\Hello,

How I can solve this algebraic to find unknowns ABCD?

I want to gain ABCD automatically without input the coefficients in rule by hand.

Because I should run the code for many input data

Thanks

 

>  restart; >  l:=0.5;a:=0.1; rho:=2700;h:=.0005; E:=72.4*10^9;v:= 0.3; n:=6; m:=1; AD:=10; mu:=(2*a*2.35)/l; nu:=sin(mu*l/(2*a))/sinh(mu*l/(2*a)); omega[m,n]:= 3067.173621; (1) >    >  E:=1:k[1,1]:=-5.660173062*10^10:k[1,2]:=-2.8552873062*10^10:k[1,3]:=-8.68528173062*10^10:k[1,4]:=-7.6788528173062*10^10:k[1,5]:=-1.52568528173062*10^10:k[2,1]:=-15.660173062*10^10:k[2,2]:=-21.8552873062*10^10:k[2,3]:=-18.68528173062*10^10:k[2,4]:=-71.6788528173062*10^10:k[2,5]:=-10.52568528173062*10^10: k[3,1]:=-5.65257260173062*10^10:k[3,2]:=-27.8552552873062*10^10:k[3,3]:=-81.6854428173062*10^10:k[3,4]:=-9.67858528173062*10^10:k[3,5]:=-3.52568528173062*10^10: k[4,1]:=-51.111660173062*10^10:k[4,2]:=-21.811552873062*10^10:k[4,3]:=-18.68528173062*10^10:k[4,4]:=-17.6788528173062*10^10:k[4,5]:=-11.52568528173062*10^10: k[5,1]:=-6.660173062*10^10:k[5,2]:=-61.852873062*10^10:k[5,3]:=-82.68528173062*10^10:k[5,4]:=-72.6788528173062*10^10:k[5,5]:=-21.52568528173062*10^10 (2)   >    >  S:=(Matrix([[rho*h*omega[m,n]^2+k[1, 1],k[1,2],k[1,3],k[1,4]],[k[2,1],rho*h*omega[m,n]^2+k[2,2],k[2,3],k[2,4]],[k[3,1],k[3,2],k[3,3]+rho*h*omega[m,n]^2,k[3,4]],[k[4,1],k[4,2],k[4,3],k[4,4]+rho*h*omega[m,n]^2]])).(Vector(1..4,[[A],[B],[C],[D]]))=-E*(Vector(1..4,[k[1,5],k[2, 5],k[3,5],k[4,5]])); (3) >

 

Download solve.mw

 

 

 

restart; interface(rtablesize = 10): _EnvHorizontalName := 'x': _EnvVerticalName := 'y': eqPA := (y-b0)/(x-a0) = k: solPA := y=solve(eqPA, y): #k coefficient directeur de PA eqPB := (y-b0)/(x-a0) = -1/k: solPB :=y= solve(eqPB, y):#PB perpendicalaire à PA xA := solve(subs(y = 0, eqPA), x): yB := solve(subs(x = 0, eqPB), y): eqAB := x/xA+y/yB = 1; x k y k eqAB := --------- + --------- = 1 a0 k - b0 b0 k + a0 t := solve(a*(xM+(1/2)*t*a)+b*(yM+(1/2)*t*b)+c = 0, t); 2 (a xM + b yM + c) t := - ------------------- 2 2 a + b #Recherche des coordonnées de la projection d'un point sur une droite D #M(x,y)un point quelconque du plan, M'(x',y') son symé trique dans la symétrie orthogonale d'axe D #le vecteur MM' est colinéaire du vecteur normal n de D; vec(MM')=t.vec(n), n=

Hi everybody,

Im trying to solve the following trivial pde using Maple 2018

pdsolve([diff(Y(x, t), t, t) = 0, Y(x, 0) = 0, (D[2](Y))(x, 1) = 0]);

Obviuosly the solution is Y(x, t) = 0, but Mapple 2018 is not giving any answer.

This works in Maple 2015.

Why is not working in Maple 2018?

Thanks,

Javier

 

Hello,

What means, please, this error

 

Error, (in RootFinding:-Analytic) Maple was unable to allocate enough memory to complete this computation.  Please see ?alloc
 

 

What should I to do to overcome this error?

restart;

##########  omega and theta are variables,where J[3],F[2],H[2],etc are constants.

#### I tried with "evlf" and "evlc" command but maple was not ready to provide the solution,please help me to solve this

t1:=-1/(-16.*omega^2+exp(-4*omega)+exp(4*omega)-2.)*(-(0.5817764173e-1*I)*exp((2/9)*omega*cos(theta))*omega^5*cos(theta)*J[3]-(.6981317009*I)*exp((2/9)*omega*cos(theta))*omega^4*cos(theta)*H[3]-0.4524927691e-1*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^3*G[3]-.6205615118*exp(.1111111111*omega*(2.*cos(theta)-9.))*cos(theta)*omega^3*H[2]+.6205615118*exp(.1111111111*omega*(2.*cos(theta)-9.))*cos(theta)*omega^3*F[2]+.9308422676*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^4*H[3]-.1034269187*exp(.1111111111*omega*(2.*cos(theta)-9.))*cos(theta)*omega^3*G[2]-0.7757018900e-1*exp(.1111111111*omega*(2.*cos(theta)-9.))*cos(theta)*omega^2*G[2]-0.7757018898e-1*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^4*J[3]-0.9696273622e-1*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^3*J[3]-0.4524927691e-1*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^2*J[3]-.2714956613*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^2*H[3]-0.7757018898e-1*exp(.2222222222*omega*(cos(theta)-9.))*cos(theta)*omega^4*G[3]+0.8726646261e-1*exp((2/9)*omega*cos(theta))*omega^3*J[3])*cos((2/9)*omega*sin(theta));

t2:=int(int(t1,omega=0..infinity),theta=0..2*Pi);

 

 

Dear Sir,

I have a question: I have an analytic function depending on a real parameter that is of the form  F_y (z) with y>0 and z is complex.

 

I search the zeros  of F_y(z), that is the complex z staisfying F_y(z)=0. I used  the Maple function

RootFinding:-Analytic(F, z, re = -5 .. 0, im = -100 .. 100, );

but he displays me an error message  "Error, (in RootFinding:-Analytic) the function, -(100+I*z)^(1/2)*(80-I*z)^2*cosh((1/150)*(100-I*z)^(1/2)*Pi)*sinh((1/150)*(100+I*z)^(1/2)*Pi)+(100-I*z)^(1/2)*(80+I*z)^2*cosh((1/150)*(100+I*z)^(1/2)*Pi)*sinh((1/150)*(100-I*z)^(1/2)*Pi)+(2*I)*y*z^2*cosh((1/150)*(100-I*z)^(1/2)*Pi)*cosh((1/150)*(100+I*z)^(1/2)*Pi), depends on more than one variable: {y, z}. "

 

Can you help me to resolve my problem?

 

 

Best regards,

Zayd.

Hi,

I am trying to find p and q from this simultaneous equation as a function of system parameters. I do not know the parameters and I need an expression. But Maple simply just gives p=0 and q=0 as an answer

Eq1:=61*q*L__1^2*C*e*eta/(16*omega__n^2)+5*q*L__1^2*C*e^3*eta^3/(8*omega__n^4)+3*C*p^3*gamma__1*(1/4)+3*q*C*p^2*R__n/(4*omega__n)+q*L__1^2*C*e^4*eta^4/(16*omega__n^5)+145*q*L__1^2*C/(64*omega__n)+3*q^3*C*R__n/(4*omega__n)+3*p*C*q^2*gamma__1*(1/4)+q*R*C/(4*omega__n)+19*q*L__1^2*C*e^2*eta^2/(8*omega__n^3):
Eq2:=-3*C*p^3*R__n/(4*omega__n)-3*p*C*q^2*R__n/(4*omega__n)-p*L__1^2*C*e^4*eta^4/(16*omega__n^5)-5*p*L__1^2*C*e^3*eta^3/(8*omega__n^4)-19*p*L__1^2*C*e^2*eta^2/(8*omega__n^3)-61*p*L__1^2*C*e*eta/(16*omega__n^2)-145*p*L__1^2*C/(64*omega__n)-p*R*C/(4*omega__n)+3*q*C*p^2*gamma__1*(1/4)+3*q^3*C*gamma__1*(1/4):
sys := { Eq1 , Eq2 };solve( sys, {p,q} );

Is there any way to help Maple to try other conditions, I know the only solution should not be just p=0 and q=0.

Thanks,

Baharm31

 

 

Hi,

I do not really understand the difference between annrow operator and unapply.
From the help pages it seems that unapply "creates" an arrow operator and thus that they could be two different ways to do the same thing.


restart:

f := x[1]+y[1]:

a := indets(f):                  # just because f can be more complex than the f above
g := (op(a)) -> f;              # generates an error, "operators not of a symbol type"
h := unapply(f, (op(a)))   # ok, but with a strange output
     h := (x__1, y__1) -> x__1+y__1

So it seems that Maple has transform by itself the indexed x[1] and y[1] into symbols x__1 and y__1.

Could you explain me what happened exactly ?

TIA

Assume we have an expression in several variables, x,y,z,..., where all of them are function of one parameter, t, for an example consider the following simple expression;

f := 2*y(t)*(diff(x(t), t))^2+3*(diff(x(t), t$3))-3*x(t)*(diff(y(t), t));

Is there any command or a way to ask Maple to give the highest order of derivation of x or y with respect to t in the expression? For example in the above example, the answer for x is 3 and for y is 1. If we remove the second term, then the answer for x should be 1.

Hello!

I am truing to simplify kretchmann variable in the following worksheet:

M >  # Obtaining Ricci and Kretchmann; with(DifferentialGeometry):with(Tensor): >  DGsetup([t, r, theta, phi], M); g := evalDG(-(1-2*M*mu/r)^(1/mu)*dt &t dt+(1-2*M*mu/r)^(-1/mu)*`&t`(dr, dr)+r^2*(1-2*M*mu/r)^(1-1/mu)*(`&t`(dtheta, dtheta)+sin(theta)^2*`&t`(dphi, dphi))); C := Christoffel(g): (1.1) Rie := CurvatureTensor(C): R := RicciScalar(g,Rie); h := InverseMetric(g): kretchmann := ContractIndices(RaiseLowerIndices(g, Rie, [1]), RaiseLowerIndices(h, Rie, [2, 3, 4]), [[1, 1], [2, 2], [3, 3], [4, 4]]); (1.2) M >  # simplification M >  simplify(normal(R),symbolic) (1.3) M >  simplify(kretchmann,size,symbolic) (1.4) M >

 

Download RicciScalarKretchmann.mw

The problem is that I cannot obtain a good form of it. With Mathematica FullSimplify[] function I got the following form (LaTeX code incoming): $K =& 4 M^2 \Bigl(A-B r+C r^2\Bigr)(r-2 M \mu)^{\frac{2}{\mu}-4}r^{-\frac{2}{\mu}-4},\
    A =&M^2 (\mu (3 \mu+2)+7) (\mu+1)^2,\,B = 8 M (\mu+2) (\mu+1),\, C = 12$, i.e. terms $(r-2 M \mu)$ and $r$ got fully factorized. However, I could never achieve the same form in Maple. Any help?

I am sorry if this is a silly and many-times-answered question, but I tried consulting with Maple help and googling solutions without any success.

Regards,
Nick

Exercises solved online with Maple exclusively in space. I attach the explanation links on my YouTube channel.

Part # 01

https://www.youtube.com/watch?v=8Aa2xzU8LwQ

Part # 02

https://www.youtube.com/watch?v=qyGT28CeSz4

Part # 03

https://www.youtube.com/watch?v=yf8rjSPbv5g

Part # 04

https://www.youtube.com/watch?v=FwHPW7ncZTg

Part # 05

https://www.youtube.com/watch?v=bm3frpukb0I

Link for download the file:

Vector_Exercises-Force_in_space.mw

Lenin AC

Ambassador of Maple

 

 

 

 

>  restart: with(LinearAlgebra): with(plots): with(geometry): with(plottools): # On appelle alpha la moitié de l'angle de rotation de la roue menée par tour de roue menante. alpha=Pi/n en raduans soit Pi/5=36° pour 5 rainures.. On a alors les relations suivantes entre l'entaxe E, le rayon de la roue ùenante R1 et le rayon de la roue menée R2 : R1=E.sin(alpha), R2=E*cos(alpha) Intersection du cercle (O,R2) avec la droite tan(phi)x-r/cos(phi), on obtient les coordonnées de P3 sol:=allvalues(solve([tan(phi)*x-r/cos(phi)=y,y^2+x^2=R2^2],[x,y])): # Intersection de 2 cercles sol1:=allvalues(solve([(x-E)^2+y^2=(R-a)^2,y^2+x^2=R2^2],[x,y])): # Coordonnées des points du pourtour de l'élément de croix #point(O,0,0): phi:=Pi/5: R2:=5: r:=1/4: E:=R2/cos(phi): R:=R2*tan(phi): a:=0.5: [(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)]; geometry[point](P1,(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)): geometry[point](P2,(R2/2)*cos(phi)+r*sin(phi),(R2/2)*sin(phi)-r*cos(phi)): >  xP2:=(R2/2)*cos(phi)+r*sin(phi): yP2:=(R2/2)*sin(phi)-r*cos(phi): xP1:=(R2/2-r)*cos(phi): yP1:=(R2/2-r)*sin(phi): # Equation paramétrique du segment OP1 (t varie de 0 à 1) ; non pris en compte x1:=t*(0-xP1)+xP1: y1:=t*(0-yP1)+yP1: n1:=5: dt:=1/(n1-1): #t varie entre 0 et 1 >  for i to n1 do tau:=(i-1)*dt: xx[i]:=evalf(subs(t=tau,x1)): yy[i]:=evalf(subs(t=tau,y1)): #print(i,xx[i],yy[i]); od: # Equation paramétrique du quart de cercle P1P2 de la rainure (t varie de 0 à 1) x2:=R2/2*cos(phi)+r*cos(t): #attention au sens de rotation du parcours de l'objet y2:=R2/2*sin(phi)+r*sin(t): n2:=6: dt:=Pi/2/(n2-1): #arc de Pi/2 for i to n2 do tau:=phi-Pi+(i-1)*dt: xx[i]:=evalf(subs(t=tau,x2)): yy[i]:=evalf(subs(t=tau,y2)): od: for i to n2 do Vector[row]([i,xx[i],yy[i]]) od: >  droite:=plot((tan(phi)*x-r/cos(phi),x=0..3),linestyle=dot,color=blue): xP3:=evalf(subs(op(1,sol[1]),x)): yP3:=evalf(subs(op(1,sol[1]),y)): xP2:=evalf(subs(op(1,sol[2]),x)): yP2:=evalf(subs(op(1,sol[2]),y)): xP4:=evalf(subs(op(1,sol1[1]),x)): yP4:=evalf(subs(op(1,sol1[1]),y)): xP5:=E-(R-a): yP5:=0: x3:=t*(xP3-xP2)+xP2: y3:=t*(yP3-yP2)+yP2: n3:=10: dt:=1/(n3-1): #t varie entre 0 et 1 >  for i to n3 do tau:=(i-1)*dt: xx[i+n2]:=evalf(subs(t=tau,x3)): yy[i+n2]:=evalf(subs(t=tau,y3)): od: >  for i to n3 do Vector[row]([i,xx[i],yy[i]]) od: >  x4:=xP5+R-a+(R-a)*cos(t):#attention au sens de rotation du parcours de l'objet y4:=(R-a)*sin(t): n4:=11: eta:=arcsin(yP4/(R-a)): dt:=(-eta)/(n2-1)/2:#arc de Pi/2 >  for i to n4 do tau:=(Pi+eta)+(i-1)*dt: #recherche de tau ? xx[i+n2+n3]:=evalf(subs(t=-tau,x4)): yy[i+n2+n3]:=evalf(subs(t=-tau,y4)): od: >  for i to n4 do Vector[row]([i,xx[i],yy[i]]) od: >  n:=n2+n3+n4; for i to n do Vector[row]([i,xx[i],yy[i]]) od: >  figure:=NULL: for i from 0 to n do xx[0]:=0: yy[0]:=0: figure:=figure,[xx[i],yy[i]]: od: >  figure:=[figure]: >  polygonplot(figure,scaling=constrained,color=yellow);#,view=[-0.1..5,-0.1..3] >  for i to n do X[i]:=xx[i]: Y[i]:=yy[i] od: >  d1:=plottools[disk]([xP1,yP1],0.05,color=blue):############ non définis d2:=plottools[disk]([xP2,yP2],0.05,color=red):############ non définis >  n := 27; ##### ci-dessus, il faut arrêter le trait à l'axe Ox ############### il suffit de prendre le symétrique puis 4 rotations de Pi/5 ############# symOX:= pt -> [pt[1],-pt[2]]: rot:=proc(t,pt) [pt[1]*cos(t)-pt[2]*sin(t),pt[1]*sin(t)+pt[2]*cos(t)]; end: figure1:=[ op(figure),op(map(symOX,figure))]: croix:=op(figure1): >  for i to 4 do croix:=croix,op(map( pt -> rot(i*2*Pi/5,pt),figure1) ): od: croix:=[croix]: polygonplot(croix,scaling=constrained,color=yellow); #How arrange this drawning and animate it. Thank you. >

 

Download GenovaDrive.mw

[Just in case the above is not a 100%-correct presentation of the original code, I've left it below in its origianally posted form.--Carl Love as Moderator]

restart:with(LinearAlgebra):with(plots):with(geometry):with(plottools): On appelle alpha la moitié de l'angle de rotation de la roue menée par tour de roue menante. alpha=Pi/n en raduans soit Pi/5=36° pour 5 rainures.. On a alors les relations suivantes entre l'entaxe E, le rayon de la roue ùenante R1 et le rayon de la roue menée R2 : R1=E.sin(alpha), R2=E*cos(alpha) Intersection du cercle (O,R2) avec la droite tan(phi)x-r/cos(phi), on obtient les coordonnées de P3 sol:=allvalues(solve([tan(phi)*x-r/cos(phi)=y,y^2+x^2=R2^2],[x,y])): Intersection de 2 cercles sol1:=allvalues(solve([(x-E)^2+y^2=(R-a)^2,y^2+x^2=R2^2],[x,y])): Coordonnées des points du pourtour de l'élément de croix #point(O,0,0): phi:=Pi/5:R2:=5:r:=1/4:E:=R2/cos(phi):R:=R2*tan(phi):a:=0.5: [(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)]: geometry[point](P1,(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)): geometry[point](P2,(R2/2)*cos(phi)+r*sin(phi),(R2/2)*sin(phi)-r*cos(phi)): xP2:=(R2/2)*cos(phi)+r*sin(phi):yP2:=(R2/2)*sin(phi)-r*cos(phi): xP1:=(R2/2-r)*cos(phi):yP1:=(R2/2-r)*sin(phi): Equation paramétrique du segment OP1 (t varie de 0 à 1) ; non pris en compte x1:=t*(0-xP1)+xP1: y1:=t*(0-yP1)+yP1: n1:=5: dt:=1/(n1-1):#t varie entre 0 et 1 for i to n1 do tau:=(i-1)*dt: xx[i]:=evalf(subs(t=tau,x1)): yy[i]:=evalf(subs(t=tau,y1)): #print(i,xx[i],yy[i]); od: Equation paramétrique du quart de cercle P1P2 de la rainure (t varie de 0 à 1) x2:=R2/2*cos(phi)+r*cos(t):#attention au sens de rotation du parcours de l'objet y2:=R2/2*sin(phi)+r*sin(t): n2:=6: dt:=Pi/2/(n2-1):#arc de Pi/2 for i to n2 do tau:=phi-Pi+(i-1)*dt: xx[i]:=evalf(subs(t=tau,x2)): yy[i]:=evalf(subs(t=tau,y2)): od: for i to n2 do Vector[row]([i,xx[i],yy[i]]) od: droite:=plot((tan(phi)*x-r/cos(phi),x=0..3),linestyle=dot,color=blue): sol[1]: xP3:=evalf(subs(op(1,sol[1]),x)):yP3:=evalf(subs(op(1,sol[1]),y)): xP2:=evalf(subs(op(1,sol[2]),x)):yP2:=evalf(subs(op(1,sol[2]),y)): xP4:=evalf(subs(op(1,sol1[1]),x)):yP4:=evalf(subs(op(1,sol1[1]),y)): xP5:=E-(R-a):yP5:=0: x3:=t*(xP3-xP2)+xP2: y3:=t*(yP3-yP2)+yP2: n3:=10: dt:=1/(n3-1):#t varie entre 0 et 1 for i to n3 do tau:=(i-1)*dt: xx[i+n2]:=evalf(subs(t=tau,x3)): yy[i+n2]:=evalf(subs(t=tau,y3)): od: for i to n3 do Vector[row]([i,xx[i],yy[i]]) od: x4:=xP5+R-a+(R-a)*cos(t):#attention au sens de rotation du parcours de l'objet y4:=(R-a)*sin(t): n4:=11: eta:=arcsin(yP4/(R-a)): dt:=(-eta)/(n2-1)/2:#arc de Pi/2 for i to n4 do tau:=(Pi+eta)+(i-1)*dt:#recherche de tau ? xx[i+n2+n3]:=evalf(subs(t=-tau,x4)): yy[i+n2+n3]:=evalf(subs(t=-tau,y4)): od: for i to n4 do Vector[row]([i,xx[i],yy[i]]) od: n:=n2+n3+n4; for i to n do Vector[row]([i,xx[i],yy[i]]) od: figure:=NULL: for i from 0 to n do xx[0]:=0:yy[0]:=0: figure:=figure,[xx[i],yy[i]]: od: figure:=[figure]: polygonplot(figure,scaling=constrained,color=yellow):#,view=[-0.1..5,-0.1..3] for i to n do X[i]:=xx[i]: Y[i]:=yy[i] od: d1:=plottools[disk]([xP1,yP1],0.05,color=blue):############ non définis d2:=plottools[disk]([xP2,yP2],0.05,color=red):############ non définis n := 27 ##### ci-dessus, il faut arrêter le trait à l'axe Ox ############### il suffit de prendre le symétrique puis 4 rotations de Pi/5 ############# symOX:= pt -> [pt[1],-pt[2]]: rot:=proc(t,pt) [pt[1]*cos(t)-pt[2]*sin(t),pt[1]*sin(t)+pt[2]*cos(t)]; end: figure1:=[ op(figure),op(map(symOX,figure))]: croix:=op(figure1): for i to 4 do croix:=croix,op(map( pt -> rot(i*2*Pi/5,pt),figure1) ): od: croix:=[croix]: polygonplot(croix,scaling=constrained,color=yellow); How arrange this drawning and animate it. Thank you.

I am working with the Physics package in Maple 2018. I have a spacetime with a metric g and the vector field l is the tangent to a null curve. I have two problems:

(1) I want to check in maple that indeed g(l,l)=0. For this I wrote:

SumOverRepeatedIndices(g_[mu,nu]l[~mu]l[~nu])

Firstly, when I try to execute this command, the program keeps on evaluating for a long time. So, I have to `interrupt the current operation', undo this command, save and reopen the file, execute the entire worksheet again and then somehow it works. However, as you can see in step(13) of the attached worksheet, Maple returns an expression for this command but doesn't cancel out terms and show that it is indeed zero. How can I do this?

(2) I wish to find $(\nabla_l l)^\nu$. Is there a way to directly do this? Or should I do something like:

Define(L[mu,~nu]=D_mu l[~nu])

SumOverRepeatedIndices(l[~mu]L[mu,~nu])

Thanks a lot.

Edit: I actually carried out these steps too. First, I defined $L_\mu^\nu$ in step (14). Then, when I ask for the non-zero components of L, I find that the L_\mu,\nu components are given. How can I get the non-zero L_\mu^\nu components? Also, in step (16), I did the `SumOverRepeatedindices' but it returned only a symbolic result without evaluationg.

try2.mw

How I can simplify result? For factor or using rule.

Thanks

 

Hello,

I am trying to work with the Physics package to calculate Christoffel symbols. I have just started using Maple today and am coming up against a few hurdles.

I want to change the signature of the metric from (+---) to (-+++). I am typing the following in math mode:

Setup( signature = '-+++')

However, this is returning the following error:

"Error, invalid sum/difference" with my command rewritten within a red box together with the last '+" within another red box. Can you please help me out with this?

Thank you.