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Prove that all integrable functions form a vector space

  1. Oct 11, 2014 #1
    This isn't a homework problem, a classmate asked for a challenging proof to try and do and this was the one we were given. We started by trying to derive some rules from un-integratable functions but realized that that would take a long time and a lot of work. After some thinking we came up with the idea of using Taylor Series and proof by definition to prove it. My question I guess is does that work? Here's the proof we came up with:

    Let the set of all integrable functions be call V. All integrable functions can be expressed as a Taylor Series, all integrable functions can be defined as Taylor Series, when can then be integrated.

    Let [itex]\oplus[/itex] be an operation defined by:
    Let u, v, w be integrable functions defined as a Taylor Series [itex]\sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n}[/itex]
    [tex]
    u = \sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n}\\
    v = \sum_{n = 0}^{\infty} \frac{g^{n}(a)}{n!}(x-a)^{n}\\
    w = \sum_{n = 0}^{\infty} \frac{h^{n}(a)}{n!}(x-a)^{n}\\
    1. u \oplus v = v \oplus u: \sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n} \oplus \sum_{n = 0}^{\infty} \frac{g^{n}(a)}{n!}(x-a)^{n} = \sum_{n = 0}^{\infty} \frac{g^{n}(a)}{n!}(x-a)^{n} \oplus \sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n}\\
    \begin{align} 2. u \oplus (v \oplus w) = w \oplus (u \oplus w&): \sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n} \oplus (\sum_{n = 0}^{\infty} \frac{g^{n}(a)}{n!}(x-a)^{n} \oplus \sum_{n = 0}^{\infty} \frac{h^{n}(a)}{n!}(x-a)^{n})\\ &= \sum_{n = 0}^{\infty} \frac{h^{n}(a)}{n!}(x-a)^{n} \oplus (\sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n} \oplus \sum_{n = 0}^{\infty} \frac{g^{n}(a)}{n!}(x-a)^{n})\\ &= \sum_{n = 0}^{\infty} \frac{h^{n}(a)}{n!}(x-a)^{n} \oplus \sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n} \oplus \sum_{n = 0}^{\infty} \frac{g^{n}(a)}{n!}(x-a)^{n} \end{align}\\
    4. u \oplus -u = 0: \sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n} + -\sum_{n = 0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^{n} = 0 \\
    \text{Closed Under Addition} \oplus \text{, closed as:} \int{\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\text{dx}}\\
    \text{Let } \odot \text{ be an operator defined by:} \mathbf{c} \odot \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} = \mathbf{c}\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\\ \text{ where } \mathbf{c} \text{ is a constant }
    \forall u, v\epsilon \mathbf{V} \text{ and c,d}\epsilon\mathbf{R}\\
    \mathbf{1) } \mathbf{ c} \odot \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} = \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} \odot \mathbf{c}\\
    \mathbf{2) } \mathbf{ c } \odot (\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} + \sum_{\text{n=0}}^{\infty} \frac{g^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}) = \mathbf{c}\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} + \mathbf{c}\sum_{\text{n=0}}^{\infty} \frac{g^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\\
    \mathbf{3) } \mathbf(c+d) \odot \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} = \mathbf{c}\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} + \mathbf{d}\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\\
    \mathbf{4) } \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} \odot \mathbf{c}) \odot \mathbf{d} = \mathbf{c} \odot (\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} \odot \mathbf{d})\\
    \mathbf{5) } 1 \times \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}} = \sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\\
    \text{Scalar Multiplication } \odot \text{ closed as: } \mathbf{c } \odot \int{\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\text{dx}} = \int{\mathbf{c}\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\text{dx}} = \mathbf{c} \int{\sum_{\text{n=0}}^{\infty} \frac{f^{\text{n}}\text{(a)}}{\text{n!}}\text{(x-a)}^{\text{n}}\text{dx}}\\ \\
    \text{So, if the properties of } \oplus \text{ and } \odot \text{ hold under V for polynomials of n-limit length then the set is a Vector Space.}[/tex]

    (Wow that took a while to type up...)
     
    Last edited: Oct 12, 2014
  2. jcsd
  3. Oct 12, 2014 #2
    "All integrable functions can be expressed as a Taylor Series"

    That is not in anyway true. I have no idea why you're doing anything with Taylor series, you're entirely overcomplicating this.

    Edit, why did you even go to Taylor series? Literally nothing in your proof required a Taylor series, you could just have easily used f(x), g(x) , etc. I'm genuinely curious, why?
     
  4. Oct 12, 2014 #3
    I thought as much, but we weren't really sure where to go. The other idea we had was to define the set of integrable functions and work from there. Any idea where I should start looking for a better solution? As you said, I know I am way over-complicating this :/.
     
  5. Oct 12, 2014 #4
    First off, make sure you're being precise about these things. A set is NOT a vector space. When you say "prove the set of all integrable functions forms a vector space", I have to ask with what operations and scalars? If you mean the usual, pointwise addition/multiplication, with scalars being real or complex numbers, then everything is basically trivial.

    If ##f(x)## and ##g(x)## are integrable, then surely ##f(x)+g(x)## is integrable. If ##c\in \mathbb{C}##, then ##cf(x)## is integrable. The rest of the axioms follow from pointwise addition/multiplication.

    If I was grading a linear algebra class, and someone wrote the above two lines, I would give them full credit.

    In addition, this should be in homework help. Even if it is not "for school", it is still a school-style problem.
     
  6. Oct 12, 2014 #5
    Oh wow, that's definitely much simpler then the way we were going about it. And yes, I was referring to the typical point-wise addition and multiplication operators...I wasted way more time on this problem then I probably should have, buuut proofs are something I need to definitely strengthen myself in. I wasn't sure if this was considered a homework or standard section post, but I'll post any future similar problems I have in the homework help section. Thank you though!
    One quick question, can there be cases where two non-integrable functions added together create an integrable function?
     
  7. Oct 12, 2014 #6
    Yup, in fact for every non integrable function, there is another nonintegrable function such that their sum is integrable.

    Let me ask YOU a quick question, can you give me an example of a non integrable function?
     
  8. Oct 12, 2014 #7

    pasmith

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    Homework Helper

    If [itex]f[/itex] is not integrable, then neither is [itex]-f[/itex] (because if [itex]-f[/itex] were integrable, then so would [itex]-(-f) = f[/itex] be). However [itex]f + (-f) = 0[/itex] is integrable.
     
  9. Oct 12, 2014 #8
    A few non integrable functions I can think of are [itex]f(x) = ln(ln(x)), \text{ or } g(x) = e^{e^{x}}, \text{ also } h(x) = \frac{1}{ln(x)} [/itex] are a few. So for any function, [itex] \int{(f(x) + -f(x))dx} = \int{dx} = c[/itex]. That's a simple way to clear that up!
     
  10. Oct 12, 2014 #9
    Actually ##e^{e^x}## IS integrable. You can't express its antiderivative in terms of elementary functions, but it IS integrable in the sense that it HAS a well defined "area under the curve".

    The rest can be integrable, depending on your domain. If you exclude points where the function blows up, both other functions are indeed integrable.
     
  11. Oct 12, 2014 #10
    Regarding your edit, tl;dr we're dumb. We were way over thinking trying to deal with un-integrable functions and somehow got on the topic of Taylor and Maclaurin series without looking into the idea that some functions can't be represented as Taylor Series. We then thought (without properly testing) that un-integrable functions could be represented as Taylor series as well therefore forgoing our crazy idea of defining un-integrable functions and why that wouldn't work, therefore only working with integrable ones. Our overthinking got the best of us, and we didn't even consider that fact that the definitions of addition under and scalar multiplication would lead us directly to the answer, which is kind of funny because on some actual homework we had, we had to use these rules for a few proofs. Oh well, it was kind of a crazy idea.
     
  12. Oct 13, 2014 #11

    pasmith

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    Homework Helper

    A function which is continuous on [itex][a,b][/itex] is integrable on [itex][a,b][/itex]. There are closed intervals on which each of the given functions is continuous.

    The canonical example of a non-(Riemann) integrable function is [tex]
    f : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases}
    1, & x \in \mathbb{Q}, \\
    0, & x \notin \mathbb{Q}.
    \end{cases}[/tex]
     
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