Hello
I have a list with a large number of elements and I need to partition it in chunks of a specific size. Here is my attempt
listpart := proc(cond::list,nchunk::nonnegint:=1000)
if numelems(cond) < nchunk then
newcond:=[cond]:
else
ss:=[`$`(1..numelems(cond))]:
sss:=map(`mod`,ss,nchunk):
ind:=[ListTools:-SearchAll(0,sss)]:
ind:=[0,op(ind),ifelse(ss[numelems(ss)]=ind[numelems(ind)],[],ss[numelems(ss)])]:
newcond:=[Threads:-Seq(cond[(ind[j]+1)..ind[j+1]],j=1..(numelems(ind)-1))];
end if:
return(newcond):
end proc:
It does not run as fast as I thought. I am not even sure if Threads:-Seq could be used in this case. Please tell me what I am doing wrong and how I can modify the code to get a faster response.
Many thanks
Ed
Hello all
could anyone tell how to solve following nonliner equations numerically.
f '''' - c1(g'') + R(f ' f '' - f f ''' )=0
g'' + c2(f '' -2g) -c3(f g' - f ' g)=0
f ' (-1)=0, f ' (1)=0, f(-1)=1-A, F(1) =1, g(-1)=0, g(1)=0
c1=3.2, c2=3.3, c3=3.4, R= -10 and A=1.6 are constants.
please help to find solution numerically and how to plot.
Thanks in advence
Hello,
I am trying to calculate an integral involving dirac delta as given below
int(Dirac(sin(x)),x=-3/2*Pi..3/2*Pi)
However, Maple returns the integral unevaluated as
How can I get it evaluated? the result should be 3 here.
Thanks in advance.
I try to make a procedure for this
Was it by a example of Riemann sum that x-values in a array are stored for numerical value, but in this task it seems that the array must be filled with the symbol x1,x2 , etc
Should this be needed ?
blz64.pdf
blz_65.pdf
Download vraag_excset3_task_4.mw
I can't figure out how Maple obtained this solution and looking for some ideas to try.
It is first order non-linear ode in y(x), which is separable.
ode:=diff(y(x),x)=x*ln(y(x));
dsolve([ode,y(1)=1],y(x))
But the general solution is
sol:=dsolve(ode)
Setting up manually an equation using the given condition in order to solve for _C1, produces no solution.
eq:=subs([y(x)=1,x=1],sol);
solve(eq,_C1)
Warning, solutions may have been lost
Also
coulditbe(exp(RootOf(1 + 2*Ei(1, -_Z) + 2*_C1))=1)
FAIL
So how did Maple solve for the constant of integration which results in particular solution y(x)=1 that is supposed to satisfy the condition y(1)=1?
It is clear that y(x)=1 satisfies the ODE itself. But I am asking about how it also satisfies y(1)=1
(odetst says it does satisfy the ODE and condition as well. So Maple must have done something very smart under the cover)
Next I tried
ode:=diff(y(x),x)=x*ln(y(x));
sol:=dsolve(ode,y(x));
sol:=DEtools:-remove_RootOf(sol);
sol:=subs([y(x)=1,x=1],sol)
And now
solve(sol,_C1)
Error, (in Ei) numeric exception: division by zero
Just wondering how did Maple decide that y(x)=1 satisfies y(1)=1? I do not see it.
Using Maple 2020.1. But same result on Maple 2019
Etude d'un cas particulier
a := 5: b := 7:
k := 9:
A := [a, 0]: B := [0, b]: #A et B fixes
P := [t, 0]: Q := [0, k/t]:#P et Q 2 points mobiles
cir := -a*x-b*y+x^2+y^2 = 0:
sol := solve(subs(y = 5, cir), x):
cen := [solve(diff(cir, x)), solve(diff(cir, y))]:
x0 := sol[1]: y0 := 5:
M := [x0, y0]:
R := sqrt(cen[1]^2+cen[2]^2):
beta := arctan(diff(solve(EQ(M, cen), y), x)):
Recherche des valeurs de t pour que les 2 droites soient perpendiculaires
eq := t^2*(y0-b)+t*(a*b-a*y0+b*x0-k)-x0*(a*b-k) = 0;
sol := solve(eq, t);
t := sol[1]; tp := sol[2];
P1 := [t, 0]; Q1 := [0, k/t];
PQ1 := simplify(x*(-a*b+b*t+k)+y*t*(t-a)-t*(-a*b+b*t+k)) = 0:#1ere tangente
PQ2 := simplify(x*(-a*b+b*tp+k)+y*tp*(tp-a)-tp*(-a*b+b*tp+k)) = 0:#2ième tangente
P2 := [tp, 0]; Q2 := [0, k/tp];
CIR := implicitplot(cir, x = -4 .. 8, y = -4 .. 12, color = red);
Fig := proc (alpha) local Dr1, DR1, Dr2, DR2, N, u0, v0, Po, t, tp, sol; global a, b, k, cen, R; u0 := cen[1]+R*cos(alpha); v0 := cen[2]+R*sin(alpha); N := [u0, v0]; sol := solve(t^2*(v0-b)+t*(b*u0-a*v0+a*b-k)-u0*(a*b-k) = 0, t); t := sol[1]; tp := sol[2]; Dr1 := simplify(x*(-a*b+b*t+k)+y*t*(t-a)-t*(-a*b+b*t+k)) = 0; DR1 := implicitplot(Dr1, x = -4 .. 8, y = -4 .. 12, color = brown); Dr2 := simplify(x*(-a*b+b*tp+k)+y*tp*(tp-a)-tp*(-a*b+b*tp+k)) = 0; DR2 := implicitplot(Dr2, x = -4 .. 8, y = -4 .. 12, color = pink); Po := pointplot([N[]], symbol = solidcircle, color = [black], symbolsize = 8); display([Po, DR1, DR2]) end proc;
DrPQ1 := implicitplot(PQ1, x = -4 .. 22, y = -4 .. 12, color = blue);
DrPQ2 := implicitplot(PQ2, x = -4 .. 22, y = -4 .. 12, color = blue);
Points := pointplot([A[], B[], M[], P1[], P2[], Q1[], Q2[], cen[]], symbol = solidcircle, color = [green], symbolsize = 10);
T := plots:-textplot([[A[], "A"], [B[], "B"], [M[], "M"], [P1[], "P1"], [P2[], "P2"], [Q1[], "Q1"], [Q2[], "Q2"], [cen[], "cen"]], font = [times, 10], align = {below, left});
n := 19;
display([seq(Fig(2*i*Pi/n), i = 0 .. n), Fig(beta), CIR, DrPQ1, DrPQ2, Points, T], scaling = constrained, size = [500, 500]);
I would find out the focus of the ellipse. Thank you.
Theoretically, if the multiplication sign is missed Maple needs to give reminders or warnings.But the following is not the case, why？I am surprised its output.
x:=1
x := 1
x(2+1)# Actually, I want to enter x*(2+1)
1
x(sin(y))# Actually, I want to enter x*(sin(y))
1
In preparing to sample problems, I came across this difference in an output depending upon the input type: 2d Input vs. Maple Input. Is there a typo on my part?
Download 2020_evalf_digits.mw
Hi,
How do I solve numerically this set of equations with the following ICs to plot U1(x), phi(x),diff(phi(x),x) versus x:
diff(U1(x),x)=-diff(phi(x),x)/(U1(x)-T/U1(x));
diff(phi(x),x$2)=(1+A1*phi(x)+A2*phi(x)**(3/2)+A3*phi(x)**2)-(M1/U1(x));
where
A1:=(2*k-1)/(2*k-3);
A2:=8*sqrt(2/pi)*(beta-1)*k*Gamma(k)/(3*Gamma(k-0.5)*(2*k-3)**(3/2));
A3:=(4*k**2-1)/(2*(2*k-3)**2);
M1=0.1+sqrt(T+(1/A1));
(Gamma is gamma function)
assume, for example, T=0.1, pi=3.14, beta=0.6, k=3.5
ICs:
U1(x=0)=M1, phi(x=0)=0, diff(phi(x=0),x)=0.001.
Thanks.
From help, it says
coulditbe routine returns true if there is a possible value of x1 that satisfies prop1
my question is, how to find out this condition/possible values that Maple found? This infomration is very useful, but now I do not see how to obtain it. All what coulditbe retuirn is true or false.
Context of why I am asking: Sometimes odetest do not verify its own solutions. And coulditbe can help in finding under what conditions the solution can satisfy the ode. Here is an example
restart;
ode:=diff(y(x),x) = abs(y(x))+1;
solExplicit:=dsolve(ode);
offset := odetest~([solExplicit],ode)
gives
[exp(-x)/_C1 - abs((-exp(-x) + _C1)/_C1) - 1, exp(x)*_C1 - abs(exp(x)*_C1 - 1) - 1]
Both solution fail odetest.
coulditbe~(offset,0)
gives true
So there are assumptions/conditions which makes the solution satisfy the ODE. In this case, by inspection one can see what these conditions are. They are, for one solution:
(-exp(-x) + _C1)/_C1 >0
and for the other, the condition is
exp(x)*_C1 - 1 >0
Under these assumptions, odetest would have given 0 for each odetest.
And it is this information I wanted to obtain automatically from coulditbe.
In Mathematica, Reduce is used for this. Reduce gives conditions under which something is satisfied. For example,
Reduce[ C[1] Exp[x] - Abs[C[1] Exp[x] - 1] - 1 == 0, {x, C[1]}, Reals]
Gives
C[1] >= Exp[-x]
While the above in Maple
coulditbe( C[1]*exp(x)- abs( C[1]*exp(x)-1)-1 = 0)
gives true only, but without the important information, true under what conditions.
Is there a different command in Maple which could give this information?
