An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

# Quiz problem: A Taylor approximation

mark

Suppose we estimate
$$e^3 \approx \sum_{k=1}^{5} \frac{3^k}{k!}.$$
Give an upper bound for the error in that approximation.

djett

Bound on error:
$$\frac{3^7}{6!} \approx 3.0375$$

mark

@djett Can you show how that arises from the remainder theorem? Regardless, you should certainly include the formula and indicate how it is applied.

Cromer

I'm a bit confused about how we can get that low of an error term. Taylor's remainder theorem is

$$\text{error} = \frac{f^{(N+1)}(\xi)}{(N+1)!} x^{N+1}$$

where $N$ is $5$ in our case, and $x=3$. So, we get

$$\text{error} = \frac{e^{\xi}}{6!} 3^6$$

for some $\xi \in (0,3)$. From here, I am thinking $e^{\xi} \in (1,e^3)$, and so the error is bounded by

$$\frac{e^{3}}{6!} 3^6.$$

From here, if $e<3$, then this error is less than $3^9/6!$, which is about 27. This is quite a bit larger than I would have expected, but I am having trouble seeing how to get your answer of about $3$. Can you see where I went wrong?

mark

@Invictus Be sure to include braces {} around your exponents! Also, in my experience, you can reload the page while editing to see how it looks before posting.