Checking behavior of critical points other than 2 nd derivative test method

Question

Question is to find local extrema and saddle points. Given $f(x,y)=\sqrt{56x^{2}-8y^{2}-16x-31}+1-8x$

I find critical points which are $(-2,0)$ and $(16/7,0)$.But using 2 nd derivative test to identify their behavior looks too lengthy.Any other way to proceed? Thanks in advance.

Answer 1

As far as I see, if we already found the extremum points, then we just need to evaluate the function at the point itself and a few points around it to guess/estimate the behavior. It is in my opinion the most simple way for functions with explicitly known expression.

Of course, there's a downside: if the function is badly behaved (oscillations, singularities, etc.) this method may not give anything reliable, no matter how many points we use.

For a more reliable way see @phaedo's answer.

But the OP's function is nice enough, so it makes sense to try this simple method.

Also, the saddlepoints can give a bit of trouble, depending on where we check.

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Let us pick the point from the OP $(-2,0)$ and check the function values around it:

$$f(-2,0)=\sqrt{56 \cdot 4+16 \cdot 2-31}+1+8 \cdot 2=32$$

$$f(-2.1,0)=33.59747 \ldots$$

$$f(-1.9,0)=30.39718 \ldots$$

$$f(-2,0.1)=f(-2,-0.1)=31.99767 \ldots$$

This is neither a maximum nor a minimun, so I would say, saddlepoint. Note that the function is even in $y$, so checking around $y=0$ is simple.

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Now for $(16/7,0)$:

$$f(16/7,0)=-2.285714 \ldots$$

$$f(17/7,0)=-2.290772 \ldots$$

$$f(15/7,0)=-2.291607 \ldots$$

$$f(16/7,1/7)=f(16/7,-1/7)=-2.296764 \ldots$$

This point looks like a maximum to me, as all the values around it are smaller (note the negative signs).

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Of course, as I said, this method may lead to uncertainties for less well behaved functions, and numerical errors may also play a role, so if you want to be certain, you have to use the Hessian matrix of second derivatives.

Using more than $5$ points (for example, $9$ points) should make the method more reliable.

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Here, I made a little R program which uses the nine point method, where you can input a two variable function, a point you want to check, and the distance around it where we check the values. It wouldn't know a saddle point though, it only recognizes a maximum or a minimum:

PointCheck <- function(x0,y0,d,f){
f0 <- f(x0,y0);
f1 <- f(x0-d,y0);
f2 <- f(x0-d,y0-d);
f3 <- f(x0,y0-d);
f4 <- f(x0+d,y0-d);
f5 <- f(x0+d,y0);
f6 <- f(x0+d,y0+d);
f7 <- f(x0,y0+d);
f8 <- f(x0-d,y0+d);
c1 <- f0 > f1;
c2 <- f0 > f2;
c3 <- f0 > f3;
c4 <- f0 > f4;
c5 <- f0 > f5;
c6 <- f0 > f6;
c7 <- f0 > f7;
c8 <- f0 > f8;
if(c1+c2+c3+c4+c5+c6+c7+c8 == 8){return(paste(c("(",x0,y0,") is a maximum"), collapse=" "))}
if(c1+c2+c3+c4+c5+c6+c7+c8 == 0){return(paste(c("(",x0,y0,") is a minimum"), collapse=" "))}
if(c1+c2+c3+c4+c5+c6+c7+c8 > 0 && c1+c2+c3+c4+c5+c6+c7+c8 < 8){return(paste(c("(",x0,y0,") is neither a maximum, nor a minimum"), collapse=" "))}}

The output for $(-2,0)$:

x0 <- -2;
y0 <- 0;
d <- 0.05;
f <- function(x,y){sqrt(56*x^2-7*y^2-16*x-31)-8*x+1}
PointCheck(x0,y0,d,f)

[1] "( -2 0 ) is neither a maximum, nor a minimum"

The output for $(16/7,0)$:

x0 <- 16/7;
y0 <- 0;
d <- 0.05;
f <- function(x,y){sqrt(56*x^2-7*y^2-16*x-31)-8*x+1}
PointCheck(x0,y0,d,f)

[1] "( 2.28571428571429 0 ) is a maximum"

Note, that $d$ is also a good estimate for the error of this method, so if we try to check a point closer than $d$ to our critical point, the program wouldn't recognize the difference.

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In fact, this function is a little tricky, because numerically it's hard to see if the second point is a maximum or not, because the plot for the whole range looks like this:

Now, plotting only the right side for $y=0$ gives us a more clear picture of a local maximum:

Answer 2

Based upon discussions below, the only critical point is (16/7, 0)

If you do not want to calculate second-order partial derivatives, you can examine the sign of the first order (FO) Taylor series remainder:

Using FO partial derivatives, the FO series at (16/7, 0) is -16/7, and the remainder is therefore:

$f(x,y)-(-16/7)=23/7-8x+\sqrt{56x^{2}-8y^{2}-16x-31}$

Furthermore:

$$ 23/7-8x+\sqrt{56x^2-8y^2-16x-31}=\frac{-(23/7-8x)^2+56x^{2}-8y^{2}-16x-31}{-23/7+8x+\sqrt{56x^{2}-8y^{2}-16x-31}} $$

Near (16/7, 0) the denominator is positive, and the numerator simplifies to $-(8/49) (16 - 7 x)^2 - 8 y^2 $ which is always negative

Conclusion: the remainder is always negative which proves that the critical point is a local maximum

Note that this method is an actual proof and does not require to calculate second-order partial derivatives at all!

Answer 3

There may be some symmetries that can be exploited. Let z=f(x,y).

$$(z+8x-1)^2=56x^2-8y^2-16x-31$$

$$z^2+64x^2+1 +16xz-2z-16x=56x^2-8y^2-16x-31$$

$$z^2+8x^2+16xz-2z+8y^2 +32=0$$

This is the equation for a hyperboloid of two sheets. The form of the equation suggests a 45 degree tilt from 1 or more coordinate axes and cylindrical symmetry.

Geometrically, vertices are expected at two different points and no saddle nodes.

In addition, symmetry of the equation suggests these vertices occur on the axis of the hyperboloids and are evenly distributed about the origin.

So these can be found by taking the above equation as a constraint and minimizing: $$g(x,y,z)=x^2+y^2+z^2$$

Resulting in $y=0$ and an equation in the multiplier.

Given the geometry of the hyperboloid, we'd expect extrema to occur at the vertices or at the boundaries being considered.

Knowing the geometry guarantees smooth behavior, no saddle points, and ways to disambiguate the extrema without taking a second derivative.