MaPal93

How to change the length or thickness of...

Asked: MaPal93 170 Product: Maple

May 28 2024

1 3

I noticed that in a legend with:

Three elements
linestyle = [solid, longdash, dash]
thickness = [4, 4, 4]

The three different linestyles are distinguishable in the plots (of course, since the curves span the whole plot area) but indistinguishable in the legend box. Since the legend box is too narrow, the symbols for a solid line, a longdash line, and a dash line are equivalent, resulting in confusion regarding which element is associated to each linestyle. Note that I don't want to use three different colors to achieve that or change my font sizes.

How can I "show more" of the linestyle in the legend box? "Longer" line symbols in the legend box would allow to dinstinguish a solid from a longdash from a dash even when all three are quite thick.
Alternatively, can I reduce the thickness of the line symbols in the legend box (leaving unaltered the thickness of the lines in the actual plot)? "Slimmer" line symbols in the legend box would allow to dinstinguish a solid from a longdash from a dash even when all three are quite short.

helping fsolve()?...

Asked: MaPal93 170 Product: Maple 2023

May 27 2024

1 22

I am using fsolve() to solve a highly nonlinear system of 6 equations in 6 variables: lambda_d1, lambda_i1, lambda_d2, lambda_i2, lambda_d3, lambda_i3.

fsolve() doesn't "solve"! I usually help fsolve() with some initial conditions and with the expected signs of the solution but in this case it's not enough. I noticed that if I comment out the expected signs line (that is, if I don't impose my 6 lambdas to be strictly positive), the fsolve() works.
How do I help fsolve() to pin down only positive solutions at each iteration? I have no reasons to believe that there aren't any positive solutions for all 6 lambdas...

Worksheet: fsolve_help.mw

thank you.

alias() in alias(), how to do it?...

Asked: MaPal93 170 Product: Maple

May 09 2024

1 0

I want to define an alias L_2 in terms of another alias L_1, the latter being a RootOf(). I noticed that doing the following does not work: alias_in_alias.mw. How to fix it? Should I define the two alias sequantially? That is, L_1 first and then (in a follow up alias()) L_2?

Sorry for the super complicated expressions. You could perhaps show me with a simpler example. Thanks!

Verify a solution of a 2-equations syste...

Asked: MaPal93 170 Product: Maple

May 08 2024

1 1

I am trying to verify a solution by checking whether eval() returns 0 but it's taking me forever. If I recall correctly it once returned [0,0] so I am quite confident that the first and only positive root of that polynomial solves my system. I am now running again the same calculation but somehow eval() is stuck in "Evaluating...". I am not sure if it matters here, but parameters gamma, sigma_v, and sigma_d are strictly positive while -1<rho<+1 (rho is a correlation coefficient).

How else can I verify such solution?

EDIT: THIS IS NOT A DUPLICATE QUESTION AND SHOULDN'T BE TAGGED AS SUCH

>

restart;

>	local gamma:

Equations:

>

eq1 := (gamma*sigma__v^2*(-1 + rho__v) - 2*lambda__2)*(rho__v*sigma__v^2 + sigma__v^2)/((2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2)*((gamma*sigma__v^2*(-1 + rho__v) - 2*lambda__2)^2*(2*rho__v*sigma__v^2 + 2*sigma__v^2)/(2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2)^2 + (-lambda__1*(gamma*sigma__v^2*(-1 + rho__v) - 4*lambda__2)/(2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2) + 1)^2*sigma__d^2 + gamma^2*lambda__2^2*sigma__v^4*(-1 + rho__v)^2*sigma__d^2/(2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2)^2)):

eq2 := (gamma*sigma__v^2*(-1 + rho__v) - 2*lambda__1)*(rho__v*sigma__v^2 + sigma__v^2)/((2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2)*((gamma*sigma__v^2*(-1 + rho__v) - 2*lambda__1)^2*(2*rho__v*sigma__v^2 + 2*sigma__v^2)/(2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2)^2 + (-(gamma*sigma__v^2*(-1 + rho__v) - 4*lambda__1)*lambda__2/(2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2) + 1)^2*sigma__d^2 + gamma^2*lambda__1^2*sigma__v^4*(-1 + rho__v)^2*sigma__d^2/(2*gamma*(lambda__1 + lambda__2)*(-1 + rho__v)*sigma__v^2 - 8*lambda__1*lambda__2)^2)):

Eq1, Eq2 := eq1 - lambda__1, eq2 - lambda__2:

Solution:

>

Gamma := gamma*sigma__v*sigma__d:
L__2 := RootOf(-8*(rho + 1)^4*_Z^4 + 12*(rho + 1)^3*Gamma*(rho - 1)*_Z^3 - 5*(rho + 1)^2*(-4/5 + Gamma^2*rho^2 + 2*(-2/5 - Gamma^2)*rho + Gamma^2)*_Z^2 - 4*(rho + 1)*Gamma*(rho^2 - 1)*_Z + Gamma^2*(rho + 1)*(rho - 1)^2);
l__2 := L__2*sigma__v*(rho__v + 1)/sigma__d:
quartic_solution := lambda__2 = simplify(allvalues~([l__2]))[1]:

(1)

Check: TOO SLOW

>	simplify(eval([eval(Eq1, lambda__1 = lambda__2), eval(Eq2, lambda__1 = lambda__2)], quartic_solution));

Download solution_check.mw

Signs of first and second derivatives of...

Asked: MaPal93 170 Product: Maple 2024

May 06 2024

1 2

I have a complicated bivariate function f(Gamma,rho) that is a RootOf of a quartic. I know that it is strictly positive (one of the four roots at least) for Gamma=0..10 and rho in (-1,+1), with bounds excluded.

I need to find the signs of its first and second derivatives (wrt to Gamma and wrt to rho: 4 derivatives in total).

I encounter numerical issues when I plot3d the derivatives using D[]() vs. fdiff() (numerical function evaluations of the RootOf). I was hoping for the two commands to produce the same output, but they don't it seems. What's going on?

Script:

>	restart; _quartic := RootOf(-8(rho + 1)^4_Z^4 + 12(rho + 1)^3Gamma(rho - 1)_Z^3 - 5(rho + 1)^2(-4/5 + Gamma^2rho^2 + 2(-2/5 - Gamma^2)rho + Gamma^2)_Z^2 - 4(rho + 1)Gamma(rho^2 - 1)_Z + Gamma^2(rho + 1)(rho - 1)^2); convert(_quartic,radical): f(Gamma,rho) := simplify(%):

RootOf((8*rho^3+24*rho^2+24*rho+8)*_Z^4+(-12*Gamma*rho^3-12*Gamma*rho^2+12*Gamma*rho+12*Gamma)*_Z^3+(5*Gamma^2*rho^3-5*Gamma^2*rho^2-5*Gamma^2*rho+5*Gamma^2-4*rho^2-8*rho-4)*_Z^2+(4*Gamma*rho^2-4*Gamma)*_Z-Gamma^2*rho^2+2*rho*Gamma^2-Gamma^2)

(1)

Synthetic representation of derivatives

>

der1_Gamma := diff(_quartic, Gamma):
der1_rho := diff(_quartic, rho):

Diff('f(Gamma,rho)', Gamma) = collect~(normal(eval(der1_Gamma, _quartic = 'f(Gamma,rho)')), 'f(Gamma,rho)');
Diff('f(Gamma,rho)', rho) = collect~(normal(eval(der1_rho, _quartic = 'f(Gamma,rho)')), 'f(Gamma,rho)');

der2_Gamma := diff(der1_Gamma, Gamma):
der2_rho := diff(der1_rho, rho):

Diff('f(Gamma,rho)', Gamma$2) = collect~(normal(eval(der2_Gamma, _quartic = 'f(Gamma,rho)')), 'f(Gamma,rho)');
Diff('f(Gamma,rho)', rho$2) = collect~(normal(eval(der2_rho, _quartic = 'f(Gamma,rho)')), 'f(Gamma,rho)');

(2)

Signs of derivatives: fdiff (numerical function evaluations of the RootOf) vs. D[]()

>	restart; with(plots):

>	_quartic := RootOf(-8(rho + 1)^4_Z^4 + 12(rho + 1)^3Gamma(rho - 1)_Z^3 - 5(rho + 1)^2(-4/5 + Gamma^2rho^2 + 2(-2/5 - Gamma^2)rho + Gamma^2)_Z^2 - 4(rho + 1)Gamma(rho^2 - 1)_Z + Gamma^2(rho + 1)(rho - 1)^2):

>	plot3d(_quartic, Gamma=0..10, rho=-1..+1, labels=[Gamma,rho,Lambda(Gamma,rho)],axesfont=["helvetica","roman",20],labelfont=["helvetica","roman",30]);

Define it as a f and test it for Gamma=1 and rho=0.5

>	f := (Gamma,rho) -> RootOf(-8(rho + 1)^4_Z^4 + 12(rho + 1)^3Gamma(rho - 1)_Z^3 - 5(rho + 1)^2(-4/5 + Gamma^2rho^2 + 2(-2/5 - Gamma^2)rho + Gamma^2)_Z^2 - 4(rho + 1)Gamma(rho^2 - 1)_Z + Gamma^2(rho + 1)(rho - 1)^2): evalf(f(1.0,0.5));

(3)

Value at zero:

>	f(0,0): allvalues(%): fl := select(is, [allvalues(f(0,0))], positive)[];evalf(%);

(4)

Value at infinity (commented out because too slow)

>	#limit(f(x,y), {x = infinity, y = 0}): #fh := select(is, [allvalues(%)], positive)[];evalf(%);

Derivative at zero:

>	allvalues([D[1](f)(0,0)]): Dfl := %[1][];

(5)

Derivative at a point, evaluated, vs numerical derivative at a point:

>	D[1](f)(1,0.5): evalf(%); fdiff(f(x,y), x, {x = 1.0, y = 0.5}); fdiff(f, [1], [1.0,0.5]); D[2](f)(1,0.5): evalf(%); fdiff(f(x,y), y, {x = 1.0, y = 0.5}); fdiff(f, [2], [1.0,0.5]);

(6)

Can make a function out of fdiff

>	fDfG := (Gamma,rho) -> fdiff(f, [1], [Gamma,rho]); fDfr := (Gamma,rho) -> fdiff(f, [2], [Gamma,rho]);

(7)

Check for numerical values close to thresholds:

>

Digits := 15:
evalf('D[1]'(f)(0.1e-8,0.5));fdiff(f, [1], [0.1e-8,0.5]);
evalf('D[1]'(f)(0.1e-7,0.5));fdiff(f, [1], [0.1e-7,0.5]);
evalf('D[1]'(f)(0.1e-5,0.5));fdiff(f, [1], [0.1e-5,0.5]);
evalf('D[1]'(f)(0.00001,0.5));fdiff(f, [1], [0.00001,0.5]);
evalf('D[1]'(f)(0.001,0.5));fdiff(f, [1], [0.001,0.5]);

evalf('D[2]'(f)(1,-0.99));fdiff(f, [2], [1,-0.99]);
evalf('D[2]'(f)(1,-0.97));fdiff(f, [2], [1,-0.97]);
evalf('D[2]'(f)(1,-0.1));fdiff(f, [2], [1,-0.1]);
evalf('D[2]'(f)(1,0.98));fdiff(f, [2], [1,0.98]);
evalf('D[2]'(f)(1,-0.99));fdiff(f, [2], [1,-0.99]);

(8)

Compare with D (vertical range here to prevent effect of large values from fdiff near zero):

>	d1G := plot3d([D[1](f), fDfG], 0..10, -0.95..+0.95, view=-0.3..0, color = [red, blue]); d1r := plot3d([D[2](f), fDfr], 0..10, -0.95..+0.95, color = [red, blue]);

Second derivatives:

>	evalf('D[1,1]'(f)(1.0,0.5)); fdiff(f, [1, 1], [1.0,0.5]); evalf('D[2,2]'(f)(1.0,0.5)); fdiff(f, [2, 2], [1.0,0.5]); fD2fG := (Gamma,rho) -> fdiff(f, [1, 1], [Gamma]); fD2fr := (Gamma,rho) -> fdiff(f, [2, 2], [Gamma]);

(9)

>	d2G:= plot3d([D[1,1](f), fD2fG], 0..10, -0.9..+0.9, color = [red, blue]); d2r:= plot3d([D[2,2](f), fD2fr], 0..10, -0.9..+0.9, color = [red, blue]);

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

>	d1d2G := plot3d([fDfG, fD2fG], 0.1e-6 .. 10, -0.98 .. +0.98, axesfont=["helvetica","roman",20],labelfont=["helvetica","roman",30], size=[1000,1000]); d1d2r := plot3d([fDfr, fD2fr], 0.1e-6 .. 10, -0.98 .. +0.98, axesfont=["helvetica","roman",20],labelfont=["helvetica","roman",30], size=[1000,1000]);

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