taylor series
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) = 0, 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 Mathematicians
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
€
R1(x)(x − a) = f (x) − f (a) − ′ f (x − a)
€
f (x) = f (a) + ′ f (a)(x − a) + R1(x)(x − a)
€
f (x) = f (a) + ′ f (a)(x − a) +′ ′ f (a)
2(x − a)2 + R2(x)(x − a)2
€
R2(x)(x − a)2 = f (x) − f (a) − ′ f (a)(x − a) −′ ′ 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:
€
f :R→R
€
a∈R
€
Rk :R→R
€
f (x) = f (a) + ′ f (a)(x − a) +′ ′ f (a)
2!(x − a)2 + ...+
f k (a)
k!(x − a)k + Rk (x)(x − a)k
€
Pk (x) = f (a) + ′ f (a)(x − a) +′ ′ f (a)
2!(x − a)2 + ...+
f k (a)
k!(x − a)k
+Works for Linear Approximations
€
f (x) = c0 + c1(x)
€
P1(x) = c0 + c1(x − a) + c1(a)€
f (a) = c0 + c1(a)
€
′ f (a) = c1
€
P1(x) = c0 + c1(a) + c1(x − a) = c0 + c1(x)
€
f (x) = f (a) + ′ f (a)(x − a) +′ ′ f (a)
2!(x − a)2 + ...+
f k (a)
k!(x − a)k + Rk (x)(x − a)k
+Works for Quadratic Approximations
€
f (x) = c0 + c1(x) + c2(x 2)
€
f (a) = c0 + c1(a) + c2(a2)
€
′ f (a) = c1 + 2c2(a)
€
′ ′ f (a) = 2c2
€
P2(x) = c0 + c1(a) + c2(a2) + c1 + 2c2(a)[ ](x − a) +2c2
2[x − a]2 =
€
c0 + c1(a) + c2(a2) + c1(x − a) + 2c2(x − a) + c2(x)2 − c2(2ax) + c2(a)2 =
€
€
P2(x) = c0 + c1(x) + c2(x 2)
€
f (x) = f (a) + ′ f (a)(x − a) +′ ′ f (a)
2!(x − a)2 + ...+
f k (a)
k!(x − a)k + Rk (x)(x − a)k
+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
If f and g have the same value and all of the same derivatives at a point, then they 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, so 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…
€
f (x) = c0 + c1(x − a) + c2(x − a)2 + ...+ cn (x − a)n
€
c0 = f (a)
€
′ f (x) = c1 + 2c2(x − a) + 3c3(x − a)2 + ...+ ncn (x − a)n −1
€
′ ′ f (x) = 2c2 + 3∗2c3(x − a) + 4∗3c4 (x − a)2 + ...+ n∗(n −1)cn (x − a)n −2
€
c1 = ′ f (a)
€
c2 =′ ′ f (a)
2!
€
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)=0 sin’(0)=1 sin’’(0)=0 sin’’’(0)=-1 sin’’’’(0)=0 sin’’’’’(0)=1 sin’’’’’’(0)=0 … etc.
€
sin(x) =(−1)n
(2n +1)!x 2n +1
n =0
∞
∑€
Pn (x) =f n (a)
n!(x − a)n
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 (a,x)
€
sin(x) =(−1)n
(2n +1)!x 2n +1
n =0
∞
∑
€
f :R→R
€
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
€
Rk (x) =f k +1(z)
(k +1)!(x − a)k +1
€
f k +1(z) ≤1
€
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 entire since it is equal to its Taylor series
€
Rk (x) ≤ 0
€
f (x) = Pk (x)
+Maclaurin Series Examples
Note:
€
log(1− x) = −x n
n!n =1
∞
∑
€
log(1+ x) = (−1)n +1 x n
n!n =1
∞
∑
€
1
1− x= x n
n =0
∞
∑
€
1+ x =(−1)n (2n)!
(1 − 2n)(n!)2(4)n x n
n =0
∞
∑
€
ex =x n
n!n =0
∞
∑
€
sin(x) =(−1)n
(2n +1)!x 2n +1
n =0
∞
∑
€
cos(x) =(−1)n
(2n)!x 2n
n =0
∞
∑
€
e ix = 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 Earth)
+Special Relativity
Let . If v ≤ 100 m/s
Then according to Taylor’s Inequality (Lagrange)
€
m =m0
1 − v 2 c 2
€
KE = mc 2 − m0c2
€
KE =m0c
2
1− v 2 c 2− m0c
2 = m0c2 1−
v 2
c 2
⎛
⎝ ⎜
⎞
⎠ ⎟
−1/ 2
−1 ⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
€
KE1(v) =1
2m0v
2
€
(1+ x)−1/ 2 =1+
−1
2
⎛
⎝ ⎜
⎞
⎠ ⎟
1x +
−1
2
⎛
⎝ ⎜
⎞
⎠ ⎟−3
2
⎛
⎝ ⎜
⎞
⎠ ⎟
2!x 2 +
−1
2
⎛
⎝ ⎜
⎞
⎠ ⎟−3
2
⎛
⎝ ⎜
⎞
⎠ ⎟
3!x 3 + ...
€
x =−v 2
c 2
+Lagrange Remainder
Lagrange formula for the remainder: Let be k+1 times
differentiable on (a,x) and continuous on [a,x]. Then
for some z in (a,x)€
f :R→R
€
Rk (x) =f k +1(z)
(k +1)!(x − a)k +1
+Special Relativity
Let . If v ≤ 100 m/s
Then according to Taylor’s Inequality (Lagrange)
€
m =m0
1 − v 2 c 2
€
KE = mc 2 − m0c2
€
KE =m0c
2
1− v 2 c 2− m0c
2 = m0c2 1−
v 2
c 2
⎛
⎝ ⎜
⎞
⎠ ⎟
−1/ 2
−1 ⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
€
R1(x) ≤1
2
3m0c2
4(1 −1002 /c 2)5 / 2
1004
c 4 < (4.17 ×10−10)m0
€
KE1(v) =1
2m0v
2
€
(1+ x)−1/ 2 =1+
−1
2
⎛
⎝ ⎜
⎞
⎠ ⎟
1x +
−1
2
⎛
⎝ ⎜
⎞
⎠ ⎟−3
2
⎛
⎝ ⎜
⎞
⎠ ⎟
2!x 2 +
−1
2
⎛
⎝ ⎜
⎞
⎠ ⎟−3
2
⎛
⎝ ⎜
⎞
⎠ ⎟
3!x 3 + ...
€
x =−v 2
c 2
+The End