Question

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.

Answer 1

@dan815 @oldrin.bataku

Answer 2

@dan815

Answer 3

bout time

Answer 4

i'm going to have dinner now. have a sus

Answer 5

u need to multiply food by human to equal eat right now

Answer 6

is linear algebra fun?

Answer 7

does it expand the way you think about life?

Answer 8

the orthogonal complement of W is the set of even functions

Answer 9

here's the proof:

we need to find\[g \in C[-1,1]\]such that \[<f,g> = 0\]\[\int\limits_{-1}^{1} f(x)g(x)\: dx=0\]\[\int\limits_{-1}^{0} f(x)g(x)\: dx + \int\limits_{0}^{1}f(x)g(x)\:dx = 0\]\[-\int\limits_{1}^{0}f(-x)g(-x) \: dx + \int\limits_{0}^{1} f(x)g(x)\: dx = 0\]\[\int\limits_{1}^{0}f(x)g(-x)\:dx+\int\limits_{0}^{1}f(x)g(x)\:dx=0\]\[-\int\limits_{0}^{1}f(x)g(-x)\:dx + \int\limits_{0}^{1}f(x)g(x)\:dx = 0\]\[\int\limits_{0}^{1}f(x)(g(x)-g(-x))\:dx=0\]it must be\[g(x) - g(-x) = 0\]\[g(x) = g(-x)\]it's clear that g is even

Answer 10

i seee! interesting.

Answer 11

oh ok. i see what u did with the integrals.

Answer 12

http://dyinglovegrape.wordpress.com/2012/08/05/orthogonal-complement-of-even-functions/
:)

Answer 13

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, :)

Answer 14

the problem states that f(x) is odd

Answer 15

the diff. is just the sign

Answer 16

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)

Answer 17

yea i had problems with manipulating the integrals...thanks for that.

Answer 18

good like btw.

Answer 19

good link*