@Markiyan Hirnyk

Thank you for the response, but I do not know what is  Lucida Sans Unicode font and how to use it to solve my problem.

For example, the symbol for white Queen  ♕  can be coded in Unicode as  U+2655 , but how to do this in Maple I do not know. 

@Carl Love 

Indeed the data are nondecreasing, but the growth rate decreases all the time . The last three values of y-coordinate is not changed, i.e. the growth rate is equal to 0. So it is quite natural that in the future growth rate becomes negative, ie, function decreases. The plot shows that the second model  much better approximates the data in the middle and on the right end of the range. The first value returned by DataFit for the second model is almost 5 times less. 

with(DirectSearch): with(plots):

xy := Matrix([[0.2e-1, .158], [0.2e-1, .159], [0.3e-1, .161], [0.3e-1, .164], [0.3e-1, .166], [0.4e-1, .169], [0.6e-1, .173], [0.8e-1, .178], [.1, .185], [.11, .187], [.14, .193], [.19, .2], [.28, .21], [.38, .223], [.44, .233], [.58, .244], [.82, .256], [1.4, .278], [1.71, .281], [1.78, .282], [1.78, .282], [1.81, .282]]):

X := xy[() .. (), 1]:

Y := xy[() .. (), 2]:

N:=nops(X):

f:=a+b*x^c:

g := a+b*x^c+d*x^e:

sol1 := DataFit(f, X, Y, x, evaluationlimit=20000):

sol2 := DataFit(g, X, Y, x, evaluationlimit=20000):

f1 := eval(f, sol1[2]);

g1 := eval(g, sol2[2]);

Er1:=sol1[1];  # error for first model

Er2:=sol2[1];  # error for second model

p1, p2, p3 := pointplot(X, Y), plot(f1, x = 0 .. 2), plot(g1, x = 0 .. 2, color=blue):

display(p1, p2, p3);  # red - first model, blue - second model

@Carl Love

This error occurs in Maple 12 Classic Worksheet. 

@Carl Love

This error occurs in Maple 12 Classic Worksheet. 

@Carl Love 

ISC:=proc(T::list, Eq::`=`)

local f:= unapply((lhs-rhs)(Eq), indets(Eq,name)[]);

     `or`(seq(f(i[]) >= 0, i= T)) and `or`(seq(f(i[]) <= 0, i= T))

end proc:

ISC([[0,0],[10,10],[10,0]], x+y=-1);

Error, (in ISC) invalid input: or expects 2 arguments, but received 3

@Carl Love 

ISC:=proc(T::list, Eq::`=`)

local f:= unapply((lhs-rhs)(Eq), indets(Eq,name)[]);

     `or`(seq(f(i[]) >= 0, i= T)) and `or`(seq(f(i[]) <= 0, i= T))

end proc:

ISC([[0,0],[10,10],[10,0]], x+y=-1);

Error, (in ISC) invalid input: or expects 2 arguments, but received 3

M. Hirnyk, your code does not work in Maple 16:

trigsubs(sin(a+b)^2 - sin(a-b)^2)[];

   Error, (in trigsubs) expecting the sum or difference of two functions but got sin(a+b)^2-sin(a-b)^2

M. Hirnyk, your code does not work in Maple 16:

trigsubs(sin(a+b)^2 - sin(a-b)^2)[];

   Error, (in trigsubs) expecting the sum or difference of two functions but got sin(a+b)^2-sin(a-b)^2

If formally integrated in each interval, we get

int(abs(x-2), x)=-x^2/2+2*x+C1, x<=2

int(abs(x-2), x)=x^2/2-2*x+C2, x>=2

But the function  int(abs(x-2), x)   is continuous throughout the real axis and, in particular, at the point  x=2  because the function  abs(x-2)  is continuous. Therefore, at the point  x=2  we obtain  2+C1=-2+C2  or  C2=C1+4

@Carl Love 

Thanks for the explanation!

@Carl Love 

Thanks for the explanation!

It's interesting, why my simple version takes less time than your multithreaded version?

t:=time(): PP({i $ i=1..20}): time()-t;

t:=time(): PowerSet({i $ i=1..20}): time()-t;

                        10.640

                        14.766

It's interesting, why my simple version takes less time than your multithreaded version?

t:=time(): PP({i $ i=1..20}): time()-t;

t:=time(): PowerSet({i $ i=1..20}): time()-t;

                        10.640

                        14.766

@Carl Love

Regarding this example, you are right. But brute force method is more universal. Try to solve by  fsolve  the same question, for example,  for a series  

sum(ln(n)/n^2, n = 1 .. infinity);

@Carl Love

Regarding this example, you are right. But brute force method is more universal. Try to solve by  fsolve  the same question, for example,  for a series  

sum(ln(n)/n^2, n = 1 .. infinity);