MaplePrimes Questions

Dear All,

Colud you please tell me how to obtain the answer f(b)-f(a), by manipulating the following expression in maple:

int(diff(f(x), x), x = a .. b)

Thank you very much. Best regards, Jand

Hello, I have a similary issue with the following post :

http://www.mapleprimes.com/questions/200835-Algsubs-With-Subscripts-And-Powers

In a constraint equation (in mechanics), I need to make these algebraic substitutions :

--> sqrt(xA1² + yA1²) =rF1 
--> arctan(abs(yA1)/abs(xA1)) = aF1
--> sqrt(xC1^2+yC1^2) = rBTP
--> arctan(abs(yC1)/abs(xC1))=a1

I try to do it with this code :

ListAlgsubs:={sqrt(xA1^2+yA1^2)=rF1,sqrt(xA2^2+yA2^2)=rF2,sqrt(xA3^2+yA3^2)=rF3, sqrt(xA4^2+yA4^2)=rF4,arctan(abs(yA1)/abs(xA1))=aF1,-arctan(abs(yA2)/abs(xA2))=aF2,pi+arctan(abs(yA3)/abs(xA3))=aF3,pi-arctan(abs(yA4)/abs(xA4))=aF4,sqrt(xC1^2+yC1^2)=rBTP,arctan(abs(yC1)/abs(xC1))=a1};
for i from 1 to nops(ListAlgsubs)
do Branch1Cons:=applyrule(op(i,ListAlgsubs),Branch1Cons)
end do:
Branch1Cons:= simplify(Branch1Cons);

I have tried with algsubs or applyrule but it wasn't successfull in both cases.

Do you have an idea how can I do to simplify my expressions ? and namely conduct a good use of applyrule function

I join an extract of my code in order to facilitate the troubleshooting.

example.mw

Thanks a lot for your help.

How can I extract the coefficients of all monomials in a multivariate polynomial?

For example if f=ax^2+bxy^3+2 then I want

coeff(f,x^2)=a

coeff(f,xy^3)=b

coeff(f,1/2)=4

coeff(f,1/10)=20 and...

Dears, When I run calculation in Maple I found an error in matrices. See the file

 

Hi, I have 10 equation system and 10 unknown variables. I however, want to reduce the equations to 2 with two unknowns. I'm wondering how this could be done in maple. The variables are Y, q, yd, y*, yx, H,  pd, w, P and Pv. I intend to solve the equations for Y and w. 

Thanks in advance for your help. The maple file has also been attached.

 

Regards

Small_Open_Economy.mw

 

 

restart; Digits := 10; F[0] := 0; F[1] := 0; F[2] := (1/2)*A; T[0] := 1; T[1] := B; M := 2; S := 1; Pr := 1

for k from 0 to 12 do F[k+3] := (-3*(sum((k+1-r)*(k+2-r)*F[r]*F[k+2-r], r = 0 .. k))+2*(sum((r+1)*F[r+1]*(k+1-r)*F[k+1-r], r = 0 .. k))+M*(k+1)*F[k+1]-T[k])*factorial(k)/factorial(k+3); T[k+2] := (-3*Pr*(sum((k+1-r)*F[r]*T[k+1-r], r = 0 .. k))-S*T[k])*factorial(k)/factorial(k+2) end do:

(1/630)*x^7*A*B+(1/8064)*x^9*A*B-(121/1209600)*x^10*A^2*B+(19/369600)*x^11*A*B-(11/725760)*x^12*A^2*B+(97/19958400)*x^12*A*B^2+(1/12)*x^4*A-(1/24)*x^4*B+(1/120)*x^5*A^2+(1/180)*x^6*A-(1/720)*x^6*B-(1/630)*x^7*A^2-(13/40320)*x^8*A^3+(11/20160)*x^8*A-(11/40320)*x^8*B-(19/60480)*x^9*A^2-(1/45360)*x^9*B^2+(391/3628800)*x^10*A+(37/604800)*x^10*A^3-(23/1814400)*x^10*B-(41/39916800)*x^11*B^2+(229/13305600)*x^11*A^4-(439/7983360)*x^11*A^2+(197/21772800)*x^12*A-(883/159667200)*x^12*B+(29/1520640)*x^12*A^3+(1/2)*A*x^2-(1/6)*x^3-(1/120)*x^5-(1/1680)*x^7-(11/362880)*x^9-(23/2661120)*x^11

(1)

print(expand(t)):

1-(20747/79833600)*x^12*A*B+(29/1680)*x^7*A^2*B-(451/241920)*x^10*A^3*B-(2507/14515200)*x^12*A^3*B+(2921/13305600)*x^11*A*B^2-(33/4480)*x^8*A*B+(761/403200)*x^10*A*B+(1/48)*x^6*A*B-(1/8)*x^4*A*B+(977/887040)*x^11*A^2*B+(1349/4838400)*x^12*A^2*B^2-(1/1152)*x^9*A^2*B-(11/7560)*x^9*A*B^2-(37/44800)*x^10*A^2+(223/604800)*x^10*B^2+(47/633600)*x^11*A-(7913/19958400)*x^11*B+(193/6652800)*x^11*B^3+(1409/1478400)*x^11*A^3-(4813/53222400)*x^12*B^2-(167/221760)*x^12*A^2+(3/40)*x^5*A+(1/30)*x^5*B+(1/240)*x^6*B^2-(1/560)*x^7*A-(23/2520)*x^7*B-(43/4480)*x^8*A^2-(1/896)*x^8*B^2+(61/13440)*x^9*A+(31/22680)*x^9*B-(1/6)*B*x^3+B*x+(2573/95800320)*x^12-(1/2)*x^2+(1/24)*x^4-(13/720)*x^6+(11/8064)*x^8-(2143/3628800)*x^10

(2)

solve({limit(numapprox:-pade(t, x, [2, 2]), x = infinity) = 0., limit(numapprox:-pade(diff(f, x), x, [2, 2]), x = infinity) = 1}, {A, B});

{A = -.7359903327, B = 1.324616408}, {A = -0.7307377025e-1+2.009578912*I, B = .3744177908+.5971332133*I}, {A = .6936483785+.1660915631*I, B = .1622123331+.9257041678*I}, {A = -2.182873922*I, B = .8203849935*I}, {A = .3431199285*I, B = 1.783825109*I}, {A = -.6936483785+.1660915631*I, B = -.1622123331+.9257041678*I}, {A = 0.7307377025e-1+2.009578912*I, B = -.3744177908+.5971332133*I}, {A = .7359903327, B = -1.324616408}, {A = 0.7307377025e-1-2.009578912*I, B = -.3744177908-.5971332133*I}, {A = -.6936483785-.1660915631*I, B = -.1622123331-.9257041678*I}, {A = 2.182873922*I, B = -.8203849935*I}, {A = -.3431199285*I, B = -1.783825109*I}, {A = .6936483785-.1660915631*I, B = .1622123331-.9257041678*I}, {A = -0.7307377025e-1-2.009578912*I, B = .3744177908-.5971332133*I}

(3)

solve({limit(numapprox:-pade(t, x, [3, 3]), x = infinity) = 0., limit(numapprox:-pade(diff(f, x), x, [3, 3]), x = infinity) = 1}, {A, B});

{A = 4.154051132, B = 17.13248053}, {A = .5466914672+.2697341397*I, B = .1291930705+.9494499975*I}, {A = .4506017673+.3824137679*I, B = -.2437153257+1.192091322*I}, {A = .5458260296+.5776530367*I, B = .3085138074+1.260130057*I}, {A = .3007754662+.5799020019*I, B = 0.8347381159e-1+1.033103936*I}, {A = .3916946210+1.036293227*I, B = .9202208108+1.239552889*I}, {A = .1349186305+.5994923360*I, B = 1.926737919+1.099451808*I}, {A = .5141206762+2.582294380*I, B = -.7917198503+.5287783790*I}, {A = 1.669898274*I, B = 1.659206265*I}, {A = 3.170666197*I, B = -.6372670837*I}, {A = -.5141206762+2.582294380*I, B = .7917198503+.5287783790*I}, {A = -.1349186305+.5994923360*I, B = -1.926737919+1.099451808*I}, {A = -.3916946210+1.036293227*I, B = -.9202208108+1.239552889*I}, {A = -.3007754662+.5799020019*I, B = -0.8347381159e-1+1.033103936*I}, {A = -.5458260296+.5776530367*I, B = -.3085138074+1.260130057*I}, {A = -.4506017673+.3824137679*I, B = .2437153257+1.192091322*I}, {A = -.5466914672+.2697341397*I, B = -.1291930705+.9494499975*I}, {A = -4.154051132, B = -17.13248053}, {A = -.5466914672-.2697341397*I, B = -.1291930705-.9494499975*I}, {A = -.4506017673-.3824137679*I, B = .2437153257-1.192091322*I}, {A = -.5458260296-.5776530367*I, B = -.3085138074-1.260130057*I}, {A = -.3007754662-.5799020019*I, B = -0.8347381159e-1-1.033103936*I}, {A = -.3916946210-1.036293227*I, B = -.9202208108-1.239552889*I}, {A = -.1349186305-.5994923360*I, B = -1.926737919-1.099451808*I}, {A = -.5141206762-2.582294380*I, B = .7917198503-.5287783790*I}, {A = -1.669898274*I, B = -1.659206265*I}, {A = -3.170666197*I, B = .6372670837*I}, {A = .5141206762-2.582294380*I, B = -.7917198503-.5287783790*I}, {A = .1349186305-.5994923360*I, B = 1.926737919-1.099451808*I}, {A = .3916946210-1.036293227*I, B = .9202208108-1.239552889*I}, {A = .3007754662-.5799020019*I, B = 0.8347381159e-1-1.033103936*I}, {A = .5458260296-.5776530367*I, B = .3085138074-1.260130057*I}, {A = .4506017673-.3824137679*I, B = -.2437153257-1.192091322*I}, {A = .5466914672-.2697341397*I, B = .1291930705-.9494499975*I}

(4)

 

Download D.T.M.mw

restart:

with(student):

with(plots):

with(plots):

Digits := 19:

inf := 28.5:

equ1 := diff(f(eta), eta, eta, eta)+3*(diff(f(eta), eta, eta))*f(eta)-2*(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+theta(eta) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+3*(diff(diff(f(eta), eta), eta))*f(eta)-2*(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+theta(eta) = 0

(1)

equ2 := diff(theta(eta), eta, eta)+3*Pr*f(eta)*(diff(theta(eta), eta))+S*theta(eta) = 0;

diff(diff(theta(eta), eta), eta)+3*Pr*f(eta)*(diff(theta(eta), eta))+S*theta(eta) = 0

(2)

FNS := f(eta), theta(eta);

f(eta), theta(eta)

(3)

s := 0:

BC := f(0) = s, (D(f))(0) = 0, (D(f))(inf) = 1, theta(0) = 1, theta(inf) = 0;

f(0) = 0, (D(f))(0) = 0, (D(f))(28.5) = 1, theta(0) = 1, theta(28.5) = 0

(4)

CODE := [M = 2, Pr = 1, S = 1]:

S1 := dsolve({BC, subs(CODE, equ1), subs(CODE, equ2)}, {f(eta), theta(eta)}, type = numeric):

S1(0)

[eta = 0., f(eta) = 0., diff(f(eta), eta) = 0., diff(diff(f(eta), eta), eta) = .7424080874401649594, theta(eta) = 1.000000000000000000, diff(theta(eta), eta) = .9438662130843066161]

(5)

NULL

NULL

 

Download shooting_method.mw

Thank you so much for your time. Here's the real problem

f'''(η) + 3f(η)f''(η) - 2[f'(η)] 2 + θ(η) - m*f'(η) = 0

θ''(η) + 3*Pr*f(η)θ'(η) + s*θ(η) = 0

Boundary conditions are:

at η=0: f(η)=f'(η)=0; θ(η)=1;

as η→∞ f'(η)=1; θ(η)=0;

Where m = magnetic parameter (in this case taken as 2)

S = shrinking parameter (in this case taken as 1)

Pr = taken as 1 too

I haven't been able to solve this using differential transforms method (i.e getting the values of f''(0) and θ'(0) denoted by A and B respectively) but shooting method works just fine. :( I seriously need help with this. Thanks you in advance.
I've attached my codes above and i'm hoping someone helps me out real soon. thanks very one.

I am planning on getting Maple 18 Student Editon and I am wondering if the calculus palette is in Maple 18 student edition.

Thanks

Nick

Hi,

sorry for that question, i'm a beginner in maple but i think my question is not as simple.

So, i want to solve numerically this first equation :

but the second term is present only if k*(U*t-x)>μc*m*g  (stick-slip problem)

initial conditions : U=1m/s; x=0m; k=10 for example

My problem is simple, i don't know how to use conditional statement for such an ode in maple.

I've tried > Xr := U*t-x;
> k := 10; m := 1; g := 10; mu := .2;
> if k*Xr > mu*m*g then ode1 := m*(diff(x(t), t, t)) = k.Xr-mu*m*g else ode1 := m*(diff(x(t), t, t)) = k.Xr end if;

But, of course, too optimistic.

Thank to anyone who will solve that (isuppose) simple problem

Philippe

How do I convert a hardware float to a software float? I have a procedure that is passed floats as arguments. They could be hardware or software floats. I need to set an attribute on these floats (with setattribute). Trying to use setattribute on a hardware float produces utterly bizarre results, although, strangely, not an error. But I'd be happy converting them to software floats and then setting the attribute. I tried (separately)

UseHardwareFloats:= false;

convert(x, float, 16);

SFloat(op(x));

And, FWIW, here's the procedure, although I doubt that it'll provide any more useful information than I've already given. Note that in this procedure I've redefined log to be the base-10 logarithm. (Not that that makes the slightest difference wrt my Question.)

LogView:= proc(viewspec::{range, identical(DEFAULT)}, Range::range)
local aV, bV, aP, bP, aL, bL;
     (aP,bP):= op(map(x-> 10^x, Range)); # The actual range of the ORIGINAL plot structure
     if viewspec = 'DEFAULT' then
          setattribute(op(1,Range), aP) .. setattribute(op(2,Range), bP)
     else
          (aV,bV):= op(viewspec); #The user-specified view in the pre-log units
          if bV <= 0 then error "upper view limit must be positive; received %1.", bV fi;
          (aL,bL):= op(Range); #Actual range of the log plot structure
          setattribute(`if`(aV >= aP, log(aV), aL - (aP-aV)/(min(bP,bV)-aP)*(log(min(bP,bV))-aL)), aV) 

          .. setattribute(`if`(bV <= bP, log(bV), bL + (bV-bP)/(bP-max(aP,aV))*(bL-log(max(aP,aV)))), bV) 

     end if
end proc;

In the specific use instance that I'm dealing with, the first parameter, viewspec, is DEFAULT, and the second, Range, is HFloat(-43.42...)..HFloat(43.42...).

I think the ideal solution would be to use parameter coercion to convert the arguments.

Hi, I have a function as following:

50*tan(8*Pi/180)=(a1/(d2-d1))*ln(d2/d1)+(a2/(d3-d2))*ln(d3/d2)+((0.6-a1-a2)/(d4-d3))*ln(d4/d3)

where d1=0.01..0.06, d2>=d1, d3>=d2, d4>=d3, a1>0,a2>0, a1+a2<0.6.

I want to get  all the solutions of d1,d2,d3,d4,a1,a2, which satify the equation and the constrains, is there any way to solve this problem by using maple?

 

 

 

Maybe a bit strange question (just exploring Maple and a new topic), but is there a way to use CylindricalAlgebraicDecompose with a rational functions, eg 1/x, where eg is x an integer? I tried to use it but always ended up with:

 

Error, invalid input: RegularChains:-SemiAlgebraicSetTools:-CylindricalAlgebraicDecompose expects its 1st argument, F, to be of type Or(list(list(Or(polynom(rational), polynom(rational) = polynom(rational), polynom(rational) <> polynom(rational), polynom(rational) <= polynom(rational), polynom(rational) < polynom(rational)))), list(polynom(rational))), but received [1/x = x, polynom(rational)]

Hello, i got the following problem. I have a function, say u(h,e) whith multiple exponents ((u(h,e))^(1/a))^a. As long as Maple does not know that u(h,e) is positiv it won't make the simplification and return u(h,e) as expected. From my view i have two possibilites to solve this problem:

1) Define u(h,e) to be positive. Sounds intuitive, but i didn't manage to find the right definition

2) Give maple the function e.g. u:=e+h^2 and simplify the epression in uneval quotes with assumption positive for e and h.

Whereas I have solution 2, the problem is that i would need excessive amounts of uneval quotes '' at the start of my worksheet, not to run out of them.

 

I definitly need my epressions in unevaluated forms and would be happy to hear any ideas.

Greetings, Fabian

 

Hi! I just started to use Maple and I play with some of its functions. I have seen in examples related to RegularChain package an output like  [regular_chain, [[-1, -1], [0, 0]]] and I wonder how I should read it. Thanks for any help and sorry if my question is very basic.

Hi,

i make an attempt to plot the solution to

Here is my code :

> with(plots); with(DEtools);
> ode1 := diff(x(t), t) = v(t); ode2 := diff(v(t), t) = -(.8*9.8)*v(t)/abs(v(t))-cos(t)^2;
> MODEL := {ode1, ode2}; VARS := {v(t), x(t)}; DOMAIN := t = 0 .. 150; RANGE := x = -1 .. 1, v = -5 .. 5; COLORS := [BLACK, BLUE]; IC1 := [x(0) = .5, v(0) = .25]; IC2 := [x(0) = 2.5, v(0) = 3];
> DEplot(MODEL, VARS, DOMAIN, RANGE, [IC1, IC2], stepsize = .1, linecolor = COLORS, scene = [t, x]);
>

and the message cannot evaluate the solution further right of .16015784, maxfun limit exceeded (see ?dsolve,maxfun for details)

Any other attemp has failed.

Have you got somme ideas

Thanks

Phil

for a to z1/T1 do ics[a*T1] := [g0(0, a*T1) = r, g0(1, a*T1) = s] end, this loop runs correctly for T1:=1, but gives the too many levels of recursion error for T1<1. In this loop i am inserting equations g0 in a list ics.

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