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a²x, a?x? Po(x) = 1, p1(x) = 1 + ax, p2(x) = 1 + ax + 2! P3 (x) = 1 + ax + 2! %3D 3! a?? ax? p4 (x) = 1+ ax + 2! atx + 4! Pn (x) = as 3! k! k=0

a²x, a?x? Po(x) = 1, p1(x) = 1 + ax, p2(x) = 1 + ax + 2! P3 (x) = 1 + ax + 2! %3D 3! a?? ax? p4 (x) = 1+ ax + 2! atx + 4! Pn (x) = as 3! k! k=0

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Your answer is correct. Find the Maclaurin polynomials of orders n = 0,1,2,3, and 4, and then find the nth Maclaurin polynomials, p„(x) for the function in sigma notation for f(x) = eax Choose the correct answer. po(x) = 1, p1(x) = 1 + ax, p2(x) = 1 + ax + 2! ax? a?x? + 2! P3 (x) = 1 + ax + 3! p4 (x) = 1+ ax + + 2! 3! 4! Pn(x) = k! k=0 ax? ax? po(x) = 1, p1(x) = 1 + ax, p2(x) = 1 + ax + p3 (x) = 1+ ax + 2 3 a a?x? at xA P4 (x) = 1+ ax + 2 + + 3 Pa(x) = x" 4 k. k-0 a²x? a?x? po(x) = 1, p1(x) = 1 – ax, p2(x) =1- ax + P3 (x) = 1 – ax + 2 2 3 a²x? a?x? Pa (x) = E(-1* k p4 (x) = 1 – ax + - - 3 4 k0 ax? Po(x) = 1, p1(x) = 1 – ax, p2(x) = 1 – ax + p3 (x) = 1 – ax + 2! 2! 3! P4(x) = 1 – ax + 2! Pa(x) = E(-1)*a*x* k! 3! 4! k=0 Po (x) = 1, p1(x) = 1 + ax, p2(x) = 1+ ax + a²x², p3 (x) = 1 + ax + a²x² + a°x³. P4(x) = 1+ ax + a²x² + a°x³ + a*x*, pa(x) = },a*x* k=0
Transcribed Image Text:Your answer is correct. Find the Maclaurin polynomials of orders n = 0,1,2,3, and 4, and then find the nth Maclaurin polynomials, p„(x) for the function in sigma notation for f(x) = eax Choose the correct answer. po(x) = 1, p1(x) = 1 + ax, p2(x) = 1 + ax + 2! ax? a?x? + 2! P3 (x) = 1 + ax + 3! p4 (x) = 1+ ax + + 2! 3! 4! Pn(x) = k! k=0 ax? ax? po(x) = 1, p1(x) = 1 + ax, p2(x) = 1 + ax + p3 (x) = 1+ ax + 2 3 a a?x? at xA P4 (x) = 1+ ax + 2 + + 3 Pa(x) = x" 4 k. k-0 a²x? a?x? po(x) = 1, p1(x) = 1 – ax, p2(x) =1- ax + P3 (x) = 1 – ax + 2 2 3 a²x? a?x? Pa (x) = E(-1* k p4 (x) = 1 – ax + - - 3 4 k0 ax? Po(x) = 1, p1(x) = 1 – ax, p2(x) = 1 – ax + p3 (x) = 1 – ax + 2! 2! 3! P4(x) = 1 – ax + 2! Pa(x) = E(-1)*a*x* k! 3! 4! k=0 Po (x) = 1, p1(x) = 1 + ax, p2(x) = 1+ ax + a²x², p3 (x) = 1 + ax + a²x² + a°x³. P4(x) = 1+ ax + a²x² + a°x³ + a*x*, pa(x) = },a*x* k=0
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