I don't understand that. The OP'sQuote:
Originally Posted by CaptainBlack
Youris
under my assumption about the unspecified index set for the
s
![\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}](http://latex.codecogs.com/png.latex?\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases})
My f is given by![f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}](http://latex.codecogs.com/png.latex?f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases})
They are orthogonal to each other.
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I don't understand that. The OP'sQuote:
Originally Posted by CaptainBlackYour
My f is given by
They are orthogonal to each other.
I'm lost,Quote:
Originally Posted by OpalgI don't understand that. The OP's![f(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}](http://latex.codecogs.com/png.latex?f(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases})
My f is given by
They are orthogonal to each other.
is in
but arn't you looking for a
function not expressible in terms of these functions?
CB
Sorry to keep harping on about this. There was a typo (now corrected) in my previous comment, which made nonsense of it. What I should have said isQuote:
Originally Posted by CaptainBlackI'm lost,
![\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere,} \end{cases}\qquad f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}](http://latex.codecogs.com/png.latex?\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere,} \end{cases}\qquad f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases})
The difference is thathas a 1 in its definition where
OK that makes sense, I still preferQuote:
Originally Posted by OpalgSorry to keep harping on about this. There was a typo (now corrected) in my previous comment, which made nonsense of it. What I should have said is
The difference is that
, which is obviously not in the space spanned by the
(though not orthogonal to
CB
Parseval's Theorem: An orthonormal sequence {xn} in H is complete iff:Quote:
Originally Posted by OpalgFor (b), I would start by taking f to be a function inthat is orthogonal to all the functions
Parseval's theorem then tells you thatis orthogonal to all the functions
.
llxll^2 = (sum n=1 to inf) l(x,xn)l^2.. for every x in H.
Using your premise above, how does Parseval's theorem tell youis incomplete? ie, how do you show "
There is more than one result that carries Parseval's name. The result that you state as Parseval's Theorem is what I call Parseval's Identity. For me, Parseval's Theorem is the result which says that if f, g are functions inQuote:
Originally Posted by HartlwParseval's Theorem: An orthonormal sequence {xn} in H is complete iff:
llxll^2 = (sum n=1 to inf) l(x,xn)l^2.. for every x in H.
Using your premise above, how does Parseval's theorem tell you
with Fourier transforms
then
It follows that if you have a (nonzero) functionsuch that
for all n, then
for all n.
That wraps it up. Thanks, very nice.Quote:
Originally Posted by OpalgThere is more than one result that carries Parseval's name. The result that you state as Parseval's Theorem is what I call Parseval's Identity. For me, Parseval's Theorem is the result which says that if f, g are functions in
It follows that if you have a (nonzero) function
I note belatedly that f is given by you in post #3, which also invokes Parseval's theorem.
The theorem that if (,
Finally, I found many versions of Parsevals theorem, with various (formula, theorem, identity) interchanged names, including for Fourier Series and one that states the Fourier transform is bijective, but none with your version. Could you please give a source for your version, preferably internet?
Parseval's theorem - Wikipedia, the free encyclopedia (see especially the section headed Equivalence of the norm and inner product forms.Quote:
Originally Posted by HartlwThat wraps it up. Thanks, very nice.
I note belatedly that f is given by you in post #3, which also invokes Parseval's theorem.
The theorem that if (
Finally, I found many versions of Parsevals theorem, with various (formula, theorem, identity) interchanged names, including for Fourier Series and one that states the Fourier transform is bijective, but none with your version. Could you please give a source for your version, preferably internet?