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Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
Taylor series
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Taylor series

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  • 1. + Taylor Series John Weiss
  • 2. + Approximating Functions   f(0)= 4   What is f(1)?   f(x) = 4?   f(1) = 4?
  • 3. + Approximating Functions   f(0)= 4, f’(0)= -1   What is f(1)?   f(x) = 4 - x?   f(1) = 3?
  • 4. + Approximating Functions   f(0)= 4, f’(0)= -1, f’’(0)= 2   What is f(1)?   f(x) = 4 – x + x2?   (same concavity)   f(1) = 4?
  • 5. + Approximating Functions   f(x) = sin(x)   What is f(1)?   f(0) = 1, f’(0) = 1   f(x) = 0 + x?   f(1) = 1?
  • 6. + Approximating Functions   f(x) = sin(x)   f(0) = 1, f’(0) = 1, f’’(0) = 0, f’’’(0) = -1,…   What is f(1)? i.e . What is sin(1)?
  • 7. + Famous Dead People   James Gregory (1671)   Brook Taylor (1712)   Colin Maclaurin (1698-1746)   Joseph-Louis Lagrange (1736-1813)   Augustin-Louis Cauchy (1789-1857)
  • 8. + Approximations  Linear Approximation f (x) = f (a) + f ʹ′(a)(x − a) + R1 (x)(x − a) R1 (x)(x − a) = f (x) − f (a) − f ʹ′(x − a)  Quadratic Approximation f ʹ′ (a)€ f (x) = f (a) + f ʹ′(a)(x − a) + (x − a) 2 + R2 (x)(x − a) 2 2 f ʹ′ (a) 2 R2 (x)(x − a) = f (x) − f (a) − f ʹ′(a)(x − a) − (x − a) 2 2€
  • 9. + Taylor’s Theorem   Letk≥1 be an integer and f : R →R be k times differentiable at a ∈ R .   Then there exists a function R : R →R such that k f ʹ′ (a) f k (a)f (x) = f (a) − f ʹ′(a)(x − a) + (x − a) 2 + ...+ (x − a) k + Rk (x)(x − a) k 2! k! € €   Note: Taylor Polynomial of degree k is: € f ʹ′ (a) f k (a) Pk (x) = f (a) − f ʹ′(a)(x − a) + (x − a) 2 + ...+ (x − a) k 2! k!
  • 10. + Works for Linear Approximations f (x) = c 0 + c1 (x) f (a) = c 0 + c1 (a) f ʹ′(a) = c1 f (x)(a)fʹ′= c + c (x − a) f = f (a) = c (a) + 0 11 1€ f (x) = f (a) + c1€ € € a) (x − € f (x) = c 0 + c1 (a) + c1 (x − a) = c 0 + c1 (x) €€ €
  • 11. + Works for Quadratic Approximations f (x) = c 0 + c1 (x) + c 2 (x 2 ) f (a) = c 0 + c1 (a) + c 2 (a 2 ) f ʹ′(a) = c1 + 2c 2 (a) f ʹ′ (a) = 2c 2 € = c + c (a) + c (a 2 ) + [c + 2c (a)](x − a) + 2c 2 [x − a]2 = € f (x) 0 1 2 1 2 € 2 c 0€ c1(a) + c 2 (a 2 ) + c1(x − a) + 2c 2 (x − a) + c 2 (x) 2 − c 2 (2ax) + c 2 (a) 2 = + f (x) = c 0 + c1 (x) + c 2 (x 2 )€
  • 12. + f(x) = sin(x) Degree 1
  • 13. + f(x) = sin(x) Degree 3
  • 14. + f(x) = sin(x) Degree 5
  • 15. + f(x) = sin(x) Degree 7
  • 16. + f(x) = sin(x) Degree 11
  • 17. + Implications Any smooth functions with all the same derivatives at a point MUST be the same function!
  • 18. + Proof: If f and g are smooth functions that agree over some interval, they MUST be the same function   Let f and g be two smooth functions that agree for some open interval (a,b), but not over all of R   Define h as the difference, f – g, and note that h is smooth, being the difference of two smooth functions. Also h=0 on (a,b), but h≠0 at other points in R   Without loss of generality, we will form S, the set of all x>a, such that f(x)≠0   Note that a is a lower bound for this set, S, and being a subset of R, S is complete so S has a real greatest lower bound, call it c.   c, being a greatest lower bound of S, is also an element of S, since S is closed   Now we see that h=0 on (a,c), but h≠0 at c. So, h is discontinuous at c, but then h cannot be smooth   Thus we have reached a contradiction, and so f and g must agree everywhere!
  • 19. + Suppose f(x) can be rewritten as a power series…   f (x) = c 0 + c1 (x − a) + c 2 (x − a) 2 + ...+ c n (x − a) n   c 0 = f (a)   f ʹ′(x) = c1 + 2c 2 (x − a) + 3c 3 (x f − a) 2 + ...+ nc n (x − a) n −1 k (a) ck =€ k!   c1 = f ʹ′(a)€   f ʹ′ (x) = 2c 2 + 3∗2c 3 (x − € + 4 ∗ 3c 4 (x − a) 2 + ...+ n ∗(n −1)c n (x − a) n −2 a)€ f ʹ′ (a)   c2 =€ 2!€ f k (a)   ck = k!€
  • 20. + Entirety (Analytic Functions) A function f(x) is said to be entire if it is equal to its Taylor Series everywhere   Entire   Not Entire   sin(x)   log(1+x)
  • 21. + Proof: sin(x) is entire   Maclaurin Series   sin(0)=1   sin’(0)=0 ∞   sin’(0)=-1 (−1) n 2n +1 sin(x) = ∑ x   sin’(0)=0 n =0 (2n +1)!   sin’(0)=1   sin’(0)=0   sin’(0)=-1   … etc. €
  • 22. + Proof: sin(x) is entire ∞ (−1) n 2n +1 sin(x) = ∑ x n =0 (2n +1)!   Lagrange formula for the remainder:   Let f : R →R be k+1 times differentiable on (a,x) and continuous on [a,x]. Then f k +1 (z) k +1 Rk (x) = (x − a) (k +1)!€ for some z in (x,a)€
  • 23. + Proof: sin(x) is entire   First, sin(x) is continuous and infinitely differentiable over all of R   If we look at the Taylor Polynomial of degree k f k +1 (z) k +1 Rk (x) = (x − a) (k +1)!   Note though f k +1 (z) ≤ 1 for all z in R k +1 (x − a) Rk (x) ≤€ (k +1)! €€
  • 24. + Proof: sin(x) is entire   However, as k goes to infinity, we see Rk (x) ≤ 0   Applyingthe Squeeze Theorem to our original € equation, we obtain that as k goes to infinity f (x) = Tk (x) and thus sin(x) is complete€
  • 25. + Maclaurin Series Examples ∞ ∞ xn xn   log(1 − x) = −∑ log(1+ x) = ∑ (−1) n +1 n =1 n! n =1 n! ∞ ∞ 1 (−1) n (2n)!   = ∑ xn 1+ x = ∑ xn 1 − x n =0 n =0 (1 − 2n)(n!) 2 (4) n ∞€ xn€   ex = ∑ n =0 n! ∞€ ∞ € (−1) n (−1) n 2n   sin(x) = ∑ x 2n +1 cos(x) = ∑ x n =0 (2n +1)! n =0 (2n)!€ ix   Note: e = cos(x) + isin(x)€ €
  • 26. + Applications   Physics   Special Relativity Equation   Fermat’s Principle (Optics)   Resistivity of Wires   Electric Dipoles   Periods of Pendulums   Surveying (Curvature of the Eart)
  • 27. + Special Relativity m0   m=   KE = mc 2 − m0c 2 1− v2 c2 m0c 2 ⎡⎛ v 2 ⎞ −1/ 2 ⎤   KE = − m0c 2 = m0c 2 ⎢⎜1 − 2 ⎟ −1⎥ 2 1−v c 2 € ⎢⎝ c ⎠ ⎣ ⎥ ⎦€   If v ≤ 100 m/s€   Then according to Taylor’s Inequality 1 3m0c 2 100 4 −10 R1 (x) ≤ 2 2 4 < (4.17 × 10 )m0 2 4(1 −100 /c ) c

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