MaplePrimes Questions

Hi everybody,

I have differential equation of the form eq := diff(x(t), t) = x(t) + 1  and I want to find the name of the dependent variable (more generally this issue concerns a differential system in N dependent variables).

How should I use “indets” in order to obtain {t, x(t)} ?

(indets(eq, name) returns {t}, while indets(eq, function) returns {t, x(t), d/dt x(t)}.

Thanks in advance

Perhaps I am being stupid, but I have a problem with functional operators, more specifically with looping through a collection of them. Consider the following code (just a study case):

nonIdMaps := []:
for f in [x -> x,x -> 2*x,x -> 3*x] do
   f,f(y);
   if f(y) <> y then nonIdMaps := [nonIdMaps[],f] end if
end do;
nonIdMaps,map(f -> f(y),nonIdMaps);

There are two things I do not understand about this output: 1.) Why the f's rather than the specific maps as given in the list? 2.) Why two times 3*y in the last line, rather than [2*y,3*y]? (this issue being of course quite likely a consequence of the first one). This behaviour can be compared with the following:

nonIdMaps := select(f -> f(y) <> y,[x -> x,x -> 2*x,x -> 3*x]):
nonIdMaps,map(f -> f(y),nonIdMaps);

the output of which I do understand, being as I expected it to be. I guess that there are some scoping rules I do not understand, or what?

how would you interpret the solutions to this:

>int(sin(y/2)^2/(x*(x-y)*y^2), x=omega__ir*t..omega__c*t) assuming t::positive, omega__ir::positive, omega__c::positive, omega__c>omega__ir;
 

which leads to the expression shown in the screen shot. In particular, I'm interested in the condition for the solution to be "undefined"

thanks.

If I make a single plot it will always be located in the centre of the worksheet. Is there an option to place it at the left border?

 

Dear friends,

I have to solve the BVP of the following type:

(x-1)*y''(x)+y'(x)=1+x, y(0)=1, y(1)=1

Expressing the highest derivative (y''(x)), we get a singularity at x=1.

So, as recommended, the method bvp[middefer] (or bvp[midrich]) is used:

dsolve({(x-1)*diff(y(x), x$2)+diff(y(x),x) = 1+x, y(0)=0, y(1)=1}, type = numeric, method = bvp[middefer], y(x), 'output' = Array([seq(k/10, k=0..10)]), 'abserr'=1.0e-3, 'maxmesh'=100)

Maple generates the following error:

Error, (in dsolve/numeric/bvp) unable to achieve requested accuracy of 0.10e-2 with maximum 100 point mesh (was able to get 0.82e3), consider increasing `maxmesh` or using larger `abserr`
But as I increase maxmesh, the achievable accuracy (reported in "(was able to get XXX)") decreases:

for

maxmesh = 200: 0.19e4

maxmesh = 500: 0.54e4

maxmesh = 1000: 0.12e5

maxmesh = 5000: 0.73e5

and so on.....

How to solve this equation with desirable accurancy and usage of Array as output?

i want to solve an ode , but maple return an integral in result, how can i have an answer?


 

restart:

eq:=1/(x*y^(2/3))*8.620689655172415*10^(-16)*(-3.11*10^23*x^2*y^(7/6)-3.92*10^19*y^(25/6)+2.14545039999999*10^29*(0.0108*exp(-45.07/y)+exp(-19.98/y^(1/3)-0.00935317203476387*y^2)))/(x+0.015*y^(1.2));

0.8620689655e-15*(-0.3110000000e24*x^2*y^(7/6)-0.3920000000e20*y^(25/6)+0.2317086432e28*exp(-45.07/y)+0.2145450400e30*exp(-19.98/y^(1/3)-0.935317203476387e-2*y^2))/(x*y^(2/3)*(x+0.15e-1*y^1.2))

(1)

eq:=subs(y=y(t),eq):

 

ans:=dsolve(diff(y(t),t)=eq);

t+Intat((12500/1724137931)*x*_a^(2/3)*(3*_a^(6/5)+200*x)/(49*_a^(25/6)+388750*x^2*_a^(7/6)-2896358040*exp(-(4507/100)/_a)-268181300000*exp(-(1/100000000000000000)*(935317203476387*_a^(7/3)+1998000000000000000)/_a^(1/3))), _a = y(t))+_C1 = 0

(2)

 

 

 


 

Download dsolve.mw

 Dear Maple users,

In Maple 18, I want to  label y axis as $\hat{\sigma}_y$.

labels = [x, sigma[y]] works fine but I have no idea how to put a hat on sigma[y] .

regards,

 

Using Tools -> Check for Updates, I have just tried to update my Maple installation from 2017.0 to 2017.1. The download itself went without a hitch, but then nothing happened. In view of the fact that I clicked a button titled 'Download and Install', I would certainly expect the installation to have started automatically after download. Are anyone else experiencing the same odd behaviour?

PS: Related question of mine is Automatic update of help pages?

Limit of function f(x)=[arcsin(x)]/tan(πx/2).  At point x=1

I can't take the inverse laplace transform of one kind of functions.

Would you like to help me?

Thanks.

hello,

i went to plot a complex numerical solution , i used odeplot but did not work

please help

Thank you very much 


 

restart; with(plots)

NULL

L := 1;

1

(1)

R := 0.57e-1;

0.57e-1

(2)

V := evalf(Pi*R^2*L);

0.1020703453e-1

(3)

pcy := m/V;

97.97164858*m

(4)

Am := 0.6e-1;

0.6e-1

(5)

``

k := evalf(2*Pi/(1.56*T*T));

4.027682890/T^2

(6)

h := 10; -1; zz := h+Am*cos(omega*t)

10+0.6e-1*cos(omega*t)

(7)

g := 9.81;

9.81

(8)

pe := 1025;

1025

(9)

T := 3.57;

3.57

(10)

omega := 2*Pi/T;

1.759995884

(11)

m := (1/2)*pe*evalf(Pi*R^2*L);

5.231105195

(12)

``

``

eq := m*(diff(z(t), `$`(t, 2)))-pe*g*L*R^2-pe*L*R^2*(diff(z(t), `$`(t, 2)))+I*R^3*pe*L*cosh(k*(zz+h))*cos(omega*t)*exp(-I*omega*t)/cosh(k*h)

1.900880195*(diff(diff(z(t), t), t))-32.66950725+(0.1607410765e-1*I)*cosh(6.320462130+0.1896138639e-1*cos(1.759995884*t))*cos(1.759995884*t)*exp(-(1.759995884*I)*t)

(13)

csi := z(0) = 0, (D(z))(0) = 0;

z(0) = 0, (D(z))(0) = 0

(14)

sol := dsolve({csi, eq}, numeric, maxfun = 0):

NULL

sol2 := simplify(evalf(expand(sol))):

fg1 := evalc(Re(sol)):

odeplot(rhs(sol), [t, z(t)], t = 0 .. 3, labels = ["t", "z(t)"], color = blue, thickness = 1, legend = ["z(t) "], axes = boxed)

Error, invalid input: rhs received sol, which is not valid for its 1st argument, expr

 

complexplot(sol, [t, z(t)], t = 0 .. 3, labels = ["t", "z(t)"], color = blue, thickness = 1, legend = ["z(t) "], axes = boxed)

Error, (in plots:-complexplot) invalid input: `plots/complexplot` expects its 2nd argument, r, to be of type {range, name = range}, but received [t, z(t)]

 

NULL

NULL

``


 

Download mapleprime.mwmapleprime.mw

I created a loop to calculate, using random numbers, the probability of a die roll. I got 0, which I originally attributed to the fact that the probability might have just been that low, but in comparing it to a binomial distribution, it shouldn't be 0.

I tried editing the loop to do a coin toss instead, where the probability is 0.5, and that worked fine. So I don't think my loop structure is necessarily wrong. But perhaps I wrongly input getting a 0.167 probability somehow.

Any advice would be appreciated!

dierollfail.mw

https://drive.google.com/file/d/0Bxs_ao6uuBDUNmd5ZVJtX29GT3c/view?usp=sharing
https://drive.google.com/file/d/0Bxs_ao6uuBDUSUdHSTcwSEQtS3M/view?usp=sharing

would like to return K map of P1

Summation expression for logic only consider 1 but how about wildcard x ?

if consider wildcard x as 1 too, then will use below

source = [[0,0,1,0],[0,0,1,1],[0,1,0,1],[0,1,1,0],[0,1,1,1],[1,0,0,0],[1,0,0,1],[1,0,1,0],[1,0,1,1],[1,1,0,0],[1,1,0,1],[1,1,1,0],[1,1,1,1]];
i use Quine Mccluskey algorithm

got result below

wildcard is 5 or x
[[0, 5, 1, 5], [5, 0, 1, 5], [5, 5, 1, 0], [1, 0, 5, 5], [1, 5, 0, 5], [1, 5, 5, 0], [5, 5, 1, 1], [5, 1, 5, 1], [5, 1, 1, 5], [1, 5, 5, 1], [1, 5, 1, 5], [1, 1, 5, 5]]
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB
 
table1 = [[0,0,0,0],
[0,0,0,1],
[0,0,1,0],
[0,0,1,1],
[0,1,0,0],
[0,1,0,1],
[0,1,1,0],
[0,1,1,1],
[1,0,0,0],
[1,0,0,1],
[1,0,1,0],
[1,0,1,1],
[1,1,0,0],
[1,1,0,1],
[1,1,1,0],
[1,1,1,1]];
 
loand(lonot(tt[0]),tt[2])
loand(lonot(tt[1]),tt[2])
loand(lonot(tt[3]),tt[2])
loand(lonot(tt[1]),tt[0])
loand(lonot(tt[2]),tt[0])
loand(lonot(tt[3]),tt[0])
loand(tt[2],tt[3])
loand(tt[1],tt[3])
loand(tt[1],tt[2])
loand(tt[0],tt[3])
loand(tt[0],tt[2])
loand(tt[0],tt[1])
 
loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));
def lonot(z):
    if z == 1:
        return 0
    else:
        return 1
def loand(a, b):
    if a == 1 and b == 1:
        return 1
    else:
        return 0
def loor(a, b):
    if a == 0 and b == 0:
        return 0
    else:
        return 1
#A'C + B'C + AB' + CD + A'BD
for tt in table1:
    print loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));
 
finally i use python to verify
return
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1
 
seems correct if wildcard is 1 too, but
can boolean simplify function simplify this
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB
 
to
 
C + A + B.D   which is
P1 = D + Q0 + Q1.N in png file ?

Hi, friends

I need to calculate the double integral over a non-rectangular domain.

Say, the domain is the triangle (in red)

When I enter

int(int((x^2+y^2)*`if`((y-x-1/2 <= 0) and (y+2*x-2<=0) and (y+x/2-1/2>=0), 1, 0), x=0..1, numeric = true), y=0..1, numeric = true);

or 

int(int((x^2+y^2)*eval(`if`((y-x-1/2 <= 0) and (y+2*x-2<=0) and (y+x/2-1/2>=0), 1, 0)), x=0..1, numeric = true), y=0..1, numeric = true);

an error occur (Error, (in int) cannot determine if this expression is true or false: y-x <= 1/2 and y+2*x <= 2 and 0 <= y+(1/2)*x-1/2)

For me, it is desirable to write boundary conditions in the int operator itself, not as a separate expession.

 

Regarding my recent question http://www.mapleprimes.com/questions/221909-How-To-Extract-Data-From-Implicit-Function I would like to share an interesting observation. Here the code of the program:

restart;
R0 := ln(y)+Re(Psi(1/2+(2*(p^2+(1/2)*sqrt(2*I+4*ksi_fs^2*p^2)*tanh(sqrt(2*I+4*ksi_fs^2*p^2)*x)/(tau+0.5e-2*a)))/y))+gamma+2*ln(2)
tau:= 10.000:ksi_fs:=10:p:=0.037:
R0p:= unapply(R0, [a,x]):
R0f:= proc(a,x)
local r:= fsolve(R0p(a,x), y= 0..1);
   `if`(r::float, r, Float(undefined))
end proc:
M:= Matrix(
   (100,100),
   (i,j)-> R0f(i, 1 + (j-1)*(0.5-0)/(100-1)),
   datatype= float[8]
);

After approximately 2 hours of calculations I get a message window

But I repeat this calculations on another computer with the same Windows 7 64 bit and Maple 17 I don't get such error and I obtain desired data.

So can Maple be sensitive to the hardware? 

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