The only problem with it is that it was published in 1994 and doesn't include recent developments.

Alec

The only problem with it is that it was published in 1994 and doesn't include recent developments.

Alec

There is no such a site.

Wikipedia, in general, is not very good as a reference in math. In this particular case, however, it is pretty good.

You don't have to give a reference to wikipedia (other than saying that you used it to find other links) - but you can use the links given there as a reference.

Alec

There is no such a site.

Wikipedia, in general, is not very good as a reference in math. In this particular case, however, it is pretty good.

You don't have to give a reference to wikipedia (other than saying that you used it to find other links) - but you can use the links given there as a reference.

Alec

Can you read? Or click on the links provided?

Alec

Can you read? Or click on the links provided?

Alec

See acer's post above. It has a procedure.

What you did, can be written as a procedure as follows,

pade:=proc(f,x,a,m,n)
local L,M,S,p,q,k;
Digits:=Digits+4;
L := add(p[k]*x^k, k=0..m);
M := 1 + add(q[k]*x^k, k=1..n);
S:=fsolve({seq(coeftayl(M*f-L,x=a,k),k=0..m+n)});
evalf(eval(L/M,S),Digits-4)
end;

For example,

pade(exp(x),x,0,4,2);
                                       2                  3
  (1. + 0.6666666667 x + 0.2000000000 x  + 0.03333333333 x

                           4    /
         + 0.002777777778 x )  /  (
                              /
                                             2
        1. - 0.3333333333 x + 0.03333333333 x )

pade(sin(x),x,Pi,2,2);

                     1.187777303 - 0.3780812583 x
                --------------------------------------
                                                     2
                1. - 0.3959257678 x + 0.06301354304 x

Alec

PS Note, however, that this procedure is not as good as numapprox:-pade, because the system of equations may have no solutions, as well as infinite number of solutions. For example,

pade(1+x^5,x,0,4,3);
Error, (in fsolve) invalid input: 
select expects 2 or more arguments, but received 1

pade(1+x,x,0,3,2);
                                               2         3
           1. + (q[1] + 1.) x + (q[2] + q[1]) x  + q[2] x
           -----------------------------------------------
                                            2
                        1. + q[1] x + q[2] x

The case of infinite number of solutions can be fixed, say, by substituting 0s to free variables, but what to do if there are no solutions? -Alec

See acer's post above. It has a procedure.

What you did, can be written as a procedure as follows,

pade:=proc(f,x,a,m,n)
local L,M,S,p,q,k;
Digits:=Digits+4;
L := add(p[k]*x^k, k=0..m);
M := 1 + add(q[k]*x^k, k=1..n);
S:=fsolve({seq(coeftayl(M*f-L,x=a,k),k=0..m+n)});
evalf(eval(L/M,S),Digits-4)
end;

For example,

pade(exp(x),x,0,4,2);
                                       2                  3
  (1. + 0.6666666667 x + 0.2000000000 x  + 0.03333333333 x

                           4    /
         + 0.002777777778 x )  /  (
                              /
                                             2
        1. - 0.3333333333 x + 0.03333333333 x )

pade(sin(x),x,Pi,2,2);

                     1.187777303 - 0.3780812583 x
                --------------------------------------
                                                     2
                1. - 0.3959257678 x + 0.06301354304 x

Alec

PS Note, however, that this procedure is not as good as numapprox:-pade, because the system of equations may have no solutions, as well as infinite number of solutions. For example,

pade(1+x^5,x,0,4,3);
Error, (in fsolve) invalid input: 
select expects 2 or more arguments, but received 1

pade(1+x,x,0,3,2);
                                               2         3
           1. + (q[1] + 1.) x + (q[2] + q[1]) x  + q[2] x
           -----------------------------------------------
                                            2
                        1. + q[1] x + q[2] x

The case of infinite number of solutions can be fixed, say, by substituting 0s to free variables, but what to do if there are no solutions? -Alec

Also, with either assuming, or assigning k, the answer given by Maple in the second integral doesn't depend on the power of y in i2 (which is obviously wrong, too) , say

assume(k>0);
f:= cos(k * y * cos(theta))/Pi :
i1 := int(f , theta = 0..Pi) :
i2 := int(y^10 * exp(-y) * i1, y = 0..infinity) ;

                                     1
                          i2 := ------------
                                       2 1/2
                                (1 + k~ )

The power of the BesselJ also doesn't seem to matter,

int(y^5*exp(-y)*BesselJ(0,k*y)^7,y=0..infinity);

                                  1
                             ------------
                                    2 1/2
                             (1 + k~ )

Alec

It seems as if GNFS is the latest one.

The second latest which is widely used is debatable - depending on the point of view what "latest" means. It is either quadratic sieve, or ECM.

Alec

It seems as if GNFS is the latest one.

The second latest which is widely used is debatable - depending on the point of view what "latest" means. It is either quadratic sieve, or ECM.

Alec

Actually, Maple 12.02 gives the answer which is wrong for not positive k, even in the original integral,

f:= cos(k * y * cos(theta))/Pi :
i1 := int(f , theta = 0..Pi) :
i2 := int(y * exp(-y) * i1, y = 0..infinity) ;

                                     1
                       i2 := ------------------
                              3 /     1  \(3/2)
                             k  |1 + ----|
                                |      2 |
                                \     k  /

It is not defined for k=0 and is negative for negative k while it should be positive.

Alec

It can be patched as

f:=sprintf("%a",eval(`convert/Sum/from`[`GAMMA/IC`])):

`convert/Sum/from`[`GAMMA/IC`]:=parse(cat(f[1..504],
"-gamma-ln(z)+Sum((-1)^k/(k+1)!*z^(k+1)/(k+1),k = 0 .. infinity)",
f[546..-1])): 

convert(Ei(1,z),Sum);

                            /infinity                     \
                            | -----        _k1  (1 + _k1) |
                            |  \       (-1)    z          |
           -gamma - ln(z) + |   )     --------------------|
                            |  /      (1 + _k1)! (1 + _k1)|
                            | -----                       |
                            \_k1 = 0                      /

Alec

The source of the bug is in `convert/Sum/from`[`GAMMA/IC`]:

elif `tools/type`(a = 0) then 
return `convert/Sum/from`['GAMMA'](('GAMMA')(z))

That converts Γ(0,z) to Γ(z). In particular,

`convert/Sum/from`[`GAMMA/IC`]('GAMMA'(0,z));

                               GAMMA(z)

convert(GAMMA(0,z),Sum);

                               GAMMA(z)

Alec

Is MySpace page also planned?

Barack Obama has a page there, by the way.

Alec