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admin-thoth Articles January 22, 2019 | 3

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

Kays says:
January 22, 2019 at 5:19 PM
Let $k$ be a positive integer and $\{\}$ be the fractional par function, then calculate in closed-form the following Multiple Integral :
$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\ln^k(xyz)\bigg\{\bigg(\frac{x}{y}\bigg)^k\bigg\}\bigg\{\bigg(\frac{y}{z}\bigg)^k\bigg\}\bigg\{\bigg(\frac{z}{x}\bigg)^k\bigg\}\mathrm dx\ \mathrm dy\ \mathrm dz$
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Kays says:
January 22, 2019 at 5:20 PM
$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\ln^k(xyz)\bigg\{\bigg(\frac{x}{y}\bigg)^k\bigg\}\bigg\{\bigg(\frac{y}{z}\bigg)^k\bigg\}\bigg\{\bigg(\frac{z}{x}\bigg)^k\bigg\}\mathrm dx\ \mathrm dy\ \mathrm dz$
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admin-thoth says:
January 23, 2019 at 8:23 AM
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 (??).
Log in to Reply

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