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# Taylor series

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### Transcript

• 1. + Taylor Series John Weiss
• 2. + Approximating Functions &#xF06E;&#x202F; f(0)= 4 &#xF06E;&#x202F; What is f(1)? &#xF06E;&#x202F; f(x) = 4? &#xF06E;&#x202F; f(1) = 4?
• 3. + Approximating Functions &#xF06E;&#x202F; f(0)= 4, f&#x2019;(0)= -1 &#xF06E;&#x202F; What is f(1)? &#xF06E;&#x202F; f(x) = 4 - x? &#xF06E;&#x202F; f(1) = 3?
• 4. + Approximating Functions &#xF06E;&#x202F; f(0)= 4, f&#x2019;(0)= -1, f&#x2019;&#x2019;(0)= 2 &#xF06E;&#x202F; What is f(1)? &#xF06E;&#x202F; f(x) = 4 &#x2013; x + x2? &#xF06E;&#x202F; (same concavity) &#xF06E;&#x202F; f(1) = 4?
• 5. + Approximating Functions &#xF06E;&#x202F; f(x) = sin(x) &#xF06E;&#x202F; What is f(1)? &#xF06E;&#x202F; f(0) = 1, f&#x2019;(0) = 1 &#xF06E;&#x202F; f(x) = 0 + x? &#xF06E;&#x202F; f(1) = 1?
• 6. + Approximating Functions &#xF06E;&#x202F; f(x) = sin(x) &#xF06E;&#x202F; f(0) = 1, f&#x2019;(0) = 1, f&#x2019;&#x2019;(0) = 0, f&#x2019;&#x2019;&#x2019;(0) = -1,&#x2026; &#xF06E;&#x202F; What is f(1)? i.e . What is sin(1)?
• 7. + Famous Dead People &#xF06E;&#x202F; James Gregory (1671) &#xF06E;&#x202F; Brook Taylor (1712) &#xF06E;&#x202F; Colin Maclaurin (1698-1746) &#xF06E;&#x202F; Joseph-Louis Lagrange (1736-1813) &#xF06E;&#x202F; Augustin-Louis Cauchy (1789-1857)
• 8. + Approximations &#xF06E;&#x202F;Linear Approximation f (x) = f (a) + f &#x2B9;&#x2032;(a)(x &#x2212; a) + R1 (x)(x &#x2212; a) R1 (x)(x &#x2212; a) = f (x) &#x2212; f (a) &#x2212; f &#x2B9;&#x2032;(x &#x2212; a) &#xF06E;&#x202F;Quadratic Approximation f &#x2B9;&#x2032; (a)&#x20AC; f (x) = f (a) + f &#x2B9;&#x2032;(a)(x &#x2212; a) + (x &#x2212; a) 2 + R2 (x)(x &#x2212; a) 2 2 f &#x2B9;&#x2032; (a) 2 R2 (x)(x &#x2212; a) = f (x) &#x2212; f (a) &#x2212; f &#x2B9;&#x2032;(a)(x &#x2212; a) &#x2212; (x &#x2212; a) 2 2&#x20AC;
• 9. + Taylor&#x2019;s Theorem &#xF06E;&#x202F; Letk&#x2265;1 be an integer and f : R &#x2192;R be k times differentiable at a &#x2208; R . &#xF06E;&#x202F; Then there exists a function R : R &#x2192;R such that k f &#x2B9;&#x2032; (a) f k (a)f (x) = f (a) &#x2212; f &#x2B9;&#x2032;(a)(x &#x2212; a) + (x &#x2212; a) 2 + ...+ (x &#x2212; a) k + Rk (x)(x &#x2212; a) k 2! k! &#x20AC; &#x20AC; &#xF06E;&#x202F; Note: Taylor Polynomial of degree k is: &#x20AC; f &#x2B9;&#x2032; (a) f k (a) Pk (x) = f (a) &#x2212; f &#x2B9;&#x2032;(a)(x &#x2212; a) + (x &#x2212; a) 2 + ...+ (x &#x2212; a) k 2! k!
• 10. + Works for Linear Approximations f (x) = c 0 + c1 (x) f (a) = c 0 + c1 (a) f &#x2B9;&#x2032;(a) = c1 f (x)(a)f&#x2B9;&#x2032;= c + c (x &#x2212; a) f = f (a) = c (a) + 0 11 1&#x20AC; f (x) = f (a) + c1&#x20AC; &#x20AC; &#x20AC; a) (x &#x2212; &#x20AC; f (x) = c 0 + c1 (a) + c1 (x &#x2212; a) = c 0 + c1 (x) &#x20AC;&#x20AC; &#x20AC;
• 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 &#x2B9;&#x2032;(a) = c1 + 2c 2 (a) f &#x2B9;&#x2032; (a) = 2c 2 &#x20AC; = c + c (a) + c (a 2 ) + [c + 2c (a)](x &#x2212; a) + 2c 2 [x &#x2212; a]2 = &#x20AC; f (x) 0 1 2 1 2 &#x20AC; 2 c 0&#x20AC; c1(a) + c 2 (a 2 ) + c1(x &#x2212; a) + 2c 2 (x &#x2212; a) + c 2 (x) 2 &#x2212; c 2 (2ax) + c 2 (a) 2 = + f (x) = c 0 + c1 (x) + c 2 (x 2 )&#x20AC;
• 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 &#x202F; Let f and g be two smooth functions that agree for some open interval (a,b), but not over all of R &#x202F; Define h as the difference, f &#x2013; g, and note that h is smooth, being the difference of two smooth functions. Also h=0 on (a,b), but h&#x2260;0 at other points in R &#x202F; Without loss of generality, we will form S, the set of all x&gt;a, such that f(x)&#x2260;0 &#x202F; 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. &#x202F; c, being a greatest lower bound of S, is also an element of S, since S is closed &#x202F; Now we see that h=0 on (a,c), but h&#x2260;0 at c. So, h is discontinuous at c, but then h cannot be smooth &#x202F; 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&#x2026; &#xF06E;&#x202F; f (x) = c 0 + c1 (x &#x2212; a) + c 2 (x &#x2212; a) 2 + ...+ c n (x &#x2212; a) n &#xF06E;&#x202F; c 0 = f (a) &#xF06E;&#x202F; f &#x2B9;&#x2032;(x) = c1 + 2c 2 (x &#x2212; a) + 3c 3 (x f &#x2212; a) 2 + ...+ nc n (x &#x2212; a) n &#x2212;1 k (a) ck =&#x20AC; k! &#xF06E;&#x202F; c1 = f &#x2B9;&#x2032;(a)&#x20AC; &#xF06E;&#x202F; f &#x2B9;&#x2032; (x) = 2c 2 + 3&#x2217;2c 3 (x &#x2212; &#x20AC; + 4 &#x2217; 3c 4 (x &#x2212; a) 2 + ...+ n &#x2217;(n &#x2212;1)c n (x &#x2212; a) n &#x2212;2 a)&#x20AC; f &#x2B9;&#x2032; (a) &#xF06E;&#x202F; c2 =&#x20AC; 2!&#x20AC; f k (a) &#xF06E;&#x202F; ck = k!&#x20AC;
• 20. + Entirety (Analytic Functions) A function f(x) is said to be entire if it is equal to its Taylor Series everywhere &#xF06E;&#x202F; Entire &#xF06E;&#x202F; Not Entire &#xF06E;&#x202F; sin(x) &#xF06E;&#x202F; log(1+x)
• 21. + Proof: sin(x) is entire &#xF06E;&#x202F; Maclaurin Series &#xF06E;&#x202F; sin(0)=1 &#xF06E;&#x202F; sin&#x2019;(0)=0 &#x221E; &#xF06E;&#x202F; sin&#x2019;(0)=-1 (&#x2212;1) n 2n +1 sin(x) = &#x2211; x &#xF06E;&#x202F; sin&#x2019;(0)=0 n =0 (2n +1)! &#xF06E;&#x202F; sin&#x2019;(0)=1 &#xF06E;&#x202F; sin&#x2019;(0)=0 &#xF06E;&#x202F; sin&#x2019;(0)=-1 &#xF06E;&#x202F; &#x2026; etc. &#x20AC;
• 22. + Proof: sin(x) is entire &#x221E; (&#x2212;1) n 2n +1 sin(x) = &#x2211; x n =0 (2n +1)! &#xF06E;&#x202F; Lagrange formula for the remainder: &#xF06E;&#x202F; Let f : R &#x2192;R be k+1 times differentiable on (a,x) and continuous on [a,x]. Then f k +1 (z) k +1 Rk (x) = (x &#x2212; a) (k +1)!&#x20AC; for some z in (x,a)&#x20AC;
• 23. + Proof: sin(x) is entire &#xF06E;&#x202F; First, sin(x) is continuous and infinitely differentiable over all of R &#xF06E;&#x202F; If we look at the Taylor Polynomial of degree k f k +1 (z) k +1 Rk (x) = (x &#x2212; a) (k +1)! &#xF06E;&#x202F; Note though f k +1 (z) &#x2264; 1 for all z in R k +1 (x &#x2212; a) Rk (x) &#x2264;&#x20AC; (k +1)! &#x20AC;&#x20AC;
• 24. + Proof: sin(x) is entire &#xF06E;&#x202F; However, as k goes to infinity, we see Rk (x) &#x2264; 0 &#xF06E;&#x202F; Applyingthe Squeeze Theorem to our original &#x20AC; equation, we obtain that as k goes to infinity f (x) = Tk (x) and thus sin(x) is complete&#x20AC;
• 25. + Maclaurin Series Examples &#x221E; &#x221E; xn xn &#xF06E;&#x202F; log(1 &#x2212; x) = &#x2212;&#x2211; log(1+ x) = &#x2211; (&#x2212;1) n +1 n =1 n! n =1 n! &#x221E; &#x221E; 1 (&#x2212;1) n (2n)! &#xF06E;&#x202F; = &#x2211; xn 1+ x = &#x2211; xn 1 &#x2212; x n =0 n =0 (1 &#x2212; 2n)(n!) 2 (4) n &#x221E;&#x20AC; xn&#x20AC; &#xF06E;&#x202F; ex = &#x2211; n =0 n! &#x221E;&#x20AC; &#x221E; &#x20AC; (&#x2212;1) n (&#x2212;1) n 2n &#xF06E;&#x202F; sin(x) = &#x2211; x 2n +1 cos(x) = &#x2211; x n =0 (2n +1)! n =0 (2n)!&#x20AC; ix &#xF06E;&#x202F; Note: e = cos(x) + isin(x)&#x20AC; &#x20AC;
• 26. + Applications &#xF06E;&#x202F; Physics &#xF06E;&#x202F; Special Relativity Equation &#xF06E;&#x202F; Fermat&#x2019;s Principle (Optics) &#xF06E;&#x202F; Resistivity of Wires &#xF06E;&#x202F; Electric Dipoles &#xF06E;&#x202F; Periods of Pendulums &#xF06E;&#x202F; Surveying (Curvature of the Eart)
• 27. + Special Relativity m0 &#xF06E;&#x202F; m= &#xF06E;&#x202F; KE = mc 2 &#x2212; m0c 2 1&#x2212; v2 c2 m0c 2 &#x23A1;&#xF8EE;&#x239B;&#xF8EB; v 2 &#x239E;&#xF8F6; &#x2212;1/ 2 &#x23A4;&#xF8F9; &#xF06E;&#x202F; KE = &#x2212; m0c 2 = m0c 2 &#x23A2;&#xF8EF;&#x239C;&#xF8EC;1 &#x2212; 2 &#x239F;&#xF8F7; &#x2212;1&#x23A5;&#xF8FA; 2 1&#x2212;v c 2 &#x20AC; &#x23A2;&#xF8EF;&#x239D;&#xF8ED; c &#x23A0;&#xF8F8; &#x23A3;&#xF8F0; &#x23A5;&#xF8FA; &#x23A6;&#xF8FB;&#x20AC; &#xF06E;&#x202F; If v &#x2264; 100 m/s&#x20AC; &#xF06E;&#x202F; Then according to Taylor&#x2019;s Inequality 1 3m0c 2 100 4 &#x2212;10 R1 (x) &#x2264; 2 2 4 &lt; (4.17 &#xD7; 10 )m0 2 4(1 &#x2212;100 /c ) c