Job 16 Taylor Formula

4.

(1) Solution: According to Taylor unfold
\[
\cos x =1-\frac{x^2}{2} + O (x^3)
\qquad \mbox{and} \qquad
\LN (1+x) = x-\frac{1}{2} x^2 +o (x^2),
\]
So
\[
\lim_{x\to 0} \frac{\cos x \ln (1+x)-x}{x^2}
= \lim_{x\to 0} \frac{(1+o (x)) (X-\frac12 x^2 +o (x^2))-X}{x^2}
=\lim_{x\to 0} \frac{x-\frac12 x^2-x+o (x^2)}{x^2}=-\frac12.
\]

Note: Because the denominator is $x ^2$, we only need to expand the molecule to $x ^2$, and other high-order items can be written as $o (x^2) $. Here $\cos x$ only used the previous expansion, because $\cos x$ 's second expansion is $x ^2$ and $\ln (1+X) $ The first expansion is $x $, so at this time to multiply is $x ^3=o (x^2) $, so we do not need to consider. Similarly, because the first item of $\cos x$ is $1$, $\ln (1+x) $ needs to be expanded to $x ^2$ item.



(2) $\frac{7}{360}$, courseware example



5. Proof: According to the book on the 144-page theorem 3.3.1, because the function $f (x) $ at $a $ point exists in the second derivative, then
\[
F (x) =f (a) +f ' (a) (x-a) +\frac{f "(A)}{2} (X-a) ^2+o ((x-a) ^2).
\]
To make $x =a+h$ and $x =a-h$
\[
F (a+h) =f (a) +f ' (a) h+\frac{f ' (a)}{2}h^2+o (h^2)
\]
And
\[
F (a-h) =f (a)-F ' (a) h+\frac{f ' (a)}{2}h^2+o (h^2).
\]
So
\[
\begin{aligned}
\lim_{h \to 0} \frac{f (a+h) +f (a-h) -2f (a)}{h^2}
&= \lim_{h \to 0} \frac{f (a) +f ' (a) h+\frac{f ' (a)}{2}h^2+f (a)-F ' (a) h+\frac{f ' (a)}{2}h^2-2f (a) +o (h^2)}{h^2}
\\
&=\lim_{h\to 0} \frac{f "(a) h^2 +o (h^2)}{h^2}
\\
&=f ' (a).
\end{aligned}
\]


Another method of proof:

Job 16 Taylor Formula