## 1365 Reputation

18 years, 151 days

## One question....

But what are you simplifying?

## Surfing the net....

You're meant to present the file on the forum. Use the green arrow to upload the worksheet.

## Rushed....

@mmcdara I apologize. In my haste it seems I have not included the requirement on the generators to be mutually disjoint. Minimal base would have least count. You can always subdivide non trivial basis elements, but that would increase the count.

Also I should point out that intersection and union are both n-ary operators, rather than simply binary. Every combination of sets is considered for intersection.

## Basis....

@Carl Love I believe so. It would be a minimal basis of disjoint elements, such that every intersection of elements of S is presentable as a union of elements of said basis.

An example:   F({{1, 2, 3}, {1, 4, 5}, {2, 3, 4}}) = {{}, {1}, {4}, {5}, {2, 3}}; and F({{1, 2, 3, 6}, {1, 4, 5, 6}, {2, 3, 4, 6}}) = {{1}, {4}, {5}, {6}, {2, 3}};

So it seems I forgot to include these criterions for T:
3. every intersection of elements of S is presentable as a union of elements of T uniquely
4. T is minimal

(I think the above means we can skip 1.)

## Apologies....

You have already spotted the essential difference. The unevaluate is needed when using the \$.

## evala@Norm...

@goli Apply this code to your expression:

```A := RootOf(6*_Z^3+(27+3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z^2+(3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2-9*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+90*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-18*l^4+6*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2-81+45*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z-324-3*l^8+l^10*RootOf(_Z^2*l^2+3*_Z^4-3)^2+108*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)-3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^6+sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^8*RootOf(_Z^2*l^2+3*_Z^4-3)^2-63*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+30*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2+45*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2+351*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-108*l^4, index = 1):
L := proc(A, lambda)
collect(
(`@`(numer, ((L) -> map2(op, 1, op(2, L))), evala,  Factors, evala, Norm))(lambda - convert(A, RootOf)), lambda)
end proc(A, lambda);
F := map(unapply, L, lambda);
factor(evala(F(A))):
subs(l = 1/2, %):
map(Testzero @ proc(x) evalf(x, 60); fnormal(%); end, %);
if member(true, %,'i') then L[i] else FAIL fi;
```

## Try series....

How effective is series approximation?

## ?...

@Carl Love Visibly, one more definition is required to solve numerically...

## Unexpected....

@Carl Love Unexpected. I get unevaluated with maple 2017.

## Did you simply forget the dot at the end...

@Glowing Obviously, integers have infinite precision. Your "error" in the code is to use "1014" instead of "1014.".

## There are no bugs here though......

```restart;
P1 := 1007;
P2 := 1014;
P3 := 1014.1;
P2 - P1;
P3 - P1;
evalf(P2 - P1, 2);
evalf(P3 - P1, 2);

restart;
P1 := 1007.;
P2 := 1014.;
P3 := 1014.1;
P2 - P1;
P3 - P1;
evalf(P2 - P1, 2);
evalf(P3 - P1, 2);
```

## Print....

@taro You can replace identical(w^sigma) with identical(w)^Non(integer) as convert/parfrac will not accept non integer powers of the variable. Also, You are correct. If You negate the first element in the list then you will see the type of which  occurences will be substituted. In this case it is Non(Non(identical(w^sigma))) = identical(w^sigma). The items in the second element in the list will present without substitution, preserved. Note, the substitutions are applied only to the face of the expression passed and the result returned. If you want to see how it works out You may want to use print:

frontend(proc(E,a,v) print('args'); convert('args'); end ,[e_n_1b2,parfrac, w^sigma],[{Non}(identical(w^sigma)), {}]);

## Distributive....

@HS I expect it is because in your equation mod is applied only to the rhs: "Q2 = P2 mod p". If you want to campare mod p then apply modulus to all elements compared. "(Q2 = P2) mod p" will apply to all elements.

## Brackets....

Does evalb((Q2 = P2) mod p); work?

## One question:...

Are we to assume that k stands for K and i stands for I?
Also notice you are using e^(-k*h), e^(-k*z). That should be exp(-k*h) and exp(-k*z).

 3 4 5 6 7 8 9 Last Page 5 of 18
﻿