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In [[mathematics]], '''trigonometric interpolation''' is [[interpolation]] with [[trigonometric polynomial]]s. Interpolation is the process of finding a function which goes through some given data points. For trigonometric interpolation, this function has to be a trigonometric polynomial, that is, a sum of [[trigonometric function|sines and cosines]] of given periods. This form is especially suited for interpolation of [[periodic function]]s.
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An important special case is when the given data points are equally spaced, in which case the solution is given by the [[discrete Fourier transform]].
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==Formulation of the interpolation problem==
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A trigonometric polynomial of degree ''K'' has the form
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:<math> p(x) = a_0 + \sum_{k=1}^K a_k \cos(kx) + \sum_{k=1}^K b_k \sin(kx). \, </math>
This expression contains 2''K'' + 1 coefficients, ''a''<sub>0</sub>, ''a''<sub>1</sub>, … ''a''<sub>''K''</sub>, ''b''<sub>1</sub>, …, ''b''<sub>''K''</sub>, and we wish to compute those coefficients so that the function passes through ''N'' points:
:<math> p(x_n) = y_n, \quad n=0,\ldots ,N-1. \, </math>
Since the trigonometric polynomial is periodic with period 2π, the ''N'' points can be distributed and ordered in one period as
:<math> 0 \leq x_0 < x_1 < x_2 < \ldots  < x_{N-1} < 2  \pi. \, </math>
(Note that we do ''not'' in general require these points to be equally spaced.) The interpolation problem is now to find coefficients such that the trigonometric polynomial ''p'' satisfies the interpolation conditions.


==Solution of the problem==
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Under the above conditions, there exists a solution to the problem for ''any'' given set of data points {''x''<sub>''k''</sub>, ''y''<sub>''k''</sub>} as long as ''N'', the number of data points, is not larger than the number of coefficients in the polynomial, i.e., ''N''&nbsp;≤&nbsp;2''K''+1  (a solution may or may not exist if ''N''>2''K''+1 depending upon the particular set of data points). Moreover, the interpolating polynomial is unique if and only if the number of adjustable coefficients is equal to the number of data points, i.e., ''N''&nbsp;=&nbsp;2''K''&nbsp;+&nbsp;1. In the remainder of this article, we will assume this condition to hold true.
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The solution can be written in a form similar to the [[Lagrange polynomial|Lagrange formula for polynomial interpolation]]:
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:<math> p(x) = \sum_{k=0}^{2K} y_k \prod_{m=0,m\ne k}^{2K} \frac{\sin\frac12(x-x_m)}{\sin\frac12(x_k-x_m)}. \, </math>
This can be shown to be a trigonometric polynomial by employing the [[trigonometric identity#Multiple-angle formulae|multiple-angle formula]] and other identities for sin&nbsp;½(''x''&nbsp;−&nbsp;''x''<sub>''m''</sub>).
 
==Formulation in the complex plane==
 
The problem becomes more natural if we formulate it in the [[complex plane]]. We can rewrite the formula for a trigonometric polynomial as
:<math> p(x) = \sum_{k=-K}^K c_k e^{ikx}, \, </math>
where ''i'' is the [[imaginary unit]]. If we set ''z'' = ''e''<sup>''ix''</sup>, then this becomes
:<math> q(z) = \sum_{k=-K}^K c_k z^{k}, \, </math>
with
:<math> q(e^{ix}) \triangleq p(x). \, </math>
This reduces the problem of trigonometric interpolation to that of polynomial interpolation on the [[unit circle]]. Existence and uniqueness for trigonometric interpolation now follows immediately from the corresponding results for polynomial interpolation.
 
For more information on formulation of trigonometric interpolating polynomials in the complex plane see [http://www.physics.arizona.edu/~restrepo/475A/Notes/sourcea.pdf , p135  Interpolation using Fourier Polynomials].
 
==Equidistant nodes and the discrete Fourier transform==
 
The special case in which the points ''x''<sub>''n''</sub> are equally spaced is especially important. In this case, we have
:<math> x_n = 2 \pi \frac{n}{N}, \qquad 0 \leq n < N.</math>
 
The transformation that maps the data points ''y''<sub>''n''</sub> to the coefficients ''a''<sub>''k''</sub>, ''b''<sub>''k''</sub> is obtained from the [[discrete Fourier transform]] (DFT) of order&nbsp;N.
 
:<math> Y_k = \sum_{n=0}^{N-1} y_n \ e^{-i 2 \pi \frac{nk}{N}} \, </math>
 
:<math> y_n = p(x_n) = \frac{1}{N} \sum_{k=0}^{N-1} Y_k \ e^{i 2 \pi \frac{nk}{N}} \, </math>
 
 
 
(Because of the way the problem was formulated above, we have restricted ourselves to odd numbers of points.  This is not strictly necessary; for even numbers of points, one includes another cosine term corresponding to the [[Nyquist frequency]].)
 
 
 
The case of the cosine-only interpolation for equally spaced points, corresponding to a trigonometric interpolation when the points have [[Even and odd functions|even symmetry]], was treated by [[Alexis  Clairaut]] in 1754.  In this case the solution is equivalent to a [[discrete cosine transform]].  The sine-only expansion for equally spaced points, corresponding to odd symmetry, was solved by [[Joseph Louis Lagrange]] in 1762, for which the solution is a [[discrete sine transform]].  The full cosine and sine interpolating polynomial, which gives rise to the DFT, was solved by [[Carl Friedrich Gauss]] in unpublished work around 1805, at which point he also derived a [[Cooley–Tukey FFT algorithm|fast Fourier transform]] algorithm to evaluate it rapidly. Clairaut, Lagrange, and Gauss were all concerned with studying the problem of inferring the [[orbit]] of [[planet]]s, [[asteroid]]s, etc., from a finite set of observation points; since the orbits are periodic, a trigonometric interpolation was a natural choice.  See also Heideman ''et al.'' (1984).
 
==References==
 
* Kendall E. Atkinson, ''An Introduction to Numerical Analysis'' (2nd edition), Section 3.8. John Wiley & Sons, New York, 1988. ISBN 0-471-50023-2.
* M. T. Heideman, D. H. Johnson, and C. S. Burrus, "[http://nctuocw.sjtu.edu.cn/new/course/fourier/supplement/heideman-johnson-etal1985.pdf Gauss and the history of the fast Fourier transform]," ''IEEE ASSP Magazine'' '''1''' (4), 14&ndash;21 (1984).
 
[[Category:Interpolation]]
[[Category:Trigonometry]]

Latest revision as of 05:35, 25 December 2014

The standard of the games being produced for our mobile phones nowadays has improved tremendously. Online video video game titles upgrade your concentration stage to an awesome extent. Some theorize that it is the sense of relief when the horror is gone that people enjoy and mistakenly take for enjoyment of the game. After the illustrious Total War series perhaps began to stagnate, given the release of Medieval 2: Total War and its subsequent expansions, the games were in need of a reboot and rethink. 

This is the reason why action rpg games online has made its presence felt in the gaming world. This new Westward game is a departure from the gritty art style in the previous games. I would like to believe, that with a corporation as big as Microsoft that, they have had plans for it years to the future. This article will be devoted to the pros of the Honor System and the cons of the Token system. However, I've found myself getting frustrated lately by the amount of quests that they are putting out, and the fact that most of my neighbors have decided that they no longer are interested, because of the "work" that is now involved in "playing".

• Four control types: Touch, Relative, Tilt, Joypad. It is great to see Dominion grow with expansions like this, adding new themes and mechanics to an already-strong core game. With people believing that life is fun when you are out partying and drinking do to shows like Jersey Shore, they are dealing with a High Risk lifestyle. A large number of the cards in this set support that theme, actively tampering with your opponents' hands and decks, and forcing them to trash valuable cards.

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