vv

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These are answers submitted by vv

Actually the equation  chi(i) = a (a>=0) may have up to 4 solutions. (The first one for i<=0.)

restart

p := proc (S) options operator, arrow; .32*exp(S)/(1.2+1.901885*exp(S)-S^3) end proc

proc (S) options operator, arrow; .32*exp(S)/(1.2+1.901885*exp(S)-S^3) end proc

(1)

eta := proc (S) options operator, arrow; (D(p))(S)*S/p(S) end proc

proc (S) options operator, arrow; (eval(diff(p(x), x), x = S))*S/p(S) end proc

(2)

f := proc (k) options operator, arrow; A*k^alpha end proc

proc (k) options operator, arrow; A*k^alpha end proc

(3)

w := proc (k) options operator, arrow; f(k)-(D(f))(k)*k end proc

proc (k) options operator, arrow; f(k)-(eval(diff(f(x), x), x = k))*k end proc

(4)

beta := 1

1

(5)

A := 28

28

(6)

alpha := .33

.33

(7)

chi := proc (i) options operator, arrow; i*((1+beta)/beta+eta(i))/(1+eta(i)) end proc

proc (i) options operator, arrow; i*((1+beta)/beta+eta(i))/(1+eta(i)) end proc

(8)

#ss:=simplify(chi(i));

#plot(ss, i=-2..18)

#sort(select(u -> Im(u)=0,[solve(ss,i),solve(1/ss,i)]));

#select(u -> (u>6), [solve(diff(ss,i),i)]);

ii:=[-1.059280978, -0.7553094684, 0, 3.612421353, 6.399747910, 9.270348450, 9.270348450, 50]:

chin:=proc(a)
  local b:= evalf(RootOf(200000*_Z^5 - 200000*_Z^4*a - 200000*_Z^4 + 400000*_Z^3*a - 760754*exp(_Z)*_Z + 380377*exp(_Z)*a - 240000*_Z^2 + 240000*_Z*a - 480000*_Z + 240000*a, _Z, ii[2*n-1] .. ii[2*n]));
`if`(b::numeric,b,undefined)
end:

NULL

Hn:=k -> chin(w(k)):
Vn := k -> ln((w(k) - Hn(k))*(A*alpha)^beta*(p(Hn(k))*Hn(k))^(alpha*beta)):

for n from 1 to 4 do
  plot(Vn, 1 .. 2, title=Plot||n);
od;

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

 

 

 

 

 


Download Worksheet_1_vv.mw

If you really want to create the (global) m||i variables even when they exist, just replace

M := [seq(m || i, i = 2 .. l)];

with

M := [seq(evaln(m || i), i = 2 .. l)];


 

It's a quadratic equation. Just use solve.

solve((a*y-x*(a+1))/(x^(2)-b*a^(2)*(y-x)^(2))=A, y);

 

With a little help, Maple can compute the coefficients.

1. Expand to a series around z=0 the function 1/(1 + z)^2.

2. Substitute here z = exp(x-5*t) and expand around t=0.

3. Compute the sum for the coefficient of t^k.

 

One obtains the coefficient

 

c := k -> (-1)^k*(polylog(-1 - k, -exp(x)) + polylog(-k, -exp(x)))*5^k/k!;

proc (k) options operator, arrow; (-1)^k*(polylog(-1-k, -exp(x))+polylog(-k, -exp(x)))*5^k/factorial(k) end proc

(1)

add(map(combine,simplify(c(k)))*t^k,k=0..6); # check

-exp(x)*(2+exp(x))/(1+exp(x))^2+10*exp(x)*t/(1+exp(x))^3+(50*exp(2*x)-25*exp(x))*t^2/(1+exp(x))^4+(125/3)*exp(x)*(4*exp(2*x)-7*exp(x)+1)*t^3/(1+exp(x))^5+(625/12)*exp(x)*(8*exp(3*x)-33*exp(2*x)+18*exp(x)-1)*t^4/(1+exp(x))^6+(625/12)*exp(x)*(16*exp(4*x)-131*exp(3*x)+171*exp(2*x)-41*exp(x)+1)*t^5/(1+exp(x))^7+(3125/72)*exp(x)*(32*exp(5*x)-473*exp(4*x)+1208*exp(3*x)-718*exp(2*x)+88*exp(x)-1)*t^6/(1+exp(x))^8

(2)

 

@Michael_Watson It seems that you want an asymptotic expansion. Note that you missed each time a paranthesis.

restart;

L/(z*L*sqrt((z/L)^2 + 1)): % = asympt(%,L,2);

1/(z*(z^2/L^2+1)^(1/2)) = 1/z+O(1/L^2)

(1)

L/(z*z*sqrt(1+(L/z)^2)): % = asympt(%,L,2) assuming z>0;

L/(z^2*(1+L^2/z^2)^(1/2)) = 1/z+O(1/L^2)

(2)

L/(z*sqrt(z^2 + L^2)): % = asympt(%,L,2);

L/(z*(L^2+z^2)^(1/2)) = 1/z+O(1/L^2)

(3)


Download Lz.mw

EQSYS:=proc(S1::set(`=`), S2::set(`=`), {vars:={x,y,z}, param:={t}}) 
  local A:=eliminate(S1,param)[2], B:=eliminate(S2,param)[2];
  evalb( eval(B,solve(A,vars)) union eval(A,solve(B,vars)) = {0});
end:

sys1:={16*x-2*y-11*z=0, 14*x-y-10*z=3}:
sys2:={(x-2)/3=(y-5)/2, (y-5)/2=(z-2)/4}:
sys2a:={ (x-2)/3=t, (y-5)/2=t, (z-2)/4=t }:
sys3:={x+y-z=2, x-z=77}:

EQSYS(sys1, sys2);
                              true

EQSYS(sys1, sys2a);
                              true

EQSYS(sys1, sys3);
                             false

 

You may use inert operators

ex := L/(z*sqrt(z^2 + L^2)):
subs(L=(L%/z), numer(ex)) %* (1/denom(ex*z));

But, why do you need this?

It makes sense to use Maple only for numeric purposes here. 
For the two examples, a few numeric computations will suggest the general result, but the proof must be given by hand.

a) The limit  (M⊗N)(a,b) = sqrt(a*b). The proof is simple, as mmcdara showed.

M := (a,b) -> (a+b)/2:
N := (a,b) -> 2*a*b/(a+b):
MN:=proc(M::procedure, N::procedure, a, b, n::nonnegint, nmax::nonnegint:=n)
  local an:=a, bn:=b, ab:=Vector(1), k; 
  ab[1]:=[a,b]; 
  for k to nmax do  an, bn := evalf(M(an,bn)), evalf(N(an,bn));
    if k>=n then ab(k+1-n):=[an,bn] fi 
  od;
 seq(ab);
end:
MN(M,N,1,2, 2,5);             # the sequence [a1,b1],[a2,b2], ...

    [1.416666666, 1.411764706], [1.414215686, 1.414211438], [1.414213562, 1.414213562], [1.414213562, 1.414213562]

MN(M,N, 1,2, 10) = sqrt(2.);  # [a10,b10]

            [1.414213562, 1.414213562] = 1.414213562

b)  The limit M⊗N is the logarithmic mean; this can be proven similarly to the Gauss' proof for the AGM mean.

L := (a, b) -> (b-a)/ln(b/a):  # the logarithmic mean
a:=2.:  b:=8.:
MN( (a,b) -> (a+sqrt(a*b))/2, (a,b) -> (b+sqrt(a*b))/2,  a, b, 50 ) = L(a,b);

        [4.328085124, 4.328085124] = 4.328085123

The correct spherical parametrization is:

plots:-spacecurve([5^(1/2)*(t^2+1)*abs(t), arctan(4*t^2,2*t^3-2*t), arccos(1/5*signum(t)*5^(1/2))],
t = -1 .. 1, coords = spherical);

(Note the order [r, theta, phi] in plot3d)

It seems that the types `*`  and  `&*`  have problems for polynomial expressions.

restart;
type(x^2*y,  identical(x^2) &* name );
#                             false

type(x^2*y,  '`*`'({identical(x^2) , name}) );
#                             false

type(x^a*y,  identical(x^a) &* name );
#                              true

type(x^a*y,  '`*`'({identical(x^a) , name}) );
#                              true

1.0/3  is automatically simplified to 0.333...   with the current value of Digits (due to the so called Floating-Point Contagion).

Compare:

Digits:=3; evalf[8](1.0/3);   # Floating-Point Contagion
                          Digits := 3

                             0.333

Digits:=3; evalf[8](1/3);  # No contagion, 1/3 is exact
                          Digits := 3

                           0.33333333

 

The integral is computed by ScalarPotential  (if it exists, i.e. the form is exact).
Here is an example and a calling procedure (it works for more than 2 variables).

int1:=proc(F::list, X::list(name))
  uses VC=VectorCalculus;
  VC:-SetCoordinates('cartesian'[X[]]);
  VC:-ScalarPotential(VC:-VectorField(F))
end:

int1([2*x*y^7, 7*x^2*y^6], [x,y]);
      x^2*y^7

If you want inert integrals, it's easy to modify ScalarPotential.

It seems that the multiplication of float matrices with dimensions > 2000 crashes Maple 2021.

restart:
n:=2001:  # n:=2000 is ok
A:=Matrix(n, (i,j)->(i+2*j)/1000., datatype=float[8]):
A^2: #crash

A basis always exists, but the coefficients are probably huge and many primes are needed.

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