MaplePrimes Questions

How I can do ?

Thank you.

 

Substitution of . 5,6,7) into Eqs. 1–(4), gives the new equation as functions of the generalized coordinates,
u_m,n(t);  v_m,n ( t), and w_m,n ( t). These expressions are then inserted in the Lagrange equations (see Eq. 8)) a set of N second-order coupled ordinary differential equations with both quadratic   and cubic nonlinearities.

In Eq (8) q are generalized coordinate such as uvw  and q = {`u__m,n`(t), `v__m,n`(t), `w__m,n`(t)}^T.

\where the elements of the vector,q_i are the time-dependent generalized coordinates.

L_Maple
 

U = (1/2)*(int(int(int(E*(`∂`(u(x, y, t))/`∂`(x)+(1/2)*(`∂`(w(x, y, t))/`∂`(x))^2+`∂`(w(x, y, t))/`∂`(x)*(`∂`(w__0(x, y, t))/`∂`(x))-z*(diff(w(x, y, t), x, x))+v(x, y, t)*(`∂`(v(x, y, t))/`∂`(y)+(1/2)*(`∂`(w(x, y, t))/`∂`(y))^2+`∂`(w(x, y, t))/`∂`(y)*(`∂`(w__0(x, y, t))/`∂`(y))-z*(diff(w(x, y, t), y, y))))*(`∂`(u(x, y, t))/`∂`(x)+(1/2)*(`∂`(w(x, y, t))/`∂`(x))^2+`∂`(w(x, y, t))/`∂`(x)*(`∂`(w__0(x, y, t))/`∂`(x))-z*(diff(w(x, y, t), x, x)))/(-nu^2+1)+E*(`∂`(nu(x, y, t))/`∂`(y)+(1/2)*(`∂`(w(x, y, t))/`∂`(y))^2+`∂`(w(x, y, t))/`∂`(y)*(`∂`(w__0(x, y, t))/`∂`(y))-z*(diff(w(x, y, t), y, y))+v(x, y, t)*(`∂`(u(x, y, t))/`∂`(x)+(1/2)*(`∂`(w(x, y, t))/`∂`(x))^2+`∂`(w(x, y, t))/`∂`(x)*(`∂`(w__0(x, y, t))/`∂`(x))-z*(diff(w(x, y, t), x, x))))*(`∂`(v(x, y, t))/`∂`(y)+(1/2)*(`∂`(w(x, y, t))/`∂`(y))^2+`∂`(w(x, y, t))/`∂`(y)*(`∂`(w__0(x, y, t))/`∂`(y))-z*(diff(w(x, y, t), y, y)))/(-nu^2+1)+E*(`∂`(u(x, y, t))/`∂`(y)+`∂`(v(x, y, t))/`∂`(x)+`∂`(w(x, y, t))/`∂`(x)*(`∂`(w(x, y, t))/`∂`(y))+`∂`(w__0(x, y, t))*`∂`(w(x, y, t))/(`∂`(x)*`∂`(y))+`∂`(w__0(x, y, t))*`∂`(w(x, y, t))/(`∂`(x)*`∂`(y))-2*z*(diff(w(x, y, t), x, y)))^2/(2*(1+nu))+E*l^2*(diff(w(x, y, t), x, y))^2/(1+nu)+E*l^2*(diff(w(x, y, t), x, y))^2/(1+nu)+E*l^2*(diff(w(x, y, t), y, y)-(diff(w(x, y, t), x, x)))^2/(2*(1+nu))+E*l^2*(diff(v(x, y, t), y, y)-(diff(u(x, y, t), x, x)))^2/(8*(1+nu))+E*l^2*(diff(v(x, y, t), x, y)-(diff(u(x, y, t), y, y)))^2/(8*(1+nu)), z = -(1/2)*h .. (1/2)*h), y = 0 .. b), x = 0 .. a))

U = (1/2)*(int(int((1/12)*(-E*(-v(x, y, t)*(diff(diff(w(x, y, t), y), y))-(diff(diff(w(x, y, t), x), x)))*(diff(diff(w(x, y, t), x), x))/(-nu^2+1)-E*(-v(x, y, t)*(diff(diff(w(x, y, t), x), x))-(diff(diff(w(x, y, t), y), y)))*(diff(diff(w(x, y, t), y), y))/(-nu^2+1)+4*E*(diff(diff(w(x, y, t), x), y))^2/(2+2*nu))*h^3+E*(`∂`(u(x, y, t))/`∂`(x)+(1/2)*`∂`(w(x, y, t))^2/`∂`(x)^2+`∂`(w(x, y, t))*`∂`(w__0(x, y, t))/`∂`(x)^2+v(x, y, t)*(`∂`(v(x, y, t))/`∂`(y)+(1/2)*`∂`(w(x, y, t))^2/`∂`(y)^2+`∂`(w(x, y, t))*`∂`(w__0(x, y, t))/`∂`(y)^2))*(`∂`(u(x, y, t))/`∂`(x)+(1/2)*`∂`(w(x, y, t))^2/`∂`(x)^2+`∂`(w(x, y, t))*`∂`(w__0(x, y, t))/`∂`(x)^2)*h/(-nu^2+1)+E*(`∂`(nu(x, y, t))/`∂`(y)+(1/2)*`∂`(w(x, y, t))^2/`∂`(y)^2+`∂`(w(x, y, t))*`∂`(w__0(x, y, t))/`∂`(y)^2+v(x, y, t)*(`∂`(u(x, y, t))/`∂`(x)+(1/2)*`∂`(w(x, y, t))^2/`∂`(x)^2+`∂`(w(x, y, t))*`∂`(w__0(x, y, t))/`∂`(x)^2))*(`∂`(v(x, y, t))/`∂`(y)+(1/2)*`∂`(w(x, y, t))^2/`∂`(y)^2+`∂`(w(x, y, t))*`∂`(w__0(x, y, t))/`∂`(y)^2)*h/(-nu^2+1)+E*(`∂`(u(x, y, t))/`∂`(y)+`∂`(v(x, y, t))/`∂`(x)+`∂`(w(x, y, t))^2/(`∂`(x)*`∂`(y))+2*`∂`(w__0(x, y, t))*`∂`(w(x, y, t))/(`∂`(x)*`∂`(y)))^2*h/(2+2*nu)+2*E*l^2*(diff(diff(w(x, y, t), x), y))^2*h/(1+nu)+E*l^2*(diff(diff(w(x, y, t), y), y)-(diff(diff(w(x, y, t), x), x)))^2*h/(2+2*nu)+E*l^2*(diff(diff(v(x, y, t), y), y)-(diff(diff(u(x, y, t), x), x)))^2*h/(8+8*nu)+E*l^2*(diff(diff(v(x, y, t), x), y)-(diff(diff(u(x, y, t), y), y)))^2*h/(8+8*nu), y = 0 .. b), x = 0 .. a))

(1)

T = rho*h*(int(int((`∂`(u(x, y, t))/`∂`(t))^2+(`∂`(v(x, y, t))/`∂`(t))^2+(`∂`(w(x, y, t))/`∂`(t))^2, y = 0 .. b), x = 0 .. a))

T = rho*h*(int(int(`∂`(u(x, y, t))^2/`∂`(t)^2+`∂`(v(x, y, t))^2/`∂`(t)^2+`∂`(w(x, y, t))^2/`∂`(t)^2, y = 0 .. b), x = 0 .. a))

(2)

F = (1/2)*c*(int(int((`∂`(u(x, y, t))/`∂`(t))^2+(`∂`(v(x, y, t))/`∂`(t))^2+(`∂`(w(x, y, t))/`∂`(t))^2, y = 0 .. b), x = 0 .. a))

F = (1/2)*c*(int(int(`∂`(u(x, y, t))^2/`∂`(t)^2+`∂`(v(x, y, t))^2/`∂`(t)^2+`∂`(w(x, y, t))^2/`∂`(t)^2, y = 0 .. b), x = 0 .. a))

(3)

W = int(int(w(x, y, t)*f__1(x, y, t)*cos(omega*t), y = 0 .. b), x = 0 .. a)

W = int(int(w(x, y, z)*f__1(x, y, z)*cos(omega*t), y = 0 .. b), x = 0 .. a)

(4)

u(x, y, t) = sum(sum(`u__m,n`(t)*sin(m*Pi*x/a)*sin(n*Pi*y/b), n = 1 .. N), m = 1 .. M)

u(x, y, t) = -(1/4)*(cos(Pi*y*N/b)*cos(Pi*x/a)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)-cos(Pi*y*N/b)*cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y/b)+cos(Pi*x/a)*sin(Pi*y*N/b)*cos(Pi*y/b)*sin((M+1)*Pi*x/a)-cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y*N/b)*cos(Pi*y/b)-cos(Pi*y*N/b)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)-cos(Pi*x/a)*sin(Pi*y*N/b)*sin((M+1)*Pi*x/a)-cos(Pi*x/a)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)+cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y*N/b)+sin(Pi*x/a)*sin(Pi*y/b)*cos((M+1)*Pi*x/a)-sin(Pi*y*N/b)*cos(Pi*y/b)*sin((M+1)*Pi*x/a)+sin(Pi*y*N/b)*sin((M+1)*Pi*x/a)+sin(Pi*y/b)*sin((M+1)*Pi*x/a))*`u__m,n`(t)/((cos(Pi*x/a)-1)*(cos(Pi*y/b)-1))+(1/4)*(-cos(Pi*y*N/b)*sin(Pi*y/b)*sin(Pi*x/a)-sin(Pi*y*N/b)*cos(Pi*y/b)*sin(Pi*x/a)+sin(Pi*y*N/b)*sin(Pi*x/a)+sin(Pi*y/b)*sin(Pi*x/a))*`u__m,n`(t)/((cos(Pi*x/a)-1)*(cos(Pi*y/b)-1))

(5)

v(x, y, t) = sum(sum(`v__m,n`(t)*sin(m*Pi*x/a)*sin(n*Pi*y/b), n = 1 .. N), m = 1 .. M)

v(x, y, t) = -(1/4)*(cos(Pi*y*N/b)*cos(Pi*x/a)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)-cos(Pi*y*N/b)*cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y/b)+cos(Pi*x/a)*sin(Pi*y*N/b)*cos(Pi*y/b)*sin((M+1)*Pi*x/a)-cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y*N/b)*cos(Pi*y/b)-cos(Pi*y*N/b)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)-cos(Pi*x/a)*sin(Pi*y*N/b)*sin((M+1)*Pi*x/a)-cos(Pi*x/a)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)+cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y*N/b)+sin(Pi*x/a)*sin(Pi*y/b)*cos((M+1)*Pi*x/a)-sin(Pi*y*N/b)*cos(Pi*y/b)*sin((M+1)*Pi*x/a)+sin(Pi*y*N/b)*sin((M+1)*Pi*x/a)+sin(Pi*y/b)*sin((M+1)*Pi*x/a))*`v__m,n`(t)/((cos(Pi*x/a)-1)*(cos(Pi*y/b)-1))+(1/4)*(-cos(Pi*y*N/b)*sin(Pi*y/b)*sin(Pi*x/a)-sin(Pi*y*N/b)*cos(Pi*y/b)*sin(Pi*x/a)+sin(Pi*y*N/b)*sin(Pi*x/a)+sin(Pi*y/b)*sin(Pi*x/a))*`v__m,n`(t)/((cos(Pi*x/a)-1)*(cos(Pi*y/b)-1))

(6)

w(x, y, t) = sum(sum(`w__m,n`(t)*sin(m*Pi*x/a)*sin(n*Pi*y/b), n = 1 .. N), m = 1 .. M)

w(x, y, t) = -(1/4)*(cos(Pi*y*N/b)*cos(Pi*x/a)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)-cos(Pi*y*N/b)*cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y/b)+cos(Pi*x/a)*sin(Pi*y*N/b)*cos(Pi*y/b)*sin((M+1)*Pi*x/a)-cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y*N/b)*cos(Pi*y/b)-cos(Pi*y*N/b)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)-cos(Pi*x/a)*sin(Pi*y*N/b)*sin((M+1)*Pi*x/a)-cos(Pi*x/a)*sin(Pi*y/b)*sin((M+1)*Pi*x/a)+cos((M+1)*Pi*x/a)*sin(Pi*x/a)*sin(Pi*y*N/b)+sin(Pi*x/a)*sin(Pi*y/b)*cos((M+1)*Pi*x/a)-sin(Pi*y*N/b)*cos(Pi*y/b)*sin((M+1)*Pi*x/a)+sin(Pi*y*N/b)*sin((M+1)*Pi*x/a)+sin(Pi*y/b)*sin((M+1)*Pi*x/a))*`w__m,n`(t)/((cos(Pi*x/a)-1)*(cos(Pi*y/b)-1))+(1/4)*(-cos(Pi*y*N/b)*sin(Pi*y/b)*sin(Pi*x/a)-sin(Pi*y*N/b)*cos(Pi*y/b)*sin(Pi*x/a)+sin(Pi*y*N/b)*sin(Pi*x/a)+sin(Pi*y/b)*sin(Pi*x/a))*`w__m,n`(t)/((cos(Pi*x/a)-1)*(cos(Pi*y/b)-1))

(7)

diff(`∂`(T(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`), t)-`∂`(T(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`)+`∂`(U(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`)+`∂`(U(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`)+`∂`(F(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`) = `∂`(W(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`), j = 1, () .. (), N

(D(`∂`))(T(x, y, t))*(diff(T(x, y, t), t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`)-`∂`(T(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`)+2*`∂`(U(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`)+`∂`(F(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`) = `∂`(W(x, y, t))/`∂`(`#mscripts(mi("q"),mi("j"),none(),none(),mo("."),none(),none())`), j = 1, () .. (), N

(8)

NULL


 

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How come maple doesn't support symbolic matrix calculus?

For example, I'd like to differentiate or simplify expressions with relation to general vectors and matrices. Besides simpifying the expressions, in runtime (e.g. optimizing a function in matlab) my vector and matrix size change according to the data, so deriving specific expressions element-wise would only be valid for a specific case.

A simple example (A is a matrix, x is a vector):

f(x) = ||Ax||^2

df/dx = 2A'Ax

Recently, I stumbled into this

http://www.matrixcalculus.org/

where you can try my example above: norm2(A*x)^2

I was wondering how come that mature symbolic softwares on the market (maple and mathematica) can't do it?

An hour or two ago, I answered a question in which it was a question of plotting a complex-valued function of 2 real variables. But the question itself and also my answer to it disappeared somewhere. Therefore, I send my answer here below.

There are two options for plotting:
1. Graphs of real and imaginary parts (as 2 surfaces in 3D).
2. Graph of the absolute value of this function (one surface in 3d) .

restart;
f:=(1+cosh(2*x))*exp(-4*I*t):
plot3d([Re,Im](f), x=0..1, t=0..1, color=[red,blue]);
plot3d(sqrt(add([Re,Im](f)^~2)), x=0..1, t=0..1, color=green);

 

I define the spherical Bessel function with j([n, z]) := sqrt(2)*sqrt(Pi/z)*BesselJ(n + 1/2, z)/2.  Then I attempt to plot with

Digits := 20;
plot(j([1, z]), z = 0 .. 25);

I get an empty plot and an error message: 

Warning, expecting only range variable z in expression j([1, z]) to be plotted but found name j

which makes no sense to me.

(I am using Maple 2019.)

Hi

I would like to solve the non-linear system to obtain rho1 and rho2 .

But when i use solve i get the following error
Error, (in solve) cannot solve for an unknown function with other operations in its arguments

 

system.mw

many thanks

>SIRvector := [op(SIRvector), op(evalf(solve(eval({Yeqnx(1), Yeqny(1), Yeqnz(1)}, SIRvector), {JJ(1), RR(1), SS(1)})))];

I have a picture. I tried to find the formula of the curve
My code

t := proc (x) options operator, arrow; b*x^10+c*x^9+d*x^8+e*x^7+f*x^6+g*x^5+h*x^4+i*x^3+j*x^2+k*x+l end proc

And solve
solve([t(-2) = 2, t(0) = -2, t(1) = 0, t(2.5) = -3, t(3.5) = 0, t(-1) = 0, eval(diff(t(x), x), x = -2) = 0, eval(diff(t(x), x), x = 0) = 0, eval(diff(t(x), x), x = 2.5) = 0, eval(diff(t(x), x), x = 1) = 0, t(2.8) = 0], [b, c, d, e, f, g, h, i, j, k, l])

[[b = -0.2688965311e-1, c = .1687428810, d = -0.3166922031e-1, e = -1.490599337, f = 1.885667707, g = 3.845330330, h = -6.939129440, i = -2.523473874, j = 7.112020606, k = 0., l = -2.]]

I ploted

plot(-.2688965311*x^10+.1687428810*x^9-.3166922031*x^8-1.490599337*x^7+1.885667707*x^6+3.845330330*x^5-6.939129440*x^4-2.523473874*x^3+7.112020606*x^2-2, x = -3.5 .. 4.5)

The result too far with picture. How can I repair?

I want to made a comparison via plots of RK-4, NSFD and LWM.

Restart

eqn1 := diff(x(t), t) = x(t)*(2-(1/20)*x(t))-x(t-tau)*y(t)/(x(t)+10)-10;

eqn2 := diff(y(t), t) = y(t)*(x(t)/(x(t)+10)-2/3);

bc := x(0) = 40, y(0) = 16,tau=0;

sol := dsolve({bc, eqn1, eqn2}, numeric);

plots[odeplot](sol, 0 .. 10);

I got the plot Vertically x and horizontally t but i need vertically y and horizontally x, Please help me, how can i write the code for ploting, thanks in advance.

I am trying to calculate the contour integral of exp(-z/A)/z^{2+n}, for constant A and integer n. Here is the Maple code (forgive me, very beginner Maple user here!):


 

restart;

f := exp(-z/A)/z^(2 + n);
                                 /  z\
                              exp|- -|
                                 \  A/
                         f := --------
                               (2 + n)
                              z       


op~([singular(f, z)]);
                            [z = 0]
poles := [0];
                          poles := [0]
C := z -> is(abs(z) < 1);
assume(2 <= n);
(2*Pi)*I*add(residue(f, z = a), a = select(C, poles));
                              /   /  z\       \
                              |exp|- -|       |
                              |   \  A/       |
                2 I Pi residue|--------, z = 0|
                              | (2 + n)       |
                              \z              /

You can do this by hand by simply expanding the exponential, but how do I make Maple actually do the calculation? (instead of just answering with "residue...".

hi, i am trying to solve a PDE f(x,z,t) with mixed boundary conditions, while Maple just gives u(x,z,t)=0 which is incorrect, so i believe somewhere must be wrong, someone has an idea?  

Governing equation: diff(u(x, z, t), t) = a*(diff(u(x, z, t), x, x))+b*(diff(u(x, z, t), z, z)),    0<x<M, 0<z<L, t>0

IC: heat source is a point at (0, z0):  u(x,z,0)=c*dirac(x-0)*dirac(z-z0) , where c is a temperature at t=0.

boundarys of domain are cauchy boundaries: du(0,z,t)/dx=0;  du(M,z,t)/dx=0; du(x,0,t)/dz=0;  du(x,L,t)/dz=0

the code is: 

PDE := diff(u(x, z, t), t) = a*(diff(u(x, z, t), x, x))+b*(diff(u(x, z, t), z, z));
IBC := u(x, z, 0) = c*Dirac(x-0)*Dirac(z-z0), (D[1](u))(0, z, t) = 0, (D[1](u))(M, z, t) = 0, (D[2](u))(x, 0, t) = 0, (D[2](u))(x, L, t) = 0;

pdsolve([PDE, IBC], u(x, z, t)) assuming 0<x<M, 0<z<L

 

 

thanks in advance!! 

Hi, 

I was able to determine a cubic spline fit, F(v), to x1 and y1. Now I have vector x2 which I would like to use F(v) to calculate y2 as another Vector[row]. I am having trouble accomplishing this task. Any help is greatly appreciated. Thanks.
 

restart

 x1 := Vector[row]([0.8e-1, .28, .48, .68, .88, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.4, 2.6, 2.8, 3, 3.2, 3.4, 3.6, 3.8, 4, 4.2]);

 y1 := Vector[row]([-10.081, -10.054, -10.018, -9.982, -9.939, -9.911, -9.861, -9.8, -9.734, -9.659, -9.601, -9.509, -9.4, -9.293, -9.183, -9.057, -8.931, -8.806, -8.676, -8.542, -8.405, -8.265]);

 

m := ArrayTools[Dimensions](x1);

maxx := rhs(m[1]);

 

F := proc (v) options operator, arrow; CurveFitting:-Spline(x1, y1, v, degree = 3) end proc;

 

x2 := Vector[row]([seq(log10(2*10^x1[k]), k = 1 .. maxx)])

 

y2:=?

 

Pts1 := plot(x1, y1, style = point, symbol = diamond, gridlines = true, color = red);

plt_sp := plot(F(v), v = x1[1] .. x1[maxx], color = blue);

plots:-display(Pts1, plt_sp)``

"# How to calculate Vector y2 using spline fit F with x2"? "    x1:=Vector[row]([0.08,0.28,0.48,0.68,0.88,1,1.2,1.4,1.6,1.8,2,2.2,2.4,2.6,2.8,3,3.2,3.4,3.6,3.8,4,4.2]):    y1:=Vector[row]([-10.081,-10.054,-10.018,-9.982,-9.939,-9.911,-9.861,-9.8,-9.734,-9.659,-9.601,-9.509,-9.4,-9.293,-9.183,-9.057,-8.931,-8.806,-8.676,-8.542,-8.405,-8.265]):    m:=ArrayTools[Dimensions](x1):  maxx:=rhs(m[1]):      F:=v->CurveFitting:-Spline(x1,y1, v,degree=3):    x2:=Vector[row]([seq(log10(2*10^(x1[k])),k=1..maxx)]):                   #` PLOT RESULTS`   Pts1:=plot(x1,y1,style=point,symbol = diamond, gridlines=true, color = red):       plt_sp:=plot(F(v),v=x1[1]..x1[maxx],color = blue):     plots:-display(Pts1,plt_sp);     "

 

``

``


 

Download splfit.mw

I have a similar matrix.

Build through matrixplot, not exactly what I need to get. I need a way to plot without zero values on the graph.
 

how to compute example 1 of linear schrodinger equation?

[Edit: uploaded .pdf file of M.M. Mousa and S.F. Ragab, Z. Naturforsch. 63a, 140 – 144 (2008) removed for copyright reasons]

I am trying to get the function of the curve in this picture. But, I don't know how to start. How can I get function of the graph in this picture?
I guess f(x) = (x + 1) (x - 1) (x - 3), if -1 <= x <=3.5 and f(x) = -(x + 1) (x + 2.5), if -2.5 <=x <=-1.

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