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Apparently inconsistent behaviour of the BesselJ() function.

Examples: BesselJ(-3, 0)  ... gives 0 (correct)

but BesselJ(-3.0, 0), BesselJ(-3, 0.0)  and BesselJ(-3, 0.0) all give Float(infinity) (wrong! - should be 0.0)

The problem seems to occur for all negative integer values of the first argument (the order) when the second argument is 0 or 0.0.

The following integral
f := u-> int(-1/(x*sqrt(-1+u^2*(x+1)^2*x^2)), x = (1/2)*(-u-sqrt(u^2-4*u))/u .. (1/2)*(-u+sqrt(u^2-4*u))/u);
arised in an applied research. I was asked about its properties:
plot on RealRange(4,infinity), limit(f(u),u=4,right), limit(f(u),u=infinity).
Unfortunately, I lost a file. As far as I remember it, I have had a problem with
the latter one only:

limit(f(u), u = infinity);

MultiSeries:-limit(f(u), u = infinity);

asympt(f(u), u, 2);

Error, (in asympt) unable to compute series

Hope my colleagues will make progress with it. The assumed value is Pi/2.

I'm trying to determine that f(x) = (a/2)*e^(-a|x|) is a pdf for which I have tried to calcuate the integral from -infinity to +infinity but I am no getting a result that converges(even the wolfram alpha widget said the integral doesn't converge). How do I correcly implement this?

hello everyone. I have an undergradute project i'm currently working on and I'm stuck where I have to use the Differential Transforms Method to solve a problem with boundary conditions at infinity


Digits := 5;

F[0] := 0; F[1] := 0; F[2] := (1/2)*A; T[0] := 1; T[1] := B; M := 2; S := 1;

for k from 0 to 10 do F[k+3] := (2*(sum((r+1)*F[r+1]*(k+1-r)*F[k+1-r], r = 0 .. k))-T[k]-3*(sum((k+1-r)*(k+2-r)*F[r]*F[k+2-r], r = 0 .. k))-M*(k+1)*F[k+1])*factorial(k)/factorial(k+3);

T[k+2] := (-3*(sum((k+1-r)*F[r]*T[k+1-r], r = 0 .. k))-S*T[k])*factorial(k)/factorial(k+2)

end do; f := 0; t := 0;

for k from 0 to 10 do

f := f+F[k]*x^k;

t := t+T[k]*x^k end do;


but the problem is that i cant seem to evaluate

or higer diagonal pade-approximant. any help will be greatly appreciated. thank you.

How to find asymptotic behaviour of a function.

For example at infinity

sinh(x) behaves as 1/2*exp(x)

1/sinh(x)  behaves as 2*exp(-x)

exp(-x)*(exp(-x)+1) behaves as exp(-x)

so that it works with a more complex expression.

I got a problem with solving a second order ODE. 

The ODE is :

-V(xi)+(1/2)*xi*(diff(V(xi), xi))+(1/4)*(diff(V(xi), xi, xi))=-(1/2)*k2*(diff(H(xi), xi))-k1*n*X/E+1+k2

where k1,k2,n,X,E  all are constant.

the condition is :

V(xi) tends to 2*xi^2 as xi tends to infinity.

I used 'dsolve' to solve the equation firstly, and got a solution with two constant C1 and C2, I want to use the condition to elimilate C2, so I used limit(sol,xi=infinity)=2*xi^2. But when I used the command 'limit', I can't get the answer.

Could any one help me? 

Many thanks!!!

Surely this is a bug.

> 0^0;
> sum( 0^m, m=0..infinity );

The limit of

G := N -> sum(1/N*numtheory[sigma](k)/k,k=1..N ) 

as N -> infinity should be Pi^2/6. However

eval( G(N), N=infinity );

returns 0.

Can anyone explain this puzzling behaviour?

Is it possible to numerically calculate  the integral

int((-12*y^2+1)*ln(abs(Zeta(x+I*y)))/(4*y^2+1)^3, [y = 0 .. infinity, x = 1/2 .. infinity])

in Maple?

The code

int((-12*y^2+1)*ln(abs(Zeta(x+I*y)))/(4*y^2+1)^3, [y = 0 .. infinity, x = 1/2 .. infinity],numeric,epsilon=0.1)

has been executed on my comp  without any output since this morning.





Maybe someone can give me a nice answer without Maple.

I am given a fourier series:
ln|cosx|=Co - sum( (-1)^k/k * cos2kx,k=1..infinity)
and am asked what this tells me about the chevychev series for ln(u).



Hi Everybody.


Doing some calculation in quantum mechanics, I stuble on that integral:

I see immediately that the answer is 1/2.  But Maple 18 cannot give an answer other than a limit that he cannot evaluate.  I even try assumption that p and hbar are realcons.  I get infinity.

Any idea?

Thank you in advance for your help.


Mario Lemelin
Maple 18 Ubuntu 13.10 - 64 bits
Maple 18 Win 7 - 64 bits messagerie : téléphone :  (819) 376-0987

The following limit does not return a value. Then the evalf gives a wrong answer.

The answer should be "undefined" or -infinity .. infinity.

limit(exp(n)/(-1)^n, n = infinity) assuming n::posint; evalf(%);

                       /exp(n)              \
                  limit|------, n = infinity|
                       |    n               |
                       \(-1)                /


The same happens if you delete the assumption.


A similar problem occurs with

limit(sin(Pi/2+2*Pi*n), n = infinity) assuming n::posint;
                            -1 .. 1
without the assumption this would be appropriate.

I've been given a question:

Let pn denote the nth prime number. Then p1 = 2, p2 = 3, p3 = 5, p4 = 7, p5 = 11, . . . .

It is known that the infinite sum 1/p1 + 1/p2 + 1/p3 + · · · + 1/pn + · · · = infinity.

Find the smallest positive integer N so that 1/p1 + 1/p2 +1/p3 + · · · + 1/pN−1 + 1/pN > e. [Hint : ithprime(n) generates the nth prime number.]

How do I start off?

Many thanks!

Would really appreciate the help in doing this question!

Pi= infinity Σ n=0 ((120n^2+An+B)/(16^n(512n^4+1024n^3+712n^2+194n+15))) for some positive integers A,B.

Hint: Set A=0 to find B, then find A.

I'm so clueless as to how to do this. THANK YOU!

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