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Golden ratio...

vv 13805
August 03 2020
1 0

For f : (0, oo) --> (0, oo), a solution is:

phi := sqrt(5)/2 + 1/2:
f := x -> phi^(1-phi)*x^phi;

Try to check it.

Another workaround: rtf or latex first...

vv 13805
August 03 2020
1 1

Save the worksheet as rtf (or LaTeX) and then produce the pdf from here.
A pdf obtained from rtf is very small!

 

f(p)+g(p)=h(p)...

vv 13805
August 02 2020
0 8

1.  y = f(p) + g(p)   means actually  y = h(p)

so, it's in the d'Alembert pattern y = x*f(p) + g(p)   for f=0, g=h.

2. dsolve(ode) solves the ode using the y = g(y') pattern [independent of x], so, a RootOf form.

The situation is almost the same as in y...

vv 13805
August 01 2020
0 0

The situation is almost the same as in your previous question except that now Expand works properly in Maple 2015
(Maple 2020 works directly -- your "Method 1").

Expand...

vv 13805
July 31 2020
1 6

In Maple 2020 the result is directly   -4.211444465*10^(-43)*y + 4.534504493*10^(-43)

In earlier versions try:

Int(q*AAA/sqrt(p^2 + q^2 + 1), [p = 0 .. L, q = 0 .. L]):
evalf(IntegrationTools:-Expand(%));

 

mpl...

vv 13805
July 31 2020
0 0

The simplest solution is to open the worksheet in Maple 2020 and then export it as mpl (i.e. text file).
Now you can open the .mpl file. You may want to split it into several execution groups (1D input).

Another solution (since you are already using 1D input)  is to simply remove the output from worksheet (Ctrl + D) and reexecute.

Euclid...

vv 13805
July 28 2020
0 3

Am am sure it is based on the Euclidean algorithm for polynomials. (The modulus is in general a prime number).

A hybrid solution...

vv 13805
July 26 2020
1 5

Your system is numerically unstable.
Supposing that the constants are "exact", it is possible to make a mix of symbolic and numeric approaches.
The ideea is to convert the constants to rationals (for a precision of 15 digits) then solve the system symbolically wrt the variable ionic (because this is possible), then solve the univariate equation in ionic with very high precision.

Here are the results. The solution is unique [almost]. Some variables are negative.


 

> 

restart;
> 

Digits:=15;

15

 (1)
> 

eq1:=K_1=(((n_Cl-x)*u_Cl)*(n_H*u_H))/(n_HCl*m);
eq2:=K_2=(((n_Na-x)*u_Na)*(n_OH*u_OH))/(n_NaOH*m);
eq3:=K_w=(n_H*u_H/m)*(n_OH*u_OH/m);
eq4:=(n_NaCl-x)=(n_Na-x)+n_NaOH;
eq5:=(n_NaCl-x)=(n_Cl-x)+n_HCl;
eq6:=(n_Na-x)+n_H=(n_Cl-x)+n_OH;
eq7:=2*ionic=(n_H/m)+((n_Cl-x)/m)+((n_Na-x)/m)+(n_OH/m);
eq8:=u_H=0.4077*ionic^2-0.3152*ionic+0.9213;
eq9:=u_Na=0.0615*ionic^2-0.2196*ionic+0.8627;
eq10:=u_OH=0.1948*ionic^2-0.1803*ionic+0.8887;
eq11:=m=r*V;
eq12:=u_Cl=(1.417625986641341e-01)*exp(-ionic/2.199955601666953e-02)+2.369460669647978e-01*exp(-ionic/3.756472377688394e-01)+5.859738096037875e-01;
 

K_1 = (n_Cl-x)*u_Cl*n_H*u_H/(n_HCl*m)

  

K_2 = (n_Na-x)*u_Na*n_OH*u_OH/(n_NaOH*m)

  

K_w = n_H*u_H*n_OH*u_OH/m^2

  

n_NaCl-x = n_Na-x+n_NaOH

  

n_NaCl-x = n_Cl-x+n_HCl

  

n_Na-x+n_H = n_Cl-x+n_OH

  

2*ionic = n_H/m+(n_Cl-x)/m+(n_Na-x)/m+n_OH/m

  

u_H = .4077*ionic^2-.3152*ionic+.9213

  

u_Na = 0.615e-1*ionic^2-.2196*ionic+.8627

  

u_OH = .1948*ionic^2-.1803*ionic+.8887

  

m = r*V

  

u_Cl = .1417625986641341*exp(-45.4554627939891*ionic)+.2369460669647978*exp(-2.66207201719228*ionic)+.5859738096037875

 (2)
> 

#constants
K_1:=10^(8);K_2:=10^(-0.2);K_w:=10^(-13.995);n_NaCl:=1.2*10^(-4);V:=5*10^(-8);r:=997.0;x:=0;
 

100000000

  

.630957344480193

  

0.101157945425990e-13

  

0.120000000000000e-3

  

1/20000000

  

997.0

  

0

 (3)
> 

eq:={seq(eq||i,i=1..12)}:
> 

Eq:=convert(eq,rational):
> 

nops(Eq);

12

 (4)
> 

X:=indets(Eq,name);nops(%);

{ionic, m, n_Cl, n_H, n_HCl, n_Na, n_NaOH, n_OH, u_Cl, u_H, u_Na, u_OH}

  

12

 (5)
> 

S:=eliminate(Eq, X minus {ionic}):
> 

f:=simplify(S[2])[]:
> 

############
> 

Digits:=200;

200

 (6)
> 

############
> 

plot(f, ionic=0..3);

 
> 

ionicS:=fsolve(f):
> 

eval(S[1], ionic=ionicS):
> 

evalf[15]([ionic=ionicS, %[]]);

[ionic = 2.40722164287621, m = 0.498500000000000e-4, n_Cl = 0.120000000000000e-3, n_H = -0.570189355755727e-11, n_HCl = -0.203222877739516e-18, n_Na = 0.120000004599272e-3, n_NaOH = -0.459927245018502e-11, n_OH = -0.110262131059512e-11, u_Cl = .586364294954344, u_H = 2.52504946683014, u_Na = .690449163557180, u_OH = 1.58348862197850]

 (7)
> 

 

> 

 

> 

 

> 

###### Precision check
> 

eval(f,ionic=ionicS);

0.23962833294454762502864124840136613066328936070999815539241196442017321814523037838085545177856951081566864562952416179056576403619594972564687659275843824010627201071429444585847063334975635851082636e-66

 (8)
> 

eval(f,ionic=ionicS-1e-190);

-0.21177153923974396361906170327470731797368197252746086982804407355632808153584734689408100550931080518334716557509197798241249396698817057004042718885026979469391788946875771652742342222284717170262391e-65

 (9)
> 

eval(f,ionic=ionicS+1e-190);

0.19949058717633589783634383929413730377718839279107346436418296037979420410590429000206216360565911775404414748657886469064599856013312814660102476347139983488847144891965012617717680226367217800405131e-65

 (10)
> 

 

 

assuming...

vv 13805
July 25 2020
0 1 
solve( sum((1/(1+x+x^2))*y^n, n=1..infinity) = 1, parametric=true, real) assuming abs(y)<1;

    [{x = x, y = (x^2 + x + 1)/(x^2 + x + 2)}]

assuming is necessary, otherwise the series is divergent.

 

answer...

vv 13805
July 23 2020
4 3

Maple cannot compute the integral directly.
We shall use residues.

> 

restart;
> 

Int( (cosh(a*x)-1)/sinh(b*x)/x, x=0..infinity) assuming a>0, b>0;
ans := -ln(cos(a*Pi/(2*b))); # abs(a)<b

Int((cosh(a*x)-1)/(sinh(b*x)*x), x = 0 .. infinity)

  

-ln(cos((1/2)*a*Pi/b))

 (1)
> 

f:=(cosh(a*x)-1)/sinh(b*x)/x;
# f is even, so we'll assume  0<a<b

(cosh(a*x)-1)/(sinh(b*x)*x)

 (2)
> 

singular(f,x);

{x = 0}, {x = I*Pi*_Z1/b}

 (3)
> 

r_n := residue(f, x=n*I*Pi/b) assuming n::integer;

-I*(cos(a*n*Pi/b)-1)/((-1)^n*n*Pi)

 (4)
> 

J:=Pi*I*sum(r_n, n=1..infinity) assuming a>0,b>a; # f is even

I*Pi*(((1/2)*I)*ln(1+exp(I*a*Pi/b))/Pi+((1/2)*I)*ln(1+exp(-I*a*Pi/b))/Pi-I*ln(2)/Pi)

 (5)
> 

evalc(J) assuming  a>0,b>a;

-Pi*((1/2)*ln((1+cos(a*Pi/b))^2+sin(a*Pi/b)^2)/Pi-ln(2)/Pi)

 (6)
> 

ANSWER:=expand(simplify(eval(%, a=2*b*c/Pi))) assuming 0<c,c<Pi/2;

-ln(cos(c))

 (7)
> 

ANSWER := eval(ANSWER, c = Pi*a/b/2);

-ln(cos((1/2)*a*Pi/b))

 (8)


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Implication...

vv 13805
July 22 2020
2 0

 

 

 

Probably you mean:

 

Show that  A  implies  B, where  x >0  and

 

> 

A := y = (sqrt(x) + 10)^(1/3) - (sqrt(x) - 10)^(1/3);

y = (x^(1/2)+10)^(1/3)-(x^(1/2)-10)^(1/3)

 (1)
> 

B := y^3 = 20 - 3*(x - 100)^(1/3)*y;

y^3 = 20-3*(x-100)^(1/3)*y

 (2)
> 

###############################################
> 

factor( eval( (lhs-rhs)(B), A));;

-3*((x^(1/2)+10)^(1/3)-(x^(1/2)-10)^(1/3))*((x^(1/2)+10)^(1/3)*(x^(1/2)-10)^(1/3)-(x-100)^(1/3))

 (3)
> 

combine(%) assuming x>0;

-3*((x^(1/2)+10)^(1/3)-(x^(1/2)-10)^(1/3))*(((x^(1/2)+10)*(x^(1/2)-10))^(1/3)-(x-100)^(1/3))

 (4)
> 

expand(%);

0

 (5)
> 

# QED
# Note: The implication holds for x :: real

 

diff/f...

vv 13805
July 19 2020
3 7 
restart;
`diff/f` := (u,x) -> f[op(procname)-1](u)*diff(u,x):

Examples

diff(f[4](x), x, x);
    f[2](x)

diff(x^2 + f[4](f[5](2*x)),  x,  x);
    2 + 4*f[2](f[5](2*x))*f[4](2*x)^2 + 4*f[3](f[5](2*x))*f[3](2*x)

 

proc...

vv 13805
July 19 2020
5 0 
f:=proc(x,y)
   local lm := op(procname);
   if not type(procname, 'indexed') or nops([lm])<>2 then error "Two indices needed" fi;
   x^lm[1] + y^lm[2]
end proc:

 

procedure...

vv 13805
July 18 2020
1 1

https://mapleprimes.com/questions/230075-Rewriting-General-Formula-#answer271028

Answer...

vv 13805
July 16 2020
1 2

It's possible in any dimension using quadratic forms. Here is an example when the quadric has a center.

[Your question is again more about maths than programming].

> 

restart;
> 

with(LinearAlgebra):
> 

n:=3;
f := 7*x[1]^2 + 4*x[1]*x[2] - 4*x[1]*x[3] + 4*x[2]^2 - 2*x[2]*x[3] + 4*x[3]^2 - 4*x[1] + 5*x[2] + 4*x[3] - 18

3

  

7*x[1]^2+4*x[1]*x[2]-4*x[1]*x[3]+4*x[2]^2-2*x[2]*x[3]+4*x[3]^2-4*x[1]+5*x[2]+4*x[3]-18

 (1)
> 

plots:-implicitplot3d(f, x[1]=-5..5, x[2]=-5..5, x[3]=-5..5, style=surface, numpoints=64000, scaling=constrained);

 
> 

A:=VectorCalculus:-Hessian(1/2*f,[seq(x[i],i=1..n)]);

Matrix(3, 3, {(1, 1) = 7, (1, 2) = 2, (1, 3) = -2, (2, 1) = 2, (2, 2) = 4, (2, 3) = -1, (3, 1) = -2, (3, 2) = -1, (3, 3) = 4})

 (2)
> 

b:=eval(Vector( [ seq(diff(f,x[i]),i=1..n)]), [seq(x[i]=0,i=1..n)]);

Vector(3, {(1) = -4, (2) = 5, (3) = 4})

 (3)
> 

c:=eval(f,[seq(x[i]=0,i=1..n)]);

-18

 (4)
> 

X:=Vector([seq(x[i],i=1..n)]);

Vector(3, {(1) = x[1], (2) = x[2], (3) = x[3]})

 (5)
> 

solve([ seq(diff(f,x[i]),i=1..n)],{seq(x[i],i=1..n)}); # the center

{x[1] = 11/27, x[2] = -26/27, x[3] = -29/54}

> 

X0:= Vector[column]( eval([seq(x[i],i=1..n)],%) );

Vector(3, {(1) = 11/27, (2) = -26/27, (3) = -29/54})

 (6)
> 

J,Q := JordanForm(A,output=['J','Q']);

J, Q := Matrix(3, 3, {(1, 1) = 3, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 9, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 3}), Matrix(3, 3, {(1, 1) = 5/6, (1, 2) = 2/3, (1, 3) = 1/2, (2, 1) = -1/3, (2, 2) = 1/3, (2, 3) = 0, (3, 1) = 4/3, (3, 2) = -1/3, (3, 3) = 1})

> 

T:=Matrix(GramSchmidt([seq( Q[..,j],j=1..n)],normalized)); # T is orthogonal

Matrix(3, 3, {(1, 1) = (5/93)*sqrt(93), (1, 2) = (1/3)*sqrt(6), (1, 3) = -(1/31)*sqrt(62), (2, 1) = -(2/93)*sqrt(93), (2, 2) = (1/6)*sqrt(6), (2, 3) = (7/62)*sqrt(62), (3, 1) = (8/93)*sqrt(93), (3, 2) = -(1/6)*sqrt(6), (3, 3) = (3/62)*sqrt(62)})

 (7)
> 

fnew:=simplify( (T.X+X0)^+. A. (T.X+X0) + b.(T.X+X0) + c ); # ==> ellipsoid

-602/27+3*x[1]^2+9*x[2]^2+3*x[3]^2

 (8)


 

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