290 Reputation

14 years, 240 days

operators...

"using operators instead of expressions seems to use a weaker tolerance".

Yeah, it looks natural. All of that functions was created via codegen stuff so they are optimized => less operations made => precision of result is higher. But problem not in tolerance i.e. smth that checked and smth else under check fails. There could be another stuff that i had with newton under c on double type.Namely, because of round-off error newton algorythm could be divergent i.e. stuck on pairs (or even triples and generally on arbitrary number, i suspect) of points x1, x2 such that

x2= newton_step(x1) and x1= newton_step(x2).

To overcome such situation after "critical times of guesses" i fixed delta=|x2-x1| and

after "other critical times of guesses" i calculate new delta and if i see that new delta wasn't twice less than old one then i changed guess to random perturbated:

x2+=unit_rand()*(x1-x2). Sequence of guesses still inside bounds and more probably covergent to only one point.

I just wonder, wheather maple developers did smth like that, because such behavior doesn't depend on neither tolerance cryterium nor precision of calculations as well. It depends only on precision in that sense if precision is higher then such situation is less probable to occur but still probable.

Aha...

Found cool stuff about that there. So, levinu is applied in your case but not fact that in others. But you can directly calculate only by levinu. Other question what Chezaro divergence order of levinu itself. At least to understand what convergent or divergent, cannot find appropriate literature.

Sum: don't understand...

From all of that stuff i don't understand only one thing: in which sense maple Sum calculated? In Chezaro sense? If in Chezaro then with which order? Can i regulate this order? I mean for every a: limit(n^(a/n), n=infinity)=1

Documentation...

example:

f := proc(x) x - cos(x) end proc:
fsolve( f, 1.0 );

- there 1.0 as initial value. But in same help page

starting_values

-

(optional) equation, set(equation), or list(value); specify initial value used by solver. You cannot specify the variables option if you specify this option.

1.0 interpretated as list(value)? Should be somehow fsolve( f, [1.0] ); i guess.

Errors in function produce errors in documentation. So arguments processing is top priority for all of that tasks.

PS...

P.S. Don't see full of both of those images above

wrong...

gc doesn't help :( Nothing help with eval environment!

wrong...

gc doesn't help :( Nothing help with eval environment!

found...

`Diff(...,y)+eval(...,y=0)  is REALLY case! Hasta la vista, series!`

found...

`Diff(...,y)+eval(...,y=0)  is REALLY case! Hasta la vista, series!`

ahhh :(...

i  recalled, i tried multiseries before, but many functions are not supported and i see that for H even...

Error, (in MultiSeries:-multiseries) unable to compute series.

And for smaller:

s:=multiseries(LerchPhi(exp(-2*y*a)*b/(b+1), 1, a), y, 4, 'exact_order');
s := LerchPhi(b/(b+1), 1, a)+(-8*b^4*a/((b+1)^4*(a+4))-2*b*a/((b+1)*(a+1))-6*b^3*a/((b+1)^3*(a+3))-4*b^2*a/((b+1)^2*(a+2)))*y+(32*b^4*a^2/((b+1)^4*(a+4))+18*b^3*a^2/((b+1)^3*(a+3))+8*b^2*a^2/((b+1)^2*(a+2))+2*a^2*b/((b+1)*(a+1)))*y^2+(-36*b^3*a^3/((b+1)^3*(a+3))-(32/3)*b^2*a^3/((b+1)^2*(a+2))-(256/3)*b^4*a^3/((b+1)^4*(a+4))-(4/3)*a^3*b/((b+1)*(a+1)))*y^3+O(y^4)

is OK but how to convert it to polynom without that O(y^4) trail?
eval(s, O=proc() 0; end proc); - nothing. Again long and windy road.

ahhh :(...

i  recalled, i tried multiseries before, but many functions are not supported and i see that for H even...

Error, (in MultiSeries:-multiseries) unable to compute series.

And for smaller:

s:=multiseries(LerchPhi(exp(-2*y*a)*b/(b+1), 1, a), y, 4, 'exact_order');
s := LerchPhi(b/(b+1), 1, a)+(-8*b^4*a/((b+1)^4*(a+4))-2*b*a/((b+1)*(a+1))-6*b^3*a/((b+1)^3*(a+3))-4*b^2*a/((b+1)^2*(a+2)))*y+(32*b^4*a^2/((b+1)^4*(a+4))+18*b^3*a^2/((b+1)^3*(a+3))+8*b^2*a^2/((b+1)^2*(a+2))+2*a^2*b/((b+1)*(a+1)))*y^2+(-36*b^3*a^3/((b+1)^3*(a+3))-(32/3)*b^2*a^3/((b+1)^2*(a+2))-(256/3)*b^4*a^3/((b+1)^4*(a+4))-(4/3)*a^3*b/((b+1)*(a+1)))*y^3+O(y^4)

is OK but how to convert it to polynom without that O(y^4) trail?
eval(s, O=proc() 0; end proc); - nothing. Again long and windy road.

@Markiyan Hirnyk e2  is name of you...

@Markiyan Hirnyk
e2  is name of your expression. It's OK.

@Markiyan Hirnyk e2  is name of you...

@Markiyan Hirnyk
e2  is name of your expression. It's OK.

what is this?...

@Markiyan Hirnyk
e2 - don't get it. And trying for me is little bit hard because input expression in general case not known. I noticed that sometimes i need radical, simetimes need power, but most of them i need size. And symbolic just to do for 'size' things be little bit easier...

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