vv

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These are answers submitted by vv

restart;
with(LinearAlgebra): with(plots):
n:=3:
f:=-x^2 + 2*y^2 + 2*z^2 - 6*x + 4*x*y - 4*x*z - 8*y*z + 4*z - 12:
f:=eval(f, [x=x[1],y=x[2],z=x[3]]);
L:=10:
quad:=implicitplot3d(f, x[1]=-L..L, x[2]=-L..L, x[3]=-L..L, style=surface, scaling=constrained):
A:=VectorCalculus:-Hessian(1/2*f,[seq(x[i],i=1..n)]):
b:=eval(Vector( [ seq(diff(f,x[i]),i=1..n)]), [seq(x[i]=0,i=1..n)]):
c:=eval(f,[seq(x[i]=0,i=1..n)]):
X:=Vector([seq(x[i],i=1..n)]):
solve([ seq(diff(f,x[i]),i=1..n)],{seq(x[i],i=1..n)}); # the center
X0:= Vector[column]( eval([seq(x[i],i=1..n)],%) ):
J,Q:=Eigenvectors(A):
T:=Matrix(GramSchmidt([seq( Q[..,j],j=1..n)],normalized)): # T is orthogonal
fnew:=simplify( (T.X+X0)^+. A. (T.X+X0) + b.(T.X+X0) + c ); # ==> Hyperboloid of Two Sheets
col:=[red,yellow,blue]:
ax:=seq(arrow(X0, T[..,j], length=10, width=0.3, color=col[j]), j=1..n): 
display(quad, ax, orientation=[175,63,21], caption="Hyperboloid of Two Sheets");

restart;
f := (x,y)->(x^2+y^2)^x: f(0,0):=1:
'D[1]'(f)(0,0) = limit((f(x,0)-f(0,0))/x, x=0);
'D[2]'(f)(0,0) = limit((f(0,x)-f(0,0))/x, x=0);

G:=GroupTheory:-GaloisGroup(x^5 + 20*x + 32, x);
                Gal(x^5+20*x+32,x)
IdentifySmallGroup(G);
                             10, 1

G has been identified as [n,d] = [10,1],  see ?IdentifySmallGroup

To identify it as a human, just access the wiki page  List of small groups - Wikipedia

==> [10,1]  corresponds to the non-abelian group D10 (i.e. D5 with Maple notation).

 

arctan(x,y) = - arctan(-x,y)  implies for x=0 that arctan(0,y) = 0,  
but this is false for y<0.
[I have used your swapped notation x <--> y ]

 

Replace fn:=solve(...)  with

seq(['q'=q, 'x'=solve(V__out/120 = 1/sqrt((-m*x^2 + m + 1)^2 + (q*(x - 1/x))^2), x, useassumptions)], q in Q__s) assuming x::positive;

 

Correct syntax:

GroupOrder(ds[2]);

The result is of course 5, because it is generated by a cycle of length 5.

sum(-5*3^(-k-1)*(x-2)^k, k=0..infinity, formal);
                             
# or, say something about the domain
sum(-5*3^(-k-1)*(x-2)^k, k=0..infinity) assuming abs(x-2)<3;
                            
# Note that tha radius of convergence is R=3.

              5/(x - 5)

              5/(x - 5)

Not all ODEs have series solutions around 0.
To illustrate the problem, let's modify a bit your second ODE.

ode1:=x^2*diff(diff(y(x),x),x)+x*diff(y(x),x)+y(x) = 0:
dsolve(ode1);

             y(x) = _C1 sin(ln(x)) + _C2 cos(ln(x))

MultiSeries:-series(sin(ln(x)), x=0);

          -sin(ln(1/x))

 

restart;
m:=166571820085905498772938593207359906487315774156514145089564134\
67009247675308229223573508773729914408724747749233779284765414\
62530835296763930087228227919982503096221081674037688617693027\
1167988018683:
for x from isqrt(m) while not issqr(x^2-m) do od:
y:=isqrt(x^2-m):  'm' = (x-y) %* (x+y);

m = 4081321110693270343633073697474256143563558458718976746753830538032062222085722974121768604305614359 %* 4081321110693270343633073697474256143563558458719886006565783193342137709249520153612225643475208637

# This works only if the two primes are "very" close. (x is close to sqrt(m).)
x-isqrt(m);
                               25

 

You must use signum, not sign.(sign is used mainly for polynomials in Maple).

Take the two sub-sequences for which sin(n)>0  and sin(n)<0 [it cannot be 0].
The limits are 11/2 and 25/2, so, R=2/25.

Using random polynomials there are big chances to have duplicates (there are only 500 such polynomials). So:

restart;
A:={seq(-2..2)}:  A0:=A minus {0}:
n:=54:
a:=Iterator:-CartesianProduct(A0,A$3, rank=500-n+1):
seq(add(v[i]*x^(i-1)+x^4,i=1..4), v=a);

 

The radius of convergence needs limsup and root test here.

a := n -> ((2 + (-1)^n)/(5 - (-1)^n))^n:
an0:=simplify(abs(a(n))^(1/n)) assuming n::even:
R0:=limit(1/an0,n=infinity):
an1:=simplify(abs(a(n))^(1/n)) assuming n::odd:
R1:=limit(1/an1,n=infinity):
R:=min(R0,R1);

      R := 4/3

Optimization:-Minimize(x->TF(x, 0, 0.03, 55.11566060, Pi/2, Pi/4), 0 .. 0.03);

                         [4.281548719743726*10^(-7), [0.15749994509751696e-1]]
 

restart;

sum(Beta(k,1/2)/(2*k+1)^2,k=1..infinity);  # Bug!

Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation

 

S:=Sum(Beta(k,1/2)/(2*k+1)^2,k=1..infinity);

Sum(Beta(1/2, k)/(2*k+1)^2, k = 1 .. infinity)

(1)

Sa:=evalf[15](S);

.336137623291124

(2)

s:=value(convert(S, GAMMA));

(1/4)*Pi^(1/2)*MeijerG([[1], [5/2, 5/2]], [[3/2, 1, 1], []], -1)

(3)

evalf[15](s); # Maple cannot compute

 

exact:=identify(Sa,BasisPolyConst=[Catalan]);

4-4*Catalan

(4)

evalf[100](S-exact); # numerical check

0.

(5)

 

 

Adding to what user acer said:

Gcd in Algebraic has a distinct syntax; it is not inert any more.
If you want the result mod 3 then use

 Gcd(f, g, characteristic=3);

Note that without characteristic=3, the gcd is computed in the ring Z[x]; if mod 3 is included (as you did), the mod 3 is applied to this polynomial.

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