## 15379 Reputation

12 years, 110 days

## Numeric solution...

Probably your system can only be solved numerically. To do this, specify the values of all 15 parameters and set the initial conditions. I took them from 1 to 15 and arbitrarily selected 3 initial conditions (see code below):

 > restart; odeA := {m*diff(x(t), t, t) = -m*A*sin(2*Pi*f*t) - k*x(t) + 0.5*q(t)*(N__f*epsilon__0*L*(-2*x(t)/(G^2 - x(t)^2) + 2*(G__1^2 - x(t)^2)*x(t)/(G^2 - x(t)^2)^2)*(G^2 - x(t)^2)/(2*tan(alpha)*(G__1^2 - x(t)^2)) + N__f*epsilon__0*L*(-2*x(t)/(G^2 - x(t)^2) + 2*(G__2^2 - x(t)^2)*x(t)/(G^2 - x(t)^2)^2)*(G^2 - x(t)^2)/(2*tan(alpha)*(G__2^2 - x(t)^2)))/(N__f*epsilon__0*L*ln((G__1^2 - x(t)^2)/(G^2 - x(t)^2))/(2*tan(alpha)) + N__f*epsilon__0*L*ln((G__2^2 - x(t)^2)/(G^2 - x(t)^2))/(2*tan(alpha)) + C__p) - d*diff(x(t), t), diff(q(t), t) = (q(t)/(N__f*epsilon__0*L*ln((G__1^2 - x(t)^2)/(G^2 - x(t)^2))/(2*tan(alpha)) + N__f*epsilon__0*L*ln((G__2^2 - x(t)^2)/(G^2 - x(t)^2))/(2*tan(alpha)) + C__p) + V__bias)/R1};
 (1)
 > indets(odeA);
 (2)
 > params:={A, C__p, G, G__1, G__2, L, N__f, R1, V__bias, alpha, d, f, k, m,  epsilon__0};
 (3)
 > n:=nops(params); Sol:=dsolve({eval(odeA,params=~{\$ 1..n})[],x(0)=1,q(0)=2,D(x)(0)=0}, numeric);
 (4)
 > plots:-odeplot(Sol,[[t,x(t)],[t,q(t)]], t=0..7,color=[red,blue]);
 >

## explicit...

You do not need to use  plots:-implicitplot  command because one variable is easily expressed through another one  mu=ln(1+x)/x . Immediately we get a smooth curve without any additional options. Here are 2 variants for plotting:

restart;
plot([ln(1+x)/x, x, x = -1 .. 5], color = black, labels = [mu, x], view = [0 .. 5, -1 .. 5]);  # x as a function of mu
plot(ln(1+x)/x, x = -1 .. 5, color = black, labels = [x, mu], view = [-1 .. 5, 0 .. 5]); # mu as a function of x 

## Recursive sequence...

I do not understand what the problem is. Maple handles this task easily:

 > restart; F := rsolve({16*s(n+1) = 2+12*s(n)-2*s(n-1), s(1) = 1, s(2) = 5/8}, s); s:=unapply(F, n); seq(s(n), n=1..10); plot([seq([n,s(n)], n=1..10)], style=point, symbol=solidcircle, color=red,size=[1000,300], view=[0..10,0..1]); limit(s(n), n=infinity);
 (1)
 > # Without rsolve restart; s:=proc(n) option remember; if n=1 then return 1 elif n=2 then return 5/8 else 1/8+3*s(n-1)*(1/4)-(1/8)*s(n-2) fi; end proc:
 > seq(s(n), n=1..10); plot([seq([n,s(n)], n=1..10)], style=point, symbol=solidcircle, color=red,size=[1000,300], view=[0..10,0..1]);
 >

Unfortunately, Optimization:-Maximize command in this example returns an erroneous result (I use Maple 2018.2), since it returns only a local extremum. For the correct solution, it is useful to plot graphs and use  initialpoint  option:

 > restart;
 > Optimization:-Maximize(2*x^2 + 2*y^2 + y, {2*x + y <= 6, y^2 - x <= 2});   # This is an incorrect result plots:-inequal({2*x + y <= 6, y^2 - x <= 2},x=-3..5,y=-3..3); solve({2*x + y = 6, y^2 - x = 2}); plot3d(2*x^2 + 2*y^2 + y, y=-5/2..2,x=(6-y)/2..y^2-2, axes=normal); Optimization:-Maximize(2*x^2 + 2*y^2 + y, {2*x + y <= 6, y^2 - x <= 2}, initialpoint=[x=3,y=-2]);  # OK eval(2*x^2 + 2*y^2 + y,[x = 17/4, y = -5/2]); # Symbolic result evalf(%);
 (1)
 >

Edit.

## eval...

As a workaround use the  eval  command:

restart;
f(x):=a+b;
a:=5;
b:=2;
f(x):=eval(f(x));

Edit. In addition to the acer's advice, you can also use indexed names:

restart;
f[x]:=a+b;
a:=5;
b:=2;
f[x]:=f[x];


gamma  is a protected constant in Maple, and  S  cannot be differentiated by a constant. Execute it first

local gamma;

## Proof...

We have

f = 2*x^5-x^3*y+2*x^2*y^2-x*y^3+2*y^5 = 2*(x^5+y^5) - x*y*(x^2+y^2-2*x*y) = 2*(x^5+y^5) - x*y*(x-y)^2

If  x<0  and  y<0  then both summands are negative.

## Procedure...

A simple procedure called  lim allows you to get the result in the desired form. I did not find in Maple the possibility of directly exporting this result to MathType, but you can first convert it to latex, and then find the conversion of latex to MathType on the Internet:

 > restart; lim:=proc(Expr,Point) 'limit'(Expr,Point)=limit(Expr,Point); end proc:
 > # Example of use: lim((x^2-4)/(x-2), x=2);
 (1)
 > 'lim'((x^2-4)/(x-2), x=2);
 (2)
 > latex(%);
 \lim _{x\rightarrow 2}{\frac {{x}^{2}-4}{x-2}}=4
 >

## Check for an arbitrary n...

You can easily verify this identity for any  n :

restart;
F:=unapply(rsolve({F(n+1) = F(n)+F(n-1),F(1)=1,F(2)=1},F(n)),n);
combine(expand(F(n+1)*F(n+2) - F(n)*F(n+3)));

Edit. The proof by mathematical induction:

restart;
simplify(subs(n=k+1,F(n+1)*F(n+2) - F(n)*F(n+3)), {F(k+1)*F(k+2) - F(k)*F(k+3)=a,F(k+2)=F(k)+F(k+1),F(k+3)=F(k+1)+F(k+2),F(k+4)=F(k+2)+F(k+3)});
eval(%, a=(-1)^k);
simplify(%-(-1)^(k+1));


## if...

f:=(i,j)->((x-GaußKnoten[j])/(GaußKnoten[i]-GaußKnoten[j]));
n:=3;
*(seq(seq(if(i<>j,f(i,j),NULL), j=1..n), i=1..n));


Edit. In order to explicitly multiply all this, use the  expand  command (code continuation):

expand(%);

## Plot and animation...

Using the  plots:-odeplot  command, we can easily not only plot the curve, but also animate it:

plots:-odeplot(F, [x(t), y(t), z(t)], t = 0 .. 0.7, color = red, thickness = 2, axes = normal, scaling = constrained)

plots:-odeplot(F, [x(t), y(t), z(t)], t = 0 .. 0.7, color = red, thickness = 2, axes = normal, scaling = constrained, frames = 15);

## convert(... , diff)...

Maybe the following more detailed form of writing will be clearer:

diff(w(y(t), t), t);
convert(%, diff);

## plots:-inequal...

This is easy to do if you use the plots:-inequal  command and Cartesian coordinates:

with(plots):
P1 := plot([-sin(t), t, t = 0 .. 2*Pi], coords = polar, color = red):
P2 := plot([cos(t), t, t = 0 .. 2*Pi], coords = polar, color = blue):
Shade := inequal({(x-1/2)^2+y^2<1/4,x^2+(y+1/2)^2<1/4}, x=-0.5..1, y=-1..0.5, color=yellow, nolines):
display(P1, P2, Shade, scaling = constrained);


Below is another method based on approximating a region on the plane by a polygon:

with(plots): with(plottools):
P1 := plot(-sin(t), t = 0 .. 2*Pi, coords = polar, color = red):
P2 := plot(cos(t), t = 0 .. 2*Pi, coords = polar, color = blue):
P:=polygon([seq([-sin(t)*cos(t),-sin(t)*sin(t)],t=-evalf(Pi/4)..0,0.1),seq([cos(t)*cos(t),cos(t)*sin(t)],t=-evalf(Pi/2)..-evalf(Pi/4),0.1)],color=yellow,style=surface):
display(P1, P2, P, scaling = constrained);


This method is automated in my post  https://www.mapleprimes.com/posts/145922-Perimeter-Area-And-Visualization-Of-A-Plane-Figure-  and is applicable to any flat region bounded by a piecewise smooth non-selfintersecting curve (the curve can be specified in Cartesian or polar coordinates or parametrically).

Edit.

## eval, alias...

As alternatives to  subs , you can also use  eval  or  alias :

restart;
s1 := RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S):

s2 := -(D1*RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S)*D6-S*D6+RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S))/D2/D6/RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S):

eval(s2, s1=s);
alias(s=s1):
s2;


If  subs  and  eval  are fairly close commands, then  alias  is essentially different from them. If in the first two cases  s  is just a symbol, then as for  alias  Maple remembers what  s  means. This may be more convenient in subsequent calculations. See help on these commands.

## Plotting...

restart;
lambda:=x+I*y:
evalc(abs(1+1/3*lambda+1/18*lambda^2-1/324*lambda^3+1/1944*lambda^4)-1);
plots:-implicitplot(%, x=-15..15, y=-15..15, gridrefine=5, scaling=constrained, size=[550,550]);

Edit. It would be interesting to show that the curves on the right (bottom and top) are exact circles (this is my assumption) and learn their equations.

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