Mariusz Iwaniuk

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10 years, 258 days

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These are answers submitted by Mariusz Iwaniuk

QDifferenceEquations packed is only do algebraic calculation.

From wikipedia: qgamma = (1-q)^(1-z)*(product((1-q^(n+1))/(1-q^(n+z)), n = 0 .. infinity))

 

qgamma := proc (z, q) options operator, arrow; (1-q)^(1-z)*(product((1-q^(n+1))/(1-q^(n+z)), n = 0 .. 1000)) end proc

z := .8; q := .9

evalf(qgamma(z, q)) =1.156991553

qgamma.mw

eq1 := 1 = abs(1/(I*Pi*z*(1+I*Pi*a)));
eq2 := -(5/9)*Pi = argument(1/(I*Pi*z*(1+I*Pi*a)));

fsolve([eq1, eq2], {a = 0 .. 1, z = 0 .. 1})

{a = 0.5612662116e-1, z = 0.3134740437}

PS: Mathematica gives symbolic answer:

 

 

Using int(func(p),p,numeric).

See file attached :)

evalfandintPerformance.mw

evalfandintPerformance1.mw

Ode_nonlinear.mw

 


 

 

 

From maple Help:"Note that isolate does not perform integration or differentiation to isolate for expr"

Use a collect function like this : collect(yours equation, diff)

Model_Maple.mw

@spalinowy 

Simple code to compute the solution? No.

On basis this webpage:

http://www.maplesoft.com/support/help/Maple/view.aspx?path=examples/pdsolve_boundaryconditions

Answer in the file:

Pde_solve.mw

Integral_A.mw
Integral_B.mw
 

restart

f := proc (y) options operator, arrow; Pi*(2*ln(y+sqrt(y^2-4))-2*ln(2))^2 end proc

proc (y) options operator, arrow; Pi*(2*ln(y+sqrt(y^2-4))-2*ln(2))^2 end proc

(1)

Int(f(y), y = 2 .. exp(1)+exp(-1))

Int(Pi*(2*ln(y+(y^2-4)^(1/2))-2*ln(2))^2, y = 2 .. exp(1)+exp(-1))

(2)

with(IntegrationTools):

simplify(IntegrationTools:-Change(Int(Pi*(2*ln(y+(y^2-4)^(1/2))-2*ln(2))^2, y = 2 .. exp(1)+exp(-1)), y+sqrt(y^2-4) = x, x), size)

2*Pi*(Int((-ln(x)+ln(2))^2*(x^2-4)/x^2, x = 2 .. exp(1)+exp(-1)+(exp(2)-2+exp(-2))^(1/2)))

(3)

simplify(value(2*Pi*(Int((-ln(x)+ln(2))^2*(x^2-4)/x^2, x = 2 .. exp(1)+exp(-1)+(exp(2)-2+exp(-2))^(1/2)))), symbolic)

4*(((exp(2)+1)*(exp(4)-2*exp(2)+1)^(1/2)+2*exp(2)+exp(4)+1)*ln(exp(2)+(exp(4)-2*exp(2)+1)^(1/2)+1)^2+(-2*(exp(2)+1)*(ln(2)+2)*(exp(4)-2*exp(2)+1)^(1/2)+(-4*exp(2)-2*exp(4)-2)*ln(2)-4*exp(4)-4)*ln(exp(2)+(exp(4)-2*exp(2)+1)^(1/2)+1)+((exp(2)+1)*ln(2)^2+(4*exp(2)+4)*ln(2)-4*exp(1)+5*exp(2)+5)*(exp(4)-2*exp(2)+1)^(1/2)+(2*exp(2)+exp(4)+1)*ln(2)^2+(4*exp(4)+4)*ln(2)-4*exp(1)+2*exp(2)-4*exp(3)+5*exp(4)+5)*Pi*exp(-1)/(exp(2)+(exp(4)-2*exp(2)+1)^(1/2)+1)

(4)

Digits := 20:

evalf(4*(((exp(2)+1)*(exp(4)-2*exp(2)+1)^(1/2)+2*exp(2)+exp(4)+1)*ln(exp(2)+(exp(4)-2*exp(2)+1)^(1/2)+1)^2+(-2*(exp(2)+1)*(ln(2)+2)*(exp(4)-2*exp(2)+1)^(1/2)+(-4*exp(2)-2*exp(4)-2)*ln(2)-4*exp(4)-4)*ln(exp(2)+(exp(4)-2*exp(2)+1)^(1/2)+1)+((exp(2)+1)*ln(2)^2+(4*exp(2)+4)*ln(2)-4*exp(1)+5*exp(2)+5)*(exp(4)-2*exp(2)+1)^(1/2)+(2*exp(2)+exp(4)+1)*ln(2)^2+(4*exp(4)+4)*ln(2)-4*exp(1)+2*exp(2)-4*exp(3)+5*exp(4)+5)*Pi*exp(-1)/(exp(2)+(exp(4)-2*exp(2)+1)^(1/2)+1))

7.0080014290760108080

(5)

int(f(y), y = 2 .. exp(1)+exp(-1), numeric)

7.0080014290760108047

(6)

``


 

Download Integral_B.mw

 

restart

f := proc (y) options operator, arrow; Pi*(2*ln(y+sqrt(y^2-4))-2*ln(2))^2 end proc

proc (y) options operator, arrow; Pi*(2*ln(y+sqrt(y^2-4))-2*ln(2))^2 end proc

(1)

simplify(convert(f(y), arctan), symbolic)

16*Pi*(arctanh((-1+y+(y^2-4)^(1/2))/(y+(y^2-4)^(1/2)+1))-arctanh(1/3))^2

(2)

simplify(int(16*Pi*(arctanh((-1+y+(y^2-4)^(1/2))/(y+(y^2-4)^(1/2)+1))-arctanh(1/3))^2, y = 2 .. exp(1)+exp(-1)), symbolic)

32*Pi*((((2+2*exp(3)+4*exp(2)+4*exp(4)+exp(1)+exp(5)+2*exp(6))*arctanh(1/3)+(1/2)*exp(1)+exp(6)+(1/2)*exp(5)+1)*(exp(4)-2*exp(2)+1)^(1/2)+(2+exp(5)+2*exp(6)+exp(7)+2*exp(8)+exp(1)+exp(3)+2*exp(2))*arctanh(1/3)+1+(1/2)*exp(1)-(1/2)*exp(3)-(1/2)*exp(5)-exp(6)+(1/2)*exp(7)+exp(8)-exp(2))*arctanh((-(exp(4)-2*exp(2)+1)^(1/2)-exp(2)+exp(1)-1)/((exp(4)-2*exp(2)+1)^(1/2)+exp(2)+exp(1)+1))+((exp(3)+2*exp(4)+(1/2)*exp(5)+exp(6)+2*exp(2)+(1/2)*exp(1)+1)*(exp(4)-2*exp(2)+1)^(1/2)+1+(1/2)*exp(3)+(1/2)*exp(1)+exp(2)+(1/2)*exp(5)+exp(6)+(1/2)*exp(7)+exp(8))*arctanh(((exp(4)-2*exp(2)+1)^(1/2)+exp(2)-exp(1)+1)/((exp(4)-2*exp(2)+1)^(1/2)+exp(2)+exp(1)+1))^2+((exp(3)+2*exp(4)+(1/2)*exp(5)+exp(6)+2*exp(2)+(1/2)*exp(1)+1)*arctanh(1/3)^2+((1/2)*exp(1)+exp(6)+(1/2)*exp(5)+1)*arctanh(1/3)-(3/4)*exp(1)+(1/2)*exp(2)+(1/2)*exp(6)-(1/2)*exp(3)-(3/4)*exp(5)+(1/2)*exp(4)+1/2)*(exp(4)-2*exp(2)+1)^(1/2)+(1+(1/2)*exp(3)+(1/2)*exp(1)+exp(2)+(1/2)*exp(5)+exp(6)+(1/2)*exp(7)+exp(8))*arctanh(1/3)^2+(1+(1/2)*exp(1)-(1/2)*exp(3)-(1/2)*exp(5)-exp(6)+(1/2)*exp(7)+exp(8)-exp(2))*arctanh(1/3)+1/2-(3/4)*exp(1)+(1/4)*exp(3)+(1/4)*exp(5)-(3/4)*exp(7)+(1/2)*exp(8))*exp(-1)/((exp(3)+2*exp(2)+2*exp(4)+exp(1)+2)*(exp(4)-2*exp(2)+1)^(1/2)+exp(1)+exp(5)+2*exp(6)+2)

(3)

Digits := 20:

evalf(32*Pi*((((2+2*exp(3)+4*exp(2)+4*exp(4)+exp(1)+exp(5)+2*exp(6))*arctanh(1/3)+(1/2)*exp(1)+exp(6)+(1/2)*exp(5)+1)*(exp(4)-2*exp(2)+1)^(1/2)+(2+exp(5)+2*exp(6)+exp(7)+2*exp(8)+exp(1)+exp(3)+2*exp(2))*arctanh(1/3)+1+(1/2)*exp(1)-(1/2)*exp(3)-(1/2)*exp(5)-exp(6)+(1/2)*exp(7)+exp(8)-exp(2))*arctanh((-(exp(4)-2*exp(2)+1)^(1/2)-exp(2)+exp(1)-1)/((exp(4)-2*exp(2)+1)^(1/2)+exp(2)+exp(1)+1))+((exp(3)+2*exp(4)+(1/2)*exp(5)+exp(6)+2*exp(2)+(1/2)*exp(1)+1)*(exp(4)-2*exp(2)+1)^(1/2)+1+(1/2)*exp(3)+(1/2)*exp(1)+exp(2)+(1/2)*exp(5)+exp(6)+(1/2)*exp(7)+exp(8))*arctanh(((exp(4)-2*exp(2)+1)^(1/2)+exp(2)-exp(1)+1)/((exp(4)-2*exp(2)+1)^(1/2)+exp(2)+exp(1)+1))^2+((exp(3)+2*exp(4)+(1/2)*exp(5)+exp(6)+2*exp(2)+(1/2)*exp(1)+1)*arctanh(1/3)^2+((1/2)*exp(1)+exp(6)+(1/2)*exp(5)+1)*arctanh(1/3)-(3/4)*exp(1)+(1/2)*exp(2)+(1/2)*exp(6)-(1/2)*exp(3)-(3/4)*exp(5)+(1/2)*exp(4)+1/2)*(exp(4)-2*exp(2)+1)^(1/2)+(1+(1/2)*exp(3)+(1/2)*exp(1)+exp(2)+(1/2)*exp(5)+exp(6)+(1/2)*exp(7)+exp(8))*arctanh(1/3)^2+(1+(1/2)*exp(1)-(1/2)*exp(3)-(1/2)*exp(5)-exp(6)+(1/2)*exp(7)+exp(8)-exp(2))*arctanh(1/3)+1/2-(3/4)*exp(1)+(1/4)*exp(3)+(1/4)*exp(5)-(3/4)*exp(7)+(1/2)*exp(8))*exp(-1)/((exp(3)+2*exp(2)+2*exp(4)+exp(1)+2)*(exp(4)-2*exp(2)+1)^(1/2)+exp(1)+exp(5)+2*exp(6)+2))

7.0080014290760108096

(4)

int(f(y), y = 2 .. exp(1)+exp(-1), numeric)

7.0080014290760108047

(5)

``


 

Download Integral_A.mw

 

From https://www.easycalculation.com/analytical/learn-angle-between-two-curves.php

ANGLE.mw

Angle between two curves is 47,56 degree.

 

With help of Mathematica:

InverseLAP.mw

 

 

You must play with values: Digits,ep,maxmesh, initmesh,abserr

to obtain greater accuracy

 

mplprimes.mw

int(sqrt(a^2+cos(x)), x = 0 .. Pi) assuming a>0

int(sqrt(a^2+cos(x)), x = 0 .. Pi) assuming a<0

(1/2*(2*a^2+2))*sqrt(2)*EllipticF(2/sqrt(2*a^2+2), (1/2)*sqrt(2*a^2+2))-2*sqrt(2)*EllipticF(2/sqrt(2*a^2+2), (1/2)*sqrt(2*a^2+2))+2*sqrt(2)*EllipticE(2/sqrt(2*a^2+2), (1/2)*sqrt(2*a^2+2))

Mathematica solution:

How about:

G := 6.67*10^(-8); c := 3*10^10:

sys := diff(y(x), x) = x^2*(z(x)^(1/2.762)+z(x))/c^2, diff(z(x), x) = -G*(z(x)^(1/2.762)+z(x))*(y(x)+x^3*z(x)/c^2)/(c^2*(x^2-G*x*y(x)/c^2))

ep := 10^(-14):

sol := dsolve([sys, y(ep) = 0, z(ep) = 1.4*10^35], [y(x), z(x)], type = numeric, range = 0 .. 10^8)

with(plots):

odeplot(sol, [x, z(x)])

 

in file more: ode.mw

Maple has trouble with Dirac() function to solve this PDE,but if I use alternative representations of Dirac() function then Maple can solve.

 

PDE.mw

 

Using fracdiff, Laplace Transform and Numeric Inverse Laplace Transfrom.

This method is not perfect.

Solution for z=0

Mariusz Iwaniuk

FracPDE_3.mw

 

 

 

Maybe if you want find solution with series.

eq := diff(y(x), x)-(Q-x*p0*exp(alpha-beta*y(x))/(1+exp(alpha-beta*y(x))))^2 = 0

Order := 5

sol := dsolve([eq, y(0) = 0], y(x), type = 'series'); convert(sol, polynom)

y(x) = Q^2*x-Q*p0*exp(alpha)*x^2/(1+exp(alpha))+(1/3)*p0*exp(alpha)*(2*Q^3*beta+p0*exp(alpha))*x^3/(1+exp(alpha))^2+(1/4)*Q^2*p0*exp(alpha)*beta*(Q^3*exp(alpha)*beta-Q^3*beta-4*p0*exp(alpha))*x^4/(1+exp(alpha))^3

 

EDITED:

Using Substitution y(x) = -(ln(v(x))-alfa)/beta I have:

diff(v(x), x)+v(x)*((-p0*x+Q)*v(x)+Q)^2*beta/(1+v(x))^2 = 0

but Maple can't find general solution.

Only  for the some values give a solution.

For p0=1,Q=0,beta=1

-(1/3)*x^3-ln(exp(alfa-y(x)))+(1/2)*exp(-2*alfa+2*y(x))+2*exp(-alfa+y(x))+_C1 = 0

for more information file included.

ode1.mw

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