Inflection point of a surface

1. Mar 27, 2014

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?

Last edited: Mar 27, 2014
2. Mar 27, 2014

micromass

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

3. Mar 27, 2014

Staff: Mentor

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) = x1/3, 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.

4. Mar 27, 2014

Jhenrique

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)...

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?

5. Mar 27, 2014

Staff: Mentor

No, f'' = 0 is not a necessary condition. One example I gave above, f(x) = x1/3, should convince you that there is an inflection point at (0, 0) even though f''(0) ≠ 0. (f''(0) is not even defined.)

6. Mar 27, 2014

Jhenrique

Words from Wolframapage:
http://mathworld.wolfram.com/InflectionPoint.html

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

7. Mar 27, 2014

Staff: Mentor

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.

8. Mar 27, 2014

Staff: Mentor

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) x1/3, (0, 0) is the inflection point, since the concavity changes sign at 0, even though g''(0) isn't zero.

9. Mar 27, 2014

Staff: Mentor

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

10. Mar 27, 2014

pasmith

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 $f$.

11. Mar 27, 2014

pasmith

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

12. Mar 27, 2014

AlephZero

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.

Last edited: Mar 27, 2014
13. Mar 27, 2014

Staff: Mentor

That thought came to my mind as well.

14. Mar 27, 2014

Staff: Mentor

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