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Wikipedia:Babel
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Mathematics
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![{\displaystyle \sum _{k=1}^{\infty }{x^{k} \over k!}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99025c99cb17dfaa4e0a5cdb7d8db9866496e0a9) | This user loves problem solving. |
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Hardware
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Software
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![](//upload.wikimedia.org/wikipedia/commons/thumb/3/39/Lambda_lc.svg/43px-Lambda_lc.svg.png) | This user can program in Scheme. |
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Topology: an application to number theory
Statement There exist infinitely many primes in
.
Proof Consider the collection
of subsets of
that consists of all sets of the form
, as well as of any (possibly trivial) union of such sets. Then, it is trivial to see that
induces a topology on
. Furthermore, observe that for all such
, which implies that
is closed, as the complement of a union of open sets.
Suppose, for the sake of contradiction, that there were finitely many primes
in
. Then
would be closed, as a finite union of closed sets. However,
, so its complement is not open, a contradiction. Consequently, there exist infinitely many primes.
Counting in two ways: an application to trigonometry
Statement For any
we have
.
Proof Consider a sequence
of unfair coins, each having probability
,
, of turning up heads when tossed. Denote by
the event that, upon sequentially tossing the coins, a head eventually turns up; this can happen in one of countably many ways: a head comes up immediately, or a tail comes up first followed by a head, or two tails come up followed by a head, and so on. For a head to first come up on the
-th toss, all previous tosses must have resulted in tails, which event happens with probability
. Therefore,
. We may, however, calculate
with another method. For
not to happen, each toss must result in a tail, for
. The associated probability is
. Since
, the result follows.
Application This can be applied whenever we can construct a sequence whose range lies in the unit interval
. One particularly interesting application is obtained by setting
, for
, thus establishing the trigonometric identity
.
Counting in two ways: an application to number theory
Lemma If two positive integers
are chosen at random, each number being equally likely to have been selected, then the probability that
are coprime is
Proof Let us adopt the notion that
is the set of positive integers. Recalling that
denotes the greatest common divisor, we wish to show that
.
First, we claim that
, for any positive integer
. Indeed, it is easy to see that
. If
, being uniformly distributed over
, are both restricted to
, then
are uniformly distributed over
, so the right factor equals
, and thus:
, and the claim follows.
Now, observe that
, so
, and the result follows.
Statement If
denotes the sequence of primes in
, then
Proof Due to the lemma, it suffices to show that
, where
are random variables uniformly distributed over
. Let us find yet another (easier) way of computing
. Two integers are coprime whenever they share no prime factors. Therefore,
are coprime if, for every
,
are not both in
, an event whose probability is
. After multiplying the expressions for
, since the events are independent, the result follows.
Isometric embeddings: an application to computational geometry
Lemma Consider the map
defined as
, with
being the normal dot product in
. Then for any two
we have
Proof Indeed, observe that
In fact, equality is achieved for
(in this case, we adopt the notion that
), and the lemma follows.
Statement Given
points
, it is possible to compute the points' Manhattan diameter,
, in runtime that is linear in
; specifically, we can achieve a runtime of
.
Proof Let
and
be as in the lemma. Then
which is easily computed in
.
Uncategorized Problems
Prove that if
is singular with respect to Lebesgue measure on
and
on
for some
irrational, then
. Stanford Qualifying Exams.