@Alejandro Jakubi 

I think that's strange because it works with an assigned n, say n=11,

infolevel[sum]:=3:
sum((-1)^(k+1)/k^11,k=1..infinity);
sum:   "process the input arguments"
sum:   "definite sum"
ClosedForm:   simple definite summations
ClosedForm:   infinite summation
Indefinite:   "process the input arguments"
ClosedForm:   "handle floats"
ClosedForm:   "indefinite summation"
ClosedForm:   "indefinite summation failed"
ClosedForm:   definite summation using hypergeometric functions
ClosedForm:   definite summation using hypergeometric functions
sum/hypergeom:   Function    -1/(x+1)^11*(-1)^(x+1)    satisfies the criteria

                            1023
                            ---- Zeta(11)
                            1024

If the definite summation using hypergeometric functions works for n=11 (and other n), it should work with an unassigned name n assuming n>1.

Another humorous thing is that the summation works after changing a sign,

sum((-1)^k/k^n,k=1..infinity) assuming n>1;

                          (1 - n)
                        (2        - 1) Zeta(n)

Even without assumptions,

sum((-1)^k/k^n,k=1..infinity);

                            polylog(n, -1)

Alec

What she wrote as an example, is slightly different,

f:=n->piecewise(n::even,f3(n/2)); 

The Array could be constructed without using seq, as in

Statistics:-ColumnGraph(Array(0..20,f));

Also, to answer to her first question - the gaps can be eliminated by adjusting width and distance options - something like the following in the original example,

Statistics:-ColumnGraph(Array(0..20,f3),width=0.99,distance=0.01);

Alec

What she wrote as an example, is slightly different,

f:=n->piecewise(n::even,f3(n/2)); 

The Array could be constructed without using seq, as in

Statistics:-ColumnGraph(Array(0..20,f));

Also, to answer to her first question - the gaps can be eliminated by adjusting width and distance options - something like the following in the original example,

Statistics:-ColumnGraph(Array(0..20,f3),width=0.99,distance=0.01);

Alec

I've already said about that in the previous reply - if the solution between 0 and 1 doesn't exist, f returns unevaluated, and in all those examples that you mentioned, it is easy to see from the plots given by g, that the solution doesn't exist - the lhs is negative on that interval. For example, take a look at

plots:-display(g(10, 1/3, 2, 1),plot(0,0..1));

If you want to find a solution outside that interval, the range 0..1 should be removed from the fsolve in f, and modified this way procedure, call it f1, would give a solution outside of that interval (if it exists),

f1(10, 1/3, 2, 1);

                             1.404823325

I don't think that there is a simple formula other than with the RootOf for the symbolic solution - try to solve the expression that f(10, 1/3, 2, 1) returns inside of the unevaluated fsolve,

solve(1-1/(423360/1968329+m)-1/(423360/1968329+4*m)-1/(423360/1968329+9*m)-1/(423360
/1968329+16*m)-1/(423360/1968329+25*m)-1/(423360/1968329+36*m)-1/(423360/1968329+49*m)-1/(423360
/1968329+64*m)-1/(423360/1968329+81*m)-1/(423360/1968329+100*m));

Maple returns it as a RootOf of a polynomial of 10th degree. The same is the general situation - the solution is a root of a polynomial of degree n, which can be explicitely written if you are interested in it, but that doesn't help with finding the root - it is generally the same equation as you had originally with fractions - it is the numerator of their sum written as one fraction.

Some asymptotics, certainly, could be found, say for n approaching infinity - perhaps even the entire series - but that's it.

Alec

I've already said about that in the previous reply - if the solution between 0 and 1 doesn't exist, f returns unevaluated, and in all those examples that you mentioned, it is easy to see from the plots given by g, that the solution doesn't exist - the lhs is negative on that interval. For example, take a look at

plots:-display(g(10, 1/3, 2, 1),plot(0,0..1));

If you want to find a solution outside that interval, the range 0..1 should be removed from the fsolve in f, and modified this way procedure, call it f1, would give a solution outside of that interval (if it exists),

f1(10, 1/3, 2, 1);

                             1.404823325

I don't think that there is a simple formula other than with the RootOf for the symbolic solution - try to solve the expression that f(10, 1/3, 2, 1) returns inside of the unevaluated fsolve,

solve(1-1/(423360/1968329+m)-1/(423360/1968329+4*m)-1/(423360/1968329+9*m)-1/(423360
/1968329+16*m)-1/(423360/1968329+25*m)-1/(423360/1968329+36*m)-1/(423360/1968329+49*m)-1/(423360
/1968329+64*m)-1/(423360/1968329+81*m)-1/(423360/1968329+100*m));

Maple returns it as a RootOf of a polynomial of 10th degree. The same is the general situation - the solution is a root of a polynomial of degree n, which can be explicitely written if you are interested in it, but that doesn't help with finding the root - it is generally the same equation as you had originally with fractions - it is the numerator of their sum written as one fraction.

Some asymptotics, certainly, could be found, say for n approaching infinity - perhaps even the entire series - but that's it.

Alec

@hermitian 

Yes, it looked as if you used a different power there.

By the way, a seemingly strange thing is that Maple is not able to evaluate the following sum symbolically,

sum((-1)^(k+1)/k^n,k=1..infinity) assuming n>1;

                         infinity
                          -----       (k + 1)
                           \      (-1)
                            )     -----------
                           /           n
                          -----       k
                          k = 1

It can be obtained from the sum with 1 instead of (-1)^(k+1) equal to Zeta(n), by a very simple transformation - subtracting twice its even terms, the sum of which is 1/2^n*Zeta(n), which gives the answer (1-1/2^(n-1))*Zeta(n).

Alec

@hermitian 

Yes, it looked as if you used a different power there.

By the way, a seemingly strange thing is that Maple is not able to evaluate the following sum symbolically,

sum((-1)^(k+1)/k^n,k=1..infinity) assuming n>1;

                         infinity
                          -----       (k + 1)
                           \      (-1)
                            )     -----------
                           /           n
                          -----       k
                          k = 1

It can be obtained from the sum with 1 instead of (-1)^(k+1) equal to Zeta(n), by a very simple transformation - subtracting twice its even terms, the sum of which is 1/2^n*Zeta(n), which gives the answer (1-1/2^(n-1))*Zeta(n).

Alec

Yes, there is also Worksheet package which can be used together with XMLTools, but it's extremely inconvenient to do that.

Alec

Yes, there is also Worksheet package which can be used together with XMLTools, but it's extremely inconvenient to do that.

Alec

@saeidmo 

Procedure f that I posted does exactly that - for given n, p, alpha, and c outputs m satisfying that equation.

It can be checked directly,

f(10,1/2,3,1);

                             0.8637165918
 
eval(sum(m/(k*p*c+m*i^alpha), i = 1 .. n)-m,
[n=10,p=1/2,alpha=3,c=1,k=1/sum(1/i^3,i=1..10),m=%]);

                                  0.

f(1000,1/4,7,1);

                             0.7631056622

eval(sum(m/(k*p*c+m*i^alpha), i = 1 .. n)-m,
[n=1000,p=1/4,alpha=7,c=1,k=1/sum(1/i^7,i=1..1000),m=%]);

                              0. + 0. I

If you meant that you wanted a symbolic solution and not a numerical one - that could be done in Maple, too, but the result would be in terms of RootOf of, generally speaking, the same equation, and numerical evaluation of it would be done the same way as in f (or, maybe, worse, because I specified the initial point m=1 in fsolve which is significantly better than if it started with initial point m=0, which may be the default otherwise.)

For some parameter values the solution on the interval from 0 to 1 doesn't exist (except the obvious solution m=0, which always exists) - in this case f will return unevaluated fsolve with sum in it replaced through its expression in terms of Psi.

Alec

@saeidmo 

Procedure f that I posted does exactly that - for given n, p, alpha, and c outputs m satisfying that equation.

It can be checked directly,

f(10,1/2,3,1);

                             0.8637165918
 
eval(sum(m/(k*p*c+m*i^alpha), i = 1 .. n)-m,
[n=10,p=1/2,alpha=3,c=1,k=1/sum(1/i^3,i=1..10),m=%]);

                                  0.

f(1000,1/4,7,1);

                             0.7631056622

eval(sum(m/(k*p*c+m*i^alpha), i = 1 .. n)-m,
[n=1000,p=1/4,alpha=7,c=1,k=1/sum(1/i^7,i=1..1000),m=%]);

                              0. + 0. I

If you meant that you wanted a symbolic solution and not a numerical one - that could be done in Maple, too, but the result would be in terms of RootOf of, generally speaking, the same equation, and numerical evaluation of it would be done the same way as in f (or, maybe, worse, because I specified the initial point m=1 in fsolve which is significantly better than if it started with initial point m=0, which may be the default otherwise.)

For some parameter values the solution on the interval from 0 to 1 doesn't exist (except the obvious solution m=0, which always exists) - in this case f will return unevaluated fsolve with sum in it replaced through its expression in terms of Psi.

Alec

The procedure f that I posted, does exactly that - for given n, p, alpha, and c outputs m satisfying that equation.

Alec

@Joe Riel 

That would be great!

Decimals, probably, would be easy - just add a dot, say "three dot forty five", or, maybe, "three and forty five hundredth"?

A TeX-like version "one over two", probably would be the easiest for fractions, but "one half", certainly, sounds better. Without "over", there might be a problem with cases like "one thousand two hundred seventeenth" - it could be either 1200/17, or 1000/217.

There are a few screen readers for blind people - if some of them are open source - perhaps, their code might be useful. I wonder if they can handle more complicated Maple formulas - with square roots etc. (like, say, a square root of x squared plus y squared - there is some ambiguity here though - perhaps, the "end of the square root" should be added, even if we don't use it normally. I'm afraid, my English is not good enough to deal with some cases :)

Alec

@Christopher2222 

Perhaps, ?PDEtools[declare] could be used with standard typesetting.

Or you could just stay in extended.

Alec

@Christopher2222 

Perhaps, ?PDEtools[declare] could be used with standard typesetting.

Or you could just stay in extended.

Alec