Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.1: Polynomial Functions Of Degree Greater Than
Problem 49E
Related questions
Question
ps1: Please help me provide a solution for this
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce6b7460-60e6-4169-ad21-ef1457a674bc%2Fb7b09998-805e-4c13-a608-3e0289984dd3%2Fm8x37lo_processed.png&w=3840&q=75)
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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