#
# -
#
#
# WORKSHEET#17
#
# A system of non-polynomial equations revisited
#
#
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# We reconsider the critical point analysis for the function f(x,y) from the last worksheet.
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> f:=(x,y)->sin(y-x^2-1)+cos(2*y^2-x);
2 2
f := (x,y) -> sin(y - x - 1) + cos(2 y - x)
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> f1:=D[1](f);f2:=D[2](f);
2 2
f1 := (x,y) -> - 2 cos(- y + x + 1) x - sin(- 2 y + x)
2 2
f2 := (x,y) -> cos(- y + x + 1) + 4 sin(- 2 y + x) y
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> e1:=f1(x,y)=0;e2:=f2(x,y)=0;
2 2
e1 := - 2 cos(- y + x + 1) x - sin(- 2 y + x) = 0
2 2
e2 := cos(- y + x + 1) + 4 sin(- 2 y + x) y = 0
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> e1+2*x*e2;
2 2
- 2 cos(- y + x + 1) x - sin(- 2 y + x)
2 2
+ 2 x (cos(- y + x + 1) + 4 sin(- 2 y + x) y) = 0
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> simplify(");
2 2
- sin(- 2 y + x) + 8 x sin(- 2 y + x) y = 0
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> factor(");
2
sin(- 2 y + x) (- 1 + 8 x y) = 0
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> e3:=";
2
e3 := sin(- 2 y + x) (- 1 + 8 x y) = 0
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> simplify(4*y*e1+e2);
2 2
- 8 y cos(- y + x + 1) x + cos(- y + x + 1) = 0
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> e4:=factor(");
2
e4 := - cos(- y + x + 1) (- 1 + 8 x y) = 0
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# By elementary manipulation of equations we have reduced the more complicated system
# of equations e1=0, e2=0 to the much easier, but equivalent system, e3=0, e4=0. By
# inspection we see that the solutions of this system are given by either 8xy=1 or
#
# sin(-2y^2+x)=0 and cos(-y+x^2+1)=0.
#
# What does Maple say about these equations?
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> solve(e3,x);
2 1
2 y , ---
8 y
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# Maple does not give the complete solution of the equation (-1+8xy)sin(-2y^2+x)=0. But
# we can easily see that the complete solution is either 8xy=1 or x=2y^2+mPi, where m is
# any integer.
#
# As far as e4=0 is concerned we see that the complete solution is either 8xy=1 or
#
# y=x^2+1+(n+1/2)Pi, where n is an integer.
#
# Thus there are 2 types of critical points for f(x,y)=sin(y-x^2-1)+cos(2*y^2-x), namely
# those points (x,y) where either 8xy=1 or
#
# x=2y^2+mPi and y=x^2+1+(n+1/2)Pi for some integers m,n.
#
# Notice that if x=2y^2+mPi and y=x^2+1+(n+1/2)Pi for some integers m,n then
# f(x,y)=sin(y-x^2-1)+cos(2y^2-x)=sin((n+1/2)Pi)+cos(-mPi)=(-1)^n+(-1)^m, and this
# equals -2, +2 or 0. Since f(x,y) assumes values between -2 and +2 we know that those
# points where f=2 are maxima and those points where f=-2 are minima.
#
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> subs({x-2*y^2=m*Pi,-y+x^2+1=(n+1/2)*Pi},f(x,y));
- sin((n + 1/2) Pi) + cos(m Pi)
> A:=simplify(");
A := - cos(Pi n) + cos(m Pi)
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> weknow:=sin(n*Pi)=0,sin(m*Pi)=0,cos(m*Pi)=(-1)^m,cos(n*Pi)=(-1)^n,\
sin((n+1/2)*Pi)=(-1)^n,cos((n+1/2)*Pi)=0;
m n
weknow := sin(Pi n) = 0, sin(m Pi) = 0, cos(m Pi) = (-1) , cos(Pi n) = (-1) ,
n
sin((n + 1/2) Pi) = (-1) , cos((n + 1/2) Pi) = 0
> subs(weknow,A);\
n m
- (-1) + (-1)
> with(linalg):
Warning: new definition for norm
Warning: new definition for trace
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> H:=hessian(f(x,y),[x,y]);
H := 2 2 2 2
[4 sin(- y + x + 1) x - 2 cos(- y + x + 1) - cos(- 2 y + x),
2 2
- 2 sin(- y + x + 1) x + 4 cos(- 2 y + x) y]
2 2
[- 2 sin(- y + x + 1) x + 4 cos(- 2 y + x) y,
2 2 2 2
sin(- y + x + 1) - 16 cos(- 2 y + x) y + 4 sin(- 2 y + x)]
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#
> H;
H
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# This is a bit peculiar.
#
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> eval(H);
2 2 2 2
[4 sin(- y + x + 1) x - 2 cos(- y + x + 1) - cos(- 2 y + x),
2 2
- 2 sin(- y + x + 1) x + 4 cos(- 2 y + x) y]
2 2
[- 2 sin(- y + x + 1) x + 4 cos(- 2 y + x) y,
2 2 2 2
sin(- y + x + 1) - 16 cos(- 2 y + x) y + 4 sin(- 2 y + x)]
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> H[1,1];
2 2 2 2
4 sin(- y + x + 1) x - 2 cos(- y + x + 1) - cos(- 2 y + x)
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> subs({x-2*y^2=m*Pi,-y+x^2+1=(n+1/2)*Pi},eval(H));
2
[4 sin((n + 1/2) Pi) x - 2 cos((n + 1/2) Pi) - cos(m Pi),
- 2 sin((n + 1/2) Pi) x + 4 cos(m Pi) y]
[- 2 sin((n + 1/2) Pi) x + 4 cos(m Pi) y,
2
sin((n + 1/2) Pi) - 16 cos(m Pi) y + 4 sin(m Pi)]
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> subs(weknow,");
[ n 2 m n m ]
[ 4 (-1) x - (-1) - 2 (-1) x + 4 (-1) y ]
[ ]
[ n m n m 2 ]
[ - 2 (-1) x + 4 (-1) y (-1) - 16 (-1) y ]
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> det(");
n 2 m 2 m n n m
- 64 (-1) x (-1) y - (-1) (-1) + 16 (-1) x (-1) y
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> factor(");
m n 2
- (-1) (-1) (- 1 + 8 x y)
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# Assume that 8xy is not equal to 1. If m and n have the same parity then the determinant
# is negative, in which case we have a saddle point. If m and n have opposite parity then
# the critical point can be either a maximum or a minimum. To decide which merely
# examine the [1,1] entry of the hessian. If this is positive then both eigenvalues will be
# positive and if it is negative both eigenvalues will be negative.
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