Transcript
+
Taylor Series John Weiss
+Approximating Functions
f(0)= 4
What is f(1)?
f(x) = 4?
f(1) = 4?
+Approximating Functions
f(0)= 4, f’(0)= -1
What is f(1)?
f(x) = 4 - x?
f(1) = 3?
+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?
+Approximating Functions
f(x) = sin(x)
What is f(1)?
f(0) = 1, f’(0) = 1
f(x) = 0 + x?
f(1) = 1?
+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)?
+Famous Dead People
James Gregory (1671)
Brook Taylor (1712)
Colin Maclaurin (1698-1746)
Joseph-Louis Lagrange (1736-1813)
Augustin-Louis Cauchy (1789-1857)
+Approximations
Linear Approximation
Quadratic Approximation
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R1(x)(x − a) = f (x) − f (a) − ʹ′ f (x − a)
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f (x) = f (a) + ʹ′ f (a)(x − a) + R1(x)(x − a)
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f (x) = f (a) + ʹ′ f (a)(x − a) +ʹ′ ʹ′ f (a)2
(x − a)2 + R2(x)(x − a)2
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R2(x)(x − a)2 = f (x) − f (a) − ʹ′ f (a)(x − a) − ʹ′ ʹ′ f (a)2
(x − a)2
+Taylor’s Theorem Let k≥1 be an integer and be k
times differentiable at . Then there exists a function such that
Note: Taylor Polynomial of degree k is:
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f :R→R
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a∈R
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Rk :R→R
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f (x) = f (a) − ʹ′ f (a)(x − a) +ʹ′ ʹ′ f (a)2!
(x − a)2 + ...+ f k (a)k!
(x − a)k + Rk (x)(x − a)k
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Pk (x) = f (a) − ʹ′ f (a)(x − a) +ʹ′ ʹ′ f (a)2!
(x − a)2 + ...+ f k (a)k!
(x − a)k
+Works for Linear Approximations
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f (x) = c0 + c1(x)
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f (x) = f (a) + c1(x − a)
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f (x) = f (a) + c1(x − a)€
f (a) = c0 + c1(a)
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f (a) = c0 + c1(a)
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ʹ′ f (a) = c1
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ʹ′ f (a) = c1
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f (x) = c0 + c1(a) + c1(x − a) = c0 + c1(x)
+Works for Quadratic Approximations
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f (x) = c0 + c1(x) + c2(x2)
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f (a) = c0 + c1(a) + c2(a2)
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ʹ′ f (a) = c1 + 2c2(a)
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ʹ′ ʹ′ f (a) = 2c2
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f (x) = c0 + c1(a) + c2(a2) + c1 + 2c2(a)[ ](x − a) +
2c22[x − a]2 =
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c0 + c1(a) + c2(a2) + c1(x − a) + 2c2(x − a) + c2(x)
2 − c2(2ax) + c2(a)2 =
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f (x) = c0 + c1(x) + c2(x2)
+f(x) = sin(x) Degree 1
+f(x) = sin(x) Degree 3
+f(x) = sin(x) Degree 5
+f(x) = sin(x) Degree 7
+f(x) = sin(x) Degree 11
+Implications
Any smooth functions with all the same derivatives at a point MUST be the same function!
+ Proof: If f and g are smooth functions that agree over some interval, they MUST be the same function
① Let f and g be two smooth functions that agree for some open interval (a,b), but not over all of R
② 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
③ Without loss of generality, we will form S, the set of all x>a, such that f(x)≠0
④ 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.
⑤ c, being a greatest lower bound of S, is also an element of S, since S is closed
⑥ 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
⑦ Thus we have reached a contradiction, and so f and g must agree everywhere!
+Suppose f(x) can be rewritten as a power series…
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f (x) = c0 + c1(x − a) + c2(x − a)2 + ...+ cn (x − a)
n
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c0 = f (a)
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ʹ′ f (x) = c1 + 2c2(x − a) + 3c3(x − a)2 + ...+ ncn (x − a)n−1
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ʹ′ ʹ′ f (x) = 2c2 + 3∗2c3(x − a) + 4 ∗3c4 (x − a)2 + ...+ n ∗(n −1)cn (x − a)n−2
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c1 = ʹ′ f (a)
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c2 =ʹ′ ʹ′ f (a)2!
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ck =f k (a)k!
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ck =f k (a)k!
+Entirety (Analytic Functions)
Entire sin(x)
Not Entire log(1+x)
A function f(x) is said to be entire if it is equal to its Taylor Series everywhere
+Proof: sin(x) is entire
Maclaurin Series sin(0)=1 sin’(0)=0 sin’(0)=-1 sin’(0)=0 sin’(0)=1 sin’(0)=0 sin’(0)=-1 … etc. €
sin(x) =(−1)n
(2n +1)!x 2n+1
n=0
∞
∑
+Proof: sin(x) is entire
Lagrange formula for the remainder: Let be k+1 times differentiable on
(a,x) and continuous on [a,x]. Then
for some z in (x,a)
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sin(x) =(−1)n
(2n +1)!x 2n+1
n=0
∞
∑
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f :R→R
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Rk (x) =f k+1(z)(k +1)!
(x − a)k+1
+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
Note though for all z in R
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Rk (x) =f k+1(z)(k +1)!
(x − a)k+1
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f k+1(z) ≤1
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Rk (x) ≤(x − a)k+1
(k +1)!
+Proof: sin(x) is entire
However, as k goes to infinity, we see
Applying the Squeeze Theorem to our original equation, we obtain that as k goes to infinity
and thus sin(x) is complete
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Rk (x) ≤ 0
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f (x) = Tk (x)
+Maclaurin Series Examples
Note:
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log(1− x) = −xn
n!n=1
∞
∑
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log(1+ x) = (−1)n+1 xn
n!n=1
∞
∑
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11− x
= xnn=0
∞
∑
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1+ x =(−1)n (2n)!
(1− 2n)(n!)2(4)nxn
n=0
∞
∑
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ex =xn
n!n=0
∞
∑
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sin(x) =(−1)n
(2n +1)!x 2n+1
n=0
∞
∑
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cos(x) =(−1)n
(2n)!x 2n
n=0
∞
∑
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eix = cos(x) + isin(x)
+Applications
Physics Special Relativity Equation Fermat’s Principle (Optics) Resistivity of Wires Electric Dipoles Periods of Pendulums Surveying (Curvature of the Eart)
+Special Relativity
If v ≤ 100 m/s
Then according to Taylor’s Inequality
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m =m0
1− v 2 c 2
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KE = mc 2 −m0c2
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KE =m0c
2
1− v 2 c 2−m0c
2 = m0c2 1− v
2
c 2⎛
⎝ ⎜
⎞
⎠ ⎟
−1/ 2
−1⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
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R1(x) ≤12
3m0c2
4(1−1002 /c 2)1004
c 4< (4.17 ×10−10)m0
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