Taylor series expansion

$f(x)=f(a) +f^{(1)}(a) \frac{(x-a)}{1 !}$

$\ldots +f^{(2)}(a) \frac{(x-a)^2}{2 !}$

$\ldots +f^{(k)}(a) \frac{(x-a)^k}{k !}$

Find the Taylor Series approximation of $\tan ^{-1}(x)$ up to $x^5$.

$f(x) =\tan ^{-1}(x) \quad f(0)=0$

$f^{\prime}(x) =\frac{1}{1+x^2}=\left(1+x^2\right)^{-1} \quad \therefore f^{\prime} (0)=1$

$f^{(2)}(x) =\frac{d}{d x}\left(1+x^2\right)^{-1}$

$=-\left(1+x^2\right)^{-2}(2 x)$

$=-2 x\left(1+x^2\right)^{-2}$

$\therefore f^{(2)}(x)=0 \text { at } x=0$

$f^{(3)}(x)=\frac{d}{d x}\left(-2 x\left(1+x^2\right)^{-2}\right)$

$=-2\left(1+x^2\right)^{-2}+(-2 x)(-2)(1+x^2)^{-3}(2 x)$

$=-2\left(1+x^2\right)^{-2}+8 x^2\left(1+x^2\right)^{-3}$

$\therefore f^{(3)}(0)=-2$

$f^{(4)}(x)=\frac{d}{d x}(-2(1+x^2)^{-2}+8 x(1+x^2)^{-3})$

$=4(1+x^2)^{-3}(2 x)+16 x (1+x^2)^{-3} + 8 x^2 + (1+x^2 )^{-4} 2 x$

$=8 x\left(1+x^2\right)^{-3}+16 x\left(1+x^2\right)^{-3} +8 x^3\left(1+x^2\right)^{-4} 2x$

$=24 x\left(1+x^2\right)^{-3}+8 x^3\left(1+x^2\right)^{-4}$

$\therefore \quad f^{(4)}(0)=0$

$f^{(5)}(x)=\frac{d}{d x} \left(24 x\left(1+x^2\right)^{-3} +8 x^3\left(1+x^2\right)^{-4}\right)$

$=24\left(1+x^2\right)^{-3}+24 x\left(1+x^2\right)^{-4}(2 x)+ \ldots$

$f^{(5)}(0)=24$

$f(0)=0$
$f^{(1)}(0)=1$
$f^{(2)}(0)=0$
$f^{(3)}(0)=-2$
$f^{(4)}(0)=0$
$f^{(5)}(0)=24$

$\therefore$ The $5^{\text {th }}$ degree polynomial approximation for $\tan ^{-1}(x)$ is:

$x-\frac{2 x^3}{3 !}+\frac{24 x^5}{5 !}.$

$\equiv x-\frac{x^3}{3}+\frac{x^5}{5} .$

$\frac{d}{d x} \tan ^{-1}(x)=\frac{1}{1+x^2}$

The expansion of $\frac{1}{1+x^2}$

$\text { i.e }\left(1+x^2\right)^{-1}$

$1-x^2+x^4-x^6 \ldots$

$\int \frac{1}{1+x^2} d x=\int 1 d x-\int x^2 d x + \int x^4 dx \ldots +C$

$\tan ^{-1} x=x-\frac{x^3}{3}+\frac{x^5}{5} \ldots+C$

By putting $x=0$

we find that $c=0$

$x-\frac{x^3}{3}+\frac{x^5}{5}$

Consider $\sin ^{-1} x$

we knows that $\frac{d}{d x} \sin ^{-1}(x)$

$=\frac{1}{\sqrt{1-x^2}}$

$=\left(1-x^2\right)^{-\frac{1}{2}}$

$\left(1-x^2\right)^{-\frac{1}{2}}$

$=1 +\left(-\frac{1}{2}\right)\left(-x^2\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2 !}\left(-x^2\right)^2$

$+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}\left(-x^2\right)^3$

$= 1+\frac{x^2}{2}+\frac{1 \cdot 3}{8} x^4+\frac{1 \cdot 3 \cdot 5}{8 \times 3 !} x^6$

$= 1+\frac{x^2}{2}+\frac{3}{8} x^4+\frac{15^4}{48} x^6+\cdots$

$\therefore \int \frac{1}{\sqrt{1-x^2}} d x$

$\quad=\int 1 d x+\int \frac{x^2}{2} d x+\int \frac{3}{8} x^4 d x +\int \frac{15}{48} x^6 d x \ldots +c$

$\text { or } \sin ^{-1}(x)=x+\frac{x^3}{3 !}+\frac{3 x^5}{40}+\frac{15}{48 \times 7} x^7 \ldots +c$

Putting $x=0$

we get $c=0$

$\therefore$ Taylor series expansion for $\sin ^{-1}(x)$

$=x^4+\frac{x^3}{3 !}+\frac{3 x^5}{40}+\frac{5}{16 \times 7} x^7 \cdots$

What is the expansion for $\cos ^{-1}(x)$?

$f^{\prime} f^{\prime \prime} f^{\prime \prime \prime} \ldots$

$\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x$

$\text { or } \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x$

$\therefore \cos ^{-1}(x)$

$=\frac{\pi}{2}-x-\frac{x^3}{3 !}-\frac{3 x^5}{40} -\frac{5 x^7}{16 \times 7} \ldots$

Let us consider ${log (1+x)}$

$f(x)=\log (1+x)\quad \therefore f(0)=\log (1)=0$

$f^{\prime}(x)=\frac{1}{1+x} \quad\therefore f^{\prime}(0)=1$

$f^{\prime \prime}(x)=-(1+x)^{-2}\quad \therefore f^{\prime \prime}(0)=-1$

$f^{\prime\prime\prime}(x)=2(1+x)^{-3}\quad \therefore f^{\prime\prime\prime}(0)=2$

$f^{\prime\prime\prime\prime}(x)=-6(1+x)^{-4} \quad\therefore f^{\prime\prime\prime\prime}(0)=-6$

$\therefore \log (1+x)$

$x-\frac{x^2}{2}+2 \frac{x^3}{3 !}-6 \frac{x^4}{4 !} \cdots$

$= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} \cdots$

$\therefore$ The series is: $\sum_{k=1}^{\infty} \frac{x^k}{k}(-1)^{k-1}$

Since we know expansion of $\frac{1}{1+x}$

$\frac{1}{1+x} =(1+x)^{-1}$

$=1-x+x^2-x^3+x^4 \ldots$

$\int \frac{1}{1+x} d x =\int 1 d x-\int x d x+\int x^2 d x -\int_0^3 x^3 d x+\int x^4 d x \ldots+C$

$\therefore \log (1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} \cdots$

To determine the value of $C$

$\text { At } x =\log (1+x)=\log (1)=0$

$\therefore \quad C=0$

By putting $x=-x$

$\log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4} \ldots$

$\quad \log (\cos x) \quad \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

$f(x)=\log (\cos (x)) \quad \therefore f(0)=\log (1)=0$

$f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x \quad\therefore f^{\prime}(0)=0$

$f^{\prime\prime}(x)=-\left(1+\tan ^2 x\right)\quad \therefore f^{(2)}(0)=-1$

$f^{\prime\prime\prime}(x)=-2 \tan x\left(1+\tan ^2\right)$

$\quad\quad\quad=-2 \tan x-2 \tan ^3 x$

$\therefore f^{\prime\prime\prime}(0)=0$

$f^{(\prime\prime\prime\prime)}(x)=$

$\quad-2\left(1+\tan ^2 x\right)-6 \tan ^2 x\left(1+\tan ^2 x\right)$

$=-2-2 \tan ^2 x-6 \tan ^2 x-6 \tan ^4 x$

$=-2-8 \tan ^2 x-6 \tan ^4 x$

$\therefore f^{(\prime\prime\prime\prime)} (0)=-2$

$f^{(\prime\prime\prime\prime\prime)} (x)=-16(\tan x)\left(1+\tan ^2 x\right) -24 \tan ^3 x(1+\tan ^2 x)$

$=-16 \tan x -16 \tan ^3 x -24 \tan ^3 x-24 \tan ^5 x$

$\therefore f^{(5)}(x)=-16 \tan x-40 \tan ^3 x-24 \tan ^5 x$

$\therefore f^{(5)}(0)=0$

$\therefore f^{(6)}(x)=-16\left(1+\tan ^2\right)+\ldots$

$\quad f^{(6)}(0)=-16$

$\therefore$ we find that for $\log (\cos (x))$

$f(0)=0$

$f^{\prime}(0)=0$

$f^{(2)}(0)=-1$

$f^{(3)}(0)=0$

$f^{(4)}(0)=-2$

$f^{(5)}(0)=0$

$f^{(6)}(0)=-16$

$\therefore$ The series expansion is:

$-\frac{x^2}{2}-2 \frac{x^4}{4 !}-16 \frac{x^6}{6 !} \cdots$

$= -\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45} \cdots$

We can also do it as follows:

$log (\cos x) = log (1+(\cos x-1))$

Euler’s constant “e”.

$\lim_{n \rightarrow \infty} (1+ \frac{1}{n})^n$

where does $\left(1+\frac{1}{x}\right)^x$ converge?

$n =2 \quad (1+0.5)^2=2.25$

$=10 \quad (1+0.1)^{10}=2.594$

$=100 \quad (1+0.01)^{100}=2.7048$

$=1000 \quad (1+0.001)^{1000}=2.7169$

$=10000 \quad (1+0.0001)^{10000}=2.7181$

$=100000 \quad(1+0.00001)^{100000}=2.7183$

$\text { As } n \rightarrow \infty$

$\left(1+\frac{1}{n}\right)^n \rightarrow 2 \cdot 718281828459045 \ldots$

Euler’s Number.

$\left(1+\frac{1}{n}\right)^n$

What is the $k^{\text {th }}$ term?

Solution:

For a given $k$ if we consider $n>k$

$\frac{n(n-1) \cdots(n-\overline{k-1})}{k!}(\frac{1}{n})^k$

${^nc_k(\frac{1}{n})^k}$

$\frac{1\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{k-1}{n}\right)}{k !}$

$\therefore$ As $x \rightarrow \infty$ the $k^{\text {th }}$ term in $\frac{1}{k!}$

$\therefore$ The series $\left(1+\frac{1}{n}\right)^n$

$\rightarrow 1+1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{k !}+\cdots$