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Homework Help: Functions forming a vector space

  1. Aug 12, 2018 #1
    1. The problem statement, all variables and given/known data
    1.1.3
    1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
    2) How about periodic functions? obeying f(0)=f(L) ?
    3) How about functions that obey f(0)=4 ?

    If the functions do not qualify, list what go wrong.


    2. Relevant equations
    upload_2018-8-12_22-22-57.png

    3. The attempt at a solution
    1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
    2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
    3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

    Is this correct?
     
  2. jcsd
  3. Aug 12, 2018 #2

    PeroK

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    Technically, the zero function is periodic for any given period. In particular, it has the same value at the end points.
     
  4. Aug 12, 2018 #3

    Ray Vickson

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    You need to have confidence in your own reasoning. (i) Have you verified all the defining properties of vector spaces in (1)-(2)? (ii) Have you shown that some vector-space property fails in (3)?
     
  5. Aug 12, 2018 #4

    Mark44

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    What's the definition of a periodic function?
     
  6. Aug 12, 2018 #5

    fresh_42

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    The notion of the corresponding equation would have been almost shorter: ##(\alpha f + \beta g)(0)=\alpha f(0)+\beta g(0)=\alpha f(L)+ \beta g(L)=(\alpha f + \beta g)(L)##
    Why this? ##0(0)=0=0(L)##
    This is the extreme version of a periodic function, namely with period ##0##. There is no reason to exclude it.
    Yes, except that I couldn't see your reasoning why the null vector in section 2 shouldn't work.

    Edit: Please do the following little exercise: A set which is closed under addition and scalar multiplication automatically contains zero.
     
    Last edited: Aug 12, 2018
  7. Aug 12, 2018 #6
    Yes, I verified it in mind. I didn't write the verification here as it was easy.
     
  8. Aug 12, 2018 #7

    Mark44

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    Or any period.
    If f(x) = 0 for all x, then f(x) = f(x + L), for any L.

    The vector space properties in the attached image in post 1 is a bit unusual. The usual definition includes statements about closure under vector addition and closure under scalar multiplication.

    Minor point: The usual definitions are of the properties these operations. I've never seen them called features.
     
  9. Aug 12, 2018 #8

    vela

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    Typically the period is required to be positive to exclude possibilities like claiming f(x)=x is periodic since f(x)=f(x+0) for all x.
     
  10. Aug 12, 2018 #9

    fresh_42

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    Makes sense. I thought the zero might be necessary for the set of all periods, or to get vector spaces for periodic functions, but "any" sounds better than "0" although "any" includes it.
     
  11. Aug 13, 2018 #10
    Let's say that ## \alpha ## and ## \beta ## belongs to this set and a, b belong to the scalar field. Then, the set being closed under addition and scalar multiplication means that ## a \alpha + b \beta ## must belong to the set.
    Taking ## b \beta = - a \alpha ## implies that 0 belongs to the set.

    Thus, a set which is closed under addition and scalar multiplication automatically contains zero.
     
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