Your answer is correct. 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 = In7 Choose the correct answer. O po(x) = a". a² (x – In7)² | P1(x) = a' [1 + a(x – In7)]. p2(x) = a' |1+ a(x – In7) + %3D 2! a (x – In7)² a° (x – In7)³] P3 (x) = a' |1+ a(x – In7) - 2! 3! a² (x – In7)2, a (x – In7)³, at (x – In7)ʻj p4 (x) = a" |1+ a(x – In7) + - 2! 3! 4! ak+7 (x – In7)* k! Pn (x) = k-0 po(x) = 7°, P1(1) = 7°[1 + ax], p2(x) = 7° 1+ ax + 2! a²x a?x3° P3 (x) = 7° |1+ ax + 2! 3! a?x a?x? P4 (x) = 7° |1+ ax + 2! 3! 4! 7° a*x* Pr(x) = k! k-0 Po(x) = 7°. a²(x – In7)* | P1(x) = 7°[1 + a(x – In7)], p2(x) = 7ª 1+ a(x – In7) + 2!
Your answer is correct. 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 = In7 Choose the correct answer. O po(x) = a". a² (x – In7)² | P1(x) = a' [1 + a(x – In7)]. p2(x) = a' |1+ a(x – In7) + %3D 2! a (x – In7)² a° (x – In7)³] P3 (x) = a' |1+ a(x – In7) - 2! 3! a² (x – In7)2, a (x – In7)³, at (x – In7)ʻj p4 (x) = a" |1+ a(x – In7) + - 2! 3! 4! ak+7 (x – In7)* k! Pn (x) = k-0 po(x) = 7°, P1(1) = 7°[1 + ax], p2(x) = 7° 1+ ax + 2! a²x a?x3° P3 (x) = 7° |1+ ax + 2! 3! a?x a?x? P4 (x) = 7° |1+ ax + 2! 3! 4! 7° a*x* Pr(x) = k! k-0 Po(x) = 7°. a²(x – In7)* | P1(x) = 7°[1 + a(x – In7)], p2(x) = 7ª 1+ a(x – In7) + 2!
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.3: Change Of Basis
Problem 17EQ
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ps2: Please help me provide a solution for this
![Your answer is correct.
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 = In7
Choose the correct answer.
O po(x) = a".
a² (x – In7)² |
P1(x) = a' [1 + a(x – In7)]. p2(x) = a' |1+ a(x – In7) +
%3D
2!
a (x – In7)² a° (x – In7)³]
P3 (x) = a' |1+ a(x – In7) -
2!
3!
a² (x – In7)2, a (x – In7)³, at (x – In7)ʻj
p4 (x) = a" |1+ a(x – In7) +
-
2!
3!
4!
ak+7 (x – In7)*
k!
Pn (x) =
k-0
po(x) = 7°,
P1(1) = 7°[1 + ax], p2(x) = 7° 1+ ax +
2!
a²x a?x3°
P3 (x) = 7° |1+ ax +
2!
3!
a?x a?x?
P4 (x) = 7° |1+ ax +
2!
3!
4!
7° a*x*
Pr(x) =
k!
k-0
Po(x) = 7°.
a²(x – In7)* |
P1(x) = 7°[1 + a(x – In7)], p2(x) = 7ª 1+ a(x – In7) +
2!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce6b7460-60e6-4169-ad21-ef1457a674bc%2F69a56498-549f-420b-b8ac-07708352d56f%2F12r9t8_processed.png&w=3840&q=75)
Transcribed Image Text:Your answer is correct.
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 = In7
Choose the correct answer.
O po(x) = a".
a² (x – In7)² |
P1(x) = a' [1 + a(x – In7)]. p2(x) = a' |1+ a(x – In7) +
%3D
2!
a (x – In7)² a° (x – In7)³]
P3 (x) = a' |1+ a(x – In7) -
2!
3!
a² (x – In7)2, a (x – In7)³, at (x – In7)ʻj
p4 (x) = a" |1+ a(x – In7) +
-
2!
3!
4!
ak+7 (x – In7)*
k!
Pn (x) =
k-0
po(x) = 7°,
P1(1) = 7°[1 + ax], p2(x) = 7° 1+ ax +
2!
a²x a?x3°
P3 (x) = 7° |1+ ax +
2!
3!
a?x a?x?
P4 (x) = 7° |1+ ax +
2!
3!
4!
7° a*x*
Pr(x) =
k!
k-0
Po(x) = 7°.
a²(x – In7)* |
P1(x) = 7°[1 + a(x – In7)], p2(x) = 7ª 1+ a(x – In7) +
2!
![Po(x) = 7ª.
a²(x – In7)2
P1(x) = 7°[1 + a(x – In7)], p2(x) = 7" 1+ a(x – In7) +:
2!
a²(x – In7)², a°(x – In7)³]
P3 (x) = 7° |1+ a(x – In7) +
+
2!
3!
P4 (x) = 7° 1+ a(x – In7) +-
a²(x – In7)² , a°(x – In7)³ , a*(x – In7)*
2!
+
3!
4!
7°a* (x – In7)*
Pn (x) = 2
%3D
k!
po(x) = 1,
a²(x – In7)²
P1 (x) = 1+ a(x – In7), p2(x) = 1+ a(x – In7) +
2!
a (x – In7)2, a (x – In7)³
Рз (х) %3D 1 + а(х - In7) +
2!
3!
a² (x – In7)?, a (x – In7)³
at (x – In7)4
p4(x) = 1+ a(x – In7) +
2!
3!
4!
Pa(x) = a*(x – Ih7)*
k!
k=0
Po(x) = 7°.
a² (x + In7)²"
P1(x) = 7°[1 + a(x + In7)], p2(x) = 7ª 1+ a(x + In7) +
2!
a* (x + In7)?, a (x + In7)³]
P3 (x) = 7° 1+ a(x + In7) +
2!
3!
a²(x + In7)² , a°(x + In7)³ , a*(x + In7)*j
P4 (x) = 7° 1+ a(x+ In7) +
2!
3!
4!
Pn(x) = "a*(x + In7)*
k!
k=0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce6b7460-60e6-4169-ad21-ef1457a674bc%2F69a56498-549f-420b-b8ac-07708352d56f%2Fydvhwg_processed.png&w=3840&q=75)
Transcribed Image Text:Po(x) = 7ª.
a²(x – In7)2
P1(x) = 7°[1 + a(x – In7)], p2(x) = 7" 1+ a(x – In7) +:
2!
a²(x – In7)², a°(x – In7)³]
P3 (x) = 7° |1+ a(x – In7) +
+
2!
3!
P4 (x) = 7° 1+ a(x – In7) +-
a²(x – In7)² , a°(x – In7)³ , a*(x – In7)*
2!
+
3!
4!
7°a* (x – In7)*
Pn (x) = 2
%3D
k!
po(x) = 1,
a²(x – In7)²
P1 (x) = 1+ a(x – In7), p2(x) = 1+ a(x – In7) +
2!
a (x – In7)2, a (x – In7)³
Рз (х) %3D 1 + а(х - In7) +
2!
3!
a² (x – In7)?, a (x – In7)³
at (x – In7)4
p4(x) = 1+ a(x – In7) +
2!
3!
4!
Pa(x) = a*(x – Ih7)*
k!
k=0
Po(x) = 7°.
a² (x + In7)²"
P1(x) = 7°[1 + a(x + In7)], p2(x) = 7ª 1+ a(x + In7) +
2!
a* (x + In7)?, a (x + In7)³]
P3 (x) = 7° 1+ a(x + In7) +
2!
3!
a²(x + In7)² , a°(x + In7)³ , a*(x + In7)*j
P4 (x) = 7° 1+ a(x+ In7) +
2!
3!
4!
Pn(x) = "a*(x + In7)*
k!
k=0
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