Math 119 - Week 7

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Last updated 8:56 PM on 8/9/26
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7 Terms

1
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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>
2
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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

<p>h = x<sub>1 </sub>- x<sub>0 </sub></p><ul><li><p>assume evenly spaced</p></li></ul><p>Δ<sup>n</sup>y<sub>0</sub> = nth level of pascal’s triangle, with alternating + - + - </p><ul><li><p>ex.) Δ³y₀ = y₃ − 3y₂ + 3y₁ − y₀</p></li></ul><p>n = # of points - 1</p>
3
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taylor polynomials =

approximates f(x) at x = a

Tn, a

  • n = degree

  • a = point to approximate f at

<p>approximates f(x) at x = a</p><p></p><p>T<sub>n, a</sub> </p><ul><li><p>n = degree</p></li><li><p>a = point to approximate f at </p></li></ul><p></p>
4
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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

5
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d/dx(arctanx) =

1/(1 + x²)

6
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d/dx(arcsin(x))

1/√(1 - x²)

7
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d/dx(arccos(x)) =

-1/√(1 - x²)