Fo, and then find xo = In9 a?(x + In9)² 2! a (x + In9) 4! In9)? a^(x – In9)4

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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Find the Taylor polynomials of orders n = 0,1,2,3, and 4 about x = x0, and then fınd the nth Taylor polynomials, p,(x) for the
function in sigma notation for
f(x) = eax; xo = In9
Choose the correct answer.
O po(x) = 9ª,
a² (x + In9)² ]
P1(x) = 9ª[1 + a(x + In9)], p2(x) = 9ª | 1 + a(x + In9) +
2!
a² (x + In9)², a (x+ In9)³]
P3 (x) = 9ª | 1+ a(x + In9) +
+
2!
3!
a (x + In9)², a°(x + In9)³
a* (x + In9)* ]
P4 (x) = 9ª |1 + a(x + ln9) +
2!
3!
4!
9° a* (x + In9)*
Σ
Pn(x) =
k !
k=0
O po (x) = 1,
a² (x – In9)²
-
P1(x) = 1 + a(x – In9), p2(x) = 1 + a(x – In9) +
2!
a² (x – In9)?, a°(x – In9)³
P3 (x) = 1 + a(x – In9) +
2!
3!
a (x – In9)², a° (x – In9)³ , a*(x – In9)ª
-
P4(x) = 1 + a(x – In9) +
2!
3!
4!
a*(x – In9)*
-
Pn(x) =
k!
k=0
Transcribed Image Text:Find the Taylor polynomials of orders n = 0,1,2,3, and 4 about x = x0, and then fınd the nth Taylor polynomials, p,(x) for the function in sigma notation for f(x) = eax; xo = In9 Choose the correct answer. O po(x) = 9ª, a² (x + In9)² ] P1(x) = 9ª[1 + a(x + In9)], p2(x) = 9ª | 1 + a(x + In9) + 2! a² (x + In9)², a (x+ In9)³] P3 (x) = 9ª | 1+ a(x + In9) + + 2! 3! a (x + In9)², a°(x + In9)³ a* (x + In9)* ] P4 (x) = 9ª |1 + a(x + ln9) + 2! 3! 4! 9° a* (x + In9)* Σ Pn(x) = k ! k=0 O po (x) = 1, a² (x – In9)² - P1(x) = 1 + a(x – In9), p2(x) = 1 + a(x – In9) + 2! a² (x – In9)?, a°(x – In9)³ P3 (x) = 1 + a(x – In9) + 2! 3! a (x – In9)², a° (x – In9)³ , a*(x – In9)ª - P4(x) = 1 + a(x – In9) + 2! 3! 4! a*(x – In9)* - Pn(x) = k! k=0
O po(x) = 9ª,
a²x
P1 (x) = 9"[1 + ax], p2(x) = 9ª |1 + ax +
a?x? a³x³
P3 (x) = 9ª | 1 + ax +
2!
3!
a²x? a³x?_ d s
P4(x) = 9ª | 1+ ax +
2!
3!
4!
9° a*x*
Σ
Pr(x) =
k!
k=0
O po(x) = 9°,
a²(x – In9)²]
P1 (x) = 9"[1 + a(x – In9)], p2(x) = 9° |1+ a(x – In9) +
2!
a?(x – In9)? , a°(x – In9)³
P3 (x) = 9ª | 1 + a(x – In9) +
2!
3!
a(x – In9)?, a²(x – In9)³ , a*(x – In9)*
P4 (x) = 94 | 1 + a(x – In9) +
2!
3!
4!
5 9° a*(x – In9)*
Pr(x) =
k!
k-0
Transcribed Image Text:O po(x) = 9ª, a²x P1 (x) = 9"[1 + ax], p2(x) = 9ª |1 + ax + a?x? a³x³ P3 (x) = 9ª | 1 + ax + 2! 3! a²x? a³x?_ d s P4(x) = 9ª | 1+ ax + 2! 3! 4! 9° a*x* Σ Pr(x) = k! k=0 O po(x) = 9°, a²(x – In9)²] P1 (x) = 9"[1 + a(x – In9)], p2(x) = 9° |1+ a(x – In9) + 2! a?(x – In9)? , a°(x – In9)³ P3 (x) = 9ª | 1 + a(x – In9) + 2! 3! a(x – In9)?, a²(x – In9)³ , a*(x – In9)* P4 (x) = 94 | 1 + a(x – In9) + 2! 3! 4! 5 9° a*(x – In9)* Pr(x) = k! k-0
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