The Taylor series of the function ○ ƒ(x) = ₁/² x 1-u (Z− x) u(1-)Z = (x) ƒ Omle n+1 | S (x) = Σ(-1)* (x − 2)*²² n! ○ ƒ (x)= [(-1) **¹ (x−2)* Ume ○ f(x)= Σ(-1)" (x-2)¹ about the point 5=2 is
The Taylor series of the function ○ ƒ(x) = ₁/² x 1-u (Z− x) u(1-)Z = (x) ƒ Omle n+1 | S (x) = Σ(-1)* (x − 2)*²² n! ○ ƒ (x)= [(-1) **¹ (x−2)* Ume ○ f(x)= Σ(-1)" (x-2)¹ about the point 5=2 is
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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![The Taylor series of the function
©
t-u (7− x) u(1-)Z = (x) ƒ
11=0
F0
00
tru (7− x) u(1-)Z = (x) ƒ
Oml
Reset Answer
f(x)=. 1- x
n+1
00
○f(x)= Σ(-1)**¹ (x-2)"
n!
00
f(x)= [(-1)" (x-2)
PEO
< Previous Question
about the point
x=2
is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fac00414d-3d48-4d6a-a66a-e93324329768%2F3ba1f9e0-ad1b-42f9-bdf7-fdb83c21a23b%2F2r1ggv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The Taylor series of the function
©
t-u (7− x) u(1-)Z = (x) ƒ
11=0
F0
00
tru (7− x) u(1-)Z = (x) ƒ
Oml
Reset Answer
f(x)=. 1- x
n+1
00
○f(x)= Σ(-1)**¹ (x-2)"
n!
00
f(x)= [(-1)" (x-2)
PEO
< Previous Question
about the point
x=2
is
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