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acer

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19 years, 344 days
Ontario, Canada

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acer 32395
June 28 2025
0 0

@Math-dashti Why didn't you tell us what you really wanted when you first posted the Question?

query...

acer 32395
June 27 2025
0 0

Do you believe that there are explicit solutions other than,

   u = 1, v = 2*Pi*n + Pi    with n::integer
   u=-1, v=0

evala...

acer 32395
June 26 2025
0 0

@vv 

> 

restart
> 

puzzle := sum(cos(k*Pi/(2*n+1))^4, k = 1 .. n)
> 

simplify(expand(convert(puzzle, exp)))

(3/8)*n-5/16

> 

`assuming`([simplify(evala(expand(convert(puzzle, exp))))], [n::posint])

(3/8)*n-5/16

> 

The expanded exp terms turn into this kind of thing,
which can be manipulated using evala (and pose extra
difficulties for simplify).

expand(exp((n + 1)*Pi*I/(2*n + 1))) assuming n::posint;

(-1)^(n/(2*n+1))*(-1)^(1/(2*n+1))

Download sum_ex05b.mw

Playing around with the intermediate expressions led me to this gem, in which a remembered result can interfere with (or help!) a simplification. I will submit a separate bug report for this item.

> 

restart;
> 

hmm := (-(-1)^((2 - 2*n)/(2*n + 1)) + (-1)^((4 + 2*n)/(2*n + 1)))/(4*(-1)^(4/(2*n + 1)) - 4);

(-(-1)^((2-2*n)/(2*n+1))+(-1)^((4+2*n)/(2*n+1)))/(4*(-1)^(4/(2*n+1))-4)

> 

simplify(hmm);

0

> 

new := (-4*(-1)^((2 - 2*n)/(2*n + 1)) + 4*(-1)^((4 + 2*n)/(2*n + 1)) + (-5 + 6*n)*(-1)^(4/(2*n + 1)) - 6*n + 5)/(16*(-1)^(4/(2*n + 1)) - 16):
> 

whoa := evala(new - ((3*n)/8 - 5/16));

-(1/4)*((-1)^(-2*(-1+n)/(2*n+1))-(-1)^(2*(2+n)/(2*n+1)))/((-1)^(4/(2*n+1))-1)

> 

simplify(whoa);

0

> 

simplify(hmm);

0

> 

restart;
> 

hmm := (-(-1)^((2 - 2*n)/(2*n + 1)) + (-1)^((4 + 2*n)/(2*n + 1)))/(4*(-1)^(4/(2*n + 1)) - 4);

(-(-1)^((2-2*n)/(2*n+1))+(-1)^((4+2*n)/(2*n+1)))/(4*(-1)^(4/(2*n+1))-4)

> 

#simplify(hmm);
> 

new := (-4*(-1)^((2 - 2*n)/(2*n + 1)) + 4*(-1)^((4 + 2*n)/(2*n + 1)) + (-5 + 6*n)*(-1)^(4/(2*n + 1)) - 6*n + 5)/(16*(-1)^(4/(2*n + 1)) - 16):
> 

whoa := evala(new - ((3*n)/8 - 5/16));

-(1/4)*((-1)^(-2*(-1+n)/(2*n+1))-(-1)^(2*(2+n)/(2*n+1)))/((-1)^(4/(2*n+1))-1)

> 

simplify(whoa);

(-(-1)^((2-2*n)/(2*n+1))+(-1)^((4+2*n)/(2*n+1)))/(4*(-1)^(4/(2*n+1))-4)

> 

simplify(hmm);

(-(-1)^((2-2*n)/(2*n+1))+(-1)^((4+2*n)/(2*n+1)))/(4*(-1)^(4/(2*n+1))-4)

 

 

Download bug01.mw

In one variant I saw an instance of simplify(foo) fail while simplify(simplify(foo)) succeeded, though it's hard to reproduce because it too seemed to depend on previous computations. More fun than a barrel of monkeys.

Maple 2024.2...

acer 32395
June 26 2025
0 0

@Alfred_F I don't understand what you're trying to convey with, "...is made again and again ..., the simple result term (6*n-5)/16 no longer occurs; the calculation stops before it."

That is, I can't tell whether you're saying it never worked for you, or that it only worked once, or something else.

But I do see now that the simplification command that I gave (and also the one that dharr gave) might succeed in Maple 2025.1 but not in older Maple 2024.

Here is a variant that works in my Maple 2024.2.

> 

restart;
> 

kernelopts(version);

`Maple 2024.2, X86 64 LINUX, Oct 29 2024, Build ID 1872373`

> 

puzzle := sum(cos(k*Pi/(2*n + 1))^4, k = 1 .. n):
> 

simplify(convert(expand(simplify(convert(puzzle, exp))),exp))

(3/8)*n-5/16

Download sum_ex05_M2024.mw

ps. In even older versions of some years ago it may have been harder still to get this one. The `simplify` command is improving, over time. I reported the weakness (that simplify alone didn't get the simple result) yesterday.

another way...

acer 32395
June 25 2025
0 0

Maple 2025.1,

> 

restart
> 

puzzle := sum(cos(k*Pi/(2*n+1))^4, k = 1 .. n)
> 

simplify(expand(convert(puzzle, exp)))

(3/8)*n-5/16

Download sum_ex05.mw

(I'll submit a report, against simplify.)

details...

acer 32395
June 24 2025
0 0

You've forgotten to provide any details of your problem.

hmm...

acer 32395
June 20 2025

@Carl Love And the More...Delete menu item is also missing here now.

The other posting also remains.

I had passed along the details to those in charge. (The poisoning might be be tricky; I don't know.)

no worries...

acer 32395
June 20 2025
0 0

@mmcdara No worries!

The int result improved in Maple 2019. The factoring of the quartic by simplify came in Maple 2016.

version...

acer 32395
June 20 2025
0 0

@mmcdara That appears to be related to your version.

> 

restart
> 

kernelopts(version)

`Maple 2025.1, X86 64 LINUX, Jun 12 2025, Build ID 1932578`

> 

f := 2*t*piecewise(t <= 0, 0, t <= 1, t^2+Pi-4*t, t <= sqrt(2), -(sqrt(t^2-1)*t^2+2*sqrt(t^2-1)*arcsin((t^2-2)/t^2)-4*t^2+2*sqrt(t^2-1)+4)/sqrt(t^2-1), sqrt(2) < t, 0)
> 

myM := int(f, t = 0 .. x):
> 

myM assuming x>0, x<1; # see version

(1/2)*x^4+Pi*x^2-(8/3)*x^3

> 

int(f, t = 0 .. x) assuming x > 0, x < 1;

(1/2)*x^4+Pi*x^2-(8/3)*x^3

> 

simplify(%); # see version
 

(1/6)*x^2*(3*x^2+6*Pi-16*x)

Download myM_ac.mw

yeah...

acer 32395
June 19 2025

@Carl Love Yes, and the Flag button as well. I noticed this on a pair of postings from a few days ago.

I was planning on reporting it.

simplified...

acer 32395
June 19 2025
0 0

@mmcdara Fwiw the symbolic median (your Exact) can be further simplified.

In trig representation, there is no small imaginary component/artefact, when evalf'd.

> 

restart
> 

kernelopts(version)

`Maple 2025.1, X86 64 LINUX, Jun 12 2025, Build ID 1932578`

> 

f := 2*t*piecewise(t <= 0, 0, t <= 1, t^2+Pi-4*t, t <= sqrt(2), -(sqrt(t^2-1)*t^2+2*sqrt(t^2-1)*arcsin((t^2-2)/t^2)-4*t^2+2*sqrt(t^2-1)+4)/sqrt(t^2-1), sqrt(2) < t, 0)
> 

myM := int(f, t = 0 .. x):
> 

# omitting the rationale
myM=1/2 assuming x>0, x<1;

(1/2)*x^4+Pi*x^2-(8/3)*x^3 = 1/2

> 

G := solve({myM=1/2, x>0, x<1},x,explicit) assuming x>0, x<1:
> 

simplify((evalc(simplify(G)))):
evalf(%);
lprint(%%);

{x = .5120032673}

{x = 1/3*(4*(Pi^2-3)^(1/4)*(16-3*Pi+3*(Pi^2-3)^(1/2)*cos(1/3*arctan((-27*Pi^4+
96*Pi^3-54*Pi^2+864*Pi-2331)^(1/2)/(Pi^3+9*Pi-48))))^(1/4)-(16-3*Pi+3*(Pi^2-3)^
(1/2)*cos(1/3*arctan((-27*Pi^4+96*Pi^3-54*Pi^2+864*Pi-2331)^(1/2)/(Pi^3+9*Pi-48
))))^(3/4)*(Pi^2-3)^(1/4)+((-3*Pi^2*cos(1/3*arctan((-27*Pi^4+96*Pi^3-54*Pi^2+
864*Pi-2331)^(1/2)/(Pi^3+9*Pi-48)))-6*Pi*(Pi^2-3)^(1/2)+32*(Pi^2-3)^(1/2)+9*cos
(1/3*arctan((-27*Pi^4+96*Pi^3-54*Pi^2+864*Pi-2331)^(1/2)/(Pi^3+9*Pi-48))))*(16-\
3*Pi+3*(Pi^2-3)^(1/2)*cos(1/3*arctan((-27*Pi^4+96*Pi^3-54*Pi^2+864*Pi-2331)^(1/
2)/(Pi^3+9*Pi-48))))^(1/2)+(36*Pi-128)*(Pi^2-3)^(1/2))^(1/2))/(Pi^2-3)^(1/4)/(
16-3*Pi+3*(Pi^2-3)^(1/2)*cos(1/3*arctan((-27*Pi^4+96*Pi^3-54*Pi^2+864*Pi-2331)^
(1/2)/(Pi^3+9*Pi-48))))^(1/4)}

Download Median_acc.mw

incomplete...

acer 32395
June 19 2025
0 0

@jalal That's does not explicitly show us what you did in your original attachment in this thread's Question.

Why can't you show us the explicit code that constructs your Tables, in your newer approach?

Is your modified (inspired) approach still using some procedure F similar (even if without piecewise...) to that in your older worksheet's approach?

comment...

acer 32395
June 17 2025
0 0

@mmcdara It seems possible to replace (by a1..a8) all the names in ra[1] except eta, gamma, and phi (noting that phi does not even appear in the rhs's of the equations).

The result is somewhat unwieldy, though.

constructed, how?...

acer 32395
June 17 2025
0 0

How are those Tables of outputs constructed? Are they inserted Tasks? Could you provide a version of your worksheet in which the underlying Task(s) or construction mechanism is shown?

inert intermediates...

acer 32395
June 13 2025
0 0

The OP's followup example showed that he didn't want  1/`%+`(k,7)  which is the inert form for k+7.  Carl's already pointed out that OP's inconsistency in requests.

So I've redone my followup Replies to show either a fully active expression, or one where the user can utilize either a partially or fully inert/active form.

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