# Higher Order Derivative Test and Germs

• A
If ##f'(0) = 0## and ##n## is the smallest natural number such that ##f^{(n)}(0)\neq 0##, then the higher-order derivative test states the following:

1. If ##n## is even and ##f^{(n)}(0)>0##, then ##f## has a local minimum at ##0##.
2. If ##n## is even and ##f^{(n)}(0)<0##, then ##f## has a local maximum at ##0##.
3. If ##n## is odd, then ##f## has an inflection point at ##0##.
4. If ##f^{(n)}(0)=0## for all ##n##, then the higher-order derivative test is inconclusive.

I'd like to find a method of finding minima and maxima in the case where the higher-order derivative test is inconclusive, using the concept of germs. Let ##X## be the set of all functions infinitely differentiable at ##0## for which ##f^{(n)}(0)=0## for all ##n##. Let's define an equivalence relation on $X$ by saying that ##f\sim g## if there exists a sufficiently small open interval ##I## containing ##0## such that ##f(x)=g(x)## for all ##x## in ##I##. Then the set of germs of ##X## denotes the set ##Y## of equivalence classes of elements of ##X## under this equivalence relation. (Note that ##Y## has infinitely many elements, as two functions which have all their nth derivatives in common can have completely different germs.)

My question is, does there exist a nontrivial function ##F## from ##Y## to ##\mathbb{R}## such that if ##F## evaluated at a particular germ yields a positive number, then all the functions in the germ have a local minimum at ##0##, and if it yields a negative number then all the functions in the germ have a local maximum at ##0##?

andrewkirk
Homework Helper
Gold Member
I think so.
Define ##p:Y\to\mathbb R## by:
##p(y)=1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(0,\epsilon_f)\ f(k)>0##
##p(y)=-1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(0,\epsilon_f)\ f(k)<0##
otherwise ##p(y)=0##.

Similarly define ##m:Y\to\mathbb R## by:
##m(y)=1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(-\epsilon_f,0)\ f(k)>0##
##m(y)=-1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(-\epsilon_f,0)\ f(k)<0##
otherwise ##m(y)=0##.

Then define ##F:Y\to\mathbb R## by:
##F(y)=2## if ##p(y)=m(y)=1##
##F(y)=-2## if ##p(y)=m(y)=-1##
##F(y)=-1## if ##p(y)=1## and ##m(y)=-1##
##F(y)=1## if ##p(y)=-1## and ##m(y)=1##
##F(y)=0## otherwise.

Then ##y## has a local minimum at 0 iff ##F(y)=2##, a local maximum iff ##F(y)=-2##, a rising inflection point iff ##F(y)=-1## and a falling inflection point iff ##F(y)=1##.

If ##F(y)=0## then, for any ##f\in y##, roots of ##f## are dense on one or both sides of ##x=0##, so it cannot be a local extremum.

This function is slightly different from what you asked for, because it gives information about inflections as well. A function that strictly answers your question would be:
##F(y)=2## if ##p(y)=m(y)=1##
##F(y)=-2## if ##p(y)=m(y)=-1##
##F(y)=0## otherwise.

I think so.
Define ##p:Y\to\mathbb R## by:
##p(y)=1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(0,\epsilon_f)\ f(k)>0##
##p(y)=-1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(0,\epsilon_f)\ f(k)<0##
otherwise ##p(y)=0##.

Similarly define ##m:Y\to\mathbb R## by:
##m(y)=1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(-\epsilon_f,0)\ f(k)>0##
##m(y)=-1## if ##\exists f\in y\ \exists\epsilon_f>0## s.t. ##\forall k\in(-\epsilon_f,0)\ f(k)<0##
otherwise ##m(y)=0##.

Then define ##F:Y\to\mathbb R## by:
##F(y)=2## if ##p(y)=m(y)=1##
##F(y)=-2## if ##p(y)=m(y)=-1##
##F(y)=-1## if ##p(y)=1## and ##m(y)=-1##
##F(y)=1## if ##p(y)=-1## and ##m(y)=1##
##F(y)=0## otherwise.

Then ##y## has a local minimum at 0 iff ##F(y)=2##, a local maximum iff ##F(y)=-2##, a rising inflection point iff ##F(y)=-1## and a falling inflection point iff ##F(y)=1##.

If ##F(y)=0## then, for any ##f\in y##, roots of ##f## are dense on one or both sides of ##x=0##, so it cannot be a local extremum.

This function is slightly different from what you asked for, because it gives information about inflections as well. A function that strictly answers your question would be:
##F(y)=2## if ##p(y)=m(y)=1##
##F(y)=-2## if ##p(y)=m(y)=-1##
##F(y)=0## otherwise.

First of all, are you assuming that ##f(0) = 0## for all the functions in ##X##?

Second of all, this isn't quite what I was looking for. Can you make a function ##F## that depends more on the specifics of the germ? That is, if ##y_1,y_2\in Y## and ##y_1\neq y_2##, I want ##F(y_1)\neq F(y_2)## almost always, although in some rare cases they might coincidentally be equal. (Akin to how if you take two random functions and you apply the second-derivative test to both of them, you'll almost always get different values for the two second derivatives.)

andrewkirk
Homework Helper
Gold Member
First of all, are you assuming that ##f(0) = 0## for all the functions in ##X##?
Yes, that follows from the assumption '##f^{(0)}=0## for all [natural] ##n##' if we take the usual meaning of natural numbers to be all non-negative integers, since ##f^{(0)}=f##. If we take the alternative interpretation that natural numbers exclude 0, the reasoning doesn't change. We just need to replace range values of 0 by ##f(0)##.

Second of all, this isn't quite what I was looking for. Can you make a function ##F## that depends more on the specifics of the germ? That is, if ##y_1,y_2\in Y## and ##y_1\neq y_2##, I want ##F(y_1)\neq F(y_2)## almost always, although in some rare cases they might coincidentally be equal. (Akin to how if you take two random functions and you apply the second-derivative test to both of them, you'll almost always get different values for the two second derivatives.)
I doubt whether such a function is possible, because I think the domain of the function may have a higher cardinality than the range, in which case an injective function is impossible.

Yes, that follows from the assumption '##f^{(0)}=0## for all [natural] ##n##' if we take the usual meaning of natural numbers to be all non-negative integers, since ##f^{(0)}=f##. If we take the alternative interpretation that natural numbers exclude 0, the reasoning doesn't change. We just need to replace range values of 0 by ##f(0)##.
I meant positive integers, but yeah it doesn't alter your post much.
I doubt whether such a function is possible, because I think the domain of the function may have a higher cardinality than the range, in which case an injective function is impossible.
First of all, the set of all continuous functions, let alone the set of smooth functions, has the same cardinality as the set of real numbers, because a continuous function is defined by its values on the rational numbers. Second of all, I'm not saying I want an injection. Just like the function which maps the set of all smooth functions to their second derivative at ##0## isn't an injection, but still if you take two random functions the chance that they'll have the same second derivative at ##0## is very small, I want a function ##F## such that if ##y_1\neq y_2##, then the chance that ##F(y_1) = F(y_2)## is very small.

Just like the function which maps the set of all smooth functions to their second derivative at ##0## isn't an injection, but still if you take two random functions the chance that they'll have the same second derivative at ##0## is very small, I want a function ##F## such that if ##y_1\neq y_2##, then the chance that ##F(y_1) = F(y_2)## is very small.
In fact, I think if a measure is appropriately defined on this function space, then the set of functions which have the same second derivative at 0 as a given function will be of measure zero. So similarly, I want the set of all ##y## which have the same ##F## value as a given ##y_1## to be of measure 0.