# Part 3 Truncation Errors Second Term 05/06.

## Presentation on theme: "Part 3 Truncation Errors Second Term 05/06."— Presentation transcript:

Part 3 Truncation Errors Second Term 05/06

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 Second Term 05/06

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 Second Term 05/06

A more general form of approximation is in terms of Taylor Series.
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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) Second Term 05/06

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. Second Term 05/06

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. Second Term 05/06

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. Second Term 05/06

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. Second Term 05/06

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 Second Term 05/06

Taylor Series Approximation Example: More terms used implies better approximation
f(x) = 0.1x x x x + 1.2 Second Term 05/06

Taylor Series Approximation Example: Smaller step size implies smaller error
Errors Reduced step size f(x) = 0.1x x x x + 1.2 Second Term 05/06

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? Second Term 05/06

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 Second Term 05/06

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? Second Term 05/06

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. Second Term 05/06

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. Second Term 05/06

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 Second Term 05/06

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 Second Term 05/06

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. Second Term 05/06

Estimation of Truncation Errors By Geometry Series
If |tj+1| ≤ k|tj| where 0 ≤ k < 1 for all j ≤ n, then Second Term 05/06

Estimation of Truncation Errors by Geometry Series
What is |R6| for the following series expansion? Solution: Second Term 05/06

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 Second Term 05/06

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 Second Term 05/06

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. Second Term 05/06

Alternating Convergent Series Theorem
Example: Eerror estimated using the althernating convergent series theorem Actual error Another example: Actual error Second Term 05/06

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) Second Term 05/06

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): Second Term 05/06