## 11977 Reputation

7 years, 256 days

## 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)

 > restart

1 < alpha, alpha < 2      are given.

 > f := (a - b + c)^2 + (alpha - 1)*b^2/alpha + (alpha - 1)*c^2/alpha
 (1)

Find M>0 such that  f >= M*a^2,  for any a,b,c>0.

Solution.

Denote  b/a=B, c/a=C. Then we have to determine  min F, where:

 > F := (1 - B + C)^2 + (alpha - 1)*B^2/alpha + (alpha - 1)*C^2/alpha;
 (2)
 > F:=simplify( eval(F, alpha=1/(1-t)) );
 (3)
 > minimize(F, B=0..infinity, C=0..infinity, location) assuming t <= 1/2, t >= 0;
 (4)
 > M__max = simplify(eval(%[1], t=(alpha - 1)/alpha))
 (5)

## explicit, autorange...

 > restart;
 > f:=2.96736996560705*10^(-12)*p^2+1.31319840299485*10^(-13)*t^2- 8.89549693662593*10^(-7)*p+8.53128393394231*10^(-7)*t-3.65558815509970*10^(-30)*p*t-1:
 > tt:=simplify([solve](f,t)):
 > pp:=solve(tt[1]>-infinity,{p}): p1,p2 := lhs(pp[1]), rhs(pp[2]):
 > plot(tt, p=p1 .. p2, scaling=constrained);

## Workaround...

 > #Workaround (inequations work in general only with polynomials -- SemiAlgebraic!)
 > solve({(4*t1+4*t3)>0,  t1^2-4*t2 = t3^2, t3>=0},[t2,t3]);
 (1)
 >

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