0
$\begingroup$

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:

enter image description here

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 $c$.
$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.