Editing Trigonometric interpolation (section)

==Solution of the problem==
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.

===Odd number of points===
If the number of points ''N'' is odd, say ''N=2K+1'', applying the [[Lagrange polynomial|Lagrange formula for polynomial interpolation]] to the polynomial formulation in the complex plane yields that the solution can be written in the form
{{NumBlk|:|<math> p(x) = \sum_{k=0}^{2K} y_k\,t_k(x),</math>|{{EquationRef|5}}}}
where
:<math> t_k(x) = e^{-iKx+iKx_k} \prod_{\begin{align}m&=0 \\[-4mu] m &\ne k\end{align}}^{2K} \frac{e^{ix}-e^{ix_m}}{e^{ix_k}-e^{ix_m}}.</math>
The factor  <math>e^{-iKx+iKx_k}</math> in this formula compensates for the fact that the complex plane formulation contains also negative powers of <math>e^{ix}</math> and is therefore not a polynomial expression in <math>e^{ix}</math>. The correctness of this expression can easily be verified by observing that <math>t_k(x_k)=1</math> and that <math>t_k(x)</math> is a linear combination of the right powers of <math>e^{ix}</math>.
Upon using the identity
{{NumBlk|:|<math>e^{iz_1}-e^{iz_2}=2i\sin\tfrac12(z_1-z_2)\,e^{(z_1 + z_2)i/2},</math>|{{EquationRef|2}}}}
the coefficient <math>t_k(x)</math> can be written in the form
{{NumBlk|:|<math> t_k(x) = \prod_{\begin{align}m&=0 \\[-4mu] m &\ne k\end{align}}^{2K} \frac{\sin\tfrac12(x-x_m)}{\sin\tfrac12(x_k-x_m)}.</math>|{{EquationRef|4}}}}

===Even number of points===
If the number of points ''N'' is even, say ''N=2K'', applying the [[Lagrange polynomial|Lagrange formula for polynomial interpolation]] to the polynomial formulation in the complex plane yields that the solution can be written in the form
{{NumBlk|:|<math> p(x) = \sum_{k=0}^{2K-1} y_k\,t_k(x),</math>|{{EquationRef|6}}}}
where
{{NumBlk|:|<math> t_k(x) = e^{-iKx+iKx_k} \frac{e^{ix}-e^{i\alpha_k}}{e^{ix_k}-e^{i\alpha_k}} \prod_{\begin{align}m&=0 \\[-4mu] m &\ne k\end{align}}^{2K-1} \frac{e^{ix}-e^{ix_m}}{e^{ix_k}-e^{ix_m}}.</math>|{{EquationRef|3}}}}
Here, the constants <math>\alpha_k</math> can be chosen freely. This is caused by the fact that the interpolating function ({{EquationNote|1}}) contains an odd number of unknown constants. A common choice is to require that the highest frequency is of the form a constant times <math>\cos(Kx)</math>, i.e. the <math>\sin(Kx)</math> term vanishes, but in general the phase of the highest frequency can be chosen to be <math>\varphi_K</math>. To get an expression for <math>\alpha_k</math>, we obtain by using ({{EquationNote|2}}) that ({{EquationNote|3}}) can be written on the form
:<math> t_k(x) = \frac{\cos\tfrac12\Biggl(2Kx-\alpha_k+\displaystyle\sum\limits_{m=0,\,m \ne k}^{2K-1} x_m\Biggr)+\sum\limits_{m=-(K-1)}^{K-1}c_k e^{imx}}{2^N\sin\tfrac12(x_k-\alpha_k)\displaystyle\prod\limits_{m=0,\,m \ne k}^{2K-1}\sin\tfrac12(x_k-x_m)}.</math>
This yields
:<math>\alpha_k=\sum_{\begin{align}m&=0 \\[-4mu] m &\ne k\end{align}}^{2K-1} x_m - 2 \varphi_K</math>
and
:<math> t_k(x) = \frac{\sin\tfrac12(x-\alpha_k)}{\sin\tfrac12(x_k-\alpha_k)}\prod_{\begin{align}m&=0 \\[-4mu] m &\ne k\end{align}}^{2K-1} \frac{\sin\tfrac12(x-x_m)}{\sin\tfrac12(x_k-x_m)}.</math>

Note that care must be taken in order to avoid infinities caused by zeros in the denominators.