Inflection Points: Understanding the Critical Points of a Surface

[Jhenrique]

Given a function $f(x)$, the critical points are where $f'=0$ and the inflection points are where $f''=0$. Given a function $f(x,y)$, the critical points are where $\vec{\nabla}f = \vec{0}$, so I can deduce that the inflection points are where $Hf=0$ . Correct?

 

[micromass]

How did you define an "inflection point" on a surface?

 

[Mark44]

Jhenrique, the definition you are using for inflection points isn't correct. For a single-variable function, an inflection point exists at any point in the domain of the function at which the concavity changes. For example, for f(x) = x^1/3, the only inflection point is at (0, 0) even though f'' is not defined for x = 0.

Also, just because f'' is zero at some point doesn't guaranteed that there is an inflection point there. For example, let f(x) = x⁴. Then f'(x) = 4x³ and f''(x) = 12x². f''(0) = 0, but this function has no inflection points, as the concavity never changes.

 

[Jhenrique]

How did you define an "inflection point" on a surface?

A definition well defined I don't know to say, but intuitively I'd say that is a point where tangent sphere have raius equal to infinity (like in a inflection point of a curve that have a tangent circle of radius equal to infinity)...

Jhenrique, the definition you are using for inflection points isn't correct. For a single-variable function, an inflection point exists at any point in the domain of the function at which the concavity changes. For example, for f(x) = x^1/3, the only inflection point is at (0, 0) even though f'' is not defined for x = 0.

Also, just because f'' is zero at some point doesn't guaranteed that there is an inflection point there. For example, let f(x) = x⁴. Then f'(x) = 4x³ and f''(x) = 12x². f''(0) = 0, but this function has no inflection points, as the concavity never changes.

Ok. Let's say that f'' = 0 is a necessary condition for the existence of a inflection point in a curve (and not a sufficient), so, Hf = 0 is a necessary condition for the existence of a inflection point in a surface?

 

[Mark44]

Jhenrique, the definition you are using for inflection points isn't correct. For a single-variable function, an inflection point exists at any point in the domain of the function at which the concavity changes. For example, for f(x) = x^1/3, the only inflection point is at (0, 0) even though f'' is not defined for x = 0.

Also, just because f'' is zero at some point doesn't guaranteed that there is an inflection point there. For example, let f(x) = x⁴. Then f'(x) = 4x³ and f''(x) = 12x². f''(0) = 0, but this function has no inflection points, as the concavity never changes.

Ok. Let's say that f'' = 0 is a necessary condition for the existence of a inflection point in a curve (and not a sufficient), so, Hf = 0 is a necessary condition for the existence of a inflection point in a surface?

No, f'' = 0 is not a necessary condition. One example I gave above, f(x) = x^1/3, should convince you that there is an inflection point at (0, 0) even though f''(0) ≠ 0. (f''(0) is not even defined.)

 

[Jhenrique]

No, f'' = 0 is not a necessary condition. One example I gave above, f(x) = x^1/3, should convince you that there is an inflection point at (0, 0) even though f''(0) ≠ 0. (f''(0) is not even defined.)

Words from Wolframapage:

A necessary condition for x to be an inflection point is f''(x)=0

http://mathworld.wolfram.com/InflectionPoint.html

2nd place: exist inflection point in a surface? If yes, how to identify them?

 

[Mark44]

No, f'' = 0 is not a necessary condition. One example I gave above, f(x) = x^1/3, should convince you that there is an inflection point at (0, 0) even though f''(0) ≠ 0. (f''(0) is not even defined.)

Words from Wolframapage:

A necessary condition for x to be an inflection point is f''(x)=0

http://mathworld.wolfram.com/InflectionPoint.html

My example shows that the MathWorld statement is incorrect. You can have an inflection point at (c, f(c)) even when f''(c) is not equal to zero.

 

[Mark44]

From the Wikipedia page (http://en.wikipedia.org/wiki/Inflection_point), with emphasis added:

If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists,...

The converse of the above is not necessarily true; e.g., f(x) = x⁴. For this function, we have f''(0) = 0, but (0, 0) is not an inflection point.

For g(x) x^1/3, (0, 0) is the inflection point, since the concavity changes sign at 0, even though g''(0) isn't zero.

 

[Mark44]

Also, I did a quick search for surface inflection points, but didn't come up with anything. Admittedly, I didn't search too hard.

 

[pasmith]

Given a function $f(x)$, the critical points are where $f'=0$ and the inflection points are where $f''=0$. Given a function $f(x,y)$, the critical points are where $\vec{\nabla}f = \vec{0}$, so I can deduce that the inflection points are where $Hf=0$ . Correct?

No. If \nabla f = 0 then there is:

• a local minimum if both eigenvalues of Hf have strictly positive real part.
• a local maximum if both eigenvalues of Hf have strictly negative real part.
• a saddle point if one eigenvalue of Hf is strictly positive the other is strictly negative.

If one or both eigenvalues have zero real part then you have to look to higher order to determine the local behaviour of f.

 

[pasmith]

Also, I did a quick search for surface inflection points, but didn't come up with anything. Admittedly, I didn't search too hard.

I think the two-dimensional equivalent of the point of inflection is the saddle point (at least in the case where Hf is defined).

 

[AlephZero]

No, f'' = 0 is not a necessary condition. One example I gave above, f(x) = x^1/3, should convince you that there is an inflection point at (0, 0) even though f''(0) ≠ 0. (f''(0) is not even defined.)

I think the Wolfram definition considers that an inflection point is a property of the geometry of the curve, not an artefact of any particular equations or coordinate systems that are used to describe it.

The existence of $dy/dx$ depends on the arbitrary choice of an axis system.

In that interpretation, the curve still has an inflection point if rotated through $\pi/4$ and the gradient at the inflection point becomes 1.

For a sufficiently smooth curve, re-parametrizing it using its arc length gets rid of arbitrary happenstance about the existence of derivatives w.r.t. the parameter.

 

[Mark44]

I think the two-dimensional equivalent of the point of inflection is the saddle point (at least in the case where Hf is defined).

That thought came to my mind as well.

 

[Mark44]

I think the Wolfram definition considers that an inflection point is a property of the geometry of the curve, not an artefact of any particular equations or coordinate systems that are used to describe it.

The existence of $dy/dx$ depends on the arbitrary choice of an axis system.

From the perspective of the original post in this thread, and countless calculus texts, the axes are already chosen for us. "Given a function f(x)..."

In that interpretation, the curve still has an inflection point if rotated through $\pi/4$ and the gradient at the inflection point becomes 1.

Yes, I understand that, but it seems to me to be something of a stretch in the context of the question that was asked.

For a sufficiently smooth curve, re-parametrizing it using its arc length gets rid of arbitrary happenstance about the existence of derivatives w.r.t. the parameter.