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lagrange interpolation formula =
pn(x) = (L0(x))(y0) + (L1(x))(y1) + … + (Ln(x))(yn)
n = # of points - 1
Li(x) is the product of (x - xk)/(xi - xk)
k ∈ [0, n], k ≠ i
![<p>p<sub>n</sub>(x) = (L<sub>0</sub>(x))(y<sub>0</sub>) + (L<sub>1</sub>(x))(y<sub>1</sub>) + … + (L<sub>n</sub>(x))(y<sub>n</sub>)</p><ul><li><p>n = # of points - 1</p></li></ul><p></p><p>L<sub>i</sub>(x) is the product of (x - x<sub>k</sub>)/(x<sub>i</sub> - x<sub>k</sub>)</p><ul><li><p>k ∈ [0, n], k ≠ i</p></li></ul><p></p>](https://assets.knowt.com/user-attachments/4009182a-8a75-43b3-8a3f-8c05fbf8eac6.png)
newton’s interpolation formula =
h = x1 - x0
assume evenly spaced
Δny0 = nth level of pascal’s triangle, with alternating + - + -
ex.) Δ³y₀ = y₃ − 3y₂ + 3y₁ − y₀
n = # of points - 1

taylor polynomials =
approximates f(x) at x = a
Tn, a
n = degree
a = point to approximate f at

when to use newton vs lagrange interpolation =
newton:
new xi
adding more points
more points dont change previous terms
lagrange:
xi fixed, yi changes
comparing different curves over the same xi
L(x) doesnt change with different curves
d/dx(arctanx) =
1/(1 + x²)
d/dx(arcsin(x))
1/√(1 - x²)
d/dx(arccos(x)) =
-1/√(1 - x²)