This is an animation of the spread of the COVID-19 over the U.S. in the first 150 days. It was created in Maple 2020, making extensive use of DataFrames.

https://www.youtube.com/watch?v=XHXeJKTeoRw

The animation of 150 Day history includes COVID-19 data published by the NY Times and geographic data assembled from other sources. Each cylinder represents a county or in two special cases New York City and Kansas City. The cross-sectional area of each cylinder is the area in square miles of the corresponding county. The height of each cylinder is on a logarithmic scale (in particular it is 100*log base 2 of the case count for the county. The argument of the logarithm function is the number of cases per county divided by the are in square miles-so an areal density. Using a logarithmic scale facilitates showing super high density areas (e.g., NYC) along with lower density areas. The heights are scaled by a prefactor of 100 for visualization.

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.

Hello each and everyone,

I have a problem with updating the Physics Package. (At the moment I run version 709)

But when I try to update to version 710 I have the following error (see the screenshot)

I don't know what's happening.

Kind regards to all.

Jean-Michel

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?