Mariusz Iwaniuk

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These are answers submitted by Mariusz Iwaniuk

Error messages says: "system must be entered as a set/list of expressions/equations"

sys := ((D@@2)(theta1))(t) = t, ((D@@2)(theta2))(t) = t;#this is a list of equations.

# I assume a system of equations, because you did not give any.

sol1 := dsolve({sys, theta1(0) = (1/2)*Pi, theta2(0) = (1/4)*Pi, (D(theta1))(0) = 0, (D(theta2))(0) = 0}, {theta1(t), theta2(t)}, type = numeric, output = listprocedure);

sol1(0);

Workaround is  changing  integration method to _d01ajc that gives 7 correct digits.


 

NULL

`assuming`([int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity)], [x < 0])

1/2+sum((-1+x^_k0)/((_k0*x-_k0+x-1)*_k0), _k0 = 2 .. infinity)

(1)

1/2+sum(eval(op([2, 1], 1/2+sum((-1+x^_k0)/((_k0*x-_k0+x-1)*_k0), _k0 = 2 .. infinity)), x = -1), _k0 = 2 .. infinity)

ln(2)

(2)

NULL

int((1/2)*(1-(-1)^floor(u))/u^2, u = 1 .. infinity, numeric)

.6687714032

(3)

evalf(Int((1/2)*(1-(-1)^floor(u))/u^2, u = 1 .. infinity, method = _DEFAULT))

.6687714032

(4)

NULL

Int((1/2)*(1-(-1)^floor(u))/u^2, u = 1 .. infinity)

Int((1/2)*(1-(-1)^floor(u))/u^2, u = 1 .. infinity)

(5)

with(IntegrationTools)

Change(Int((1/2)*(1-(-1)^floor(u))/u^2, u = 1 .. infinity), u = 1/t)

(1/2)*(Int(1+(-1)^(1+floor(1/t)), t = 0 .. 1))

(6)

evalf(Int(1/2*(1+(-1)^(1+floor(1/t))), t = 0 .. 1, method = _d01ajc, epsilon = 0.1e-6, methodoptions = [maxintervals = 300000]))

.6931471751

(7)

evalf(ln(2))

.6931471806

(8)

``


 

Download integral_with_floor.mw

y := func^2; eval(diff(y, func), func = diff(tau(t), t));

#2*(diff(tau(t), t))

Check yours syntax,because is wrong!

Execute in command line for more information about itegration:  ?int

Probably you want as a example:

int(x*y, [x = 4 .. 16, y = 0 .. 2]);

#240

Wrong address.Your question is pure math. Ask that in math forums.

Laplace transform of yours examples not exist. This is only my opinion.

Not sure what is expected from that input.

A correct syntax is:

L1 := inttrans:-laplace(psi1(t)*(diff(z1(t), t)), t, s);
L2 := inttrans:-laplace((diff(psi1(t), t))^2, t, s);

#Return unevaluated

 

You made a  simple mistake in the equation (18).You  wrote xi(1),but  it should be: 1/xi(1).

See attached file:

feijisuan.mw

 

You forgot to deliver a inital conditions to DEplot function.

with(DEtools):

DE3 := {diff(y(x), x) = y(x)-z(x), diff(z(x), x) = z(x)-2*y(x)};

DEplot(DE3, [y(x), z(x)], x = 0 .. 3, y = 0 .. 2, z = -4 .. 4, [[y(0) = 1, z(0) = 1]], arrows = medium, numpoints = 200);

 

Download DEplot.mw

For more information execute: ?DEplot in  Maple command line.

On Maple 2018.1  definition doublefactorial is the same as 2016 nothing has changed.

From Maple help: ?factorial;

"Note: In Maple, !! is used for repeated factorials and so it does not indicate the double factorial."

then use: doublefactorial(n);

Maybe you can use alias function:

alias(`bi-factorial` = doublefactorial);

`bi-factorial`(15);

#2027025

doublefactorial(15);

#2027025

 

What version of Maple do you use?.

If you are using an older version, the latest version of Maple int correctly solves.


 

NULL

restart

infolevel[IntegrationTools] := 3

lambda[1] := 3/10; lambda[2] := 2*(1/10); alpha := 4

3/10

 

1/5

 

4

(1)

int(2*alpha^2*Z*exp(lambda[1]*Z)/((exp(lambda[1]*Z)-1+alpha)^2*(exp(lambda[2]*Z)-1+alpha)), Z = 0 .. infinity)

Definite Integration:   Integrating expression on Z=0..infinity
Definite Integration:   Using the integrators [distribution, piecewise, series, o, polynomial, ln, lookup, cook, ratpoly, elliptic, elliptictrig, meijergspecial, improper, asymptotic, ftoc, ftocms, meijerg, contour]

IntegralTransform LookUp Integrator:   Integral might be a laplace transform with expr=32/(exp(1/5*Z)+3)*Z/(exp(3/10*Z)+3)^2 and s=-3/10.

Definite Integration:   Integrating expression on T=0..infinity
Definite Integration:   Using the integrators cook
Cook LookUp Integrator:   but does not fit into the first class
Cook LookUp Integrator:   returning answer from cook pattern 1a
Definite Integration:   Returning integral unevaluated.
Definite Integration:   Integrating expression on T=0..infinity
Definite Integration:   Using the integrators cook

Cook LookUp Integrator:   but does not fit into the first class
Cook LookUp Integrator:   returning answer from cook pattern 1a
Definite Integration:   Returning integral unevaluated.
Definite Integration:   Integrating expression on T=0..infinity
Definite Integration:   Using the integrators cook
Definite Integration:   Returning integral unevaluated.

IntegralTransform LookUp Integrator:   Integral transform returned unevaluated.
LookUp Integrator:   unable to find the specified integral in the table

Definite Integration:   Method ftoc succeeded.
Definite Integration:   Finished sucessfully.

 

(1600/27)*ln(2)-(400/9)*dilog(1+(1/3)*3^(2/3))-(1000/81)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(1/3)-(1000/81)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(1/3)+(800/81)*3^(5/6)*arctan((2/3)*3^(1/6)-(1/3)*3^(1/2))-(400/81)*3^(5/6)*Pi-(400/27)*3^(1/6)*Pi+(250/81)*Pi^2+(250/27)*ln(3)^2-(400/81)*3^(1/3)*ln(3^(2/3)-3^(1/3)+1)+(800/27)*3^(1/6)*arctan((2/3)*3^(1/6)-(1/3)*3^(1/2))-((200/27)*I)*3^(1/6)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))+((200/27)*I)*3^(1/6)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-((1000/81)*I)*3^(5/6)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))+((1000/81)*I)*3^(5/6)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-((200/9)*I)*3^(1/2)*dilog(1-((1/3)*I)*3^(1/2))+((200/9)*I)*3^(1/2)*dilog(1+((1/3)*I)*3^(1/2))+(800/81)*3^(1/3)*ln(1+3^(1/3))+(200/81)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(2/3)+(2000/81)*dilog(1+(1/3)*3^(2/3))*3^(1/3)-(400/81)*dilog(1+(1/3)*3^(2/3))*3^(2/3)+(400/81)*ln(3^(2/3)-3^(1/3)+1)*3^(2/3)-(800/81)*ln(1+3^(1/3))*3^(2/3)+(200/81)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(2/3)+(200/3)*dilog(1-((1/3)*I)*3^(1/2))+(200/3)*dilog(1+((1/3)*I)*3^(1/2))-(400/9)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-(400/9)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-(4000/729)*3^(5/6)*Pi*ln(3)-(800/243)*3^(1/6)*Pi*ln(3)-(800/729)*3^(2/3)*Pi^2+(4000/729)*3^(1/3)*Pi^2+(100/9)*3^(1/2)*ln(3)*Pi

(2)

evalf(((200/27)*I)*3^(1/6)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))+((1000/81)*I)*3^(5/6)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))+((200/9)*I)*3^(1/2)*dilog(1+((1/3)*I)*3^(1/2))-(4000/729)*3^(5/6)*Pi*ln(3)-(800/243)*3^(1/6)*Pi*ln(3)+(100/9)*3^(1/2)*ln(3)*Pi-((200/27)*I)*3^(1/6)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-((1000/81)*I)*3^(5/6)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-((200/9)*I)*3^(1/2)*dilog(1-((1/3)*I)*3^(1/2))+(1600/27)*ln(2)-(400/9)*dilog(1+(1/3)*3^(2/3))-(400/9)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))-(400/9)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))+(250/81)*Pi^2+(250/27)*ln(3)^2+(200/3)*dilog(1-((1/3)*I)*3^(1/2))+(200/3)*dilog(1+((1/3)*I)*3^(1/2))-(400/81)*3^(1/3)*ln(3^(2/3)-3^(1/3)+1)+(800/27)*3^(1/6)*arctan((2/3)*3^(1/6)-(1/3)*3^(1/2))+(800/81)*3^(1/3)*ln(1+3^(1/3))+(200/81)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(2/3)+(2000/81)*dilog(1+(1/3)*3^(2/3))*3^(1/3)-(400/81)*dilog(1+(1/3)*3^(2/3))*3^(2/3)+(400/81)*ln(3^(2/3)-3^(1/3)+1)*3^(2/3)-(800/81)*ln(1+3^(1/3))*3^(2/3)+(200/81)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(2/3)-(800/729)*3^(2/3)*Pi^2+(4000/729)*3^(1/3)*Pi^2-(1000/81)*dilog(1-((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(1/3)-(1000/81)*dilog(1+((1/2)*I)*3^(1/6)-(1/6)*3^(2/3))*3^(1/3)+(800/81)*3^(5/6)*arctan((2/3)*3^(1/6)-(1/3)*3^(1/2))-(400/81)*3^(5/6)*Pi-(400/27)*3^(1/6)*Pi)``

19.31389182+0.*I

(3)

``


Maybe you can try and see what hapens ?

infolevel[IntegrationTools] := 3;
lambda[1] := 3/10; lambda[2] := 2*(1/10); alpha := 4:
int(2*alpha^2*Z*exp(lambda[1]*Z)/((exp(lambda[1]*Z)-1+alpha)^2*(exp(lambda[2]*Z)-1+alpha)), Z = 0 .. infinity, method = ftocms);

Download aquestion_ver_3.mw

 

Substitution x = t^2:

IntegrationTools:-Change(Int(1/(1+sqrt(x)), x = 0 .. 1), x = t^2);

value(%);

#2-2*ln(2)

Or using convert(expr,elementary):

Int(1/(1+sqrt(x)), x = 0 .. 1) = int(1/(1+sqrt(x)), x = 0 .. 1);

#Int(1/(1+sqrt(x)), x = 0 .. 1) = MeijerG([[0, 0, 1/2], []], [[1/2, 0], [-1]], 1)/Pi

Substitution: 1=x

convert(MeijerG([[0, 0, 1/2], []], [[1/2, 0], [-1]], x)/Pi, elementary);

#-ln(1-x)/x-2*arctanh(sqrt(x))/x+2/sqrt(x)

limit(%, x = 1);

#2-2*ln(2)

It's a bug in int !


 

restart; f := ((1/2-I*t)^(-s)-(1/2+I*t)^(-s))/(2*I); fc := evalc(f); f1 := `assuming`([simplify(int(f, t = 0 .. infinity))], [s > 1]); f2 := `assuming`([simplify(int(fc, t = 0 .. infinity))], [s > 1])

-((1/2)*I)*((1/2-I*t)^(-s)-(1/2+I*t)^(-s))

 

exp(-(1/2)*s*ln(1/4+t^2))*sin(s*arctan(2*t))

 

2^(s-1)/(s-1)

 

4^s/(2*s-2)

(1)

eval(f1, s = 3)

2

(2)

int(eval(fc, s = 3), t = 0 .. infinity, numeric)

2.

(3)

eval(f2, s = 2)

8

(4)


 

Download function_evaluation_goes_wrong_v2.mw

 

Only for: z>0,j>0,Re(z),Re(j)


 

restart

func := expand((-exp(j*z)*Ei(1, j*z)+I*Pi*exp(-j*z)+exp(-j*z)*Ei(1, -j*z))/(2*j))

-(1/2)*exp(j*z)*Ei(1, j*z)/j+((1/2)*I)*Pi/(j*exp(j*z))+(1/2)*Ei(1, -j*z)/(j*exp(j*z))

(1)

func2 := eval(func, [Ei(1, j*z) = -Ei(-j*z), Ei(1, -j*z) = -Ei(j*z)])

(1/2)*exp(j*z)*Ei(-j*z)/j+((1/2)*I)*Pi/(j*exp(j*z))-(1/2)*Ei(j*z)/(j*exp(j*z))

(2)

plot([Re(eval(func, [j = 1/2])), Re(eval(func2, [j = 1/2]))], z = 0 .. 5, color = ["red", "black"])

 

eval(Ei(1, -z) = -Ei(z)+(1/2)*ln(z)-(1/2)*ln(1/z)-ln(-z), z = j*z)

Ei(1, -j*z) = -Ei(j*z)+(1/2)*ln(j*z)-(1/2)*ln(1/(j*z))-ln(-j*z)

(3)

Ei(z) = -Ei(1, -z)+(1/2)*ln(z)-(1/2)*ln(1/z)-ln(-z)

func3 := simplify(eval(func, [Ei(1, j*z) = -Ei(-j*z)+(1/2)*ln(-j*z)-(1/2)*ln(-1/(j*z))-ln(j*z), Ei(1, -j*z) = -Ei(j*z)+(1/2)*ln(j*z)-(1/2)*ln(1/(j*z))-ln(-j*z)]))

(1/4)*(exp(j*z)*ln(-1/(j*z))-ln(1/(j*z))*exp(-j*z)+((2*I)*Pi-2*ln(-j*z)-2*Ei(j*z)+ln(j*z))*exp(-j*z)-exp(j*z)*(ln(-j*z)-2*ln(j*z)-2*Ei(-j*z)))/j

(4)

plot([Re(eval(func, [j = 1/2])), Re(eval(func3, [j = 1/2]))], z = 0 .. 5, color = ["red", "black"])

 

help("Ei # you can find formula here")

``


 

Download Ei.mw

interface(typesetting = extended);
F := n-> ifactors(ithprime(n)-2)[2, 1, 1] ;
F(11);
print(F);

#                               29
#n -> ifactors(ithprime(n) - 2)[2, 1, 1]

 

On Maple 2018.1

I a little speed up computation.The plot is now gererated by Maple about 1 minute.

If N=0 then integral is probably singular ,I changed q = 0.1e-2 to q = 0.1e-1.


 

restart; Omega := 2*Pi*N; R0 := a*tanh((a^2-mu)/(2*T_c))*ln((2*a^2+2*a*q+q^2-2*mu-I*Omega)/(2*a^2-2*a*q+q^2-2*mu-I*Omega))/q-2; T_c := 0.169064e-1; mu := .869262; N := 10; R1 := unapply(Int(R0, a = 0.1e-3 .. 100, method = _Gquad), q)

2*Pi*N

 

a*tanh((1/2)*(a^2-mu)/T_c)*ln((2*a^2+2*a*q+q^2-2*mu-(2*I)*Pi*N)/(2*a^2-2*a*q+q^2-2*mu-(2*I)*Pi*N))/q-2

 

0.169064e-1

 

.869262

 

10

 

proc (q) options operator, arrow; Int(a*tanh(29.57459897*a^2-25.70807505)*ln((2*a^2+2*a*q+q^2-1.738524-(20*I)*Pi)/(2*a^2-2*a*q+q^2-1.738524-(20*I)*Pi))/q-2, a = 0.1e-3 .. 100, method = _Gquad) end proc

(1)

Digits := 5

n := 5; plots:-pointplot({seq([q, evalf(abs(R1(q)))], q = 0.1e-1 .. 10, 1/n)}, connect = true)

 

``


 

Download PLOT_fun.mw

A simple example than yours and  we convert to piecewise to understand more:

convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, 0, 0 < x, 1)

 

 

convert(signum(0, x,1), piecewise);

#piecewise(x < 0, -1, 0 <= x, 1)

 

Using derivatives:

diff(signum(0, x),x);

convert(%, piecewise);

#signum(1, x)

#piecewise(x = 0, undefined, 0)

 

 

diff(signum(0, x, 1),x);

convert(%, piecewise);

#signum(1, x,1)

#piecewise(x = 0, undefined, 0)

 

_Envsignum0 := 0; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, 0, 0 < x, 1)

 

_Envsignum0 := 1; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, 0 <= x, 1)

 

_Envsignum0 := 2; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, 2, 0 < x, 1)

 

_Envsignum0 := -2; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, -2, 0 < x, 1)

 

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