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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 (??).

 

    \[ \boxed{f(x)=\int_1^{\infty}\frac{1}{x^2}\,\mathrm{d}x=1} \]

 

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Binomial theorem
—————-

    \[(x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k\]

Exponential function
——————–

    \[e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n\]

Cauchy–Schwarz inequality
——————————-

    \[\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\]

Bayes’ theorem
————–

    \[P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)}\]

Euler’s summation formula
————————-
_Theorem._ Euler’s summation formula. _If f has a continuous
derivative f' on the interval [y, x], where 0 < y < x,
then_

(3)   \begin{align*} \sum_{y < n \le x} f(n) = & \int_y^x f(t) dt + \int_y^x (t - [t]) f'(t) dt \notag \\ & + f(x)([x] - x) - f(y)([y] - y). \end{align*}

_Proof._ Let m = [y], k = [x]. For integers n and n - 1
in [y, x] we have

    \begin{align*} \int_{n-1}^n [t] f'(t) dt & = \int_{n-1}^n f'(t) dt \\ & = (n - 1) \bigl( f(n) - f(n - 1) \bigr) \\ & = \bigl( n f(n) - (n - 1) f(n - 1) \bigr) - f(n). \end{align*}

Summing from n = m + 1 to n = k we find

    \begin{align*} \int_{m}^k [t] f'(t) dt & = \sum_{n = m + 1}^k \bigl( n f(n) - (n - 1) f(n - 1) \bigr) - \sum_{y < n \le x} f(n) \\ & = k f(k) - m f(m) - \sum_{y < n \le x} f(n). \end{align*}

Hence,

(4)   \begin{align*} \sum_{y < n \le x} f(n) & = - \int_{m}^k [t] f'(t) dt + k f(k) - m f(m) \notag \\ & = - \int_{y}^x [t] f'(t) dt + k f(x) - m f(y). \end{align*}

Integration by parts gives us

    \begin{equation*} \int_y^x f(t) dt = x f(x) - y f(y) - \int_y^x t f'(t) dt. \end{equation*}

When this is combined with \eqref{summation} we obtain
\eqref{theorem}.

3 Responses

  1. Kays says:

    Let k be a positive integer and \{\} be the fractional par function, then calculate in closed-form the following Multiple Integral :

        \[\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\ln^k(xyz)\bigg\{\bigg(\frac{x}{y}\bigg)^k\bigg\}\bigg\{\bigg(\frac{y}{z}\bigg)^k\bigg\}\bigg\{\bigg(\frac{z}{x}\bigg)^k\bigg\}\mathrm dx\ \mathrm dy\ \mathrm dz\]

  2. 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 (??).

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