Taylor series

+

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

€

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

€

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)

€

sin(x) =(−1)n

(2n +1)!x 2n+1

n=0

∞

∑

€

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

€

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⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

€

R1(x) ≤12

3m0c2

4(1−1002 /c 2)1004

c 4< (4.17 ×10−10)m0