Find the Taylor polynomials of orders n = 0,1,2,3, and 4 about x = xo, and then find the nth Taylor polynomials, p„(x) for the function in sigma notation for f(x) = eax; xo = In2
Find the Taylor polynomials of orders n = 0,1,2,3, and 4 about x = xo, and then find the nth Taylor polynomials, p„(x) for the function in sigma notation for f(x) = eax; xo = In2
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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![Find the Taylor polynomials of orders
0,1,2,3, and 4 about x = x0, and then find
the nth Taylor polynomials, pn(x) for the function
in sigma notation for
n =
f(x) = eax; xo = In2
Choose the correct answer.
O po(x) = 2ª,
P1(x) = 2ª[1 + a(x + In2)],
a² (x + In2)?
P2(x) = 2ª | 1 + a(x + In2) +
2!
a²(x + In2)²
a° (x + In2)
+
P3 (x) = 2ª | 1 + a(x+ In2) +
2!
3!
a² (x + In2)? a (x + In2)³
+
a* (x + In2)*
P4(x) = 2ª | 1 + a(x + In2) +
2!
3!
4!
2“ a* (x + In2)*
Pn(x) = >
k!
k=0
O po(x) = 2“,
P1 (x) = 2ª[1 + a(x – In2)],
a²(x – In2)²
P2 (x) = 2ª | 1 + a(x – In2) +
2!
a? (x – In2)?
a³ (x – In2)³
P3(x) = 2° 1+ a(x – In2) +
2!
3!
a² (x – In2)?
a° (x – In2)³
a*(x – In2)4
P4(x) = 2° | 1+ a(x – In2) +
2!
3!
4!
2ª a* (x – In2)*
Pn(x) = >
k!
k=0
О ро(х) %3D 1,
P1(x) = 1 + a(x – In2),
a²(x – In2)²
P2(x) = 1 + a(x – In2) +
2!
a²(x – In2)?, a (x – In2)³
P3 (x) = 1+ a(x – In2) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3379b191-ca9f-4e3d-a8d1-2d89bef8e14f%2Fd1cfc517-b2a8-4655-834c-5f1b39a2b796%2Fx2vdsmn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Find the Taylor polynomials of orders
0,1,2,3, and 4 about x = x0, and then find
the nth Taylor polynomials, pn(x) for the function
in sigma notation for
n =
f(x) = eax; xo = In2
Choose the correct answer.
O po(x) = 2ª,
P1(x) = 2ª[1 + a(x + In2)],
a² (x + In2)?
P2(x) = 2ª | 1 + a(x + In2) +
2!
a²(x + In2)²
a° (x + In2)
+
P3 (x) = 2ª | 1 + a(x+ In2) +
2!
3!
a² (x + In2)? a (x + In2)³
+
a* (x + In2)*
P4(x) = 2ª | 1 + a(x + In2) +
2!
3!
4!
2“ a* (x + In2)*
Pn(x) = >
k!
k=0
O po(x) = 2“,
P1 (x) = 2ª[1 + a(x – In2)],
a²(x – In2)²
P2 (x) = 2ª | 1 + a(x – In2) +
2!
a? (x – In2)?
a³ (x – In2)³
P3(x) = 2° 1+ a(x – In2) +
2!
3!
a² (x – In2)?
a° (x – In2)³
a*(x – In2)4
P4(x) = 2° | 1+ a(x – In2) +
2!
3!
4!
2ª a* (x – In2)*
Pn(x) = >
k!
k=0
О ро(х) %3D 1,
P1(x) = 1 + a(x – In2),
a²(x – In2)²
P2(x) = 1 + a(x – In2) +
2!
a²(x – In2)?, a (x – In2)³
P3 (x) = 1+ a(x – In2) +
![P1(x) = 1 + a(x – In2),
a²(x – In2)?
P2(x) = 1 + a(x – In2) +
%3D
-
2!
a?(x – In2)?, a°(x – In2)³
P3(x) = 1 + a(x – In2) +
+
2!
3!
a²(x – In2)², a (x – In2)³ , a*(x – In2)“
P4(x) = 1 + a(x – In2) +
+
2!
3!
4!
Pn(x) = Sa*(r – In2)*
k!
n
k=0
O po(x) = a²,
P1(x) = a°[1+ a(x – In2)],
a²(x – In2)²
P2(x) = a² |1 + a(x – In2) +
2!
a²(x – In2)?, a (x – In2)³
P3(x) = a² |1 + a(x – In2) +
2!
3!
a² (x – In2)?, a°(x – In2)³ , a*(x – In2)*
p4(x) = a? |1+ a(x – In2) +
+
2!
3!
4!
" ak+2 (x – In2)*
Σ
Pn (x) :
k!
k=0
O po(x) = 2ª,
P1 (x) = 2ª[1 + ax],
a²x²
P2(x) = 2ª |1 + ax +
2!
a²x?
a³x³
Рз (х) —D 2" | 1+ ах +
+
2!
3!
a²x?
a³x³
a*x
P4(x) = 2ª |1 + ax
2!
3!
4!
2° ak xk
n
Pn(x) = }
k!
k=0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3379b191-ca9f-4e3d-a8d1-2d89bef8e14f%2Fd1cfc517-b2a8-4655-834c-5f1b39a2b796%2Fu1atd1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:P1(x) = 1 + a(x – In2),
a²(x – In2)?
P2(x) = 1 + a(x – In2) +
%3D
-
2!
a?(x – In2)?, a°(x – In2)³
P3(x) = 1 + a(x – In2) +
+
2!
3!
a²(x – In2)², a (x – In2)³ , a*(x – In2)“
P4(x) = 1 + a(x – In2) +
+
2!
3!
4!
Pn(x) = Sa*(r – In2)*
k!
n
k=0
O po(x) = a²,
P1(x) = a°[1+ a(x – In2)],
a²(x – In2)²
P2(x) = a² |1 + a(x – In2) +
2!
a²(x – In2)?, a (x – In2)³
P3(x) = a² |1 + a(x – In2) +
2!
3!
a² (x – In2)?, a°(x – In2)³ , a*(x – In2)*
p4(x) = a? |1+ a(x – In2) +
+
2!
3!
4!
" ak+2 (x – In2)*
Σ
Pn (x) :
k!
k=0
O po(x) = 2ª,
P1 (x) = 2ª[1 + ax],
a²x²
P2(x) = 2ª |1 + ax +
2!
a²x?
a³x³
Рз (х) —D 2" | 1+ ах +
+
2!
3!
a²x?
a³x³
a*x
P4(x) = 2ª |1 + ax
2!
3!
4!
2° ak xk
n
Pn(x) = }
k!
k=0
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