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How to calculate the projection of a function in a vector space

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    In the real linear space C(-1, 1) with inner product (f, g) = integral(-1 to 1)[f(x)g(x)]dx, let f(x) = ex and find the linear polynomial g nearest to f.

    2. Relevant equations


    3. The attempt at a solution
    I understand that the best approximation for g is equal to the projection of f. Therefore, in order to find the linear polynomial g nearest to f, I must calculate this projection. However, I do not know how to solve for the projection.

    (The solution in my book is g(x) = 1/2(e-e-1) + (3/e)x)
     
  2. jcsd
  3. Oct 8, 2014 #2

    Ray Vickson

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    This problem is almost, word-for-word, the same as one you have already described in this forum (the one about C(1,3) and the function 1/x), and you have already been given the solution. What is it you are not seeing?
     
  4. Oct 8, 2014 #3

    RUber

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    If you have a set of bases, ##\{e_i\}## and a function ##v##, the projection is ##\displaystyle \sum_i \frac{ \langle e_i, v \rangle}{\langle e_i, e_i \rangle} e_i##.
     
  5. Oct 8, 2014 #4
    I do not see the relationship between the definition of a projection and finding the specific projection of a function. I do not understand how to apply the concept.
     
  6. Oct 8, 2014 #5

    LCKurtz

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    @Cassi: When you abandon threads where you have been given help, why should we continue to waste our time with you?
     
  7. Oct 9, 2014 #6
    I did not abandon a thread. I asked a new question for clarification. How do you calculate the projection of a particular function?
    Thank you for your insightful help.
     
  8. Oct 9, 2014 #7

    RUber

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    The space that you are projecting onto has a set of basis functions.
    In this case, let's say they are the powers of x from 0 to N, since we are looking for linear polynomials N=1. Note that these are not orthonormal bases given your definition of the inner product.
    The projection of ##f(x)=e^x## onto the span of your first basis ##x^0## is ##\displaystyle \frac{\int_{-1}^1 f(x)b_0 dx}{\int_{-1}^1 b_0b_0 dx} b_0= \frac{ \int_{-1}^1 e^x (1) dx }{\int_{-1}^1 dx}(1)##, repeat this process for higher powers of x up to N.
     
  9. Oct 9, 2014 #8

    Mark44

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    If the new question pertains to a thread you have already started, ask it it the existing thread, not a new one. If the question is unrelated to the existing thread, then a new thread is appropriate.
     
  10. Oct 9, 2014 #9

    Ray Vickson

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    Several people have told you exactly how to do it, complete with explicit formulas. You need to do the actual work of applying those formulas.
     
  11. Oct 9, 2014 #10

    LCKurtz

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  12. Oct 9, 2014 #11

    pasmith

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    The only relevant difference between this problem and this problem (as continued here) is that in this problem you will end up having to minimize a function of two variables (the coefficients of [itex]1[/itex] and [itex]x[/itex]) instead of a function of one variable.

    But the principle is the same: the "nearest" linear polynomial is the linear polynomial [itex]g[/itex] which minimizes the distance between [itex]f[/itex] and [itex]g[/itex]. This is also, for the reasons given in the first thread linked above, the linear polynomial which minimizes the square of that distance. This is easier to work with and given by the formula in both the cited threads (with an appropriate adjustment to the domain of integration).
     
  13. Oct 10, 2014 #12
    Thank you, this was very helpeful. So I see that when it is a polynomial, I know repeat this with b1 = x, etc.
     
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