 07 2018
0 0

here is a numerical solution based on the procedure on this link :
https://www.mapleprimes.com/questions/123970-Finding-All-Roots-For-An-ODE-In-A-Given-Interval
 

> 

restart:
> 

allRoots := proc(f :: procedure, rng :: range, $)
local result, root;
  result := Vector(0, 'datatype = float');
  root := RootFinding:-NextZero(f, lhs(rng), 'maxdistance' = rhs(rng) - lhs(rng));
  while root <> FAIL and root <= rhs(rng) do
    result(numelems(result) + 1) := root;
    root := RootFinding:-NextZero(f, root, 'maxdistance' = rhs(rng) - root);
  end do;
  return result;
end proc:
> 

f:=proc(x) sin(sqrt(x));end proc

proc (x) sin(sqrt(x)) end proc

 (1)
> 

allRoots(f,0..500)

Vector(7, {(1) = 9.86960440100000, (2) = 39.4784176000000, (3) = 88.8264396000000, (4) = 157.913670400000, (5) = 246.740110000000, (6) = 355.305758400000, (7) = 483.610615600000})

 (2)
> 

 

NULL


 

Download allroots.mw

with some changes according to your comm...

mehdi jafari 754
September 08 2017
0 3

@Preben Alsholm 
 

> 

restart:
> 

EQ:=diff(C(t,r),t$2)=(Z(r)^2-1)-(C(t,r)^2)/(4*alpha)*(1+((1+(16*(alpha)^2/C(t,r)^4)*(Z(r)^2-1)^2-(4*alpha*c/3)+(8*alpha*axi/C(t,r)^4)))^(1/2));

diff(diff(C(t, r), t), t) = Z(r)^2-1-(1/4)*C(t, r)^2*(1+(1+16*alpha^2*(Z(r)^2-1)^2/C(t, r)^4-(4/3)*alpha*c+8*alpha*axi/C(t, r)^4)^(1/2))/alpha

 (1)
> 

pdsolve(EQ)

C(t, r) = RootOf(-6*Intat(alpha/(-6*alpha*(-12*Z(r)^2*_g*alpha+3^(1/2)*(Int(((-4*alpha*c*_g^4+48*Z(r)^4*alpha^2+3*_g^4-96*Z(r)^2*alpha^2+48*alpha^2+24*alpha*axi)/_g^4)^(1/2)*_g^2, _g))+12*_g*alpha-6*_F1(r)*alpha+_g^3))^(1/2), _g = _Z)+t+_F2(r))

 (2)
> 

 

> 

 


 

Download pdsolve.mw

dear saba, is this what you want ?

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