Taylor's Formula.

Definition. Let the function $f(z)$ be defined in a neighborhood of the point $x_0$ and have derivatives up to $(n-1)$th order inclusive in this neighborhood, and let $f^{(n)}(x_0)$ exist. Then

$$f(x)=f(x_0)+\frac{f^{\prime }(x_0)}{1!}(x-x_0)+\frac{f^{″}(x_0)}{2!}(x-x_0{)}^2+...+$$ $$+\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+o((x-x_0{)}^n)$$ при $x\to x_0.$

So $$f(x)=\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k+o((x-x_0{)}^n),x\to x_0.$$

The polynomial $P_n(x)=\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$ is the Taylor polynomial. $r_n=f(x)-P_n(x)$ is the remainder term of the $n$th order Taylor formula.

For $x_0=0$, $f(x)=\sum _{k=0}^n\frac{f^{(k)}(0)}{k!}{x}^k+o(x^n)$ - the Maclaurin formula.

Taylor formulas around the point $x_0=0$ for basic elementary functions.

$$e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+o(x^n);x\to 0$$

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...+\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}+o(x^{2n+2});x\to 0$$

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...+(-1{)}^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1});x\to 0$$

$$(1+x{)}^{\alpha }=1+\alpha x+\frac{\alpha (\alpha -1)}{2!}{x}^2+...+$$ $$+\frac{\alpha (\alpha -1)...(\alpha -(n-1))}{n!}{x}^n+o(x^n);x\to 0$$

$$\frac{1}{1+x}=\sum _{k=0}^n(-1{)}^k{x}^k+o(x^n);x\to 0$$

$$\frac{1}{1-x}=\sum _{k=0}^n{x}^k+o(x^n);x\to 0$$

$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...+\frac{(-1{)}^{n-1}{x}^n}{n}+o(x^n);x\to 0$$

$$\ln (1-x)=-\sum _{k=0}^n\frac{x^k}{k}+o(x^n)x\to 0.$$

$$tgx=x+\frac{x^3}{3}+\frac{x^5}{15}+\bar{o}(x^5)x\to 0$$

$$\arcsin x=x+\sum _{k=1}^n\frac{(2k-1)!!}{2^{k}k!(2k+1)}{x}^{2k+1}+\bar{o}(x^{2n+2}),x\to 0$$

Examples.

Expand using the Maclaurin formula to $o(x^n)$ the function

$\mathbf{1.}{\mathbf{e}}^{\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{1}}$

Solution.

$$e^{5x-1}=\frac{e^{5x}}{e}=\sum _{k=0}^n\frac{(5x{)}^k}{k!e}+o(x^n).$$

$\mathbf{2.}\sin \mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{)}$

Solution.

$$\sin (2x+3)=\sin 2x\cos 3+\sin 3\cos 2x=$$

$$\cos 3\sum _{k=0}^n(-1{)}^k\frac{(2x{)}^{2k+1}}{(2k+1)!}+\sin 3\sum _{k=0}^n(-1{)}^k\frac{(2x{)}^{2k}}{(2k)!}+o(x^{2n+1})$$

$\mathbf{3.}\cos (\frac{x}{2}+2)$

Solution.

$${\cos }^{(k)}(\frac{x}{2}+2)=\frac{1}{2^k}\cos (\frac{x}{2}+2+\frac{\pi }{2}k)$$

$$\cos (\frac{x}{2}+2)=\sum _{k=0}^n\frac{1}{2^k}\frac{\cos (2+\frac{\pi }{2}k)}{k!}{x}^k+o(x^k).$$

$\mathbf{4.}\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{x}}$

Solution.

$$\frac{1}{1-2x}=\sum _{k=0}^n{2}^k{x}^k+o(x^n).$$

$\mathbf{5.}\frac{\mathbf{1}}{\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}}$

Solution.

$$\frac{1}{3x+4}=\frac{1}{4}\frac{1}{\frac{3}{4}x+1}=\frac{1}{4}\sum _{k=0}^n(-1{)}^k{\left(\frac{3}{4}\right)}^k{x}^k+o(x^n).$$

Computing limits using Taylor formula.

Let's find $\underset{x\to 0}{lim}\frac{f(x)}{g(x)},$ where $f(0)=g(0)=0.$ Assuming that the functions $f(x)$ and $g(x)$ can be expanded using the Maclaurin formula, we will limit ourselves to the first non-zero terms in the expansion of these functions:

$f(x)=a{x}^n+o(x^n),$ $a\ne 0,$ $g(x)=b{x}^m+o(x^m),$ $b\ne 0.$

If $m=n,$ then

$$\underset{x\to 0}{lim}\frac{f(x)}{g(x)}=\underset{x\to 0}{lim}\frac{f(x)}{g(x)}=\underset{x\to 0}{lim}\frac{a{x}^n+o(x^n)}{b{x}^n+o(x^n)}=\frac{a}{b};$$

If $n>m$ , then $$\underset{x\to 0}{lim}\frac{f(x)}{g(x)}=0;$$

if $m>n$ ,then $$\underset{x\to 0}{lim}\frac{f(x)}{g(x)}=\mathrm{\infty }.$$

Examples.

Compute the limits using the Taylor formula.

1. $\underset{x\to 0}{lim}\frac{\ln (1+x)-x}{x^2}.$

Solution.

$$\underset{x\to 0}{lim}\frac{\ln (1+x)-x}{x^2}=\underset{x\to 0}{lim}\frac{x-\frac{x^2}{2}-x}{x^2}=-\frac{1}{2}.$$

2. $\underset{x\to 0}{lim}\frac{e^x-1-x}{x^2}.$

Solution.

$$\underset{x\to 0}{lim}\frac{e^x-1-x}{x^2}=\underset{x\to 0}{lim}\frac{1+x+\frac{x^2}{2}-1-x}{x^2}=\frac{1}{2}.$$

3. $\underset{x\to 0}{lim}\frac{\cos x-1+\frac{x^2}{2}}{x^4}.$

Solution.

$$\underset{x\to 0}{lim}\frac{\cos x-1+\frac{x^2}{2}}{x^4}=\underset{x\to 0}{lim}\frac{1-\frac{x^2}{2}+\frac{x^4}{4!}-1+\frac{x^2}{2}}{x^4}=\frac{1}{24}.$$

4. $\underset{x\to 0}{lim}\frac{tgx-\sin x}{x^3}.$

Solution.

$$\underset{x\to 0}{lim}\frac{tgx-\sin x}{x^3}=\underset{x\to 0}{lim}\frac{x+\frac{x^3}{3}-x+\frac{x^3}{3}}{x^3}=\frac{2}{3}.$$

5. $\underset{x\to 0}{lim}\frac{\sqrt{1+2tgx}-e^x+x^2}{arctgx-\sin x}.$

Solution.

Let's expand the numerator using the Taylor formula:

$tgx=x+\frac{x^3}{3}+o(x^3),x\to 0$

$2tgx=2x+\frac{2x^3}{3}+o(x^3),x\to 0$

$\sqrt{1+t}=(1+t{)}^{\frac{1}{2}}=1+\frac{1}{2}t-\frac{1}{8}{t}^2+\frac{1}{16}{t}^3+o(t^3),t\to 0;$

Thus,

$$\sqrt{1+2tgx}=1+\frac{1}{2}2tgx-\frac{1}{8}(2tgx{)}^2+\frac{1}{16}(2tgx{)}^3+o(t{g}^{3}x)=$$

$$=1+tgx-\frac{1}{2}t{g}^{2}x+\frac{1}{2}t{g}^3+o(t{g}^{3}x)=$$

$$=1+x+\frac{x^3}{3}-\frac{1}{2}{x}^2+\frac{x^3}{2}+o(x^3)=1+x-\frac{1}{2}{x}^2+\frac{5}{6}{x}^3+o(x^3).$$

Given that $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)$, using the Maclaurin formula, we find the numerator of the fraction as

$$\sqrt{1+2tgx}-e^x+x^2=$$ $$=1+x-\frac{1}{2}{x}^2+\frac{5}{6}{x}^3-1-x-\frac{x^2}{2}-\frac{x^3}{6}+x^2+o(x^3)=$$ $$=\frac{2}{3}{x}^3+o(x^3),x\to 0.$$

Next, we expand the denominator:

$\sin x=x-\frac{x^3}{6}+o(x^3);$

$\arcsin x=x+\frac{x^3}{6}+o(x^3).$

Hence, $\arcsin x-\sin x=\frac{x^3}{3}+o(x^3).$

Thus, the fraction can be represented as $$\frac{\frac{2}{3}{x}^3+o(x^3)}{\frac{1}{3}{x}^3+o(x^3)}.$$

From here

$$\underset{x\to 0}{lim}\frac{\sqrt{1+2tgx}-e^x+x^2}{arctgx-\sin x}=2.$$

Answer: 2.

Tags: Maclaurin formula, Taylor polynomial, Taylor's Formula, calculus, differential, higher-order differentials, mathematical analysis