Polynomial bound for complex exponential function

Question

Can you help me prove that $\vert \exp(it) - 1 -it \vert \le t^2$, where $t > 0$? Thank you in advance.

Answer 1

One can do better, $|\exp(it)-1-it|\leq\frac{t^2}2$. This estimate is stronger than in the OP, and is valid for any real $t$, not just positive. The method below works for estimating the difference between $\exp(it)$ and its Taylor polynomial of any order.

Proof: Applying FTC twice, $$\exp(it) = 1 + \int_0^ti\exp(is)ds=1+\int_0^ti\left(1+\int_0^si\exp(ir)dr\right)ds\\=1+it-\int_0^tds\int_0^s\exp(ir)dr.$$ Therefore, taking absolute values and using that $|\exp(ir)|=1$ for real $r$: $$ |\exp(it)-1-it|=\left|\int_0^tds\int_0^s\exp(ir)dr\right|\leq\int_0^tds\int_0^s|\exp(ir)|dr\\ =\int_0^tds\int_0^s1dr=\int_0^tsds=\frac{t^2}2. $$ Q.E.D

Analogous argument shows that $\displaystyle{\left|\exp(it)-\sum_{k=1}^{n-1}\frac{(it)^k}{k!}\right|\leq\frac{t^n}{n!}}$ for any real $t$ and positive integer $n$. The same estimate can be derived without using integrals from the mean value form for the remainder of the Taylor approximation. And conversely, the mean value form can be derived by this method.