## 65 Reputation

6 years, 144 days

## The issue with mtaylor for a function of...

Maple 17

Hello.

I want to check my analytical calculations and to compare my results with the series expansion for x=0 and y=0 of the function

R0 := -(1-tanh((1/2)*l*sqrt(k*y^2+x^2)*d)^2)*x/(k*y^2+x^2)+2*tanh((1/2)*l*sqrt(k*y^2+x^2)*d)*x/(l*(k*y^2+x^2)^(3/2)*d)

by means of the command

mtaylor(R0, [x, y], 5);

However, I have a message from Maple 17

Error, (in mtaylor) does not have a taylor expansion, try series()

I read here a little bit about some bugs of mtaylor function, but how to resolve this issue?

## Discrepancy between numerical and analyt...

Maple

As usual, I have a tricky question. There is an integral that Maple can take numerically

R0 := 1/(a-sqrt(b+c*cos(x)));

Now let's put the coefficients, e.g.

a := 0.9; b := 4.5; c :=0.1

and take the integral from 0 to 2*Pi

R1 := evalf(int(R0, x = 0 .. 2*Pi));

Also, there is an exact analytical result that Maple gives (I give it after simplifying it to avoid division by zero for the limit x=0 and x=2*Pi)

R2:=-4*((a^2-b+c)*EllipticK(sqrt(-2*c/(b-c)))-a^2*EllipticPi(2*c/(a^2-b+c), sqrt(-2*c/(b-c))))/((a^2-b+c)*sqrt(b-c));

As it turns out, the results are completely different. In the first case -5.145818656, while for the second case -3.612771378+0.I

Moreover, If we change the coefficients to a := 0.9; b := 4.5; c := -4 then I obtain Float(undefined)+3.662506136*I and -2.362349457+3.662506117*I , respectively.

My question: how to avoid this descepancy?

## How to solve an integro-functional equat...

Maple

Hello everybody.

My goal is to solve the following integro-functional equation:

```(int(p^2/(f(p)-f(p+q)+omega), p = a .. b))/omega^2 = ln(omega^2+q^2)
```

(int(p^2/(f(p)-f(p+q)+omega), p = a .. b))/omega^2 = ln(omega^2+q^2)

where where is the unknown function and a, b are some numerical values as well as ω and q are real positive variables.

I would be grateful for any ideas.

## Divergence or convergence?...

Maple

Hello everybody.

I have a short question. I have an expression related to the calculation of the Gaussian curvature

`-24.*y^6*(y-1.)^7*(y+1.)^7*(4.+6.*y^8-22.*y^6+30.*y^4-18.*y^2)/(4.*y^18-28.*y^16+84.*y^14+4.-136.*y^12+116.*y^10-24.*y^8-52.*y^6+56.*y^4-24.*y^2)^2`

that is diverged for y=1 as (y-1)^(-2).

At the same time substitution of something like y:=1.00001 gives the value almost zero.

What is the trick?

## Error "Found wrong extra argument(s)" du...

Maple

Hello.

I would like to solve numerically highly nonlinear and cumbersome the second order differential equation.

Applying the numerical procedure I got an error "Error, (in dsolve) found wrong extra argument(s): range = 0 .. 4*Pi, type = numerical". The similar problem has been described earlier here however I can't realize my problem.

Below is my code

```restart;

A1 := 8*Pi^3*R^2*n(x)^4*m+(2*Pi*sin((1/2)*x)*m*omega0*p+Pi*sin((1/2)*x)*m*omega0+3*Pi^2*(diff(n(x), x, x)))*n(x)^3+(-2*sin((1/2)*x)^2*m^2*omega0^2*p^2+2*cos((1/2)*x)^2*m^2*omega0^2*p^2-2*sin((1/2)*x)^2*m^2*omega0^2*p+2*cos((1/2)*x)^2*m^2*omega0^2*p)*n(x)^2+(-4*(diff(n(x), x, x))*sin((1/2)*x)^2*m^2*omega0^2*p^2-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p^2-4*(diff(n(x), x, x))*sin((1/2)*x)^2*m^2*omega0^2*p-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p)*n(x)+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p^2+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p;

R := 1; m := 1; p := 10; omega0 := 1000;

A2 := A1;

with(plots):
A3 := dsolve({A2, n(0) = n(4*Pi), (D(n))(0) = (D(n))(4*Pi)}, type = numerical, range = 0 .. 4*Pi):

odeplot(A3);

```

I appreciate for any help and suggestion.

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