Step 1 Express the function in the Taylor format. The Taylor format is, ƒ(x+1) = f(x) + f(x) • h+ ƒ^(x) + E + ƒ˜¨(x) = ² + ƒ˜¨¯( x) • H fe=0; % Enter the original equation here to evaluate f(e). A good program is like a puzzle is assembled. So its a good idea to assemble the derivatives. Let us call these terms term1, term2, term3 etc... => term0/1^? f(0) =? f(0) =? => term1/h^? f(0) =? => term2/h^? f(0) =? => term3/h^? f(0) =? => term4/h^? f(x) =??; f(x) = ??; f"(x) =??; f(x) =??; f(x) = ??; % write the expressions for terms in terms of xe. That leaves you free to % assign any xe at step e. terme=e* (h^e.)/factorial(e) % Assign f(x) term1-0* (h^1)/factorial (1) % Assign (1/1!)*f'(x) term2=0*(h^2)/factorial(2) *h^1|x=0 h^2|x=0 % Assign (1/2!) *f*'(x) term3=0*(h^3)/factorial (3) % Assign (1/3!)*f***(x) *h^3|x=0 term4=8* (h^4)/factorial (4) % Assign (1/4!) *f**** (x)*h^4|x=0

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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MATLAB PLEASE

EXAMPLE 4.1 Taylor Series
Approximation
of a Polynomial
Problem Statement. Use zero-through fourth-order Taylor
series expansions to approximate the function
f(x) = -0.1x4 -0.15x³ -0.5x²-0.25x + 1.2
from x; = 0 with h = 1. That is, predict the function's value at
Xi+1 = 1.
Transcribed Image Text:EXAMPLE 4.1 Taylor Series Approximation of a Polynomial Problem Statement. Use zero-through fourth-order Taylor series expansions to approximate the function f(x) = -0.1x4 -0.15x³ -0.5x²-0.25x + 1.2 from x; = 0 with h = 1. That is, predict the function's value at Xi+1 = 1.
Step 1
Express the function in the Taylor format. The Taylor format is,
h²
f(xn+1) = f(xi) + f'(x₂) * h + ƒ¨(x;) * 27 + ƒ¨¨(x;) *
=>term1/h^?
=> term2/h^?
=> term3/h^?
hª
fe=0;
% Enter the original equation here to evaluate f (0).
A good program is like a puzzle is assembled. So its a good idea to assemble the derivatives. Let us call these terms term 1, term2, term3 etc...
f(x)=??;
=> term0/1^?
f(0) =?
f'(0) =?
f'(x) = ??;
f"(x) =??;
f"(0) =?
f""(x) =??;
f(0) = ?
f(x) = ??;
f(0) = ?
=>term4/h^?
+ f(x₂) * !
% write the expressions for terms in terms of xe. That leaves you free to
% assign any xe at step 0.
terme=0* (h^0.)/factorial (0) % Assign f(x)
term1=0* (h^1)/factorial (1) % Assign (1/1!) *f'(x)
term2=0* (h^2)/factorial(2) % Assign (1/2!) ³f''(x)
term3=0* (h^3)/factorial (3) % Assign (1/3!) *f¹''(x)
term4=0* (h^4)/factorial (4) % Assign (1/4!)³f¹'*'(x)*h^4|x=0
*h^1|x=0
h^2|x=0
h^3|x=0
Transcribed Image Text:Step 1 Express the function in the Taylor format. The Taylor format is, h² f(xn+1) = f(xi) + f'(x₂) * h + ƒ¨(x;) * 27 + ƒ¨¨(x;) * =>term1/h^? => term2/h^? => term3/h^? hª fe=0; % Enter the original equation here to evaluate f (0). A good program is like a puzzle is assembled. So its a good idea to assemble the derivatives. Let us call these terms term 1, term2, term3 etc... f(x)=??; => term0/1^? f(0) =? f'(0) =? f'(x) = ??; f"(x) =??; f"(0) =? f""(x) =??; f(0) = ? f(x) = ??; f(0) = ? =>term4/h^? + f(x₂) * ! % write the expressions for terms in terms of xe. That leaves you free to % assign any xe at step 0. terme=0* (h^0.)/factorial (0) % Assign f(x) term1=0* (h^1)/factorial (1) % Assign (1/1!) *f'(x) term2=0* (h^2)/factorial(2) % Assign (1/2!) ³f''(x) term3=0* (h^3)/factorial (3) % Assign (1/3!) *f¹''(x) term4=0* (h^4)/factorial (4) % Assign (1/4!)³f¹'*'(x)*h^4|x=0 *h^1|x=0 h^2|x=0 h^3|x=0
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