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What does it mean for a function to be defined on an interval?

  1. May 11, 2014 #1
    So I was looking at one of the definitions of first order DE's.
    But I don't get what this statement means:

    let a function f(x) be defined on an interval I.
     
  2. jcsd
  3. May 11, 2014 #2

    lurflurf

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    I assume you are working with real numbers.
    A (real) interval is a convex subset of the real numbers.
    So I is some subset of R such that
    $$\text{if }x_k,a_k\in \mathbb{R} \text{ and }x_k\in I \text{ with}\sum_{k=1}^n a_k= 1 \text{ then }\sum_{k=1}^n x_k a_k \in I$$
    In other words we have a block of values for which the function is defined for all values


    In differential equations we can have problems if the function is defined on a set with gaps such as in the set of integers or rationals.

    Things like
    $$x\le 7\\
    x>57\\
    3<x\le 7 \\
    x\in \mathbb{R}\\
    x\in \emptyset\\
    x=1$$
    are what you should have in mind
     
    Last edited: May 11, 2014
  4. May 11, 2014 #3
    Ok, so basically to be defined means to not have any discontinuities, right?
     
  5. May 11, 2014 #4

    lurflurf

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    ^There can be discontinuities, there cannot be gaps.
    $$\mathrm{f}(x)=\begin{cases}\phantom{\frac{0}{0}}0 & x<0 \\
    \,\,\,\, \frac{1}{2} & x=0\\
    \phantom{\frac{0}{0}}1&x>0\end{cases}$$
    is defined on an interval, but not continuous
     
  6. May 11, 2014 #5
    Ok, so a piecewise function. As long as in the interval, all x values in the interval has one y value or vice versa?
     
  7. May 12, 2014 #6

    HallsofIvy

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    That's pretty much the definition of "function", isn't it?
     
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