Lagrange remainder bound for a fourth-degree Taylor approximation

Expert intro: When a Taylor polynomial is used for approximation, the only thing the grader really wants is a clean remainder estimate. You do not need the exact function here; you need the derivative bound, the order of the polynomial, and the point where you are approximating.

Detailed walkthrough

For a fourth-degree Taylor polynomial, the Lagrange remainder satisfies

$|R_4(0.1)|=|T_4(0.1)-f(0.1)|\le \frac{M}{5!}|0.1|^5$

where

$M=\max_{0\le x\le 0.1}|f^{(5)}(x)|.$

Given that $|f^{(5)}(x)|\le 15$ on the interval, we may take $M=15$.

So

$|T_4(0.1)-f(0.1)|\le \frac{15}{120}(0.1)^5.$

Now compute:

$(0.1)^5=10^{-5}$

and

$\frac{15}{120}=\frac18.$

Therefore

$|T_4(0.1)-f(0.1)|\le \frac18\cdot 10^{-5}<10^{-5}.$

So the approximation is within

$\boxed{\frac{1}{10^5}}$

of the exact value of $f(0.1)$.

The setup is exactly what the Lagrange error bound asks for: an interval, a derivative bound, and the factorial term from the next derivative order.

💡 Pitfall guide

Do not stop after writing the remainder formula; the point is to plug in the maximum fifth derivative and simplify the power of $0.1$. Also, make sure the polynomial degree and derivative order match: a fourth-degree Taylor polynomial uses the fifth derivative in the error term, not the fourth.

🔄 Real-world variant

If you were approximating at $x=0.2$ instead of $0.1$, the same formula would become

$|R_4(0.2)|\le \frac{M}{5!}(0.2)^5,$

so the error would grow by a factor of $2^5=32$. If the polynomial were degree 3 instead, you would need a bound on $|f^{(4)}(x)|$ and a fourth-power error term.

🔍 Related terms

Taylor polynomial, Lagrange remainder, error bound