Upper bound for Taylor polynomial for $f(x) = e^x \cdot \cos(x)$ on $[0,1]$

Question

Let $f:[0,1] \to \mathbb{R}, f(x) = e^x \cdot \cos(x)$. I want to find a sharp upper bound for the absolute value of the remainder term of the second order Taylor polynomial of $f$ around $0$, on the interval $[0,1]$.

It's pretty straightforward to see that the second order Taylor polynomial of $f$ is $P_2(x) = 1 + x, \forall x \in [0,1].$

Now, by the definition of the remainder (Lagrange remainder), for every $x \in (0,1)$ there is some $c_x \in (0,x)$ such that $$f(x) - P_2(x) = R_2(x) = \frac{f'''(c_x)}{3!} \cdot x^3 = -\frac{1}{3}e^{c_x}(\sin(c_x) + \cos(c_x)) \cdot x^3,$$ so $$|R_2(x)| = \frac{1}{3}e^{c_x} (\sin(c_x) + \cos(c_x)) \cdot x^3, $$ since $\sin(t), \cos(t) \geq 0, \forall t \in [0,1]$. Hence, we can deduce an upper bound for $R_2$: $$|R_2(x)| \leq \frac{1}{3}e(\sin(\pi/4) + \cos(\pi/4)) = \frac{\sqrt{2}}{3} e, \forall x \in [0,1], $$ since $x^3 \leq 1, \forall x \in [0,1]$, $e^{c_x} \leq e, \forall x \in [0,1]$ and $\sin(c_x) + \cos(c_x) \leq \sin(\pi/4) + \cos(\pi/4), \forall x \in [0,1]$. Can we make this sharper?

Answer 1

If you look at the $e^c(\sin c+\cos c)$ part, you can see that it's strictly increasing in the $[0,1]$ interval. Take the derivative with respect to $c$, you can see that it's always positive. That means that the maximum value is at $c=1$. And $\sin 1+\cos 1\approx 1.38<\sqrt 2$

Answer 2

You can always take the next higher degree Taylor polynomial and its remainder term to get a better remainder term in the original degree. With $$ f(x)=Re(e^{(1+i)x})\implies f^{(n)}(x)=Re((1+i)^ne^{(1+i)x}) $$ we get $$ f'(x)=(\cos(x)-\sin(x))e^x\\ f''(x)=-2\sin(x)e^x\\ f'''(x)=-2(\cos(x)+\sin(x))e^x\\ f^{(4)}(x)=-4\cos(x)e^x $$ so that $$ e^x\cos x=1+x-\frac{x^3}3-\frac{e^{c_x}\cos(c_x)x^4}6 $$ $e^c\cos(c)$ has its maximum at $c=\frac\pi4$ with value $1.5508...<1.6$ so that $$ |e^x\cos x-1-x|\le \frac{x^3}6(2+1.6x), $$ which is closer than the previous bound.