0 like 0 dislike
0 like 0 dislike
Can a special case be generalized in more ways than one? Any well known examples?

18 Answers

0 like 0 dislike
0 like 0 dislike
another way to phrase this, you have two different theorems/objects/etc, but a special case of both is the same thing.
0 like 0 dislike
0 like 0 dislike
Absolutely — any area of research gives more questions than answers :) and those questions lead to new conjectures and definitions!

An example in my field is Reed-Solomon codes, which are:

* an example of a linear code

* an example of a Maximum Distance Separable code

* an example of an Algebraic Geometry code

* an example of an evaluation code

* an example of a trivial locally recoverable code

* an example of a self-dual code (at times)

* etc.

RS codes as an object is are examples of many more general types of codes :)
0 like 0 dislike
0 like 0 dislike
I don't know if this is what you're looking for, but if you have finitely many topological spaces then the topology on the product is the topology generated by the set of products of open sets.

Many people meeting this for the first time suspect that this might be a way to define the topology on an infinite product as well. You can do it (its called the box topology) but the *product topology* has better properties. For the product topology you take the topology generated by products of opens of each factor, where all but finitely many elements of the product are the whole of the space they sit inside.
0 like 0 dislike
0 like 0 dislike
Fundamental theorem of calculus generalizes in different settings.

There is a generalization of the FTC to Lebesgue integration with the Lebesgue differentiation theorem, basically this generalizes the class of functions which FTC applies to to include measurable functions without a typical derivative.

You also have stokes theorem which relates integrals over a region to integrals over its boundary, FTC is a specific case of this when your region is an interval.
0 like 0 dislike
0 like 0 dislike
Every fixed parameter is an opportunity for generalization and there are always invisible parameters.

Got a theorem about certain functions f(x)?

Well, maybe it's true for functions f(x^(n)) as well. Or f(x)^n. Or d^n /dx^n  f(x)
0 like 0 dislike
0 like 0 dislike
Yes, this is the most likely thing to happen with a result - that it can be generalized in many ways. We’re this not the case, mathematical research would be very shallow as a whole!

An example - my most recent research considered a problem in optimal stochastic control for a one-dimensional linear system driven by Gaussian noise under fairly relaxed information constraints. There were many directions to generalize this in:

-    Consider nonlinearity
-    More general noise process
-    Make the information constraints harder
-    Consider multidimensionality

I am currently researching the extensions 2 and 4 above, simultaneously. Despite the generalizations I have considered, there are several still open.
0 like 0 dislike
0 like 0 dislike
Yes. Weakening any one of the conditions of the conjecture gives different opportunities for generalization, and any one condition can be relaxed in different ways. This is also true of mathematical objects, not just theorems. Consider something simple like a vector space: We can think of this as a field acting on an abelian group by scalar multiplication, and one immediate generalization is to ask "why only a field, why not a ring?", which leads to the theory of modules. We can also take a vector space and look at the relationships between subspaces: We can decompose a vector space into subspaces, and each of those subspaces has a basis. A subset of those basis vectors defines another subspace contained in the first. The relationships between these subspaces can be represented as a combinatorial object, and rather than studying vector spaces *per se*, we can study these combinatorial objects that satisfy the same basic properties as the subspaces of a vector space. This leads to the theory of *matroids*.
0 like 0 dislike
0 like 0 dislike
Intermediate value theorem generalizes in different ways:

- The image of a connected set by a continuous function is a connected set.

- Darboux's theorem: a derivative has the intermediate value property.

- Brouwer fixed-point theorem is somewhat a generalization too: in one dimension it is a specific case of the intermediate value theorem.
0 like 0 dislike
0 like 0 dislike
Fourier theory can be generalized to compact non-abelian groups, or to locally compact abelian groups, for example.
0 like 0 dislike
0 like 0 dislike
The rings Z/pZ for prime p are special cases of both the rings Z/qZ and the finite fields F*_q_* for prime powers q. In each context you have a notion of Gauss sum, and in each context there is a notion of zeta-function (due to Igusa and Weil). The two kinds of zeta functions do not quite match, but they have some analogous properties (like rationality).

Related questions

0 like 0 dislike
0 like 0 dislike
3 answers
SalvadorHeresi asked Jun 21, 2022
Is there any reason to believe that big unsolved conjectures are either provable / disprovable in ZFC?
SalvadorHeresi asked Jun 21, 2022
0 like 0 dislike
0 like 0 dislike
11 answers
phillyvictor asked Jun 21, 2022
Can any number theorists explain the patterns going on in this factorization? (Sorta relevant to cyclotomic polynomials.)
phillyvictor asked Jun 21, 2022
0 like 0 dislike
0 like 0 dislike
10 answers
0 like 0 dislike
0 like 0 dislike
62 answers
VivianBala asked Jun 21, 2022
Do any of you guys dream about your math work in your head? OR just math in general?
VivianBala asked Jun 21, 2022
0 like 0 dislike
0 like 0 dislike
59 answers
timor_sv asked Jun 21, 2022
Got about a free 50-60 days before joining my PhD. Any advice or suggestions on what to do in this free time ?
timor_sv asked Jun 21, 2022

33.4k questions

135k answers

0 comments

33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!