## 12915 Reputation

8 years, 355 days

## 3 half-lines...

The system is not very simple, but I agree that SolveTools:-SemiAlgebraic is very slow in general, and I had not the patience for a result.

Anyway, the result (for eq1) seems to be (and is confirmed by MMA):

a := RootOf(_Z^3 - 4*_Z^2 - 27, 5.054 .. 5.063):
sol := {z=0, y = x, a < x}, {y=0, z = x, a < x}, {x=0, y = z, a < z};

plots:-spacecurve([[t,t,0], [t,0,t], [0,t,t]], t=a..3*a, color=[red,blue,green], thickness=5);

Can the equation (x-a)^4 + (x-b)^4 = c have four different integer solutions,where  a, b, c  are integers?

The answer is NO.  Actually not even four real solutions!

Suppose the polynomial  f := (x-a)^4 + (x-b)^4 - c  has 4 real solutions. After a shift, we may suppose that one of the solutions is 0, so c = a^4 + b^4,  a, b nonzero.

 > restart;
 > f := (x-a)^4 + (x-b)^4 - a^4 - b^4;
 (1)
 > factor(f);
 (2)

 > f1:=op(-1,%);
 (3)

f1  must have two real solutions

 > d1:=discrim(f1,x); # should be >0. But
 (4)
 > d2:=discrim(%,a);  # should be >= 0 (otherwise d2<0 implies d1<0).  ==>
 (5)
 (6)
 > solve(f1,x);  # Actually:
 (7)

## ALG...

The procedure ALG works only for finite sets! Actually, the support for infinite sets in Maple is very limited, see ?SetOf.

[0,2] is a list, not an interval. [0,1) is a nonsense in Maple (for 1D input).

Note also that ALG computes the algebra generated by C, not the sigma-algebra; of course, X, C must be finite, so this is the same thing.

## Pure maths...

You cannot solve such theoretical problems using Maple. You cannot even formulate them within Maple, because a CAS does other things!

But your problem is very simple if you know basic measure theory:

If A is a Borel subset in R^n and  g : A --> R  and  h : R^n \ A  --> R are continuous
then the function f : R^n --> R,  f(x) = g(x), if x in A, and f(x) = h(x), if x in R^n \ A
is a Borel function.

In your example, A = {(0,0)}; in your previous version A = {0} x R.

## utf-8...

Use:

LEN:=proc(s::string) Python:-EvalFunction("len",s) end proc:

It works for utf-8 characters, used by Maple, see https://mapleprimes.com/posts/214347-Multibyte-Characters

For example

LEN("România"); #  7

Note that the solution proposed by mmcdara works only for characters coded with at most 2 bytes.

## workaround...

LC := proc(expr)
local i;
if type(expr, {numeric,name,string}) then return 1  # or maybe atomic
else add(i, i = map(thisproc,[op(expr)]))  +  1
end if
end proc;

## OK...

Everything is correct, even for a without assumptions (i.e. for any complex a).

a:=sqrt(3):
plot3d(y^2,  x=y/a .. 1-y/a, y=0...a/2, orientation=[-75,60]);

## du*dx etc...

I suspect that you actually want du*dx instead of dudx etc. If so, try:

ss:=convert(aa,string):
ss:=StringTools:-SubstituteAll(ss,"dudx","du*dx"):
ss:=StringTools:-SubstituteAll(ss,"dudy","du*dy"):
ss:=StringTools:-SubstituteAll(ss,"dvdx","dv*dx"):
ss:=StringTools:-SubstituteAll(ss,"dvdy","dv*dy"):
ddaa:=parse(ss):
collect(ddaa,[du,dv,dx,dy],distributed);

## Telescoping sums...

Why do you complicate things?

The identity

holds for k in Z, x in C  (actually for k,x in C).
Summing for k in 0..n-1  we obtain S1 and summing for -k in 1..n  we obtain S6.

## Workaround...

 > restart;
 > expr := [ exp(y*LambertW(ln(y))), (ln(y)/LambertW(ln(y)))^y, eval(x^(x^x), x = exp(LambertW(ln(y)))), eval(x^(x^x), x = ln(y)/LambertW(ln(y))) ]:
 > Matrix(4, (i,j)->'is'(expr[i]=expr[j])): simplify(%) assuming y>1;
 (1)

Workaround

 > Y:=solve(LambertW(ln(y))=t, y);
 (2)
 > mat:=simplify(subs(LambertW(ln(y))=t, y=Y, expr)) assuming t>0;
 (3)
 > Matrix(4, (i,j)->is(mat[i]=mat[j]) );
 (4)

## direct...

We don't need Maple for this.

ΔABC is similar to ΔEBC and ΔFDC. So, D is the midpoint of BC. DF=AE=EB=3, DE=AF=FC=4, DB=DC=5 (Pythagoras).

H being the centroid of ΔABC we have AD=BD=DC=5, and finally DH = AD/3 = 5/3.

## Change...

It's possible to use the standard changes of variables (Chebyshev). E.g.:

 > J := Int(x^(1/2)/(x^2+1)^(3/4), x);
 (1)
 > (x^2+1)/x^2 = t^4; X:=solve(%,x)[1]; T:=solve(%%,t)[1];
 (2)
 > simplify( IntegrationTools:-Change(J, x=X) ) assuming t>1;
 (3)
 > simplify(value(%)) assuming t>1;
 (4)
 > J=simplify(eval(%, t=T)) assuming x>0;
 (5)
 > simplify(diff(J - rhs(%), x)); # check
 (6)

 > restart;
 > with(plots): with(plottools): #with(DEtools); N := S(t) + In(t) + C(t); eqn1 := diff(S(t), t) = lambda - (lambda + sigma)*S(t) - (beta + zeta)*S(t)*In(t) - beta__1*S(t)*C(t), S(0) = ic1; eqn2 := diff(In(t), t) = beta*S(t)*In(t) - (lambda + gamma)*In(t), In(0) = ic2; eqn3 := diff(C(t), t) = zeta*In(t) + zeta*In(t)^2 - (rho + lambda)*C(t) - zeta*C(t)*In(t), C(0) = ic3; lambda := 0.117852; mu := 0.035378; beta := 0.11; beta__1 := 0.05; g := 1; rho := 0.1; zeta := 0.02; sigma := 0.066; ic1 := 2390000; ic2 := 753; ic3:= 358500;
 (1)
 > dsol := dsolve([eqn1, eqn2, eqn3], numeric);
 (2)
 > dsol(0.); dsol(0.001);
 (3)
 > odeplot(dsol,[[t,C(t)],[t,In(t)],[t,S(t)]], 0..0.001);

Remark. gamma in Maple denotes the Euler constant. I hope you used it as such.

## Not a bug...

 > restart;
 > S:= Sum(1/(4*n^2-4*n+4*100000000^2+1)/10^n,n=1..infinity);
 (1)
 > s:=convert(S, hypergeom);
 (2)
 > v:=value(S);
 (3)
 > #evalf(v);
 > #evalf(s);
 > evalf[100](S);
 (4)
 > evalf[500](S);
 (5)

 5 6 7 8 9 10 11 Last Page 7 of 115
﻿