Why doesn't Maple show a solution to the following odesys of second order and not show an error?
 

Differential Equations Trebuchet, Phase II_2020-06-05   (1) (2) (3) (4) (5) (6)   (7) (8) (9) (10) (11) (12) Error, (in ODEtools/info) Required a specification of the indeterminate function

 

Download Differential_Equations_Trebuchet_Phase_II_2020-06-05.mw

 

Hi,

I need taylor expasion of

A(w):=M/sqrt(M^2+2*w)-mup*sqrt(((mu*M^2+3*sigmap-2*w)-sqrt((mu*M^2+3*sigmap-2*w)^2-12*mu*sigmap*M^2))/(6*sigmap))

I used taylor(A,w,4), but I had a problem!

If I know that A(w) is a real function, how can I remove ‘csgn’ from the  ‘taylor(A,w,4)’ command:

Thanks.

12.mws

restart; with(plots):with(LinearAlgebra):unprotect(O); alias(conj = conjugate); conj z = lambda*v+a; La droite D est représentée par son équation complexe Appelons H l'affixe h le pied de la perpendicukaire abaissée de O sur (D) Les vecteurs OH et V sont orthogonaux donc z = lambda v + a h*conj(v)+conj(h)*v = 0; Le point H appartient à la droite (D) donc : h = lambda*v+a; conj(h) = conj(a)+lambda*conj(v); (lambda*v+a)*conj(v)+(conj(a)+lambda*conj(v))*v = 0; solve(%, lambda); h := simplify(subs(lambda = %, lambda*v+a)); a := 3-I*4; v := -2/3+4*I; evalc(h); H := [Re(h), Im(h)]; Représentation graphique d'un cas particulier : f := proc (x) options operator, arrow; -3*x+5 end proc: a := 3: A := [a, f(a)]:O:=[0,0]: zo := [8/3+I*f(8/3)]; ze := [2+I^(eval(diff(f(x), x), x = 2))]; Zo := [8/3, f(8/3)]; Ze := [2, f(2)]; ex := -3*x+5; V := `

Good day sirs, I write a system of DAE but giving me this code "(The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)". The code is attached below.

Thanking you in anticipating for your help.

Help!!!!.mw

 

f := x -> exp(-x)*sin(x); intvx:= 0..3;
f := proc (x) options operator, arrow; exp(-x)*sin(x) end proc
intvx := 0 .. 3
 

And this one below ( i prefer this one ) , but got in worksheet now the one above
Probably a option issue ?

 

f := x -> exp(-x)*sin(x); intvx:= 0..3;

restart;

ii := 50;

seq(ii, ii = 0 .. 5);

evalf(int(ii, ii = 0 .. 5))

 

For sequence the syntax is correct i.e the output is 0,1,2,3,4,5. is ii inside sequence command takes the value 5 assigned it then changes when ii ranges from 0 to 5?

restart;

ii:=50:
seq(ii)

output: 50

restart;

ii:=50:

seq(seq(ii),ii=1..5)

expected answer: 50,50,50,50,50

obtained: 1,2,3,4,5

please explain how seq command works

 

Trebuchet, Phase I, 2020-05-27 Ki restart; with(RealDomain); with(SolveTools); assume(h < r1); additionally(h < r2); [Im, Re, ^, arccos, arccosh, arccot, arccoth, arccsc, arccsch, arcsec, arcsech, arcsin, arcsinh, arctan, arctanh, cos, cosh, cot, coth, csc, csch, eval, exp, expand, limit, ln, log, sec, sech, signum, simplify, sin, sinh, solve, sqrt, surd, tan, tanh ] [AbstractRootOfSolution, Basis, CancelInverses, Combine, Complexity, Engine, GreaterComplexity, Identity, Inequality, Linear, Parametric, Polynomial, PolynomialSystem, RationalCoefficients, SemiAlgebraic, SortByComplexity] hz1 := g*(r3*m3*cos(phi1(t))-r1*m2*sin(phi2(t))/sin(phi2(t)-phi1(t)))/theta3; whattype(phi1(t)); whattype(phi2(t)); / r1 m2 sin(phi2(t)) \ g |r3 m3 cos(phi1(t)) + -----------------------| \ sin(-phi2(t) + phi1(t))/ hz1 := ------------------------------------------------ theta3 function function hz2 := phi1(t)+arcsin((h+r1*sin(phi1(t)))/r2); /h + r1 sin(phi1(t))\ hz2 := phi1(t) + arcsin|-------------------| \ r2 / subs(phi2 = hz2, hz1); / / | | 1 | | ------ |g |r3 m3 cos(phi1(t)) theta3 | | | | \ \ // /h + r1 sin(phi1(t))\\ \ \\ r1 m2 sin||phi1(t) + arcsin|-------------------||(t)| || \\ \ r2 // / || + ----------------------------------------------------------|| / / /h + r1 sin(phi1(t))\\ \|| sin|-|phi1(t) + arcsin|-------------------||(t) + phi1(t)||| \ \ \ r2 // /// deq := diff(phi1(t), t, t)-% = 0; / / / 2 \ | | | d | 1 | | deq := |---- phi1(t)| - ------ |g |r3 m3 cos(phi1(t)) | 2 | theta3 | | \ dt / | | \ \ / /h + r1 sin(phi1(t))\ \ \ r1 m2 sin|phi1(t)(t) + arcsin|-------------------|(t)| | \ \ r2 / / | + -----------------------------------------------------------| / /h + r1 sin(phi1(t))\ \| sin|-phi1(t)(t) - arcsin|-------------------|(t) + phi1(t)|| \ \ r2 / // \ | | | = 0 | | / ics := phi1(0) = -arcsin(h/r1), (D(phi1))(0) = 0; /h \ ics := phi1(0) = -arcsin|--|, D(phi1)(0) = 0 \r1/ dsolve({deq, ics}, phi1(t)); dsolve(deq); deq_numeric := subs(r1 = 8, r2 = 8, r3 = 1, h = 5, m2 = 1, m3 = 20, theta3 = 20, deq); / | / d / d \\ 1 | |--- |--- phi1(t)|| - -- g |20 cos(phi1(t)) \ dt \ dt // 20 | | \ / /5 \ \ \ 8 sin|phi1(t)(t) + arcsin|- + sin(phi1(t))|(t)| | \ \8 / / | + --------------------------------------------------------| = 0 / /5 \ \| sin|-phi1(t)(t) - arcsin|- + sin(phi1(t))|(t) + phi1(t)|| \ \8 / // ics_numeric := phi1(0) = 0, (D(phi1))(0) = -.63; ics_numeric := phi1(0) = 0, D(phi1)(0) = -0.63 hz1 := dsolve({ics, deq_numeric}, phi1(t), numeric); Error, (in dsolve/numeric/process_input) unknown arcsin(5/8+sin(phi1(t))) present in ODE system is not a specified dependent variable or evaluatable procedure sin(-phi1(t)-arcsin((h+r1*sin(phi1(t)))/r2)); / /h + r1 sin(phi1(t))\\ -sin|phi1(t) + arcsin|-------------------|| \ \ r2 //

Hello everyone,

first, I'd like to mention that I am relatively new to Maple and am therefore thankful for any advice you might have!

I am trying to integrate the term (k_1^2 * r) from a to infinity, see the picture below as well as the attached file. Maple seems to have some issues with that. However, if I break the integral down into more manageable parts it suddenly works! Why is that? How can I get Maple to solve this immeadiately?  I suspect the culprit lies in the term that contains (-Ei(-B*r)*r^(-1)) where Ei is the exponential integral as defined in Maple. The resulting  hypergeometric function seems suspicious.

The problem is that I have to evaluate 21 integrals of this type (k_x*k_y*r) and breaking them down manually becomes pretty cumbersome, especially as the number of terms in the expanded expressions increases. Is there a way to automate this procedure? I guess I would need to extract individual terms and automatically plug them into the integral expression. That should the last resort, however.

The specific problem (everything included for context, weird stuff happens after equation 15):

Maple_Problem.mw

As for the variables: E, t, and R are real positive numbers. a and B are already assumed as real and positive. A_0, C_0, A_2, and C_2 are real numbers (could be negative, I do not know yet since they must be determined later on). a_0 is definitely real, but it may be negative. r is the polar coordinate, so it is also real and positive, but adding this assumption did not yield a better result.

 

Thank you for your help!

 

conj := conjugate; d := a*x+b*y-c = 0; z := x+I*y; evalc(z+conj(z)); evalc(z-conj(z)); d := expand((1/2)*a*(z+conj(z))+b*(z-conj(z))/(2*I))-c; is(d = z(a-I*b)+conj(z)*(a+I*b)-2*c); varpi = a+I*b; is(d = z*conj(varpi)+conj(z)*varpi-2*c); How to perform calculations correctly ? Thank you.

How to get the inflection points for this function.?

fprime_expr:=x^sin(x)*(cos(x)*ln(x)+sin(x)/x);

i tried symbolic and numeric , but no answer 

for X:= solve( fprime_expr =0, x);

 

Hello,

While working for an assignment I had to use a piecewise function which gives a result based on a randomly generated value. The following is a much simpler version of what I've been working on, but it gives the same error.

A random value is uniformly distributed between 0 and 100. The piecewise returns a 1 if the value is between 0 and 50 and it returns 2 if the value is between 50 and 100. As far as I understand this function should only ever give 1 or 2, never something else. However when I loop this a few times Maple regularly returns a 0, which doesn't make sense to me. I've printed the values that return a 0 but none of these should break the piecewise. Can someone please explain to me what's going wrong here and how to fix it?

My code:

restart; randir := piecewise(0 <= r1() and r1() < 50, 1, 50 <= r1() and r1() < 100, 2);
for i to 1000 do r1 := rand(0. .. 100.0); if randir = 0 then print(fail[i], r1()) end if end do;
 

I've included a failure check to see when and at what values it returns a 0, and as you can see it happens very often.

I'm trying to create a graph using a matrix that has numbers and text values, how can I specify that some values of the matrix are strings and others are numbers? I'd like to create something like this

Hello

I use Maple 20018.2.

When I use "Data Set Search" and press Search I get following Error Message. I check that the network access to the internet is on enable. Does anybody has an Idea?

thank you

Murad

z1 := a1+I*b1; z2 := a2+I*b2; abs(z1) = 1; abs(z2) = 1; argument(z1) = alpha; argument(z2) = beta; On considère dans ℂ les complexes z1 et z2 de module 1 et d'argument α et β Show that (z1+z2)^2/(z1+z2) est un réel positf ou nul. Dans quel cas est-il nul ? is((z1^2+2*z1*z2+z2^2)/(z1+z2) = z1/z2+z2/z1+2);#wrong answer z1/z2 = exp(I*(alpha-beta)); z2/z1 = exp(I*(beta-alpha)); is(z1/z2+z2/z1+2 = 2*(1+cos(alpha-beta)));#wong answer Miscalculations. Thank you for your help.

I'm trying to obtain the dynamical response of a simply-supported beam with a cantilever extension, coupled to a spring-mass system. In mathematical terms, this system is ruled by three PDEs (relative to each bare part of the main structure) and one ODE (relative to the spring-mass system). I think my mathemical model is finely formulated, but Maple keeps telling me this:

Error, (in pdsolve/numeric/process_IBCs) improper op or subscript selector

I believe it is because my PDEs depend on "x" and "t", while the ODE depends solely on "t". I have tried to transform my ODE into a "PDE", making it also dependent of "x", but without imposing any boundary conditions relative to "x". However, after this Maple points a new error message:

Error, (in pdsolve/numeric) initial/boundary conditions must be defined at one or two points for each independent variable

Could someone help me finding a solution? My algorythm in shown in the attached file below.

Worksheet.mw