User:Hjilderda/sandbox


 * $$x_{1,2}={-b\pm\sqrt{b^2-4ac}\over2a}$$
 * $$f^{(2)}_0$$
 * $$f_0^{(2)}$$
 * $$f^{(1)}_0=\frac{-f_0+f_1}{h}-\frac{1}{2}h\,f^{(2)}(\xi)$$
 * $$f^{(1)}_1=\frac{-f_0+f_1}{h}+\frac{1}{2}h\,f^{(2)}(\xi)$$

(n+1)-punts eerste afgeleide:
 * $$h\,f'(x_i) \; = \sum_{j=0, i \ne j}^n \frac{(-1)^{i-j+1}}{i-j} \frac{i!(n-i)!}{j!(n-j)!} (f_i - f_j) \; + \; \frac{(-1)^{n-i}i!(n-i)!}{(n+1)!} h^{n+1} f^{(n+1)}(\xi) \; = \sum_{j=0, i \ne j}^n \frac{(-1)^{i-j}}{i-j} \frac{\binom{n}{j}}{\binom{n}{i}} (f_j - f_i) \; + \; \frac{(-1)^{n-i}}{\binom{n}{i}(n+1)} h^{n+1} f^{(n+1)}(\xi)$$

2-punts eerste afgeleide:
 * $$h\,f^{(1)}(x_0) = -f_0 + f_1 - \frac{1}{2} h^2 f^{(2)}(\xi)$$
 * $$h\,f^{(1)}(x_1) = -f_0 + f_1 + \frac{1}{2} h^2 f^{(2)}(\xi)$$

3-punts eerste afgeleide:
 * $$h\,f^{(1)}(x_0) = \frac{1}{2} (-3 f_0 + 4 f_1 - f_2) + \frac{1}{3} h^3 f^{(3)}(\xi)$$
 * $$h\,f^{(1)}(x_1) = \frac{1}{2} (-f_0 + f_2) - \frac{1}{6} h^3 f^{(3)}(\xi)$$
 * $$h\,f^{(1)}(x_2) = \frac{1}{2} (f_0 - 4 f_1 + 3 f_2) + \frac{1}{3} h^3 f^{(3)}(\xi)$$

3-punts tweede afgeleide:
 * $$h^2 f^{(2)}(x_0) = f_0 - 2 f_1 + f_2 - h^3 f^{(3)}(\xi)$$
 * $$h^2 f^{(2)}(x_1) = f_0 - 2 f_1 + f_2 - \frac{1}{12} h^4 f^{(4)}(\xi)$$
 * $$h^2 f^{(2)}(x_2) = f_0 - 2 f_1 + f_2 + h^3 f^{(3)}(\xi)$$

4-punts eerste afgeleide:
 * $$h\,f^{(1)}(x_0) = \frac{1}{6} (-11 f_0 + 18 f_1 - 9 f_2 + 2 f_3) - \frac{1}{4} h^4 f^{(4)}(\xi)$$
 * $$h\,f^{(1)}(x_1) = \frac{1}{6} (-2 f_0 - 3 f_1 + 6 f_2 - f_3) + \frac{1}{12} h^4 f^{(4)}(\xi)$$
 * $$h\,f^{(1)}(x_2) = \frac{1}{6} (f_0 - 6 f_1 + 3 f_2 + 2 f_3) - \frac{1}{12} h^4 f^{(4)}(\xi)$$
 * $$h\,f^{(1)}(x_3) = \frac{1}{6} (-2 f_0 + 9 f_1 - 18 f_2 + 11 f_3) + \frac{1}{4} h^4 f^{(4)}(\xi)$$

4-punts tweede afgeleide:
 * $$h^2 f^{(2)}(x_0) = 2 f_0 - 5 f_1 + 4 f_2 - f_3 + \frac{11}{12} h^4 f^{(4)}(\xi)$$
 * $$h^2 f^{(2)}(x_1) = f_0 - 2 f_1 + f_2 - \frac{1}{12} h^4 f^{(4)}(\xi)$$
 * $$h^2 f^{(2)}(x_2) = f_1 - 2 f_2 + f_3 - \frac{1}{12} h^4 f^{(4)}(\xi)$$
 * $$h^2 f^{(2)}(x_3) = -f_0 + 4 f_1 - 5 f_2 + 2 f_3 + \frac{11}{12} h^4 f^{(4)}(\xi)$$

4-punts derde afgeleide:
 * $$h^3 f^{(3)}(x_0) = -f_0 + 3 f_1 - 3 f_2 + f_3 - \frac{3}{2} h^4 f^{(4)}(\xi)$$
 * $$h^3 f^{(3)}(x_1) = -f_0 + 3 f_1 - 3 f_2 + f_3 - \frac{1}{2} h^4 f^{(4)}(\xi)$$
 * $$h^3 f^{(3)}(x_2) = -f_0 + 3 f_1 - 3 f_2 + f_3 + \frac{1}{2} h^4 f^{(4)}(\xi)$$
 * $$h^3 f^{(3)}(x_3) = -f_0 + 3 f_1 - 3 f_2 + f_3 + \frac{3}{2} h^4 f^{(4)}(\xi)$$

5-punts eerste afgeleide:
 * $$h\,f^{(1)}(x_0) = \frac{1}{12} (-25 f_0 + 48 f_1 - 36 f_2 + 16 f_3 - 3 f_4) + \frac{1}{5} h^5 f^{(5)}(\xi)$$
 * $$h\,f^{(1)}(x_1) = \frac{1}{12} (-3 f_0 - 10 f_1 + 18 f_2 - 6 f_3 + f_4) - \frac{1}{20} h^5 f^{(5)}(\xi)$$
 * $$h\,f^{(1)}(x_2) = \frac{1}{12} (f_0 - 8 f_1 + 8 f_3 - f_4) + \frac{1}{30} h^5 f^{(5)}(\xi)$$
 * $$h\,f^{(1)}(x_3) = \frac{1}{12} (-f_0 + 6 f_1 - 18 f_2 + 10 f_3 + 3 f_4) - \frac{1}{20} h^5 f^{(5)}(\xi)$$
 * $$h\,f^{(1)}(x_4) = \frac{1}{12} (3 f_0 - 16 f_1 + 36 f_2 - 48 f_3 + 25 f_4) + \frac{1}{5} h^5 f^{(5)}(\xi)$$

5-punts tweede afgeleide:
 * $$h^2 f^{(2)}(x_0) = \frac{1}{12} (35 f_0 - 104 f_1 + 114 f_2 - 56 f_3 + 11 f_4) - \frac{5}{6} h^5 f^{(5)}(\xi)$$
 * $$h^2 f^{(2)}(x_1) = \frac{1}{12} (11 f_0 - 20 f_1 + 6 f_2 + 4 f_3 - f_4) + \frac{1}{12} h^5 f^{(5)}(\xi)$$
 * $$h^2 f^{(2)}(x_2) = \frac{1}{12} (-f_0 + 16 f_1 - 30 f_2 + 16 f_3 - f_4) + \frac{1}{90} h^6 f^{(6)}(\xi)$$
 * $$h^2 f^{(2)}(x_3) = \frac{1}{12} (-f_0 + 4 f_1 + 6 f_2 - 20 f_3 + 11 f_4) - \frac{1}{12} h^5 f^{(5)}(\xi)$$
 * $$h^2 f^{(2)}(x_4) = \frac{1}{12} (11 f_0 - 56 f_1 + 114 f_2 - 104 f_3 + 35 f_4) + \frac{5}{6} h^5 f^{(5)}(\xi)$$

5-punts derde afgeleide:
 * $$h^3 f^{(3)}(x_0) = \frac{1}{2} (-5 f_0 + 18 f_1 - 24 f_2 + 14 f_3 - 3 f_4) + \frac{7}{4} h^5 f^{(5)}(\xi)$$
 * $$h^3 f^{(3)}(x_1) = \frac{1}{2} (-3 f_0 + 10 f_1 - 12 f_2 + 6 f_3 - f_4) + \frac{1}{4} h^5 f^{(5)}(\xi)$$
 * $$h^3 f^{(3)}(x_2) = \frac{1}{2} (-f_0 + 2 f_1 - 2 f_3 + f_4) - \frac{1}{4} h^5 f^{(5)}(\xi)$$
 * $$h^3 f^{(3)}(x_3) = \frac{1}{2} (f_0 - 6 f_1 + 12 f_2 - 10 f_3 + 3 f_4) + \frac{1}{4} h^5 f^{(5)}(\xi)$$
 * $$h^3 f^{(3)}(x_4) = \frac{1}{2} (3 f_0 - 14 f_1 + 24 f_2 - 18 f_3 + 5 f_4) + \frac{7}{4} h^5 f^{(5)}(\xi)$$

5-punts vierde afgeleide:
 * $$h^4 f^{(4)}(x_0) = f_0 - 4 f_1 + 6 f_2 - 4 f_3 + f_4 - 2 h^5 f^{(5)}(\xi)$$
 * $$h^4 f^{(4)}(x_1) = f_0 - 4 f_1 + 6 f_2 - 4 f_3 + f_4 - h^5 f^{(5)}(\xi)$$
 * $$h^4 f^{(4)}(x_2) = f_0 - 4 f_1 + 6 f_2 - 4 f_3 + f_4 - \frac{1}{6} h^6 f^{(6)}(\xi)$$
 * $$h^4 f^{(4)}(x_3) = f_0 - 4 f_1 + 6 f_2 - 4 f_3 + f_4 + h^5 f^{(5)}(\xi)$$
 * $$h^4 f^{(4)}(x_4) = f_0 - 4 f_1 + 6 f_2 - 4 f_3 + f_4 + 2 h^5 f^{(5)}(\xi)$$

- A backward difference uses the function values at $x$ and $x − h$, instead of the values at $x + h$ and $x$:
 * $$ \nabla_h[f](x) = f(x) - f(x-h). $$


 * $$\nabla^n_h[f](x) = \sum_{i = 0}^{n} (-1)^i \binom{n}{i} f(x - ih),$$

Newton's series
The Newton series consists of the terms of the Newton forward difference equation, named after Isaac Newton; in essence, it is  the Newton interpolation formula, first published in his Principia Mathematica in 1687, namely the discrete analog of the continuum Taylor expansion,

which holds for any polynomial function $f$ and for many (but not all) analytic functions (It does not hold when $f$ is exponential type $$\pi$$. This is easily seen, as the sine function vanishes at integer multiples of $$\pi$$; the corresponding Newton series is identically zero, as all finite differences are zero in this case. Yet clearly, the sine function is not zero.). Here, the expression
 * $$\binom{x}{k} = \frac{(x)_k}{k!}$$

is the binomial coefficient, and
 * $$(x)_k=x(x-1)(x-2)\cdots(x-k+1)$$

is the "falling factorial" or "lower factorial", while the empty product $(x)_{0}$ is defined to be 1. In this particular case, there is an assumption of unit steps for the changes in the values of $x, h = 1$ of the generalization below.

Note the formal correspondence of this result to Taylor's theorem. Historically, this, as well as the Chu–Vandermonde identity,
 * $$(x+y)_n=\sum_{k=0}^n \binom{n}{k} (x)_{n-k} \,(y)_k ,$$

(following from it, and corresponding to the binomial theorem), are included in the observations that matured to the system of umbral calculus.

In a compressed and slightly more general form and equidistant nodes the formula reads
 * $$f(x)=\sum_{k=0}\binom{\frac{x-a}h}{k} \sum_{j=0}^k (-1)^{k-j}\binom{k}{j}f(a+j h).$$


 * $$(-1)^{k}k e_k(x_1,\ldots,x_n) = \sum_{j=1}^k (-1)^{k-j+1} p_j(x_1,\ldots,x_n)e_{k-j}(x_1,\ldots,x_n)\Rightarrow k a_k = -\sum_{j=1}^k a_{k-j}p_j(x_1,\ldots,x_n)$$