## 12745 Reputation

8 years, 295 days

## freeze...

eq1a:=subs(rhs(eq1)=freeze(rhs(eq1)),eq1):
eq2a:=subs(rhs(eq2)=freeze(rhs(eq2)),eq2):
solve({eq1a,eq2a},{a,e}):
thaw(%);

## Parametrization...

Maple cannot obtain a rational parametrization for curve(alpha,beta)=0, because the genus is not 0.

But, being quadratic in alpha, a parametrization can be easily obtained by solving wrt alpha.

Let's take a simpler example, to see easier the results.

ec:= (u,v) -> u^2-u+3*v^3-v-2:  # the curve is ec(u,v)=0
# choose a point on it
u0:=0.6:
v0:=fsolve(ec(u0,v),v);

vv:=v0+z^2:   # choose v as you wish
solve(ec(u,v),u);   # find parametrization for u  (exact)

eval(%[1],v=vv);   # choose the desired branch
uu:=series(%,z,8):  # obtain the approx parametrization
u=uu,v=vv;   # Parametrization near (u0,v0)

## Simple but slow...

Here is a very simple one. Unfortunately it's slow for n>11.
But it is very fast for n=1  and n=2  (mod 4)    :-)

LangfordSeq:=proc(n::posint)
local A,B,C,C1,z,i:=irem(n,4),j,s,S, V:=Vector(2*n);
if i=1 or i=2 then return "Solutions exist only for n = 0 or 3 (mod 4)" fi;
C:= seq(seq(z[i,j]=z[i+n,i+j+1], j=1..2*n-i-1), i=1..n);
C1:= seq(seq(z[i,j]=0, j=2*n-i..2*n), i=1..n);
s:=Optimization:-LPSolve(0,{A,B,C,C1},assume=binary);
S:=eval(Matrix(2*n,symbol=z), s[2]);
for i to n do for j to 2*n do
if S[i,j]=1 then V[j]:=i; V[j+i+1]:=i; fi od;od:
convert(V,list);
end:

LangfordSeq(7);

## Maths...

In such problems maths always beats a CAS.

f:=x -> x^4+c*x^2+x^3+d*x-c-1

==> diff(f(x),x,x)  = 12*x^2+6*x+2*c  >= 0  for  c>= 3/8.

So, f in convex in x for c>= 3/8 ==>

g(c,d) = max(f(0), f(1)) = max(-1-c, 1+d) ,  for  c>= 3/8.

So, g(c, -c-2) = -c-1, (c>= 3/8)
and
limit(g(c, -c-2), c=infinity) = -infinity

==> g is not bounded below.

Edit. g(c,d) here represents max f(x) for x in [0,1] instead of [-1,1] as was asked.
For the [-1,1] version see the comment below.

## All you need is...

kekuncirahsia:=proc(n::posint) local r:=irem(n,9); `if`(r=0,9,r) end;

add8 := (x,y) -> convert( convert(x,decimal,octal) + convert(y,decimal,octal),  octal,decimal );

(The addition must be done in base 10).

Note that lists are added componentwise.

## proc...

Here is a general but not very efficient procedure.
Of course it would be much better to solve the equation only once, but in this case, the solutions must be verified and the special cases must be considered.

inteq:=proc(f,x,vpar)
local vab,ab,s,T,r,n:=0;
indets(f,name) minus {x}; vab:=[op(%)];
T:=combinat[cartprod]([ [seq(vpar)] \$ nops(vab) ]);
while not T[finished] do
ab:=T[nextvalue]();
s:=solve(eval(f, vab=~ab),x);
if s=x then print(vab=ab,"identity!") fi;
s:=select(type,[s],integer);
if s<>[] then n:=n+1; r[n]:=[ab,s];DocumentTools[Tabulate]([n]) fi;
#if s<>[] then n:=n+1; print(n,[vab=ab,x=s]) fi
od;
n,vab,x,eval(r);
end:

inteq(a*x + b = sqrt(c*x + d),x,-2..2);

[a, b, c, d] = [0, 0, 0, 0], "identity!"
[a, b, c, d] = [0, 1, 0, 1], "identity!"

176, [a, b, c, d], x,

## spirals...

Factorizing f7 ==>

p1-t1  or  p2-t2  or  p3-t3  is an integer multiple of Pi.

For p1-t1 = 0 (or Pi), from f1 and f2 ==> two spirals, namely

plots[spacecurve]([2.350000000*cos(t1), 2.350000000*sin(t1), t1], t1=-Pi..Pi);
plots[spacecurve]([-.350000000*cos(t1), -.350000000*sin(t1), t1], t1=-Pi..Pi);

For p2-t2 = 0 (or Pi) you get two spirals in (x,y,t2) etc.  If you want here (x,y,t1), t2 must be expressed wrt t1.

Note that the (x,y) projection is contained in 6 circles.

## cartprod...

with(combinat, cartprod):
n:=2:
A:=Array((1..3) \$ n):

T:=cartprod([[1,2,3]\$n]):
while not T[finished] do
i:=T[nextvalue]();
A[i[]]:=i;  # or ... whatever
end do;

## size...

If you have Maple >= 2015, you may use the size option:

plot( max(sin(x),0)^2*sin(16*x), x=0..8*Pi, numpoints=10000,size=[1200,300]);

## second recurrence...

# b(m)=m-1/2;
c:=m -> convert(m,symbol):
a:=m -> a(m-1)*b(m-1)/(b(m)*c(m)):    a(0):=a0:

a(8);

a(11);

## A...

(case A only).

Obviously sup f = oo  (for a1 --> oo).

f is increasing wrt a1.
f(2,a2,a3) is increasing wrt a3.
So, min f = min f(2,a2,3)
and is easy.

## Double series...

are tough! Usually the sum cannot be computed exactly.

a:=1/(i^4+j^4);

S:=Sum(Sum(a,i=1..infinity),j=1..infinity);

(this differs by yours by constant which is easy to compute).

value(S); does not work, as expected

Unfortunately Maple cannot approximate its sum dirrectly.

evalf(S); #does not work (unexpected!)

But the inner sum can be computed exactly, ==>

evalf(  sum(sum(a,i=1..infinity),j=1..infinity) );

So, the series seems to be convergent, but this is not a proof.

( It is not difficult to see that se series is convergent using  a <= 1/i^2 * 1/j^2 ).

Another way for convergence:

S is convergent iff J < oo, where

J:=Int(Int(a,i=1..infinity),j=1..infinity):

Unfortunately J cannot be computed directly.

value(J); # does not work

We must use:

int(a,i=1..infinity) assuming j::posint:
int(%,j=1..infinity);

so, J<oo and hence S<oo.

To compute a partial sum and verify numerically the sum:

0.7617308252

(close to the value obtained earlier).

## combine...

with(IntegrationTools):
f:=t^2*exp(-t)/ln(t):
J1:=Change(Int(f,t=0..1-a),t=(1-a)*x,x):
J2:=Change(Int(f,t=1+a..infinity), t=(1+a)/x,x):
simplify(limit(combine(J1+J2),a=0));

evalf(%);

2.153547692

## Matrix...

f := (i,j) -> `if`(j=1, i^2, i^3):
Matrix(10,2,f);

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