According to Taylor's Formula, what is the maximum error possible in the use of the sum $\sum _ { n = 0 } ^ { 4 } \frac { x ^ { n } } { n ! }$ to approximate $e ^ { x }$ in the interval $- 1 \leq x \leq 1$ ?
A) $\frac { e } { 240 }$
B) $\frac { e } { 48 }$
C) $\frac { e } { 480 }$
D) $\frac { e } { 24 }$
E) $\frac { e } { 20 }$
F) $\frac { e } { 120 }$
G) $\frac { e } { 12 }$
H) $\frac { e } { 60 }$

Question

According to Taylor's Formula, what is the maximum error possible in the use of the sum $\sum _ { n = 0 } ^ { 4 } \frac { x ^ { n } } { n ! }$ to approximate $e ^ { x }$ in the interval $- 1 \leq x \leq 1$ ?
A) $\frac { e } { 240 }$ 
B) $\frac { e } { 48 }$ 
C) $\frac { e } { 480 }$ 
D) $\frac { e } { 24 }$ 
E) $\frac { e } { 20 }$ 
F) $\frac { e } { 120 }$ 
G) $\frac { e } { 12 }$ 
H) $\frac { e } { 60 }$

Quizplus · Accepted Answer

The error term in Taylor's approximation is given by the next term in the series: $\frac{x^5}{5!}$. The maximum value of $e^x$ on $[-1, 1]$ is $e$, so the error is maximized at $\frac{e}{5!} = \frac{e}{120}$.

According to Taylor's Formula, What Is the Maximum Error Possible ∑n=04xnn!\sum _ { n = 0 } ^ { 4 } \frac { x ^ { n } } { n ! }∑n=04​n!xn​

According to Taylor's Formula, What Is the Maximum Error Possible $\sum _ { n = 0 } ^ { 4 } \frac { x ^ { n } } { n ! }$