## 1436 Reputation

7 years, 357 days

## One way is:...

```f := x -> convert(sin(theta), x);
map(f, [cos, tan, cot, csc, sec]);```

#[-cos(theta + Pi/2), 2*tan(theta/2)/(1 + tan(theta/2)^2), 2*cot(theta/2)/(cot(theta/2)^2 + 1), 1/csc(theta), -1/sec(theta + Pi/2)]

## Numerically....

```sol := dsolve({diff(f(x), x) = f(1/x), f(0) = 0}, numeric, delaymax = 1.0, delaypts = 2000);
plots:-odeplot(sol, [x, f(x)], x = -1 .. 1);```

Maybe you can try with another initial condition f(0)=1 ?

## Simpler....

```evalf[30](subs(z = 2 + 3*I, Zeta(-z) + 2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1)));

rtable([seq(evalf[2^m](subs(z = 2 + 3*I, Zeta(-z) + 2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1))), m = 1 .. 10)], subtype = Vector[column]);#Depending of the precise```

## Another sum...

I have another sum with Laguerre polynomials to  give exp(z).

`Sum((m + 1)!*z^k*LaguerreL(k, k + m, z)/(k + m + 1)!, k = 0 .. infinity) = exp(z)`

We can check:

```restart;
f := (z, m) -> sum((m + 1)!*z^k*LaguerreL(k, k + m, z)/(k + m + 1)!, k = 0 .. infinity)
interface(rtablesize = 100);
rtable([seq([m, evalf(f(1, m))], m = 0 .. 20)], subtype = Vector[column]);```

And for your example:

```Digits := 30;
f := (x, m) -> sum(m!*x^(m - n)*LaguerreL(m, n - m, x)^2/n!, n = 0 .. infinity);
interface(rtablesize = 100);
rtable([seq([m, evalf(f(1, m))], m = 0 .. 20)], subtype = Vector[column]);```

## Fracdiff equation in Maple....

From this book on page 152 I borrowed the code.

See attached files.

fracdiff.mw

fracdiff_for_alpha=1_test.mw

## Answer from MMA....

For first question:

`Int(exp(-(abs(x - mu)/sigma)^beta), x = -infinity .. s) = piecewise(Or(s = mu, s <= 0), sigma*GAMMA(1 + 1/beta), And(mu < s, 0 < s), ((-s + mu)*Ei(-(1 + beta)/beta, (s - mu)^beta*sigma^(-beta)) + 2*sigma*GAMMA(1/beta))/beta, sigma*GAMMA(1/beta, (-s + mu)^beta*sigma^(-beta))/beta)`

## Try...

Try this:

```eq := 2*exp(-2*t) + 4*t = 127;
solve([eq, 0 < t], {t});
limit(LambertW(-exp(-x))/2 + 127/4, x = infinity);
interface(rtablesize = 100);
rtable([evalf[100](seq(limit(LambertW(-exp(-x))/2 + 127/4, x = m), m = 2 .. 200, 5))], subtype = Vector[column]);```

## Gamma......

This integral, in most cases, cannot be expressed in terms of elementary functions,but we can expressed in terms of GAMMA function.

 (1)

 (2)

 (3)
 >
 (4)
 >
 (5)
 >
 (6)

 (7)

 Int(exp(beta2*_z1)*(1+N-_z1)^(-theta),_z1 = T .. t) = -1/beta2*exp(N*beta2+ beta2)*((1+N-t)^(-theta)*(beta2*(1+N-t))^theta*(theta*GAMMA(-theta)+GAMMA(1- theta,beta2*(1+N-t)))+(1+N-T)^(-theta)*(beta2*(1+N-T))^theta*(GAMMA(1-theta)- GAMMA(1-theta,beta2*(1+N-T))))

```
```

 >
 >
 >
 (1)
 >
 (2)
 >
 >
 (3)
 >
 >
 (4)
 >
 (5)
 >
 (6)
 >

## Try:...

Do you have any reason to think there is a closed form?

Most integrals don't have one.

Maybe the best you can do is numerical methods.

See atthached file.

ode.mw

See teory of First Order Differential Equations. Only one initial value problem can be not two.

## Try:...

See attached file:

integral.mw

## Maybe...My answers helps...

See attached file:

PDE_by_Elziki_Transform.mw

## Homework.....

For first question:

```f := x -> 36*x^6 + 2665*x^4 + 240*x - 675 + 4534*x^2 - 5836*x^3 - 516*x^5;

minimize(f(x), x = 0 .. 4, location);
#-675, {[{x = 0}, -675]}

evalf(maximize(f(x), x = 0 .. 4, location));
#703.9550742, {[{x = 3.800387934}, 703.9550742]}```

## Try...

Try:

```ode := diff(U(z), z \$ 4) + c^2*diff(U(z), z \$ 2) + k*c*diff(U(z), z \$ 2) - (3*U(z)^2 + a)*diff(U(z), z \$ 2) = 0;
Order := 5;dsolve(ode, U(z), type = 'series');

#U(z) = U(0) + D(U)(0)*z + 1/2*(D@@2)(U)(0)*z^2 + 1/6*(D@@3)(U)(0)*z^3 + (U(0)^2*(D@@2)(U)(0)/8 - c^2*(D@@2)(U)(0)/24 - k*c*(D@@2)(U)(0)/24 + (D@@2)(U)(0)*a/24)*z^4 + O(z^5)```

With initial conditions

```Order := 5;dsolve([ode, U(A) = A1, D(U)(A) = B1, (D@@2)(U)(A) = C1], U(z), type = 'series');

#U(z) = A1 + B1*(z - A) + 1/2*C1*(z - A)^2 + 1/6*(D@@3)(U)(A)*(z - A)^3 + (1/8*A1^2*C1 - 1/24*c^2*C1 - 1/24*k*c*C1 + 1/24*C1*a)*(z - A)^4 + O((z - A)^5)```

## Workaround....

As a workround using fourier transform:

(inttrans:-invfourier(int((inttrans:-fourier(sin(p*r), p, s) assuming (0 <= r))*sin(q*r)/(p*q), r = 0 .. infinity), s, p) assuming (q < p));

#-Pi*Dirac(p + q)/(2*p*q)

 1 2 3 4 5 6 7 Last Page 3 of 19
﻿