1. Part 3
Truncation Errors
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2. Key Concepts Taylor's Series
Using Taylor's series to approximate functions
The Remainder of Taylor's Series
Using the Remainder to estimate truncation errors
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3. Truncation Errors
Truncation Errors are those that result from using an approximation in place of an exact mathematical procedure.
Approximation of accelaration
Approximation
Truncation Errors
MacLaurin series of ex
Exact mathematical formulation
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4. A more general form of approximation is in terms of Taylor Series.
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5. Taylor's Theorem
Taylor's Theorem: If the function f and its first n+1 derivatives are continuous on an interval containing a and x, then the value of the function at x is given by
where the remainder Rn is defined as
(the integral form)
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6. Derivative or Lagrange Form of the remainder
The remainder Rn can also be expressed as
(the Lagrange form)
for some ξ between a and x
The Lagrange form of the remainder makes analysis of truncation errors easier.
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7. Taylor Series
Any smooth function can be approximated as a polynomial. Taylor series provides a mean to predict a function value at one point x in terms of the function and its derivatives at another point a.
Note:
This is the same expression except that a and x are replaced by xi and xi+1 respectively.
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8. Taylor Series
If we let h = xi+1 - xi, we can rewrite the Taylor series and the remainder as
h is called the step size.
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9. Taylor Series Approximation
If we take away the remainder, the Taylor series becomes the Taylor series approximation.
The remainder Rn accounts for all the terms not included in the approximation starting from term n+1.
n can be as small as zero.
The larger n is (more terms used), the better the approximation and thus the smaller the error.
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10. Some Terminologies Zero-order approximation: use only the first term
First-order approximation: use only the first two terms
Second-order approximation: use only the first three terms
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11. Taylor Series Approximation Example: More terms used implies better approximation
f(x) = 0.1x x x x + 1.2
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12. Taylor Series Approximation Example: Smaller step size implies smaller error
Errors
Reduced step size
f(x) = 0.1x x x x + 1.2
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13. The Remainder of the Taylor Series Expansion
Relationship between h and Rn
Smaller h implies smaller Rn, but if we reduce/magify h by a factor of k, how much smaller/larger Rn would be?
That is,
if h' = kh, what is Rn' in terms of Rn?
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14. Example: Relationship between h and R1
Use the first order Taylor series to approximate the function f(x) = x4 at x+h where x=1. h
True value
(x+h)4
1st Order Approx
f(x) + f'(x)h = x4+4x3h
= 1+4h
Error = R1=
True value –
1st order Approx. 1 16 5 11
0.5
5.0625 3
2.0625
0.25 2
0.125
1.5
0.0625
1.25
1.125
1.0625
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15. Pictorial View of h vs. R1
A slop of 2 shown in the log-log plot of the remainder indicates that the truncation error is propotional to h2
Theoritical result
Do they agree?
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16. The Remainder of the Taylor Series Expansion
Summary
To reduce truncation errors, we can reduce h or/and increase n.
If we reduce h, the error will get smaller quicker (with less n).
This relationship has no implication on the magnitude of the errors because the constant term can be huge! It only give us an estimation on how much the truncation error would reduce when we reduce h or increase n.
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17. Example This is the Maclaurin series expansion for ex
If we want to approximate e0.01 with an error less than 10-12, at least how many terms are needed?
Note: Maclaurin series expansion is a special case of the Taylor series expansion with xi = 0 and h = x.
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18. So we need at least 5 terms
Note: is about > e
To find the smallest n such that Rn < 10-12, we can find the smallest n that satisfies
With the help of a computer:
n=0 Rn= e-02
n=1 Rn= e-05
n=2 Rn= e-07
n=3 Rn= e-10
n=4 Rn= e-13
So we need at least 5 terms
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19. Same problem with larger step size
Note:1.72 is 2.89 > e
With the help of a computer:
n=0 Rn= e-01
n=1 Rn= e-01
n=2 Rn= e-02
n=3 Rn= e-03
n=4 Rn= e-04
n=5 Rn= e-05
n=6 Rn= e-06
n=7 Rn= e-07
n=8 Rn= e-09
n=9 Rn= e-10
n=10 Rn= e-11
n=11 Rn= e-13
So we need at least 12 terms
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20. Other methods for estimating truncation errors of a series
By Geometry Series
By Integration
Alternating Convergent Series Theorem
Note: Some Taylor series expansions may exhibit certain characteristics which would allow us to use different methods to approximate the truncation errors.
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21. Estimation of Truncation Errors By Geometry Series
If |tj+1| ≤ k|tj| where 0 ≤ k < 1 for all j ≤ n, then
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22. Estimation of Truncation Errors by Geometry Series
What is |R6| for the following series expansion?
Solution:
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23. Estimation of Truncation Errors By Integration
If we can find a function f(x) s.t. |tj| ≤ f(j) j ≥ n
and f(x) is a decreasing function x ≥ n, then
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24. Estimation of Truncation Errors by Integration
Estimate |Rn| for the following series expansion.
Solution:
We can pick f(x) = x-3 because it would provide a tight bound for |tj|. That is So
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25. Alternating Convergent Series Theorem (Leibnitz Theorem)
If an infinite series satisfies the conditions
It is strictly alternating.
Each term is smaller in magnitude than that term before it.
The terms approach to 0 as a limit.
Then the series has a finite sum (i.e., convergent) and moreover if we stop adding the terms after the nth term, the error thus produced is between 0 and the 1st non-zero neglected term not taken.
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26. Alternating Convergent Series Theorem
Example:
Eerror estimated using the althernating convergent series theorem
Actual error
Another example:
Actual error
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27. More Examples
If the sine series is to be used to compute sin(1) with an error less than 0.5x10-14, how many terms are needed?
Solution:
This series satisfies the conditions of the Alternating Convergent Series Theorem.
Solving
for odd n, we get n = 17 (Need 8 terms)
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28. More Examples
How many terms should be taken in order to compute π4/90 with an error of at most 0.5x10-8?
Solution (by integration):
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