Vtyjkyuk

I am trying to grasp the idea of Taylor Remainder Theorem. I want to know the way I understand it is right or wrong.

Like in the linear approximation

it is $$f(x) approx f(c) + f'(c)(x-c)$$
or $$f(x) = f(c) + f'(c)(x-c) + f''(zeta)(x-c)$$

which is in the picture below

The error is exactly $$f''(zeta)(x-c)$$

Then I thought of a case with a lower order of derivative which is

$$f(x) = f(c) + f'(zeta)(x-c)$$

The error is exactly $$f'(zeta)(x-c)$$

Then I noticed as there are more terms the order of derivatives for error gets higher which leads to the equation $$fracf^n+1(zeta)(n+1)!(x-a)^n+1$$

Is this the right way of understanding the theorem?

If $f$ is sufficiently often differentiable in the neighborhood of the point $cinmathbb R$ then for each $ngeq0$ its $n^rm th$ Taylor polynomial $j_c^nf$ is defined as follows:
$$j_c^nf(x):=sum_k=0^nf^(k)(c)over k!(x-c)^k .$$
$bigl($Note: Sometimes the increment variable $X:=x-c$ is used as variable for $j_c^nf$. One then writes
$$j_c^nf(X):=sum_k=0^nf^(k)(c)over k!X^k .bigr)$$
Given $x$ the value of such a polynomial can be computed exactly in finitely many steps.

Why should we introduce this polynomial? That's where "Taylor's theorem with remainder" comes in. It turns out that when $|x-c|$ is small this polynomial gives a good approximation to the true value of $f$ at $x$:
$$f(x)approx j_c^nf (x)quadbigl(|x-c|ll1) .$$
Now "good approximation" is just a colloquial description. We want error bounds! There are various ways to quantify the error
$$R_n(x):=f(x)-j_c^nf(x) .$$
One of them reads as follows: There is a point $xi$ between $c$ and $x$ such that
$$R_n(x)=f^(n+1)(xi)over (n+1)!(x-c)^n+1 .tag1$$
The formula $(1)$ only is of use if you have simple control over the values of $f^(n+1)$. This is, e.g., the case when $f=sin$. But don't think the formula
$$f(x)=j_c^nf(x)+R_n(x)$$
allows you to compute the value of some special $f(x)$ exactly in a "finitary", albeit very complicated way.