Answer:
0.0032
Step-by-step explanation:
We need to compute [tex]e^{0.4}[/tex] by the help of third-degree Taylor polynomial that is expanded around at x = 0.
Given :
[tex]e^{0.4}[/tex] < e < 3
Therefore, the Taylor's Error Bound formula is given by :
[tex]$|\text{Error}| \leq \frac{M}{(N+1)!} |x-a|^{N+1}$[/tex] , where [tex]$M=|F^{N+1}(x)|$[/tex]
[tex]$\leq \frac{3}{(3+1)!} |-0.4|^4$[/tex]
[tex]$\leq \frac{3}{24} \times (0.4)^4$[/tex]
[tex]$\leq 0.0032$[/tex]
Therefore, |Error| ≤ 0.0032