Koder search engine
Wednesday, June 22nd, 2005While reading about Google’s history of acquisitions, as well as likely future acquisitions, I discovered a search engine just for code! I tried running a few queries, and was pleased by the fact that I even got results; I wonder what metric they’re using for matching: it seems like they’re just searching for direct matches. The only problem I’ve noticed is that they don’t index small, out of the way projects; of course, this may be due to three plausible considerations: that they don’t want to accidentally index someone’s private code and get into legal issues (although, one could argue this is a non-issue: if someone posted code that could be spidered, they must not have wanted it kept private), it might be hard to distinguish code from general web content (e.g. short code excerpts on a web-page probably aren’t worth the bother of identifying), and … maybe the metric they’re using relegates small projects to the back of the list, were I didn’t look.
is a bounded sequence. We’ve seen that if 
is a point, 
, then the sequence 
converges, where
is a function with initial set that contains all points 
. Show that if 
and 
, there is a positive integer 
such that if 
and 
,
.
let 
be the least integer satisfying 
. Let 
be the collection of all 
is finite. Assume 
, there is a non-empty set 
. Notice 
, and each 
contains infinitely many elements. From here we can use the same ‘boxing in’ process that we used to show power series converge: namely, since 
, and then a rectangle of even smaller size which contains infinitely many of 
. Since we get to choose the size of the rectangles, we can make them converge to a point 
as 
. This point is in 
, and therefore has the property that 
, which is not possible. Therefore 
which satisfies: if 
and 
is used in making the rectangles to ensure that the 
; otherwise you couldn’t claim a contradiction.
be? Let’s find out…
are uniquely associated with it. Note, in this context, ![L^1[-\pi, \pi]](/cz/latexrender/pictures/46ec57933585c7eaab6c3151eb85c1e9.png "L^1[-\pi, \pi]")
, and likewise for 
. 
: it is linear and continuous. Note also that 
is also a linear, continuous operator from 
can also be characterized by 
. If 
, then 
can be defined as 
, and is a continuous linear transformation. 
, 
, consisting of all the continuous linear transformations on 
as defined before, 
. In fact, it is clear that 
.
, then 
, and 
. Since 
, also 
. Let 
, then 
.
, then 
in the sense that 
in the 
is continuous, 
, so for some subsequence 
, 
almost everywhere. Since 
almost everywhere also, 
almost everywhere, so 
almost everywhere.
such that 
for all 
. This since 
for all 
. 
is a Hilbert space and 
. If 
(or equivalently, 
), 
be a doubly invariant subspace of 
, such that 
. 
be the 
is a continuous linear transformation, and acts as an identity on its image. 
, since 
where 
and 
, so 
since 
. Therefore 
, so 
for some 
; specifically, taking 
, 
.
, so 
; then 
. Note that 
guarantees 
and suppose 
for all 
. Then 
. 
where 
. Assume 
. Since 
is orthogonal to the closure, M, of the linear span, N, of 
. 
. Then 
and 
. Since 
and 
, so 
, so 
. Therefore 
. Since 
, 
for some 
, so 
.
a.e.
periodic function with a continuous derivative 
on ![[-\pi, \pi]](/cz/latexrender/pictures/86f9a3a2cc344b78b146c202b2226bae.png "[-\pi, \pi]")
. Let the Fourier series for 
. Without using the fact that 
that 
. 
be two continuous 
in ![\displaystyle h(\theta) = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) g(\theta - t) dt, \quad \forall \theta \in [-\pi, \pi]](/cz/latexrender/pictures/1e27addeec562fd7356b68b7d2538f74.png "\displaystyle h(\theta) = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) g(\theta - t) dt, \quad \forall \theta \in [-\pi, \pi]")
.
is continuous and has F.S. 
in 
be defined as 
where 
and 
.