Fourier Analysis - Tough Question

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    Question:

    The deviation of a truncated Fourier Series from the function f(x) which it is approximating is defined by:

    Δp=π∫{(a0/2 - f(x) + pm=1∑[amcos(mx)+bmsin(mx)]}2dx

    Find an expression of the partial derivatives of Δp with respect to an and bn, and devise the values of n which give maxima and minima.

    My working:

    Eliminating the easy parts first, I know that when I expand the square in the integrand, the product of f(x) and a0/2 is part of the result, as are their squares. But these do not depend on an or bn, so can be ignored when finding the partial derivatives.
    Second, when multiplying a0/2 or f(x) by the summation, I realise that differentiating each term by a and b will reduce to either sine or cosine integrated over the interval [-π, π]. In each case, knowing that m is an integer, this will reduce to zero when integrated, therefore, the only term which is necessary in the integrand is the square of the summation term.

    When squaring this summation, we take the Cauchy Product of the summation term with itself, which came to:

    pk=1km=1∑ [amak-mcos(mx)cos((k-m)x) + bmbk-msin(mx)sin((k-m)x) + ak-mbmsin(mx)cos((k-m)x) + ambk-mcos(mx)sin((k-m)x)]

    For m being a certain integer n:

    pn=1∑ [anak-ncos(nx)cos((k-n)x) + bnbk-nsin(nx)sin((k-n)x) + ak-nbnsin(nx)cos((k-n)x) + anbk-ncos(nx)sin((k-n)x)]

    Due to the orthogonality of the components which define the Fourier Series, all terms will integrate to zero, with the exception of those where n=k-n. Thus, we need only integrate:

    y = an2cos2(nx) + bn2sin2(nx) + anbnsin(2nx)

    So:

    ∂y/∂an = 2ancos2(nx) + bnsin(2nx)

    Integrating over the interval [-π, π]:

    ∂Δp/∂an = [(an/2n)(2nx+sin(2nx))-(bn/2n)cos(2nx)]π = 2anπ

    Similarly for b:

    ∂y/∂bn = 2bnsin2(nx) + ansin(2nx)

    ∂Δp/∂bn = [(bn/2n)(2nx-sin(2nx))-(an/2n)cos(2nx)]π = 2bnπ


    Now I'm confused. In order to return maxima, that would require the second order derivative to have negative values. But if one were to differentiate the expressions I got, it would always result in a positive value of 2π, which to me seems strange in itself. Have I gone wrong somewhere?
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    (Original post by Nuclear Ghost)
    Question:

    The deviation of a truncated Fourier Series from the function f(x) which it is approximating is defined by:

    Δp=π∫{(a0/2 - f(x) + pm=1∑[amcos(mx)+bmsin(mx)]}2dx

    Find an expression of the partial derivatives of Δp with respect to an and bn, and devise the values of n which give maxima and minima.

    My working:

    Eliminating the easy parts first, I know that when I expand the square in the integrand, the product of f(x) and a0/2 is part of the result, as are their squares. But these do not depend on an or bn, so can be ignored when finding the partial derivatives.
    Second, when multiplying a0/2 or f(x) by the summation, I realise that differentiating each term by a and b will reduce to either sine or cosine integrated over the interval [-π, π]. In each case, knowing that m is an integer, this will reduce to zero when integrated, therefore, the only term which is necessary in the integrand is the square of the summation term.

    When squaring this summation, we take the Cauchy Product of the summation term with itself, which came to:

    pk=1km=1∑ [amak-mcos(mx)cos((k-m)x) + bmbk-msin(mx)sin((k-m)x) + ak-mbmsin(mx)cos((k-m)x) + ambk-mcos(mx)sin((k-m)x)]

    For m being a certain integer n:

    pn=1∑ [anak-ncos(nx)cos((k-n)x) + bnbk-nsin(nx)sin((k-n)x) + ak-nbnsin(nx)cos((k-n)x) + anbk-ncos(nx)sin((k-n)x)]

    Due to the orthogonality of the components which define the Fourier Series, all terms will integrate to zero, with the exception of those where n=k-n. Thus, we need only integrate:

    y = an2cos2(nx) + bn2sin2(nx) + anbnsin(2nx)

    So:

    ∂y/∂an = 2ancos2(nx) + bnsin(2nx)

    Integrating over the interval [-π, π]:

    ∂Δp/∂an = [(an/2n)(2nx+sin(2nx))-(bn/2n)cos(2nx)]π = 2anπ

    Similarly for b:

    ∂y/∂bn = 2bnsin2(nx) + ansin(2nx)

    ∂Δp/∂bn = [(bn/2n)(2nx-sin(2nx))-(an/2n)cos(2nx)]π = 2bnπ


    Now I'm confused. In order to return maxima, that would require the second order derivative to have negative values. But if one were to differentiate the expressions I got, it would always result in a positive value of 2π, which to me seems strange in itself. Have I gone wrong somewhere?
    Rather than drowning in complexity, I suggest you first look at something like choosing A that minimizes

    I = \int (f(x) - A \sin x)^2\, dx (limits -pi to pi).

    Then expanding the integrand gives \int f(x)^2  - 2A f(x) \sin x - A^2 \sin^2x\,dx

    Now diff w.r.t. A: we can ignore f(x)^2 since it doesn't depend on A, so we get

    \displaystyle \dfrac{\partial I}{\partial A} = 2 \int f(x) \sin x \,dx - 2A \int \sin^2 x\,dx

    and it is now easy to find A s.t. dI/dA = 0.

    The original problem follows much the same lines, only it's messier and you'll need to use orthogonality to get rid of a lot of cross terms.
 
 
 
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