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# When the Hessian=0 and fxx=0 Watch

1. How do I classify the stationary point (in this case 0,0 ) if the Hessian=0 and fxx=0 for, for example, this function: x^3 + y^3 - 3xy^2?

Is it true that a point of inflection occurs when the second derivative of a function about the stationary point changes sign? as in, if the stationary point was 0,0 then for small values around 0,0 would the Hessian have to change signs? what's the relation between points of inflection and saddle points?

I know in this case, that in order to classify the stat. pt (0,0), I can observe the function value when x=0 (f(x,y)= y^3) and y=0 ((f(x,y)=x^3) in turn and see how it behaves, but how do I infer from that whether it's a saddle pt or a max/min?
2. Well, if it's a maximum, what would have to be true about f(x, 0) for small x? (Compared with f(0,0)).
And if it's a minimum, what would have to be true about f(y, 0) for small x? (Compared with f(0,0)).+
3. (Original post by DFranklin)
Well, if it's a maximum, what would have to be true about f(x, 0) for small x? (Compared with f(0,0)).
And if it's a minimum, what would have to be true about f(y, 0) for small x? (Compared with f(0,0)).+
not sure?
4. (Original post by zomgleh)
not sure?
If you fix one variable, then you're left with a function of just one variable - i.g. f(x,0) becomes g(x)

So, consider what DFranklin is asking in regard to that.
5. (Original post by ghostwalker)
If you fix one variable, then you're left with a function of just one variable - i.g. f(x,0) becomes g(x)

So, consider what DFranklin is asking in regard to that.
Okay so when x=0 it becomes y^3 which has a pt. of inflection (since the 1st derivatives around small values of 0 don't change signs) and when y=0 it become x^3 which again has a pt of inflection, is that why the function is a saddle point?
6. (Original post by zomgleh)
Okay so when x=0 it becomes y^3 which has a pt. of inflection (since the 1st derivatives around small values of 0 don't change signs) and when y=0 it become x^3 which again has a pt of inflection, is that why the function is a saddle point?
From what I understand (never studied multivariate calculus), if it's a stationary point, and it's not a local maximum, and it's not a local minimum, then it's called a saddle point (even though it may not look like a saddle), and that's the reason.

But I could be totally wrong in this.
7. (Original post by ghostwalker)
..
I was also a little unsure about this, which is why I phrased my hint in term of "well, it isn't a minimum, and it isn't a maximum" without wanting to pin down exactly what it is.

But wiki seems pretty clear that "if it ain't a min/max, it's a saddle" - they even give a point of inflexion as an example.
8. (Original post by DFranklin)
But wiki seems pretty clear that "if it ain't a min/max, it's a saddle" - they even give a point of inflexion as an example.
Thanks for that. I was using wiki as my reference too. lol.

However, to back it up, I've found a definition in Apostol's Calculus, refering to a point "a".
"A stationary point is called a saddle point if every n-ball B(a) contains points x such that f(x) < f(a) and other points such that f(x) > f(a)".
9. (Original post by ghostwalker)
Thanks for that. I was using wiki as my reference too. lol.

However, to back it up, I've found a definition in Apostol's Calculus, refering to a point "a".
"A stationary point is called a saddle point if every n-ball B(a) contains points x such that f(x) < f(a) and other points such that f(x) > f(a)".
following that line of reasoning, what if when I fix each variable in turn, I end up with one of them with a max and another with a pt of inflection? IE-- for x^3 - 3xy^2 + y^4.

So z=f(0,y) = y^4 (min. pt)
and z=f(x,0) = x^3 (inflection)

how does that indicate that the function has a saddle pt?
10. (Original post by zomgleh)
following that line of reasoning, what if when I fix each variable in turn, I end up with one of them with a max and another with a pt of inflection? IE-- for x^3 - 3xy^2 + y^4.

So z=f(0,y) = y^4 (min. pt)
and z=f(x,0) = x^3 (inflection)

how does that indicate that the function has a saddle pt?
Fixing one of the variables, means that you're just looking at a small set of the values around the point you're interested in.

But there is sufficient information there, in the f(x,0), for instance, to tell you that any neighbourhood of (0,0) is going to contains points that are > f(0,0) and others that are < f(0,0), where f(0,0) is of course 0.
11. (Original post by ghostwalker)
Fixing one of the variables, means that you're just looking at a small set of the values around the point you're interested in.

But there is sufficient information there, in the f(x,0), for instance, to tell you that any neighbourhood of (0,0) is going to contains points that are > f(0,0) and others that are < f(0,0), where f(0,0) is of course 0.
Okay, but what about for a function like this one where the Hessian=0 again: z= x^4 - y^4? (it's supposed to be a saddle pt)
12. (Original post by zomgleh)
Okay, but what about for a function like this one where the Hessian=0 again: z= x^4 - y^4? (it's supposed to be a saddle pt)
In any neighbourhood of (0,0), there are points greater than 0, and others less than 0, the value of z at (0,0). The same process as before will suffice.
13. (Original post by ghostwalker)
In any neighbourhood of (0,0), there are points greater than 0, and others less than 0, the value of z at (0,0). The same process as before will suffice.
So do we evaluate the entire function for small values around (0,0) or do we evaluate only the functions we get by fixing x=0 and then y=0 in turn?
14. (Original post by zomgleh)
So do we evaluate the entire function for small values around (0,0) or do we evaluate only the functions we get by fixing x=0 and then y=0 in turn?
Just evaluating the functions restricted to x=0, and y=0 respectively is sufficient to determine that it's a saddle point, in this case.

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