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In [[mathematics]], the '''trigonometric [[moment problem]]''' is formulated as follows: given a finite sequence {''&alpha;''<sub>0</sub>,&nbsp;...&nbsp;''&alpha;<sub>n</sub>''&nbsp;}, does there exist a positive [[Borel measure]] ''&mu;'' on the interval [0, 2''&pi;''] such that
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:<math>\alpha_k = \frac{1}{2 \pi}\int_0 ^{2 \pi} e^{-ikt}\,d \mu(t).</math>
 
In other words, an affirmative answer to the problems means that {''&alpha;''<sub>0</sub>,&nbsp;...&nbsp;''&alpha;<sub>n</sub>''&nbsp;} are the first ''n'' + 1 ''Fourier coefficients'' of some positive Borel measure ''&mu;'' on [0, 2''&pi;''].
 
== Characterization ==
 
The trigonometric moment problem is solvable, that is, {''&alpha;<sub>k</sub>''} is a sequence of Fourier coefficients, if and only if the (''n'' + 1) &times; (''n'' + 1) [[Toeplitz matrix]]
 
:<math>
A =
\left(\begin{matrix}
\alpha_0      & \alpha_1          & \cdots  & \alpha_n    \\
\bar{\alpha_1} & \alpha_0          & \cdots  & \alpha_{n-1} \\
\vdots        & \vdots            & \ddots  & \vdots      \\
\bar{\alpha_n} & \bar{\alpha_{n-1}} & \cdots  & \alpha_0    \\
\end{matrix}\right)</math>
 
is [[positive semidefinite]].
 
The "only if" part of the claims can be verified by a direct calculation.
 
We sketch an argument for the converse. The positive semidefinite matrix ''A'' defines a [[sesquilinear]] product on '''C'''<sup>''n'' + 1</sup>, resulting in a [[Hilbert space]]
 
:<math>(\mathcal{H}, \langle \;,\; \rangle)</math>
 
of dimensional at most ''n'' + 1, a typical element of which is an equivalence class denoted by [''f'']. The Toeplitz structure of ''A'' means that a "truncated" shift is a [[partial isometry]] on <math>\mathcal{H}</math>. More specifically, let {&nbsp;''e''<sub>0</sub>,&nbsp;...''e''<sub>''n''</sub>&nbsp;} be the standard basis of '''C'''<sup>''n'' + 1</sup>. Let <math>\mathcal{E}</math> be the subspace generated by {&nbsp;[''e''<sub>0</sub>],&nbsp;...&nbsp;[''e''<sub>''n'' - 1</sub>]&nbsp;} and <math>\mathcal{F}</math> be the subspace generated by {&nbsp;[''e''<sub>1</sub>],&nbsp;...&nbsp;[''e''<sub>''n''</sub>]&nbsp;}. Define an operator
 
:<math>V: \mathcal{E} \rightarrow \mathcal{F}</math>
 
by
 
:<math>V[e_k] = [e_{k+1}] \quad \mbox{for} \quad k = 0 \ldots n-1.</math>
 
Since
 
:<math>\langle V[e_j], V[e_k] \rangle = \langle [e_{j+1}], [e_{k+1}] \rangle = A_{j+1, k+1} = A_{j, k} = \langle [e_{j+1}], [e_{k+1}] \rangle,</math>
 
''V'' can be extended to a partial isometry acting on all of <math>\mathcal{H}</math>. Take a minimal [[unitary operator|unitary]] extension ''U'' of ''V'', on a possibly larger space (this always exists). According to the [[spectral theorem]], there exists a Borel measure ''m'' on the unit circle '''T''' such that for all integer ''k''
 
:<math>\langle (U^*)^k [ e_ {n+1} ], [ e_ {n+1} ] \rangle = \int_{\mathbf{T}} z^{k} dm .</math>
 
For ''k'' = 0,...,''n'', the left hand side is
 
:<math>
\langle (U^*)^k [ e_ {n+1} ], [ e_ {n+1} ] \rangle
= \langle (V^*)^k [ e_ {n+1} ],  [ e_{n+1} ] \rangle
= \langle [e_{n+1-k}], [ e_{n+1} ] \rangle
= A_{n+1, n+1-k}
= \bar{\alpha_k}.
</math>
 
So
 
:<math>
\int_{\mathbf{T}} z^{-k} dm
= \int_{\mathbf{T}} \bar{z}^k dm
= \alpha_k.
</math>
 
Finally, parametrize the unit circle '''T''' by ''e<sup>it</sup>'' on [0, 2''&pi;''] gives
 
:<math>\frac{1}{2 \pi} \int_0 ^{2 \pi} e^{-ikt} d\mu(t) = \alpha_k</math>
 
for some suitable measure ''&mu;.
 
=== Parametrization of solutions ===
 
The above discussion shows that the trigonometric moment problem has infinitely many solutions if the Toeplitz matrix ''A'' is invertible. In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the [[partial isometry]] ''V''.
 
==References==
 
* N.I. Akhiezer, ''The Classical Moment Problem'', Olivier and Boyd,  1965.
* N.I. Akhiezer, M.G. Krein, ''Some Questions in the Theory of Moments'', Amer. Math. Soc., 1962.
 
[[Category:Probability theory]]
[[Category:Measure theory]]
[[Category:Functional analysis]]

Latest revision as of 03:33, 7 January 2015

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