## 1491 Reputation

8 years, 260 days

 >
 >
 (1)
 >
 (2)
 >
 (3)
 >
 (4)
 >
 (5)
 >
 (6)
 >
 (7)
 >
 (8)
 >
 (9)
 >
 (10)
 >
 (11)
 >
 (12)
 >
 (13)
 >
 (14)
 >
 (15)
 >
 (16)
 >
 (17)
 >

## Try numerically....

```sol := `assuming`([dsolve(Eq2 = 0)], [phi(xi) > 0]);
with(DEtools);

dsolve can't solve symbolical Abel's equation. Try numerically.

## ....

I invert that laplace transform numerically,but answer is different than :(section 4.1 page 13629, data used for FIG. 3 )

 >
 >
 (1)
 >
 (2)
 >
 (3)
 >
 (4)
 >
 >
 >
 (5)
 >
 (6)
 >
 (7)
 >
 (8)
 >
 (9)
 >
 (10)
 >
 >
 >
 >
 >

## Use assuming....

You can restrict T  using assumptions.

`plot([Re(sqrt(sin(x))), Im(sqrt(sin(x)))], x = 0 .. 4*Pi, legend = [typeset("Curve: ", Re(sqrt(sin(x)))), typeset("Curve: ", Im(sqrt(sin(x))))]);`

for: 0 < T < Pi

`simplify(int(sqrt(sin(x)), x = 0 .. T)) assuming 0 < T < Pi`
`#-sqrt(2)*(EllipticF(sqrt(2)*sqrt(sin(T))/sqrt(sin(T)+1), (1/2)*sqrt(2))-EllipticPi(sqrt(2)*sqrt(sin(T))/sqrt(sin(T)+1), 1/2, (1/2)*sqrt(2)))`

and for: Pi < T < 2*Pi

`simplify(int(sqrt(sin(x)), x = 0 .. T)) assuming Pi < T < 2*Pi`
`#sqrt(2)*(I*EllipticF(2*sqrt(sin((1/2)*T))*sqrt(-cos((1/2)*T))/(cos((1/2)*T)-sin((1/2)*T)), (1/2)*sqrt(2))-I*EllipticPi(2*sqrt(sin((1/2)*T))*sqrt(-cos((1/2)*T))/(cos((1/2)*T)-sin((1/2)*T)), 1/2, (1/2)*sqrt(2))+4*EllipticE((1/2)*sqrt(2))-2*EllipticK((1/2)*sqrt(2)))`

EDITED:

`Generally my method works if Maple does not have errors(Bugs).`

Workaround:

```f := proc (T) options operator, arrow; int(sqrt(sin(x)), x = 0 .. T) end proc;
[f((1/4)*Pi), f((1/2)*Pi), f(3*Pi*(1/4)), f(Pi)];

#[-sqrt(2)*EllipticF(sqrt(2)*sqrt(sqrt(2)/(2+sqrt(2))), (1/2)*sqrt(2))+sqrt(2)*EllipticPi(sqrt(2)*sqrt(sqrt(2)/(2+sqrt(2))), 1/2, (1/2)*sqrt(2)), -sqrt(2)*EllipticK((1/2)*sqrt(2))+2*sqrt(2)*EllipticE((1/2)*sqrt(2)), -sqrt(2)*EllipticK((1/2)*sqrt(2))+2*sqrt(2)*EllipticE((1/2)*sqrt(2))-sqrt(2)*EllipticF(cos(3*Pi*(1/8))*sqrt(2), (1/2)*sqrt(2))+2*sqrt(2)*EllipticE(cos(3*Pi*(1/8))*sqrt(2), (1/2)*sqrt(2)), -2*sqrt(2)*EllipticK((1/2)*sqrt(2))+4*sqrt(2)*EllipticE((1/2)*sqrt(2))]```

Regards,MI

## Weakness in Maple ....

Well Maple is very weak for solving symbolic sum and can't solve your example.

````assuming`([sum((d*k+a)^r, k = 1 .. n)], [n::posint, r::posint, r > 0, a > 0, d > 0]);

# returns unevaluated```

Mathematica 11.2 solution:

Translated code form MMA to Maple:

`Sum((d*k+a)^r, k = 1 .. n) = d^r*(Zeta(0, -r, (a+d)/d)-Zeta(0, -r, (d*n+a+d)/d));`

Regards,MI

## One way is:...

```sol:=identify(evalf(2*ln(3)-3*ln(2)));

# arcsinh(17/144)

simplify(convert(sol, arccoth), symbolic);

# 2*arctanh(1/17)
```

Only for inverse hyperbolic function.

## ....

```with(gfun);
l := [8, 32, 128, 512, 2048];
rec := listtorec(l, u(n), [ogf]);

# rec := [{u(1+n)-4*u(n), u(0) = 8}, ogf]```

You must only change to:  u(1) = 8

```sol:=rsolve({u(1+n)-4*u(n), u(1) = 8}, u(n));

#or:
sol:=rsolve({eval(op([1, 2], rec), 0 = 1), op([1, 1], rec) = 0}, u(n))

[seq(sol, n = 1 .. 6)]; #Check:

#[8, 32, 128, 512, 2048, 8192]```

I don't know how to change: 2*4^n = 2*2^(2 n) = 2^(2n+1) in Maple!

`simplify(2*4^n, power);# Dosen't work!`

## ....

For first example:

```restart:
Digits := 30:
convert((k-2)*(k^2+5)*(k^3-k^2+7*k+8)/(6*k*(k^2-3*k+8)), parfrac, k);```

and factor expression k^2-3k+8  by:

```(1/6)*k^3+k-1/2-5/(3*k)+convert((-3*k-6)/identify(factor(denom(op(4, sol)), complex)), parfrac, k);
```

and last 2 expression:

```identify(factor(k^3+3*k^2+11*k-3, complex));
identify(factor(k^2+2*k+9, complex))```

## simplify(expr,symbolic)...

```f := proc (x, z) options operator, arrow; ln(x^z) end proc;
simplify(f(x, 2), symbolic);
collect(expand(simplify(f(x, 1/2+I*y), symbolic)), ln(x));```

Note: When the symbolic option is specified, any branch of a multi-valued function can be chosen during the simplification process. The result of such an operation is in general not valid over the whole complex plane and can lead to incorrect results if you assume the expressions represent analytical functions.

## One way.....

```eq16 := r(t) = d[vol]*V/(KUS*V^2+L*tau);
ex := InertForm:-MakeInert(convert(rhs(eq16), fullparfrac, KUS)):
eq17 := r(t) = op(1, ex)/expand(simplify(op(2, ex)));```

## Alternative....

```restart:
with(Physics[Vectors]):
Setup(mathematicalnotation = true):
eq := r(t) = 2*t^2*_i+16*_j+(10*t-12)*_k;
v := diff(rhs(eq), t);
V := eval(v, t = 10);
simplify(Norm(V));```

## Use Physics[Vectors] Subpackage....

```restart;
with(Physics[Vectors]);
Setup(mathematicalnotation = true);
eq := r(t) = 3*cos(5*t)*_i+sin(5*t)*_j+3*sin(5*t)*_k;
Norm(rhs(eq));#Calculate Norm
plots:-spacecurve([Component(rhs(eq), 1), Component(rhs(eq), 2), Component(rhs(eq), 3)], t = 0 .. 4*Pi, color = pink);```

Have fun!

## Adding 2 new diff equations and changing...

```restart;
q1 := 9045.084972*(diff(z[1](t), t\$2))+863728.7570*z[1](t) = -1963.525491562420*sin(20*t);
q2 := 3454.915028*(diff(z[2](t), t\$2))+2.261271243*10^6*z[2](t) = -286.4745084375789*sin(20*t); icy := seq([z[i](0) = 0, (D(z[i]))(0) = 0], i = 1 .. 2, 1);

so := dsolve({q1, q2, seq(icy[i][], i = 1 .. 2), A(t) = diff(z[1](t), t\$2), B(t) = diff(z[2](t), t\$2)}, numeric); so(1);

plots:-odeplot(so, [[t, A(t)], [t, B(t)]], t = 0 .. 1, color = [red, blue], legend = ["diff(z[1](t), t\$2)", "diff(z[2](t), t\$2)"], legendstyle = [font = [times, bold, 20]]);```

Have fun!

## Seq....

```w := proc (x, y) options operator, arrow; piecewise(y <= .5, -2*tanh(y-.25), .5 < y, 2*tanh(.75-y)) end proc;
Matrix([seq([seq(w(x, y), x = 0 .. 10)], y = 0 .. 10)]);

#or;

Matrix([seq([seq(w(x, y), y = 0 .. 10)], x = 0 .. 10)]);
```

## A numeric method....

You don't gives values of constans,so I assuming.

```restart;
Digits := 20;
Theta := (1/3)*Pi; Upsilon := 1/10;#assume!
eq := ((D@@2)(u))(r) = (-(D(u))(r)^2*u(r)+((Upsilon-1)*(1/2))*(1-u(r)^2-(D(u))(r)^2)*((D(u))(r)*cot(Theta)+2*u(r)))/((D(u))(r)^2-((Upsilon-1)*(1/2))*(1-u(r)^2-(D(u))(r)^2));
sol := dsolve({eq, u(13.75) = .7787, (D(u))(13.75) = .344037}, numeric, abserr = 1.*10^(-16));
plots:-odeplot(sol, [[r, u(r)], [r, (D(u))(r)]], r = 0 .. 13.75, legend = [typeset("Curve: ", u(r)), typeset("Curve: ", (D(u))(r))]);```

 First 13 14 15 16 17 18 19 Page 15 of 19
﻿