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Riemann...

vv 13785
May 17 2025
3 3

Maple cannot compute the limit, but your expression 
Sum(Sum(j/(j^2 + k^2), j = 1..n), k = 1..n) / n
is a Riemann sum of the double integral

int(y/(x^2+y^2), [x=0..1,y=0..1]);

         

and this is the result of the limit (not difficult to justify).

Solutions...

vv 13785
May 15 2025
2 0
 

It seems that Maple does not like your problems. I do :-) 

 

Problem 1.

> 

restart
> 

sum(x^(2^k)/product(x^(2^m) + 1, m = 0 .. k), k = 0 .. infinity) assuming x>1;

sum(x^(2^k)/(product(x^(2^m)+1, m = 0 .. k)), k = 0 .. infinity)

 (1)

So, this does not work. Amplifying the fraction by  x-1

Workaround:   Amplifying the fraction by  x-1 ==>

> 

sum(x^(2^k)/product(x^(2^m) + 1, m = 0 .. k), k = 0 .. infinity) =
sum(x^(2^k)*(x - 1)/(x^(2^(k + 1)) - 1), k = 0 .. infinity);

sum(x^(2^k)/(product(x^(2^m)+1, m = 0 .. k)), k = 0 .. infinity) = sum(x^(2^k)*(x-1)/(x^(2^(k+1))-1), k = 0 .. infinity)

 (2)

The main ingredient is the identity:

> 

sum(x^(2^k)*(x - 1)/(x^(2^(k + 1)) - 1), k = 0 .. n) = (x^(2^(n + 1)) - x)/(x^(2^(n + 1)) - 1);

sum(x^(2^k)*(x-1)/(x^(2^(k+1))-1), k = 0 .. n) = (x^(2^(n+1))-x)/(x^(2^(n+1))-1)

 (3)

This can ve checked by induction. Finally:

 

> 

map(limit, %, n=infinity) assuming x>1;

sum(x^(2^k)*(x-1)/(x^(2^(k+1))-1), k = 0 .. infinity) = 1

 (4)

Problem 2.

 

> 

S:=Sum(Sum(1/i,i=1..n)/n/(n+1), n=1..infinity);

Sum((Sum(1/i, i = 1 .. n))/(n*(n+1)), n = 1 .. infinity)

 (5)
> 

S = value(S);

Sum((Sum(1/i, i = 1 .. n))/(n*(n+1)), n = 1 .. infinity) = sum((Psi(n+1)+gamma)/(n*(n+1)), n = 1 .. infinity)

 (6)

So, Maple practically also fails.

 

Here it is enough to change the summation order (legit, because the summands are positive).

 

> 

Sum(Sum(1/i/n/(n+1), n=i..infinity), i=1..infinity);

Sum(Sum(1/(i*n*(n+1)), n = i .. infinity), i = 1 .. infinity)

 (7)
> 

S = value(%);

Warning, unable to determine if the summand is singular in the interval of summation (1 <= i or not); try to use assumptions or option 'parametric'

  

Sum((Sum(1/i, i = 1 .. n))/(n*(n+1)), n = 1 .. infinity) = (1/6)*Pi^2

 (8)

It is interesting that the warning disappears if we execute the last command once more!NULL


Download test12-vv.mw

No algorithms...

vv 13785
May 04 2025
1 3 
B:=(4*exp(x/2) - 3*x - 2*exp(x/3)+1)^11 + (4*exp(x/2) + x - 2*exp(x/3)+1)^11 - 13:
A:=combine(expand(B)):
A1:=simplify(A, size): B1:=simplify(B, size):
length~([A, B, A1, B1]);
MmaTranslator:-Mma:-LeafCount~([A, B, A1, B1]);                   

                    [6209, 120, 3487, 120]

                      [1539, 38, 976, 38]

It will be very difficult (no algorithms available) to simplify A to B.  For the other software too!

x=2*t...

vv 13785
May 03 2025
4 1 
eval(A-B, x=2*t):
simplify(factor(expand(%))) assuming real;

                     0

workaround...

vv 13785
April 26 2025
2 0

Maple 2024 does not solve the problem.
Here is a workaround:

convert(series((f:=(lhs-rhs)(eq)),t,16),polynom):
solve([coeffs(%,t)]);
eval(f,%); #check

        {A[1]=0, A[2]=0, A[3]=-7/25, A[4]=1/25, A[5]=-2/5, A[6]=1/5, A[7]=3/16, A[8]=0, A[9]=0, A[10]=3/8}
        0

simplify radicals...

vv 13785
April 23 2025
3 2 
f := n -> (ln(x)^n)^(1/n);

simplify(f(2))  returns  csgn(ln(x))*ln(x)
just because the "simple" function  csgn exists.  csgn(z) equals sqrt(z^2)/z for z<>0.

simplify(f(3))  remains unsimplified because a similar "sign" function does not exist for z^(1/3);

For n = 3 or n=5,  simplify((log(x)^n)^(1/n)) assuming x::positive, x<1   give both  - ln(x) * (-1)^(1/n)
but when n=3,   (-1)^(1/3)   is further simplified to 1/2+I*sqrt(3)/2; 
When n=5, it would be too complicated:  (-1)^(1/5) = sqrt(5)/4 + 1/4 + sqrt(2)*sqrt(5 - sqrt(5))*I/4.

Directly...

vv 13785
April 13 2025
2 0

odetest cannot work with such generalized series. But it is easy to test the solution directly: 

fsol:=series(rhs(maple_sol), x=0):
# series(eval(lhs(ode), y(x)=fsol), x);  # O(x^5)
map(series, map(eval, ode, y(x)=fsol), x)

          O(x^5) = 0

Numerical solution...

vv 13785
April 06 2025
3 3
> 

restart;
> 

X,Y := t->5*cos(t), t->(5+sin(6*t))*sin(t):
> 

with(plots):
> 

P0:=plot([X,Y, 0..2*Pi], color=black):
> 

T:=D(Y)/D(X): # the sloap of the tangent
> 

K:=unapply(VectorCalculus:-Curvature(<X(t),Y(t)>),t):
> 

tt[1],tt[2] := eval( [a,b], fsolve({T(a)-T(b), K(a)-K(b)}, {a, b}, {a=0..1, b=3..5}) )[];

.5465721754, 4.440166423

 (1)
> 

tt[3],tt[4] := eval( [a,b], fsolve({T(a)-T(b), K(a)-K(b)}, {a, b}, {a=2..2.5, b=3..6}) )[];

2.168601832, 3.704639334

 (2)
> 

tt[5],tt[6] := eval( [a,b], fsolve({T(a)-T(b), K(a)-K(b)}, {a, b}, {a=2.5..3, b=3..6}) )[];

2.845163972, 5.737842021

 (3)
> 

col:=[red$2,blue$2,green$2]:
> 

curvatures=seq(K(tt[i]),i=1..6)

curvatures = (1.391345054, 1.391345047, 1.111111250, 1.111111261, 0.3791851261e-1, 0.3791851136e-1)

 (4)
> 

display(P0,
        seq( pointplot([[X(tt[i]),Y(tt[i])]], color=col[i]), i=1..6),
        seq( plot(T(tt[i])*(x-X(tt[i])) + Y(tt[i]), x=-40..40, color=col[i]), i=1..6),
        symbolsize=24, symbol=solidcircle, scaling=unconstrained, view=[-7..7,-7..7],gridlines=false
);

 
> 

 


 

Download equal-sloap-curvature-vv.mw

Solution...

vv 13785
April 05 2025
4 3

The correct syntax for rsolve is:

> 

sol := rsolve({f(n)=sin(f(n-1)), f(0)=a}, f(n));

(sin@@n)(a)

 (1)

Unfortunately Maple cannot compute the following "classical" limit directly:

 

> 

limit(sol*sqrt(n/3), n=infinity) assuming a>0,a<Pi;

limit((1/3)*(sin@@n)(a)*3^(1/2)*n^(1/2), n = infinity)

 (2)

However, it can more than this: it is able to compute the asymptotic expansion!

 

> 

asympt('rsolve'({f(0) = a, f(n) = sin(f(n - 1))}, f(n)), n);

3^(1/2)*(1/n)^(1/2)+(_C-(3/10)*3^(1/2)*ln(n))*(1/n)^(3/2)+O((1/n)^(5/2))

 (3)

So, the limit is 1.

Note however that this is a generic result; for its validity we must add a condition e.g. 0<a, a<Pi.

Simple solution...

vv 13785
April 03 2025
0 1

Using the identity theorem, a very simple solution is possible:

u := sin(2*x)/(cosh(2*y) - cos(2*x)):
f:=eval(u, [y=0,x=z]) + I*C : # C = real constant
simplify(convert(f, cot));

        C*I + cot(z)

Answer...

vv 13785
March 09 2025
5 3
The idea is to expand in power series and then change the order of summation. After that, the limit will be simple.
> 

restart;
> 

f := k*t^k/(1-t^k); # without (1-t)^2

k*t^k/(1-t^k)

 (1)
> 

f := convert(eval(f, t^k=u), FPS, u) assuming k::posint

Sum(k*u^(n+1), n = 0 .. infinity)

 (2)
> 

f := (eval(%, u=t^k)) assuming k>0,t>0

Sum(k*(t^k)^(n+1), n = 0 .. infinity)

 (3)
> 

term := op(1,%)

k*(t^k)^(n+1)

 (4)
> 

F := Sum(sum((1-t)^2*term, k=1..infinity),n=0..infinity) assuming t>0,t<1;

Sum((t-1)^2*t^(n+1)/(t^(n+1)-1)^2, n = 0 .. infinity)

 (5)
> 

limit(F, t=1,left)

Sum(1/(n+1)^2, n = 0 .. infinity)

 (6)
> 

ans = value(%);

ans = (1/6)*Pi^2

 (7)
 

 

Download lim_ser-vv.mw

identify...

vv 13785
March 06 2025
2 1

Maple can guess the result:

s:=sum( arctan(2/n^2), n=1 .. infinity);
ans:=identify(evalf(s));
evalf[200](s - ans);  # check 200 digits, --> 0

                  

PS. Why don't you express the series in Maple, i.e.  sum( arctan(2/n^2), n=1 .. infinity) ?

PS2. Of course this is not a proof.  It is easy to obtain one in Maple, but only if the user knows it mathematically.
I omit it because you probably know it.

Stoltz...

vv 13785
March 04 2025
1 0

Maple cannot compute it, but it's easy with Stoltz-Cesaro theorem:

a:=n -> 2^n/n:
limit(a(n) / (a(n)-a(n-1)), n=infinity);

                    2

By hand with Maple...

vv 13785
February 26 2025
3 1
> 

J:=Int(ln(x)*ln(1 - x), x = 0 .. 1);

Int(ln(x)*ln(1-x), x = 0 .. 1)

 (1)
> 

F2:=convert(ln(1-x), FPS);

Sum(-x^(n+1)/(n+1), n = 0 .. infinity)

 (2)
> 

J1:=Int(ln(x)*op(1,F2), x=0..1);

Int(-ln(x)*x^(n+1)/(n+1), x = 0 .. 1)

 (3)
> 

value(%) assuming n>=0; # by parts

1/((n+1)*(n+2)^2)

 (4)
> 

V1:=convert(%, parfrac);

-1/(n+2)+1/(n+1)-1/(n+2)^2

 (5)
> 

J = sum(V1, n=0..infinity);  # Easy to justify by dominated convergence

Int(ln(x)*ln(1-x), x = 0 .. 1) = 2-(1/6)*Pi^2

 (6)


Download INT1-vv.mw

worksheetdir...

vv 13785
February 23 2025
2 1

Try  (*.mw, not *.mpl):

interface(worksheetdir);

 

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