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Approximations Using Taylor Expansions
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over some open interval that contains [a, b]
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over some open interval that contains [a, b]
a( )[ ]
b
f(x) is infinitely differentiable in here
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over some open interval that contains [a, b] and pn(x) be the n'th Taylor-poly expanded around at a,
a( )[ ]
b
f(x) is infinitely differentiable in here
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over some open interval that contains [a, b] and pn(x) be the n'th Taylor-poly expanded around at a,
then there exists a "c" between a and b
a( )[ ]
bc
f(x) is infinitely differentiable in here
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over some open interval that contains [a, b] and pn(x) be the n'th Taylor-poly expanded around at a,
then there exists a "c" between a and b such that
f(b) = pn(b) + (b – a)n+1(n + 1)! f(n+1)(c)
a( )[ ]
bc
f(x) is infinitely differentiable in here
We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over some open interval that contains [a, b] and pn(x) be the n'th Taylor-poly expanded around at a,
then there exists a "c" between a and b such that
f(b) = pn(b) + (b – a)n+1(n + 1)! f(n+1)(c)
a( )[ ]
bc
f(x) is infinitely differentiable in here
We use the remainder formula to control the error when we use Taylor polynomials to approximate f(b).
An approximation is said to be accurate to n decimal places if the error is less than 0.5 x 10-n. Therefore,
Approximations Using Taylor Expansions
An approximation is said to be accurate to n decimal places if the error is less than 0.5 x 10-n. Therefore, 1 decimal place of accuracy error < 0.05
Approximations Using Taylor Expansions
An approximation is said to be accurate to n decimal places if the error is less than 0.5 x 10-n. Therefore, 1 decimal place of accuracy error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy error < 0.005
An approximation is said to be accurate to n decimal places if the error is less than 0.5 x 10-n. Therefore, 1 decimal place of accuracy error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy error < 0.0053 decimal places of accuracy error < 0.0005
An approximation is said to be accurate to n decimal places if the error is less than 0.5 x 10-n. Therefore, 1 decimal place of accuracy error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy error < 0.0053 decimal places of accuracy error < 0.0005
Example: The Mac-polynomial of ex is x + 2!1 + x2
+ .. ++ 3!x3
n! xn
pn(x) =
Find an n such that pn(1) approximates e with an accuracy to 4 decimal places.
An approximation is said to be accurate to n decimal places if the error is less than 0.5 x 10-n. Therefore, 1 decimal place of accuracy error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy error < 0.0053 decimal places of accuracy error < 0.0005
Example: The Mac-polynomial of ex is x + 2!1 + x2
+ .. ++ 3!x3
n! xn
pn(x) =
Find an n such that pn(1) approximates e1 with an accuracy to 4 decimal places.
f(n+1)(c)(n+1)! bn+1 isThe error term Rn(b) = ec
(n+1)! 1n+1
for some c between 0 and b = 1.
Approximations Using Taylor Expansions
Assuming we know that e < 3, then the errorec
(n+1)! Rn(1) = < 3(n+1)!
For n large enough such that
we will have pn(1) accurate to 4 decimal places.
0.00005 > 3(n+1)!
By trying different values for n (trial and error), we found 0.00005 > 3/9!, or n = 8 is large enough. So p8(1) e = 1 + 1 + 1/2! + 1/3! + .. + 1/8! 2.7183 approximates e accurately to at least 4 places.
> Rn(1)
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
Set b = 3o as π/60.
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =Expanding around 0, the Mac-poly of cos(x) is
+ 4!x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,cos(n+1)(c)π
60
,and at b = π/60,
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
cos(n+1)(c)π60
,and at b = π/60,
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1,
cos(n+1)(c)π60
,and at b = π/60,
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π60
,and at b = π/60,
π60 (n+1)!
(π/60)n+1
.
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π60
,and at b = π/60,
π60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)! (π/60)n+1
.
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π60
,and at b = π/60,
π60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)! (π/60)n+1
.
By trial and error n = 2 is sufficient.
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π60
,and at b = π/60,
π60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)! (π/60)n+1
.
By trial and error n = 2 is sufficient. So p2( ) π/60= 1 – (π/60)2/2
Approximations Using Taylor Expansions Example: Approximate cos(3o) to an accuracy of 3 decimal places.
1 – pn(x) =
with the error term Rn( ) = (n+1)!
Expanding around 0, the Mac-poly of cos(x) is+ 4!
x4
6!x6
8!x8
+ – 2!x2
..
Set b = 3o as π/60.
π60( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π60
,and at b = π/60,
π60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)! (π/60)n+1
.
By trial and error n = 2 is sufficient. So p2( ) π/60= 1 – (π/60)2/2 0.998629 gives us the desired result.