vv

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@Carl Love Yes, but then the results of LC and LeafCount will not be exactly the same.

@sursumCorda Yes, I see it now.

It seems that in seq(expr, 1..n),  expr is evaluated n times   but in seq(expr, n) ,  expr is evaluted n+1 times (being evaluated once befor creating the sequence).
But note that this is not actually a bug,  because the second form is not documented!

restart;
u:=10;
seq((u:=u+1), 3)
#                            u := 10
#                           12, 13, 14
u:=10;
seq((u:=u+1), 1..3)
#                            u := 10
#                           11, 12, 13

 

@sursumCorda 

evalf(Int(convert(x^(x^x), exp), x=1..6));

     1.102664999 * 10^36300
 

@Carl Love I obtain an error due to complex values for sample. But this works:

plot([LambertW(x), LambertW(-1,x)], x=-1..4, view=[-1..4, -3.5..1.5], colour=[red,blue], scaling=constrained, labels=[x,LambertW(x)], adaptive=15);

 

@Axel Vogt  mma gives:

The problem here is that LerchPhi is evaluated numerically very slowly for large arguments.
Yes, this could be seen as a bug, but there are easy workarounds. For example,

L:=LerchPhi(1/10, 1, 1/2-100000000*I);
evalf(L);   # must be interrupted

but 

evalf(convert(L,Sum)) ;  # works with any precision

@Zeineb Ok, I understand now, but you merge Maple and math notations: in piecewise I interpreted y as a second component of a point (x,y,z) in R^3, instead of a point.

Your function f : R^3 --> R should be locally integrable, but it's not!
So, the Hardy−Littlewood maximal function makes no sense.

@Thomas Richard Sometimes the situation is reversed!

f:=polylog(2, -1 - s) + polylog(2, (1 + s)/(s + 2));
 

polylog(2, -1-s)+polylog(2, (1+s)/(s+2))

(1)

simplify(f);  # nothing

polylog(2, -1-s)+polylog(2, (1+s)/(s+2))

(2)

simplify(convert(f, dilog));  # That's it!

-(1/2)*ln(s+2)^2

(3)

 

@sursumCorda So, you want an explicit solution instead of a parametric one (t3 being the parameter).
This could be very difficult (or even impossible) to obtain in general.

It would be nice to tell us how the equation was obtained.

Here is a (simpler) parametric one (Klimek G.&M. - Discovering curves and surfaces with Maple, Springer 1997.)
 

plot3d([(4+3.8*cos(y))*cos(x),
        (4+3.8*cos(y))*sin(x), 
        (cos(y)+sin(y)-1)*(1+sin(y))*log(1-Pi/10*y)+7.5*sin(y)],
      x=-Pi .. Pi, y=-Pi .. Pi, color=[sin(x+1)+cos(y), sin(y), 0.3],
      light=[85, 50,  0.6, 0.8, 0.2], orientation=[47, 55],
      style=patchnogrid, axes=none);

 

Your function of Pe also depends on several variables and some arbitrary functions!
How can you imagine that the limit could be computed?

@mmcdara For me, it's not about assume or assuming. 
For the same assumptions, assumptions := a>0, b>0, a*b < 1:
assuming
gives the same results.
Also, in Maple 2022, assumptions := a>0, b>0, b < 1/a: 
==> all answers are true (But it seems that in 2023, one answer is FAIL !!).

So, everything seems to be about weakness!

@Carl Love  Unfortunately the assume facility is very weak here.

restart;
# _EnvTry := true;  # useless
assume(a>0, b>0, a*b<1);
is( (1 + sqrt(1-a^2*b^2))/(a*b) >0 ); # FAIL   ?
is( (1 + sqrt(1-a*b))/(a*b) >0 );     # FAIL   ?
is( (1 + sqrt(1-a^2*b^2)) >0 );       # false ??
is( (1 - sqrt(1-a^2*b^2)) >0 );       # false ??
is( (1 - sqrt(1-a*b)) >0 );           # true
is( (1 + sqrt(1-a*b)) >0 );           # true

 

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