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At first, we sample f(x) in the N (N is odd) equidistant points around x^*:

    \[ f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2} \]

where h is some step.
Then we interpolate points \{(x_k,f_k)\} by polynomial

(1)   \begin{equation*}  P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j} \end{equation*}

Its coefficients \{a_j\} are found as a solution of system of linear equations:

(2)   \begin{equation*}  \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2} \end{equation*}

Here are references to existing equations: (1), (2).
Here is reference to non-existing equation (??).

Post Tag Math

Prove sin^2 (x) + cos^2 (x) = 1 without using Pythagoras Theorem

Sin2x + cos2x = sinx sinx + cosx cosx

= cos (x – x) [cos(A - B) = cosA cosB + sinA sinB]

= cos 0

= 1

OR

Sin2x + cos2x = [(1 – cos2x)/2] + [(1 + cos2x)/2]

= [(1 – cos2x + 1 + cos2x) / 2]

= 2/2

= 1

Post Tag Math

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Post Tag Math