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A necessary but not sufficient condition for a point of inflection is that $$f''(x)=0$$ If the second derivative is 0 and the point is not a point of inflection, Wikipedia tells me that is called an undulation point, which apparently means

a point on a curve where the curvature vanishes but does not change sign.

An example given is $f(x)=x^4$ at $(0,0)$.

What I do not understand is why $(0,0)$ is not classified as a local minimum? Surely for $f(x)=x^4$ the first derivative changes polarity either side of $(0,0)$.

Is there a better example of an 'undulation point' for a smooth function? Do 'undulation points' exist for more than one variable or is the condition that all partial derivatives are zero a necessary and sufficient one?

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  • $\begingroup$ Look at the "but does not change sign" part of the definition. $\endgroup$
    – John Joy
    Jan 6, 2015 at 18:35

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The point $(0,0)$ is a minimum point. It is also an undulation point. You are right that in some ways this is a poor example of an undulation point, since it also has other properties. On the other hand, this example does make the point easy to see, and it has an extremely easy formula.

A better example in some ways would be $(0,0)$ in the graph of $f(x)=x^4+x$. The point is still fairly noticeable but is not a minimum.

enter image description here

ADDED: I just took at look at the Wikipedia definition of undulation point, and it does give the example $f(x)=x^4$ in the text. However, it also has the example $y=x^4-x$ in a graph later in the article. This is the same as my example but reflected in the $y$ axis. It seems Wikipedia wanted to have it both ways.

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  • $\begingroup$ That is a better example thanks. I'm still confused as to why (0,0) is an undulation point for $f(x)=x^4$ though. Doesn't the sign of the derivative change either side of (0,0)? $\endgroup$
    – Gridley Quayle
    Jan 6, 2015 at 18:41
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    $\begingroup$ @GridleyQuayle: Yes, the derivative changes sign, but that is irrelevant to the definition of undulation point. The second derivative (and therefore the curvature) do not change sign, and that is what is relevant. $\endgroup$
    – Rory Daulton
    Jan 6, 2015 at 19:03
  • $\begingroup$ If you go to the Wikipedia page for 'Inflection Point' and go to the section 'a necessary but not sufficient condition' the example is at the end of the first paragraph. $\endgroup$
    – Gridley Quayle
    Jan 6, 2015 at 21:24
  • $\begingroup$ @GridleyQuayle: Ah, yes, I see, you are right. I shall edit my answer accordingly. I referred to the graph later in the article. $\endgroup$
    – Rory Daulton
    Jan 6, 2015 at 22:21
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    $\begingroup$ As a future visitor I thank You! $\endgroup$
    – uhu23
    Aug 17, 2016 at 21:01
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Who said it's not a local minimum? It certainly is.

I don't know how you would want to define "undulation point" in several variables. One problem is that there isn't just one "curvature".

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  • $\begingroup$ this got me thinking too, 🤔.. Since there are multiple ways to look at the curvature with respect to the variables present in multivariable calculus, can we not define undulation points separately for each variable? (ofcourse, if it has one). And if all the variables agree at some particular point that becomes the universal undulation point.. note that the word universal is just made up by me 😅 $\endgroup$
    – Stranger Forever
    Jun 13, 2020 at 2:32

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