Items tagged with piecewise piecewise Tagged Items Feed

Hi guys,

 

I would like to convert a piecewise defined function of two variables into an if structure
Example :

F := (x,y) -> piecewise(x < 0, piecewise(y < 0, 0, 1), piecewise(y < 0, 1, 0));

What I expect is something like (the indentation is just for fun)
if x < 0 then
   if y < 0 then 0 else 1 end if;
else 
   if y < 0 then 1 else 0 end if;
 end if


I tried the following :

convert(F(x,y), `if`)  # but it only converts the "outer x condition" and keeps the "inner y conditions" unchanged

And next this :

map(u -> convert(u, `if`), F(x,y)):   # fails to give m
# and
map(u -> convert(u(x,y), `if`), F(x,y)): 
# ...

without success.

Can anoybody help me please.
I am also interested in an explanation of the reasons why the previous command do not work.

Thanks all

I have 2 questions.

 

1) The help pages discuss the use of functions described by piecewise when using dsolve. The examples are clear enough but nothing is said about using the numeric option and piecewise. I get an error message for a very simple example: y' +R(t) y^2 = 0. Are I correct in assuming that dsllve numeric cannot deal with piecwise continuous functions?

2) I tried to solve a system of 2 second order constant coefficient linear ODEs using dsolve. After a very long wait (over an hour) I gave up. I tried again using "method=laplace" and got an instant answer. I did not bother to mention it here but it occurred again, this time with a fourth order and second order constant coefficient system. Very strange! I know that the laplace transform is a very useful tool; shouldn't Maple also know that?

Howdy,

I am trying to create a mathmatical model that shows a predator-prey relationship along with seasonal variations (hibernation). I made two piecewise functions where one is "on" during the winter and the other is "off" during the winter and the model worked well for one unit of time. 

My problem is that I would like the piecewise functions to apply over an arbitrary amount of time units periods without having to set the range within the function itself for 100 units of time. I was thinking that a for loop would work but i'm not sure how I would impliment that.

Thanks in advance!!

Edit: 

Here's a little bit more info on the problem.

I have three O.D.E's in an array..

ode := {r'(t),h'(t),c'(t)}

which are dependent on a list of parameters..

constants := {a,b,...,s,w,c0,h0,r0};

where s and w are my piecewise functions.

I am able to solve the three odes using dsolve for one period of s and w.

Again, I'm really not sure where to start to make my model periodic.

Hello,

I would like to build a periodic function based on the function x-->x².

For x belonging to [0,2], the function is defined like this :

 

And for the rest of the domain in R, the function should be periodic with a period of 2.

In other words, I would like to define a function which permits me to obtain this plot :

 

Sorry, for my picture which is very ugly, but it should enable to explain what I'm looking for.

1) Is a new piecewise function based on f function is the better solution to create this kind of periodic function ?

2) I didn't find or understand with the help how i can create "infinite" piecewise function. May you help me to define this kind of piecewise function ?

Thanks a lot for your help.

P.S: With the term "nested" in my title I only want to say that I want to apply another piecewise function (for example g) on a already created function (for example f). this term may be not very appropriated.

 

 

Hello,

I would like to build this piecewise function :

 

I try this manner :

f:=x->piecewise(x=>0 and x <= 1, x^2,x > 1 and x<=2, (2-x)^2);

But, it doesn't work. I receive the following error message :

Error, `>` unexpected

Do you have some ideas about my mistake?

Thank you for your help.

 



thanks. I played around, and had problems implementing your ideas for one of the systems I'm interested in.I don't see a difference between this and what you had advised me on, but it gets an error.

any idea why?
or how to fix it?

thing1 := diff(B[1](t), t) = piecewise(t <= 500, 0.3e-2-(63/10000)*B[1](t)-(3/500)*B[2](t), -(3/10000)*B[1](t)):
thing2 := diff(B[1](t), t) = piecewise(t <= 500, 0.1e-1-(1/50)*B[1](t)-(13/625)*B[2](t), -(1/1250)*B[2](t)):
sol := dsolve({thing1, thing2, B[1](0) = 0, B[2](0) = 0}, {B[1](t), B[2](t)}, numeric, output = listprocedure); plots:-odeplot(sol, [B[1](t), B[2](t)], t = 450 .. 550);

Error, (in dsolve/numeric/DAE/explicit) unable to obtain the standard form of the DAE system due to the presence of leading dependent variables/derivatives in the piecewise: piecewise(t <= 500, 1/100-(1/50)*B[1](t)-(13/625)*B[2](t), -(1/1250)*B[2](t))-piecewise(t <= 500, 3/1000-(63/10000)*B[1](t)-(3/500)*B[2](t), -(3/10000)*B[1](t))
Error, (in plots/odeplot) curve is not fully specified in terms of the ODE solution, found additional unknowns {B[1](t), B[2](t)}


I'm trying to plot a piecewise function whose return value has units, but I get an error saying that it found a "name" (i.e., the unit symbol).

Example:

> f := t -> 2*t;

> fp := piecewise(f(tt) <= 4, 1*Unit(W), f(tt) > 4 and f(tt) <= 8, 2*Unit(W), f(tt) > 8, 3*Unit(W));

> plot(fp, tt = 0..10);

 

I've tried using the option useunits, but no success. Any help will be appreciated.

Piecewise function in Maple are continuous function. Now, how I able to write the commands for following discontinuous piecewise function?

 

and when find out f(0) the result is undefined?

In this code, plot() and Norm() fail when used over a piecewise() function:

Error, invalid input: VectorCalculus:-Norm expects its 1st argument, v, to be of type {Matrix, Vector}, but received piecewise(t < 0, Vector[row](2, {(1) = -1, (2) = 0}, attributes = [coords = cartesian]), Vector[row](2, {(1) = 1, (2) = 0}, attributes = [coords = cartesian]))

Note that the same Norm() and piecewise() work fine when used without the plot() function:

Norm(vec(1));
                               1

Code works fine when piecewise() is removed, leaving a plot() / Norm() combination, for example:

So only the simultaneous combination plot() / Norm() / piecewise() fails. 

Finally, the question is:

Is there a way to plot Norm() / piecewise() combinations without workarounds like intermediate PLOT structures?

Thank you

 with small step rather than two points?

I had writen my questions into this code hw2_finished.mw And I think there's maybe something wrong with A=[0..130]

I think you can understand what I want by reading this code but if you are confused please let me know.

This is the code  hw2_final.mw

 

Let me explain it.

I am sure that the mistakes must be about the expresstion of the I1(t) and I2(t). Actually if you delete I1(t) and I2(t) , the whole code works and get the picture at the bottom. 

What I want is to put the expresstion of the I1(t) and I2(t) into 'sol:=dsolve...' and 'plots...' to get the picture of I1(t) and I2(t) with respect to t. Before the t* which subject to Phi(t*)=0 (The blue line in the picture at the bottom is Phi) I want I1(t) and after t* I want I2(t).

I1(t) = (int(sqrt(2*(H(t)+omega*cos(q(t)))), q(t) = q(t)-2*Pi .. q(t), numeric))/Pi.    what I want of this experesstion is to get  'int(sqrt(2*(H(t)+omega*cos(q(t)))' from  'q(t)-2*Pi' to 'q(t)' by numeric method.This q(t) is the solution of the ODE sys.

For example(the number I used is not true,just for example) , at the point t=20, q(t)=30-2*Pi.

so I1(t)= (int(sqrt(2*(H(t)+omega*cos(x))), x = 30-2*Pi .. 30, numeric))/Pi.The I2(t) I want is similar to I1(t).

 

How can I solve it?

Hello,

I use a map() command to get the values of function for each element of vector. Here is an example of a simple task:

restart:

A:=[1,2,8];

f1:=x->2*x;
f1_table_A:=Vector([map(f1,A)]);

I would like to do the same with piecewise function but it doesn't work. Here is an example:

restart:
with(plots):
f:=2*x:
g:=x^2:

h:=x->piecewise( 0<=x and x<= 5, f, 5<x and x <= 10, g ) ;
A:=[1,2,8]:
h_table_A:=map(h,A):

Is it possible to work with piecewise function? Maybe someone has an idea how to do it in different way?

Kind regards,

Iza

Hello,

I'm writing a code and I seem to have an issue when trying to implement a procedure. Here is the code:

 

with(plots):

Z := 75; A := 189; k := 14.6; Rm := 8*R; r0 := 10^(-8)*R; c := 137.036; ms := 105.66/(.51100)

fmtoau := 10^(-15)/(0.529177e-1*10^(-9)):

R := 1.1*fmtoau*A^(1.0/(3.0)):

f := proc (x) options operator, arrow; 1/(1+exp(k*(x-1))) end proc:

n0 := 3*Z*k^2/(4*Pi*(Pi^2+k^2)*R^3):

n := proc (r) options operator, arrow; 4*Pi*n0*f(r/R) end proc:

int(r^2*n(r), r = 0 .. Rm);

73.00000007

(1)

plot(n/n0, 0 .. 2*R);

 

v1 := unapply(int(x^2*f(x), x), x):

Vfermi := proc (r) options operator, arrow; -4*Pi*n0*R^2*(R*(v1(r/R)-v10)/r+v2Rm-v2(r/R)) end proc:

Vuniform := proc (r) options operator, arrow; piecewise(r < R, -Z*(3/2-(1/2)*r^2/R^2)/R, -Z/r) end proc:

plot([Vuniform(r), Vfermi(r), -Z/r], r = r0 .. 2*R, V = 1.2*Vfermi(r0) .. 0, legend = ["uniform charge", "Fermi distribution", "point charge"]);

 

``

plotsol1s := proc (E, K, r0, S, col) local Eqns, ICs, fnl, gnl, r, soln; global ms, c; Eqns := diff(fnl(r), r) = gnl(r)*[E+2*ms*c^2-Vuniform(r)]/c-(K+1)*fnl(r)/r, diff(gnl(r), r) = -fnl(r)*[E-Vuniform(r)]/c-(1-K)*fnl(r)/r; ICs := fnl(r0) = 1, gnl(r0) = 0; soln := dsolve({Eqns, ICs}, numeric); plots:-odeplot(soln, [r, fnl(r)], r0 .. S, color = col) end proc:

plotsol1s(-3*10^5, -1, 10^(-10), Rm, red)

Error, (in f) unable to store '[HFloat(0.0)]' when datatype=float[8]

 

``

``


Any help would be greatly appreciated.

 

Gambia Man

 

Hi, 

 

  I have the following input

 

***

restart;
with( Statistics ):


a:=2;c:=0.3;
g:= exp(-a*x) + c*a*exp(-a*x);
#f := x -> piecewise( x < 0, 0, x>0, g );
 f :=x -> piecewise( x < 0, 0, x>0, exp(-a*x) + c*a*exp(-a*x));

norm_factor:=int( f(x), x=0..infinity );
print(norm_factor);


randomize():
F := Distribution( PDF = 1/norm_factor*f ):
X := RandomVariable( F ):


N := 20;
S := convert( Sample(X,N), list );

print(`cc`,S[1]);

***

 

The code works. However, if I comment out 

 f :=x -> piecewise( x < 0, 0, x>0, exp(-a*x) + c*a*exp(-a*x));

 , then use

f := x -> piecewise( x < 0, 0, x>0, g );

 

i.e.

f := x -> piecewise( x < 0, 0, x>0, g );
 #f :=x -> piecewise( x < 0, 0, x>0, exp(-a*x) + c*a*exp(-a*x));

 

It is said "

Error, (in Statistics:-Sample) unable to construct the envelopes for _R, try to specify the initial range"

 

The norm_factors are actually the same for both inputs. What is the reason for the error message?  Suppose I still want to use something like

f := x -> piecewise( x < 0, 0, x>0, g );

,how to fix the problem?

 

Thank you very much

 

 

  restart: with(plots):
  H := a -> piecewise(a>=0,1):
  f1 := y->(H(y-1*Pi)-H(y-2*Pi))*sin(y)^2:

  g1:=(f1(y)/sqrt(4*Pi*t))*(sin((x-y)^2/4/t+Pi/4)-sin((x+y)^2/4/t+Pi/4)):
  g2:= int(g1, y= 0..100):

  g3:= diff(g2,t):
  g4:= diff(g2,x$2):
  g5:= (g3^2+g4^2)/2:
  E2:= unapply(Int(g5, x= 0..100, epsilon= 1e-4, digits= 7), t):
  CodeTools:-Usage( plot(E2, 0..20, numpoints= 50, labels= [t, E]));

The above mentioned code should give constant figure, but it takes a lot of time and not accurate.

If you can help me to improve these codes, I would be pleased.

 

Thanks!

1 2 3 4 5 6 7 Page 1 of 8