The document discusses various methods of interpolation, including Lagrange and Newton interpolation polynomials. Lagrange interpolation involves constructing a polynomial that passes through a set of n data points, represented by its values at the points. Newton interpolation similarly uses a polynomial but is based on divided differences. Both can be used to interpolate values within or extrapolate beyond the original data range. The complexity of calculating the interpolation polynomials is also addressed.

1. Interpolation
Dmytro Mitin
fb.com/dmitry.mitin
21 October 2015. Algoclub #17
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2. Interpolation of functions
Finite differences
Divided differences
Lagrange interpolating polynomial
Newton interpolating polynomial
Hermite interpolating polynomial∗
Interpolation error
Neville algorithm
Algorithmic complexity of interpolation
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3. String interpolation
int x = 5, y = 4;
System.out.println(String.format("%d * %d = %d",
x, y, x * y));
5 * 4 = 20
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4. How many roots?
c ·
(x − a)(x − b)
(c − a)(c − b)
+ a ·
(x − b)(x − c)
(a − b)(a − c)
+ b ·
(x − c)(x − a)
(b − c)(b − a)
= x
(a, b, c are fixed, a = b, b = c, c = a)
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5. How many roots?
(x − a)(x − b)
(c − a)(c − b)
+
(x − b)(x − c)
(a − b)(a − c)
+
(x − c)(x − a)
(b − c)(b − a)
= 1
(a, b, c are fixed, a = b, b = c, c = a)
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6. How many roots?
c2
·
(x − a)(x − b)
(c − a)(c − b)
+a2
·
(x − b)(x − c)
(a − b)(a − c)
+b2
·
(x − c)(x − a)
(b − c)(b − a)
= x2
(a, b, c are fixed, a = b, b = c, c = a)
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7. How many roots?
c3
·
(x − a)(x − b)
(c − a)(c − b)
+a3
·
(x − b)(x − c)
(a − b)(a − c)
+b3
·
(x − c)(x − a)
(b − c)(b − a)
= x3
(a, b, c are fixed, a = b, b = c, c = a)
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8. How many roots?
c4
·
(x − a)(x − b)
(c − a)(c − b)
+a4
·
(x − b)(x − c)
(a − b)(a − c)
+b4
·
(x − c)(x − a)
(b − c)(b − a)
= x4
(a, b, c are fixed, a = b, b = c, c = a)
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9. Extrapolation
World population
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10. Approximation
Uniform approximation
max
x∈[a,b]
|f (x) − p(x)| → min
p∈P
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11. Approximation
Mean approximation
b
a |f (x) − p(x)| → min
p∈P
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12. Interpolation
p(xk) = f (xk), k = 0, n − 1 or p(xk) = yk, k = 0, n − 1
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13. Lagrange interpolating polynomial
n = 3
p(x) = f (c)·
(x − a)(x − b)
(c − a)(c − b)
+f (a)·
(x − b)(x − c)
(a − b)(a − c)
+f (b)·
(x − c)(x − a)
(b − c)(b − a)
p(a) = f (a), p(b) = f (b), p(c) = f (c)
(a = b, b = c, c = a)
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14. Lagrange interpolating polynomial
p(x) = f (x0) ·
(x − x0)(x − x1) . . . (x − xn−1)
(x0 − x0)(x0 − x1) . . . (x0 − xn−1)
+ f (x1) ·
(x − x0)
(x − x1) . . . (x − xn−1)
(x1 − x0)
(x1 − x1) . . . (x1 − xn−1)
+ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
+ f (xn−1) ·
(x − x0)(x − x1) . . .
(x − xn−1)
(xn−1 − x0)(xn−1 − x1) . . .(((((((
(xn−1 − xn−1)
p(x0) = f (x0), p(x1) = f (x1), . . . , p(xn−1) = f (xn−1)
(x0 = . . . = xn−1)
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15. p(x) =
n−1
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xn−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xn−1)
Definition
Lagrange interpolating polynomial
p(x) =
n−1
k=0
f (xk) ·
0≤j≤n−1
j=k
(x − xj )
0≤j≤n−1
j=k
(xk − xj )
p(xk) = f (xk), k = 0, n − 1
(x0 = . . . = xn−1)
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16. Joseph-Louis Lagrange (1736–1813)
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17. Uniqueness
p(x) =
n−1
i=0
ai xi
p(xk) = f (xk), k = 0, n − 1
n−1
i=0
ai xi
k = f (xk), k = 0, n − 1
det
1 1 . . . 1
x0 x1 . . . xn−1
x2
0 x2
1 . . . x2
n−1
. . . . . . . . . . . .
xn−1
0 xn−1
1 . . . xn−1
n−1
=
0≤jk≤n−1
(xk − xj ) = 0,
x0 = . . . = xn−1
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18. Equidistant nodes
p(x) =
n−1
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xn−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xn−1)
Partial case: x1 − x0 = x2 − x1 = . . . = xn−1 − xn−2 = h,
xk = x0 + k · h, k = 0, n − 1
p(x) =
n−1
k=0
f (x0 + k h) · (x−x0)(x−x0−h)...((((((x−x0−k·h)...(x−x0−(n−1)h)
k(k−1)...(k−k)...(k−n+1)·hn−1
p(x) =
n−1
k=0
f (x0 + k h) ·
t(t − 1) . . .(t − k) . . . (t − n + 1)
k!(n − k − 1)!(−1)n−k−1
,
t =
x − x0
h
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19. Complexity
p(x) =
n−1
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xn−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xn−1)
What is complexity?
Is this formula good?
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20. p(x) =
n−1
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xn−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xn−1)
≡ pn(x)
p(x) = p1(x) +
n−1
j=1
pj+1(x) − pj (x)
p(x) = f (x0) +
n−1
j=1
j
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xj )
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
−
j−1
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xj−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xj−1)
p(x) = f (x0) +
n−1
j=1
Cj · (x − x0)(x − x1) . . . (x − xj−1)
Attention! Algebraic trick:
x = x0, x1, . . . , xj−1
j
are zeroes, deg = j
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21. Let’s rewrite formula
p(x) = f (x0) +
n−1
j=1
j
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xj )
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
−
j−1
k=0
f (xk) ·
(x − x0) . . .
(x − xk) . . . (x − xj−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xj−1)
p(x) = f (x0) +
n−1
j=1
Cj · (x − x0)(x − x1) . . . (x − xj−1)
Cj ? x = xj :
f (xj ) −
j−1
k=0
f (xk) ·
(xj − x0) . . .
(xj − xk) . . . (xj − xj−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xj−1)
= Cj · (xj − x0)(xj − x1) . . . (xj − xj−1)
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22. Let’s rewrite formula
f (xj ) −
j−1
k=0
f (xk) ·
(xj − x0) . . .
(xj − xk) . . . (xj − xj−1)
(xk − x0) . . .
(xk − xk) . . . (xk − xj−1)
= Cj · (xj − x0)(xj − x1) . . . (xj − xj−1)
Cj =
f (xj )
(xj − x0)(xj − x1) . . . (xj − xj−1)
−
j−1
k=0
f (xk) ·
1
(xk − x0) . . .
(xk − xk) . . . (xk − xj−1)
·
1
xj − xk
Cj =
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
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23. Cj =
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
p(x) = f (x0) +
n−1
j=1
Cj · (x − x0)(x − x1) . . . (x − xj−1)
p(x) = f (x0) +
n−1
j=1
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
·(x − x0)(x − x1) . . . (x − xj−1)
Definition
Newton interpolating polynomial
p(x) = f (x0) +
n−1
j=1
[f ; x0, x1, . . . , xj ]
Divided differences
· (x − x0)(x − x1) . . . (x − xj−1)
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24. Isaac Newton (1643–1727)
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25. Equidistant nodes
p(x) =
n−1
j=0
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
·(x − x0)(x − x1) . . . (x − xj−1)
Partial case: x1 − x0 = x2 − x1 = . . . = xn−1 − xn−2 = h,
xk = x0 + k · h, k = 0, n − 1
p(x) =
n−1
j=0
j
k=0
f (x0 + k h)
k!(j − k)!(−1)j−khj
·(x − x0)(x − x0 − h) . . . (x − x0 − (j − 1)h)
p(x) =
n−1
j=0
1
j!
j
k=0
(−1)j−k
Ck
j f (x0 + k h) · t(t − 1) . . . (t − j + 1)
t = x−x0
h , Ck
j = j!
k!(j−k)!
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26. Finite differences
p(x) =
n−1
j=0
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
·(x − x0)(x − x1) . . . (x − xj−1)
p(x) =
n−1
j=0
1
j!
j
k=0
(−1)j−k
Ck
j f (x0 + k h) · t(t − 1) . . . (t − j + 1)
t =
x − x0
h
, Ck
j =
j!
k!(j − k)!
∆j
f (x0, h) =
j
k=0
(−1)j−k
Ck
j f (x0 + k h)
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27. Finite differences
∆j
f (x0, h) =
j
k=0
(−1)j−k
Ck
j f (x0 + k h)
∆0
f (x0, h) = f (x0)
∆j
f (x0, h) = ∆(∆j−1
f )(x0, h)
∆f (x0, h) = f (x0 + h) − f (x0)
∆2
f (x0, h) = ∆(∆f )(x0, h) = ∆f (x0 + h) − ∆f (x0)
= f (x0 + 2h) − 2f (x0 + h) + f (x0)
∆3
f (x0, h) = f (x0 + 3h) − 3f (x0 + 2h) + 3f (x0 + h) − f (x0)
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28. Complexity
p(x) =
n−1
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
p(xk) = f (xk), k = 0, n − 1
[f ; x0, x1, . . . , xj ] =
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
How to add new node?
What is complexity?
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29. Divided differences
p(x) =
n−1
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
[f ; x0, x1, . . . , xj ] =
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
[f ; x0] = f (x0)
[f ; x0, x1] =
f (x0)
x0 − x1
+
f (x1)
x1 − x0
=
f (x0) − f (x1)
x0 − x1
[f ; x0, x1, x2] = f (x0)
(x0−x1)(x0−x2) + f (x1)
(x1−x0)(x1−x2) + f (x2)
(x2−x0)(x2−x1)
=
f (x0)−f (x1)
x0−x1
− f (x1)−f (x2)
x1−x2
x0 − x2
=
[f ; x0, x1] − [f ; x1, x2]
x0 − x2
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30. Divided differences
p(x) =
n−1
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
[f ; x0, x1, . . . , xj ] =
j
k=0
f (xk)
(xk − x0) . . .
(xk − xk) . . . (xk − xj )
Recurrence:
[f ; x0, x1, . . . , xj−1, xj ] =
[f ; x0, x1, . . . , xj−1] − [f ; x1, . . . , xj−1, xj ]
x0 − xj
, j ≥ 1
[f ; x0] = f (x0)
[f ; . . . , xr , . . . , xs, . . .] =
[f ; . . . , xr , . . . ,xs, . . .] − [f ; . . . ,xr , . . . , xs, . . .]
xr − xs
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31. Interpolation error
p(x) =
n−1
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
p(xk) = f (xk), k = 0, n − 1
q(x) =
n
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
= p(x) + [f ; x0, x1, . . . , xn] · (x − x0)(x − x1) . . . (x − xn−1)
q(xk) = f (xk), k = 0, n
f (xn) = q(xn) = p(xn) + [f ; x0, x1, . . . , xn]
·(xn − x0)(xn − x1) . . . (xn − xn−1)
f (x) = p(x) + [f ; x0, x1, . . . , xn−1, x]
·(x − x0)(x − x1) . . . (x − xn−1)
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32. Interpolation error
f (x) = p(x) + [f ; x0, x1, . . . , xn−1, x]
·(x − x0)(x − x1) . . . (x − xn−1)
f (xk) − p(xk) = 0, k = 0, n − 1
f (n−1)
(ξ) − p(n−1)
(ξ) = 0, ξ ∈ [a, b] x0, . . . , xn−1
f (n−1)
(ξ) = p(n−1)
(ξ) = [f ; x0, x1, . . . , xn−1] · (n − 1)!
[f ; x0, x1, . . . , xn−1] =
f (n−1)(ξ)
(n − 1)!
f (x) − p(x) =
f (n−1)(ξ)
(n − 1)!
· (x − x0)(x − x1) . . . (x − xn−1)
|f (x) − p(x)| ≤
1
(n − 1)!
max
[a,b]
|f (n−1)
| · (b − a)n
, x ∈ [a, b]
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33. Neville algorithm
p(x) =
n−1
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
= f (x0) + (x − x0) · [f ; x0, x1] + (x − x1) ·
. . . [f ; x0, . . . , xn−2] + (x − xn−1)[f ; x0, . . . , xn−1] . . .
[f ; x0, x1, . . . , xj−1, xj ] =
[f ; x0, x1, . . . , xj−1] − [f ; x1, . . . , xj−1, xj ]
x0 − xj
[f ; x0, . . . , xn−1]
[f ; x0, . . . , xn−2] [f ; x1, . . . , xn−1]
[f ; x0, . . . , xn−3] [f ; x1, . . . , xn−2] [f ; x2, . . . , xn−1]
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
[f ; x0, x1] [f ; x1, x2] [f ; x2, x3] . . . . . . [f ; xn−3, xn−2] [f ; xn−2, xn−1]
f (x0) f (x1) f (x2) f (x3) . . . f (xn−3) f (xn−2) f (xn−1)
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34. Eric Harold Neville (1889–1961)
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35. Neville algorithm
p(x) =
n−1
j=0
[f ; x0, x1, . . . , xj ] · (x − x0)(x − x1) . . . (x − xj−1)
[f ; x0, x1, . . . , xj−1, xj ] =
[f ; x0, x1, . . . , xj−1] − [f ; x1, . . . , xj−1, xj ]
x0 − xj
[f ; x0, x1, x2, . . . , xj ] =
[f ; x0, x2, . . . , xj ] − [f ; x1, x2, . . . , xj ]
x0 − x1
p(x) ≡ p(f ; x0, . . . , xn−1; x)
p(f ; x0, . . . , xn−1; x) = f (x0) +
f (x0) − f (x1)
x0 − x1
(x − x0)
+
x − x1
x0 − x1
n−1
j=2
[f ; x0, x2, . . . , xj ] · (x − x0)(x − x2) . . . (x − xj−1)
−
x − x0
x0 − x1
n−1
j=2
[f ; x1, . . . , xj ] · (x − x1) . . . (x − xj−1)
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36. Neville algorithm
p(f ; x0, . . . , xn−1; x) = f (x0) +
f (x0) − f (x1)
x0 − x1
(x − x0)
+
x − x1
x0 − x1
n−1
j=2
[f ; x0, x2, . . . , xj ] · (x − x0)(x − x2) . . . (x − xj−1)
−
x − x0
x0 − x1
n−1
j=2
[f ; x1, . . . , xj ] · (x − x1) . . . (x − xj−1)
p(f ; x0, . . . , xn−1; x) = f (x0) +
f (x0) − f (x1)
x0 − x1
(x − x0)
+
x − x1
x0 − x1
p(f ; x0, x2, . . . , xn−1; x) − f (x0)
−
x − x0
x0 − x1
p(f ; x1, . . . , xn−1; x) − f (x1)
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37. Neville algorithm
p(f ; x0, . . . , xn−1; x) = f (x0) +
f (x0) − f (x1)
x0 − x1
(x − x0)
+
x − x1
x0 − x1
p(f ; x0, x2, . . . , xn−1; x) − f (x0)
−
x − x0
x0 − x1
p(f ; x1, . . . , xn−1; x) − f (x1)
p(f ; x0, . . . , xn−1; x) =
x − x1
x0 − x1
p(f ; x0, x2, . . . , xn−1; x)
+
x0 − x
x0 − x1
p(f ; x1, . . . , xn−1; x)
p(f ; x0, . . . , xn−1; x) =
x − xn−1
x0 − xn−1
p(f ; x0, . . . , xn−2; x)
+
x0 − x
x0 − xn−1
p(f ; x1, . . . , xn−1; x)
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38. Neville algorithm
Recurrence for interpolating polynomials
p(f ; x0, . . . , xn−1; x) =
x − xn−1
x0 − xn−1
p(f ; x0, . . . , xn−2; x)
+
x0 − x
x0 − xn−1
p(f ; x1, . . . , xn−1; x)
p(f ; x0, . . . , xn−1; x)
p(f ; x0, . . . , xn−2; x) p(f ; x1, . . . , xn−1; x)
p(f ; x0, . . . , xn−3; x) p(f ; x1, . . . , xn−2; x) p(f ; x2, . . . , xn−1; x)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
p(f ; x0, x1; x) p(f ; x1, x2; x) . . . p(f ; xn−3, xn−2; x) p(f ; xn−2, xn−1; x)
f (x0) f (x1) f (x2) . . . f (xn−2) f (xn−1)
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39. Hermite interpolating polynomial
p(xk) = f (xk), p (xk) = f (xk), . . . , p(mk −1)
(xk) = f (mk −1)
(xk),
k = 0, n − 1
1
(x − x0)m0 . . .
(x − xk)mk . . . (x − xn−1)mn−1
=
∞
j=0
ckj (x − xk)j
= qki (x) +
∞
j=mk −i
ckj (x − xk)j
, deg qki ≤ mk − i − 1
p(x) =
n−1
k=0
(x − x0)m0
. . .
(x − xk)mk
. . . (x − xn−1)mn−1
·
mk −1
i=0
f (i)(xk)
i!
(x − xk)i
qki (x)
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40. Charles Hermite (1822–1901)
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41. What’s next?
Nonpolynomial interpolation (rational functions, . . . )
Multivariate interpolation
Fractal interpolation
. . .
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42. It’s coding time!
Problem 1.
Double interpolateAtPoint
(ListDouble nodes,
Double point,
UnaryOperatorDouble function){
// UnaryOperatorDouble function = x - x * x;
// code here
}
Problem 2.
void addNodeAndUpdateDividedDifferences
(ListDouble nodes,
Double newNode,
ListListDouble dividedDifferences,
UnaryOperatorDouble function) {
//code here
}
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43. Links
https://en.wikipedia.org/wiki/Interpolation
https://en.wikipedia.org/wiki/Finite difference
https://en.wikipedia.org/wiki/Divided differences
https://en.wikipedia.org/wiki/Polynomial interpolation
https://en.wikipedia.org/wiki/Lagrange polynomial
https://en.wikipedia.org/wiki/Newton polynomial
https://en.wikipedia.org/wiki/Hermite interpolation
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
http://mathworld.wolfram.com
/NewtonsDividedDifferenceInterpolationFormula.html
http://mathworld.wolfram.com/HermitesInterpolatingPolynomial.html
https://en.wikipedia.org/wiki/Neville’s algorithm
http://mathworld.wolfram.com/NevillesAlgorithm.html
https://reference.wolfram.com/language/ref/Interpolation.html
http://algolist.ru/maths/approx.php
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