Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

help me! 

 

I have a problem with the system, looking forward to everyone's help!

Hello!

How can I make MAPLE to create the solution of the following system?

https://math.stackexchange.com/questions/301068/how-do-you-find-a-corresponding-recurrence-relation-for-some-random-algorithm/301709

according to this link, how to parse or walk through the algorithm in maple to generate recurrence relation formula?

I have a differential equation involving several functions of the following form:

diff(h,z) = iAf + iBg,

where h, f and g are functions of the Cartesian coordinates x, y and R and the third coordinate corresponds to z = R for some fixed constant value R.  The derivative is then with respect to the coordinate z and A and B are constants, with i the usual imaginary unit.  Is there some way this equation could be solved explicitly with Maple?

Basically it spits out the subset of values for which a division by zero error will occur for the function you specify on  range you specify for each of it's arguments, but I get an ambigous error when ever exponentiation features in the function I specify, which of course dramatically reduces the application of the calculator. Division,addition,substraction and multiplication are currently the only available arithmetic operators availble for the function window that I know the error will not occur.

If some one can help it is much appriciated

 

DIVISION_BY_ZERO_CALCULATOR.mw

hello everyone,
   INGT.mw
 

L__d := 100:

L__b := 200:

L__c := L__d+(1/2)*L__b;

200

(1)

X := .3;

.3

(2)

Error, (in plot) procedure expected, as range contains no plotting variable

 

`ΔE__g` := 1.155*X+.37*X^2;

.3798

(3)

V__0 := .6*`ΔE__g`;

.22788

(4)

m__D := 0.67e-1*m[e];

0.67e-1*m[e]

(5)

m__B := (0.67e-1+0.83e-1*X)*m[e];

0.919e-1*m[e]

(6)

Sol := solve(2*cos(L__d*sqrt(E__x))+(m__D*sqrt((V__0-E__x)/E__x)/m__B-m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))-(m__D*sqrt((V__0-E__x)/E__x)/m__B+m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))*exp(-sqrt(V__0-E__x)*L__b) = 0, E__x);

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(7)

 

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(8)

E__1 := 0.1110897170e-2:

K__1 := sqrt(E__1);

0.3333012406e-1

(9)

K__2 := sqrt(V__0-E__1);

.4762027959

(10)

C := cosh((1/2)*K__2*L__b);

0.2399908351e21

(11)

beta := m__D*K__2/(m__B*K__1);

10.41631973

(12)

B := -beta*sinh((1/2)*K__2*L__b);

-0.2499821271e22

(13)

A := -B*sin(K__1*L__d)+C*cos(K__1*L__d);

-0.7112056933e21

(14)

h := proc (x) options operator, arrow; piecewise(x <= -L__c, A*exp(K__2*(x+L__c)), -L__c < x and x < -(1/2)*L__b, -B*sin(K__1*(x+(1/2)*L__b))+C*cos(K__1*(x+(1/2)*L__b)), abs(x) <= (1/2)*L__b, (1/2)*exp(K__2*x)+(1/2)*exp(-K__2*x), (1/2)*L__b < x and x < L__c, -B*sin(K__1*(x-(1/2)*L__b))+C*cos(K__1*(x-(1/2)*L__b)), L__c <= x, A*exp(K__2*(x-L__c))) end proc:

'h(x)' = h(x);

h(x) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))

(15)

L__y := 200:

L__z := 200:

P := proc (x, y, z) options operator, arrow; h(x)*cos(Pi*y/L__y)*cos(Pi*z/L__z) end proc:

'Psi(x, y, z)' = P(x, y, z);

Psi(x, y, z) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))*cos((1/200)*Pi*y)*cos((1/200)*Pi*z)

(16)

INGT := proc (x__i) `assuming`([evalf(int(int(int(P(x, y, z)^2*exp(-lambda*sqrt((x-x__i)^2+y^2+z^2)), x = -infinity .. infinity), y = -L__y .. L__y), z = -L__z .. L__z))], [0 < lambda]) end proc

evalf(INGT(2))

``

Warning,  computation interrupted

 

``


 

Download INGT.mw
 

L__d := 100:

L__b := 200:

L__c := L__d+(1/2)*L__b;

200

(1)

X := .3;

.3

(2)

Error, (in plot) procedure expected, as range contains no plotting variable

 

`&Delta;E__g` := 1.155*X+.37*X^2;

.3798

(3)

V__0 := .6*`&Delta;E__g`;

.22788

(4)

m__D := 0.67e-1*m[e];

0.67e-1*m[e]

(5)

m__B := (0.67e-1+0.83e-1*X)*m[e];

0.919e-1*m[e]

(6)

Sol := solve(2*cos(L__d*sqrt(E__x))+(m__D*sqrt((V__0-E__x)/E__x)/m__B-m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))-(m__D*sqrt((V__0-E__x)/E__x)/m__B+m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))*exp(-sqrt(V__0-E__x)*L__b) = 0, E__x);

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(7)

 

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(8)

E__1 := 0.1110897170e-2:

K__1 := sqrt(E__1);

0.3333012406e-1

(9)

K__2 := sqrt(V__0-E__1);

.4762027959

(10)

C := cosh((1/2)*K__2*L__b);

0.2399908351e21

(11)

beta := m__D*K__2/(m__B*K__1);

10.41631973

(12)

B := -beta*sinh((1/2)*K__2*L__b);

-0.2499821271e22

(13)

A := -B*sin(K__1*L__d)+C*cos(K__1*L__d);

-0.7112056933e21

(14)

h := proc (x) options operator, arrow; piecewise(x <= -L__c, A*exp(K__2*(x+L__c)), -L__c < x and x < -(1/2)*L__b, -B*sin(K__1*(x+(1/2)*L__b))+C*cos(K__1*(x+(1/2)*L__b)), abs(x) <= (1/2)*L__b, (1/2)*exp(K__2*x)+(1/2)*exp(-K__2*x), (1/2)*L__b < x and x < L__c, -B*sin(K__1*(x-(1/2)*L__b))+C*cos(K__1*(x-(1/2)*L__b)), L__c <= x, A*exp(K__2*(x-L__c))) end proc:

'h(x)' = h(x);

h(x) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))

(15)

L__y := 200:

L__z := 200:

P := proc (x, y, z) options operator, arrow; h(x)*cos(Pi*y/L__y)*cos(Pi*z/L__z) end proc:

'Psi(x, y, z)' = P(x, y, z);

Psi(x, y, z) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))*cos((1/200)*Pi*y)*cos((1/200)*Pi*z)

(16)

INGT := proc (x__i) `assuming`([evalf(int(int(int(P(x, y, z)^2*exp(-lambda*sqrt((x-x__i)^2+y^2+z^2)), x = -infinity .. infinity), y = -L__y .. L__y), z = -L__z .. L__z))], [0 < lambda]) end proc

evalf(INGT(2))

``

Warning,  computation interrupted

 

``


 

Download INGT.mw

 

I'm trying to calculate a triple integral complicated by a procedure that changes each time a variable xi, while the program takes a lot of time and it gives me the message "Warning, computation interrupted". If anyone can help me I will be very happy

Is there something I should be doing whenever I use simplify to avoid things like this, or should I stop using the "is" function all together?

 

interface(showassumed = 0):

 

sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1)

(n-k+1)*binomial(n+1, k)/(k+1) = binomial(n+1, k+1)

(1)

#And we have:
is(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1))

FAIL

(2)

#And since:
is(simplify(convert(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1), 'factorial')))

true

(3)

is(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1)) = is(simplify(convert(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1), 'factorial')))


 

Download main.mw

Hello people :) 

As the captian says, im trying to remove an old task i've made.
But i get this:

Error in Get, invalid object [_XML_reply_data_get("reference" =
"_Maplets_reference_12","parameter" =
"value",_XML_content("Task,UserTasks,Nyops",&Entity "#xc3",&Enity
"#xa6","tning"))]

And i have no idea what it is, but it won't erase my task :'D

Thanks a bunch in advance! 

Have a great weekend you all
Best regards Lucas :)

A few seconds after calling up Help starts zucking araound and the whole computer then freezes. Ctrl-Alt-Delete doesn't work, hard reset required. Very funny. Am I alone?

Hi good afternoon , im looking for maple code of HPM to solve time dependent diffusion reaction , can anybody help me . I didnt get how to solve it by maple . 

Hello

I know that anames( 'user' ) recovers the contents of the workspace but I wonder whether it would be possible to recover only the contents of each m-file read within a worksheet.  

Many thanks

Ed

 

how to convert a function to transition matrix?

how to convert a differential system to transition matrix?

 

 

 

In addition to defining a function in a standard way, a fast and convenient way to turn a formula (an expression) into a function is to use unapply command (check Maple help center for this command). The following items are about defining functions. 

 

 

  1. Define a function midpoint, which returns the average of two arguments given on input. For example, midpoint(2,3) returns 2.5, midpoint(a,b) returns 𝑎+𝑏2. 
  2. b) For two positive natural numbers 𝑎 and 𝑏 we can define the least common multiple lcm(𝑎,𝑏) via their greatest common divisor gcd (𝑎,𝑏), i.e.: lcm(𝑎,𝑏) = 𝑎𝑏gcd (𝑎,𝑏). When 𝑎 and/or 𝑏 are zero, lcm(𝑎; 𝑏)=0. Use the arrow operator and piecewise to define the function my_lcm which returns the least common multiple of two natural numbers. 
  3. c) The quadratic formula allows to write the roots of 𝑝 = 𝑎𝑥2+𝑏𝑥+𝑐 explicitly in terms of 𝑎,𝑏, and 𝑐. 

 

 Define a piecewise function of 𝑥 so that the value of the function is 0, if 𝑥<−(𝑟1+ 𝑟2+𝑟3) or 𝑥>𝑟1+ 𝑟2+𝑟3, and it is equal to the area of confined region if −(𝑟1+ 𝑟2+𝑟3)<𝑥<𝑟1+ 𝑟2+𝑟3. 

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