anonymous
  • anonymous
Let C[−1,1] be the real inner product space consisting of all continuous functions f : [−1,1] → R, with the inner product := integral of f(X)g(x)dx from x=-1 to x=1 Let W be the subspace of odd functions, i.e. functions satisfying f(−x) = −f(x). Find (with proof) the orthogonal complement of W.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@dan815 @oldrin.bataku
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
bout time

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anonymous
  • anonymous
i'm going to have dinner now. have a sus
anonymous
  • anonymous
u need to multiply food by human to equal eat right now
dan815
  • dan815
is linear algebra fun?
dan815
  • dan815
does it expand the way you think about life?
anonymous
  • anonymous
the orthogonal complement of W is the set of even functions
anonymous
  • anonymous
here's the proof: we need to find\[g \in C[-1,1]\]such that \[ = 0\]\[\int\limits_{-1}^{1} f(x)g(x)\: dx=0\]\[\int\limits_{-1}^{0} f(x)g(x)\: dx + \int\limits_{0}^{1}f(x)g(x)\:dx = 0\]\[-\int\limits_{1}^{0}f(-x)g(-x) \: dx + \int\limits_{0}^{1} f(x)g(x)\: dx = 0\]\[\int\limits_{1}^{0}f(x)g(-x)\:dx+\int\limits_{0}^{1}f(x)g(x)\:dx=0\]\[-\int\limits_{0}^{1}f(x)g(-x)\:dx + \int\limits_{0}^{1}f(x)g(x)\:dx = 0\]\[\int\limits_{0}^{1}f(x)(g(x)-g(-x))\:dx=0\]it must be\[g(x) - g(-x) = 0\]\[g(x) = g(-x)\]it's clear that g is even
anonymous
  • anonymous
i seee! interesting.
anonymous
  • anonymous
oh ok. i see what u did with the integrals.
anonymous
  • anonymous
http://dyinglovegrape.wordpress.com/2012/08/05/orthogonal-complement-of-even-functions/ :)
anonymous
  • anonymous
be carfull, because this is the oposite thing. He proved (copy paste) complement for even functions, and you need to supose that f(x) is odd, :)
anonymous
  • anonymous
the problem states that f(x) is odd
anonymous
  • anonymous
the diff. is just the sign
anonymous
  • anonymous
if f(x) is even, when you manipulate the integrals, you'll get g(x) + g(-x), and if f(x) is odd you'll get g(x) - g(-x)
anonymous
  • anonymous
yea i had problems with manipulating the integrals...thanks for that.
anonymous
  • anonymous
good like btw.
anonymous
  • anonymous
good link*

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