I'm struggling with Taylor Remainders in general, and once I link the particular question I'll elaborate on my thought process and understanding of the concept in hopes my confusions can be rectified:

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I'm going to attempt (ii) here.

My thought process:

  • The Taylor cubic should just be the Taylor Series expansion of the function up until degree $3$.
  • The function about $x = \frac{\pi}{2}$ is (I did this on paper to not let this be question too lengthy): $$P_3 = e^{\pi /2} + e^{\pi /2 }(x-{\pi / 2}) - \frac{e^{\pi /2}(x-\pi /2)}{3}$$
  • The remainder is such that:

$$P_3(x) + R(x) = e^x sin \ x$$

  • According to my lecture notes, $$R_n(x) = \frac{f^{n+1}(c)}{(n+1)!}(x-x_0)^{n+1}$$
  • In this case $n+1 = 3+1 = 4$, so the form should be in $$R_4(x) = \frac{f^{4}(c)}{(4)!}(x-\pi / 2)^{4}$$

  • However I have no clue about what $c$ should be.

Note, my understanding of remainder terms in Taylor Approximations is poor. From my understanding, it's the following:

  • Say you approximate a Taylor Series to a Taylor cubic.
  • Your function overlapped with the true function will start to have a difference $f(x) - p(x) \ne 0$. This non-zero value is your remainder.
  • Other than that though, my lecture notes had a bizarre way of expressing it in the form I've expressed and I don't know how to use my weak intuition to find $c$.

It is easier to set $x=\dfrac{\pi}2+u$ with $u\to 0$.

$\displaystyle f(x)=e^{(\frac{\pi}2+u)}\sin(\frac{\pi}2+u)=e^{\frac{\pi}2}e^u\cos(u)$

So we basically have to find the expansion of $g(u)=e^u\cos(u)$ in zero.

$\begin{cases}e^u=1+u+\dfrac{u^2}2+\dfrac{u^3}6+o(u^3)\\ \cos(u)=1-\dfrac{u^2}2+o(u^3)\end{cases}$

We multiply these two expansions and ignore terms smaller than $u^3$.

This gives : $\require{cancel} \left(1+u+\dfrac{u^2}2+\dfrac{u^3}6\right)-\dfrac 12\left(u^2+u^3+\cancel{\dfrac{u^4}2}+\cancel{\dfrac{u^5}6}\right)+o(u^3)=1+u-\dfrac{u^3}3+o(u^3)$

I used Taylor-Young remainder $R_3(u)=o(u^3)$

If you want to use another one you get to calculate $g^{(4)}=-4e^u\cos(u)$

For Taylor-Lagrange you don't have to exhibit a particular $c$, just say $R_3(u)=-\dfrac{e^c\cos(c)}6u^4$ with $c\in]0,u[$

For Taylor-Laplace this is $\displaystyle R_3(u)=\int_0^u -\frac 23e^t\cos(t)(u-t)^3dt$

In fact in Taylor-Lagrange, we are only interested in finding an upper bound for the error. So the knowledge of the value for $c$ is not that important.

For instance, with $u\ll 1$ then the rough inequalities occurs $\lvert e^c\lvert<2$ and $\lvert \cos(c)\lvert\le 1$

Thus $\lvert R_3(u)\lvert=\left\lvert-\dfrac{e^c\cos(c)}6u^4\right\lvert\le\dfrac{\lvert u^4\lvert}3$

Whenever we want more precise approximations for the error, we generally use the Taylor-Laplace with integral remainder formula, since it is exact.


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