## Equivalency on Mutliple Intervals

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jewish_scientist
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### Equivalency on Mutliple Intervals

Here is a hypothesis/ theorem/ problem/ whatever that I thought up a while ago:

If f(x) =/= g(x) on the interval [a,b] and are not piece wise functions, then f(x) =/= g(x) on the interval [c,d].

I am pretty sure it is true and I know it is true for polynomials, but I have not been able to thing of a proof. I have not even been able to think of how to approach it. What do you guys think?
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doogly
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### Re: Equivalency on Mutliple Intervals

I don't think being piecewise is a property of the function.
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SuicideJunkie
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### Re: Equivalency on Mutliple Intervals

What are a,b,c and d? And what else is missing?

If [c,d] is not contained within [a,b], there's nothing you can really say anything about it, since the two functions could cross at an arbitrarily small distance from a or b.
But if it is, then you're just concluding a subset of the assumptions which doesn't say anything either?

doogly
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### Re: Equivalency on Mutliple Intervals

Here's what is true: if two functions are *analytic*, then if they agree on some measurable interval they must be identical.

I don't think "not piecewise" is even a category of functions, but even if so, it's not equivalent to being analytic. Analytic functions are a strict but rich and exciting area.
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Tub
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### Re: Equivalency on Mutliple Intervals

Step 1) write down the requirements in a precise, mathematical notation
Step 2) write the theorem down in a precise, mathematical notation
Step 3) be understood.

I think you're thinking about something that can be proven, but what you've written is ambiguous nonsense. The first step to tackle the problem is to understand it well enough to explain it. Then you can think about solving it.

/edit: and doogly has spoiled it. oh well.

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### Re: Equivalency on Mutliple Intervals

Is |x| considered a piecewise function? It's really just x^2 / x, right? If it's not, take f = x, g = |x|.

Is x^3/x^2 considered a different function from x because it's not defined at zero? If so, take [a,b] = [1,2] and [c,d] = [-1,1].

How about the floor function? How about a function defined as a limit of other functions?

I agree that the biggest issue with the statement is that "piecewise" is not a property of a function, it's a property of how you write down a function. But the same function can be written down in many different ways. I suspect that for any definition of "piecewise" you come up with, someone can probably find a clever way to construct a counterexample in a way that doesn't fit your definition.
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Eebster the Great
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### Re: Equivalency on Mutliple Intervals

Yeah "piecewise" refers to a definition of a function, not the function itself. You can define a function piecewise. Separately, if a function can be cut up into pieces that each have some property, but the function as a whole doesn't necessarily have that property, you can say that it holds the property piecewise. For instance, the absolute value function is piecewise differentiable, as it consists of two pieces, each of which is differentiable on its own. Splines are piecewise polynomials, since each piece is a polynomial.

Given that, probably what you mean by "not piecewise" here is "analytic," because for a function to be analytic, it has to be equal to its Taylor series (i.e. it has an infinite polynomial representation) and therefore behaves sort of like a polynomial. It is certainly true that if two polynomials are identical over some nonempty interval, they are identical over all real numbers. This is also true of many other functions, including rational functions (if you ignore removable discontinuities), exponential and logarithmic functions, trigonometric functions, and indeed all functions called "elementary" in the theory of differential algebra. Again, we will have to have certain specific exceptions to ensure the functions are defined over their entire domains, but that's not hard. There are actually many more functions that are analytic than just those, but I think these are enough examples. Almost any functions you come across in practice will have the property you mentioned.

jewish_scientist
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### Re: Equivalency on Mutliple Intervals

I did mean that f(x) and g(x) are analytic functions. I have not heard of the term analytic functions before, so I did not know how to properly express my idea. Sorry for the confusion.
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ConMan
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### Re: Equivalency on Mutliple Intervals

Analytic functions are just about the most nicely-behaved functions you can find. If there's a property that "makes sense" for functions to do, then odds are that it will be true for analytic functions, but outside of them there will be some weird pathological example that breaks your property in the meanest way imaginable.
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Lothario O'Leary
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### Re: Equivalency on Mutliple Intervals

ConMan wrote:Analytic functions are just about the most nicely-behaved functions you can find. If there's a property that "makes sense" for functions to do, then odds are that it will be true for analytic functions, but outside of them there will be some weird pathological example that breaks your property in the meanest way imaginable.

In particular, in this case, even infinitely differentiable (smooth) is not enough - the "weird pathological example" is the bump function.

For analytic functions, however, your statement is essentially a special case of the identity theorem (or, rather, its contraposition).

DavCrav
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### Re: Equivalency on Mutliple Intervals

The best way to do this is that if f and g are functions with a particular property, hope that f-g also has that property. If that's the case, then your question becomes:

If f is a function with property P, and f=0 on an interval, is f=0?

I think analytic is the only class of functions for which this property holds.