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Integration by parts and fourier series

  1. Dec 19, 2008 #1
    i'm a self learner currently learning fourier series.

    Anyway, I'm having some problem with a question regarding the inner product of two complex functions. This is defined by an integral from negative infi to positive infi of the multiplication of one function and the complex conjugate of the other. Since it is an integral of two function i'm wondering if integration by parts that we learn in high school which applies for real function is workable on complex functions?

    A link to a proof will be even more appreciated!

  2. jcsd
  3. Dec 19, 2008 #2
    The proof should be quite simple. Do you know how to prove integration by parts for the reals? (If not, start by proving the product rule).

    After you have that proof, just substitute "real" for "complex". All the math in the proof is addition and multiplication, and so everything holds for both real and complex numbers.
  4. Dec 19, 2008 #3
    ok. but is there a proof of the differention of a complex function f is the differentiation of its real part plus i times its differentiated imaginary part? or simply the differentiation of the complex function can be done exactly the same way as one would do for a real function?
  5. Dec 19, 2008 #4

    Differentiation on complex functions is defined the same way.

    [tex]f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/tex]

    There are some subtle differences in the meaning of a limit in complex numbers. (In the real numbers, you can only approach from the left or right, but in C, you can approach from any direction). But those differences don't matter. The only limit identity used in the proof is that [tex]\lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)[/tex], which is true for both R and C.

    So what you're really doing is showing that

    [tex]\lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}g(x) + \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}f(x)[/tex].

    (I hope I got that all correct).
  6. Dec 19, 2008 #5


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    Tac-tics, the functions the OP is concerned about are complex-valued functions defined on the set of real numbers.

    The rule (f+g)'(x)=f'(x)+g'(x) follows immediately from the definitions (of the derivative and f+g). The rule (cf)'(x)=cf'(x) where c is a complex number follows immediately from the definitions (of the derivative and cf). What to do with f+ig should be obvious from these rules.
  7. Dec 19, 2008 #6
    Yes, and what I was getting at applies to functions C -> C, of which functions R -> C is a subset.
  8. Dec 22, 2008 #7


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    My point was just that there's no need to consider complex limits here.

    By the way, the set of functions from [itex]\mathbb R[/itex] to [itex]\mathbb C[/itex] is not a subset of the set of functions from [itex]\mathbb C[/itex] to [itex]\mathbb C[/itex].
  9. Dec 22, 2008 #8
    Ah, you're right.

    It depends on your formalism, but if you treat R as a subset of C, then every function R->C where f(x) = y, there is a function f' : C->C, f'(x + 0i) = y. So unless I missed something, regardless of your formalism, it's isomorphic to a subset of C->C.
  10. Dec 22, 2008 #9


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    No, {f| f:C->R} is a subset of {f|f:C->C}. {f|f:R->C} is not. The notation f:C->C means that f is defined for all complex numbers which is not the case for f:R->C. The range does not have to be all of C.
  11. Dec 22, 2008 #10
    I suppose you are right. But each can still be lifted trivially into C->C.
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