## 1491 Reputation

8 years, 260 days

## Use convert......

1.

convert(Ei(1, -ln(a-2)), Sum, dummy = n);

#-gamma-ln(-ln(a-2))+Sum((-1)^n*(-ln(a-2))^(n+1)/(factorial(n+1)*(n+1)), n = 0 .. infinity)

2.

convert(Ei(a, b), Sum, dummy = n);

eval(%, b = 1);

value(%);

limit(%, a = 1);

evalf(%);

#0.2193839344

## No possible....

Probably your equation have no analytical solution.

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 > rhs((3)[2][1])
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## Works fine....

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 > eq:=I*mu*A(t[2])*omega[0]*(1/2)-A(t[2])*omega[0]*(diff(B(t[2]), t[2]))+I*(diff(A(t[2]), t[2]))*omega[0]-(1/4)*A(t[2])^5*beta[2]*omega[0]^2-(1/4)*A(t[2])^3*beta[1]*omega[0]^2-(1/2)*F[0]*exp(I*sigma*t[2]-I*B(t[2]))+5*alpha[2]*A(t[2])^5*(1/16)+3*alpha[1]*A(t[2])^3*(1/8);
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 > eq2:=simplify(eq);
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 > eval(eq2,-sigma*t[2]+B(t[2])=C(t2));
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## fsolve.....

The standard way to solve systems of equations numerically is using fsolve.  For example:

```fsolve({x^2-y-3, x*y+2*y^2-4 = 0}, {x = 0.5 .. 2.5, y = 0 .. 3});

#{x = 2.000000000, y = 1.000000000}```

That's what the fsolve command will do with it.

```infolevel[fsolve]:=6:
fsolve({x^2-y-3, x*y+2*y^2-4 = 0}, {x = 0.5 .. 2.5, y = 0 .. 3});

#fsolve: trying multivariate Newton iteration
#fsolve:
#guess vector Vector(2, [2.1564724756843477,.13446305023563622])
#fsolve: norm of errors: 5.1897839975614897
#fsolve: new norm: 3.6816464474611895
#fsolve: iter = 1 |incr| = 1.4002 new values x = 2.1216 y = 1.4998
#fsolve: new norm: .43227630491510698
#fsolve: iter = 2 |incr| = .53690 new values x = 2.0189 y = 1.0655
#fsolve: new norm: 0.9718475514730430e-2
#fsolve: iter = 3 |incr| = 0.82477e-1 new values x = 2.0005 y = 1.0015
#fsolve: new norm: 0.5297608520287e-5
#fsolve: iter = 4 |incr| = 0.19416e-2 new values x = 2.0000 y = 1.0000
#fsolve: new norm: 0.1568000e-11
#fsolve: iter = 5 |incr| = 0.10595e-5 new values x = 2.0000 y = 1.0000
#fsolve: new norm: 0.
#fsolve: iter = 6 |incr| = 0.31360e-12 new values x = 2.0000 y = 1.0000
#               {x = 2.000000000, y = 1.000000000}
```

Or using code from:https://www.mapleprimes.com/questions/200377-Nonlinear-Equations-With-Newoton-Newton#answer201481 thanks for Carl Love.

restart:

F:= unapply(<x1^2-y1-3,x1*y1+2*y1^2-4>, x1, y1):#Equations

NewtonsMethod:= proc(F, x0, {maxiters::posint:= 99}, {epsilon::positive:= 10^(3-Digits)})
local
x, y,
J:= unapply(VectorCalculus:-Jacobian(F(x,y), [x,y]), x, y),
X:= x0,
newX:= <1,1>,
err:= <1+epsilon, epsilon>,
k
;
for k to maxiters while LinearAlgebra:-Norm(err)/LinearAlgebra:-Norm(newX) > epsilon do
newX:= X - LinearAlgebra:-LinearSolve(J(X[1],X[2]), F(X[1],X[2]));
err:= newX - X;
X:= newX
end do;
if k > maxiters then  WARNING("Did not converge.")  end if;
X
end proc:

NewtonsMethod(F, <0.5, 3.0>);

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## A test example:...

Maybe this helps:

A test example:

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## Is build in....

Execute:

with(Student[Basics]);

LinearSolveSteps((x+1)/(2*y*z) = 4*y^2/z+3*x/y, x);

LinearSolveSteps function have some limitation only  for linear equations give steps.

LinearSolveSteps("x^2 -2*x+1 = 0", x); #Nonlinear equation. It doesn't work.

## No solution....

Eliminating variable m1 and m2 from equations, then we have only eq. with 2 variables k1 and k2.

Ploting we see they do not intersect anywhere, even at point {0,0}.

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## ....

`You must add 4 more (intial,boundary) condition if you what to solve by numeric method.`
`e.g. :{ U(0, t) = 1, U(1, t) = 0, V(0, t) = 0, V(1, t) = 0}`

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Executed in Maple 2018 !!!

## Procedure with IF...

```Y := proc (delta, b, n)
if n = 0 then 1/(b^2+1)
elif n = 1 then arctan(1/b)
elif n = 2 then 1+(1/2)*ln(b^2+1)-b*arctan(1/b)
elif n = 3 then delta-(3/2)*b-(1/2)*b*ln(b^2+1)+(1/2)*arctan(1/b)*(b^2-1)
elif n = 4 then (1/2)*delta^2-b*delta+(11/12)*b^2-11/36+(1/12)*ln(b^2+1)*(3*b^2-1)+(1/6)*b*arctan(1/b)*(3-b^2)
elif n = 5 then (1/3)*delta^3-(1/2)*delta^2*b+(1/6)*delta*(3*b^2-1)+(25/72)*b-(25/72)*b^3+(1/12)*b*log(b^2+1)*(1-b^2)+(1/24)*arctan(1/b)*(1-6*b^2+b^4)
elif n = 6 then (1/4)*delta^4-(1/3)*delta^2*b+(1/12)*delta^2*(3*b^2-1)+(1/6)*b*delta*(1-b^2)+(137/1440)*b^4-(137/720)*b^2+137/7200+(1/240)*log(b^2+1)*(5*b^4-10*b^2+1)+(1/120)*b*arctan(1/b)*(-5+10*b^2-b^4)
elif n = 7 then (1/5)*delta^5-(1/4)*delta^4*b+(1/18)*delta^3*(3*b^2-1)+(1/12)*delta^2*b*(1-b^2)+(1/120)*delta*(5*b^4-10*b^2+1)-(49/2400)*b+(49/720)*b^3-(49/2400)*b^5+(1/720)*b*log(b^2+1)*(-3*b^4+10*b^2-3)+(1/720)*arctan(1/b)*(-1+15*b^2-15*b^4+b^6)
end if
end proc:

plot(Y(5, b, 7), b = 0 .. 10); # Example of use```

## Use a correct syntax....

u1 := proc (x, y) options operator, arrow; 1-exp(a*x)*cos(2*Pi*y) end proc;

a := -0.39323780;

evalf(int(u1(x, y)^2, y = -.5 .. 1.5, x = -.5 .. 1.5));

#5.493248990

#OR:

u := unapply(1-exp(a*x)*cos(2*Pi*y), [x, y]);

a := -0.39323780;

evalf(int(u(x, y)^2, y = -.5 .. 1.5, x = -.5 .. 1.5));

#OR:

u := 1-exp(a*x)*cos(2*Pi*y);

a := -0.39323780;

evalf(int(u^2, y = -.5 .. 1.5, x = -.5 .. 1.5));

## ....

Use  ExcelTools for import file.

## ....

I don't have Maple V (This is an obsolete version,you need upgrade).

integral := Int(2*(sin(theta)/cos(theta))^(2*p-1), theta = 0 .. (1/2)*Pi);

with(IntegrationTools):

integral2 := Change(convert(integral, tan), tan(theta) = sqrt(t));

`assuming`([value(integral2)], [0 < p and p < 1]);

integral.mw

Executed in Maple 2018

## Using build-in function pdetest....

Maple can't simplify better, probably V is not a solution of pde[1].

See attached file:

solution_ver_3.mw

## In Maple 2018....

BVP := [4*(diff(u(x, t), t))-9*(diff(u(x, t), x, x))-5*u(x, t) = 0, u(0, t) = 0, u(6, t) = 0, u(x, 0) = sin((1/6)*Pi*x)^2];

sol := pdsolve(BVP);

plot3d(eval(rhs(sol), infinity = 10), x = 0 .. 6, t = 0 .. 4);

OR:

plot3d(subs(infinity = 10, rhs(sol)), x = 0 .. 6, t = 0 .. 4);

OR:

plot3d(subs(infinity = 10, op(2, sol)), x = 0 .. 6, t = 0 .. 4)

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