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I was reading this article in wikipedia related to saddle points. When I came across this line

In one dimension, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.

Actually, I didn't get it when it said that it could also be a point of inflection. Can anyone give me an example of such a function and its saddle point which is both a stationary point and point of inflection?

rajan sthapit
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  • Images at https://www.google.com/search?q=saddle+point&hl=en&prmd=imvns&tbm=isch – whuber Jun 15 '12 at 20:32
  • Thats what I didn't get. I mean the point is a saddle point by definition. That is fine. It is also a stationary point because the derivative at that point is 0. But how come it is a point of inflection? As mentioned in the wiki they are referring to one dimesion. So any example in one dimension – rajan sthapit Jun 15 '12 at 20:59

2 Answers2

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Take the function $y=x^3$, $x=0$ is a stationary point and a point of inflection.

draks ...
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Michael R. Chernick
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For functions of one real variable $t$ they call a critical point $t_0$ a saddle point if $t\mapsto f(t)-f(t_0)$ changes sign a $t_0$, and in most cases this means that the first nonvanishing derivative of $f$ at $t_0$ is of odd order.

For functions $f$ of $n\geq2$ real variables a critical point ${\bf p}=(p_1,\ldots, p_n)$ is a saddle point if the Hessian of $f$ at ${\bf p}$ is indefinite. This means that $$f({\bf p}+{\bf X})-f({\bf p})={\bf X}'\ H\ {\bf X}+o(|{\bf X}|^2)\qquad({\bf X}\to{\bf 0})$$ where the symmetric matrix $$H=\left[{\partial^2 f(x_1,\ldots, x_n)\over \partial x_i\partial x_k}\right]_{i,k}$$ has at least one strictly positive and at least one strictly negative eigenvalue. It follows that $f({\bf p}+{\bf X})-f({\bf p})$ takes as well positive as negative values in the immediate neighborhood of ${\bf p}$.

In the case of $n\geq2$ variables one usually assumes that the critical point in question is nondegenerate, meaning that the rank of $H$ is $=n$. The saddle points in the one-variable case are degenerate critical points in this sense, as the rank of the Hessian is $=0$ there.

Christian Blatter
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  • What is the rationale behind calling inflection points saddle points? How should we class $(x,y)=(0,0)$ of $f(x,y)=x^3$ and $f(x,y)=(x-y)^3$? One common characteristic of the two definitions is that every neighbourhood of a critical point contains values strictly greater and strictly less. But don't we want some additional structure? Also, do you know any examples of $n=1$ inflection points behaving like $n\geq2$ saddle points? I mean, any theorems or any properties? Otherwise it looks like someone extend a regular definition to $n=1$ just for a textbook. – paperskilltrees Oct 18 '20 at 01:26
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    I cannot say much. It is a fact that at a point $p$ with $df(p)=0$ all kinds of things can happen when this critical point is degenerate, i.e., the Hessian has not full rank. I don't know whether the notion "saddle point" is ISO-standardized in some way. What I wrote is more or less what Wikipedia says. – Christian Blatter Oct 18 '20 at 15:28