$(1-x)^{-n} \rarr$ integer

$(1-x)^{p/q} \rarr$ rational number

In particular we were looking at $(1-x)^{-\frac{1}{2}}$.

What is the series expansion for $(1-x)^{\frac{1}{2}}$?

Solution:

We have $(1-x)^{\frac{1}{2}} \times (1-x)^{\frac{1}{2}} = (1 - x)$

$\therefore$ If we write

$(1-x)^{\frac{1}{2}}=a_0+a_1{x}+a_2 x^2$

then by multiply it with itself we should get $(1 - x).$

Let $(1-x)^{\frac{1}{2}}=a_0+a_1 x +a_2 x^2+\cdots$

$\therefore (a_0+a_1{x}+a_2 x^2+…) \times (a_0+a_1{x}+a_2 x^2+.. .)$

$= 1 - x$

$\therefore$ coeff of $x^0=a_0^2=1$

Taking the $(+)$ve root $a_0=1$

coeff of $x=2 a_0 a_1=-1$

$\therefore 2 a_1=-1$

$\therefore a_1=-\frac{1}{2}$

Coeff of $x^2, \$ $\ a_0 a_2+a_1^2+a_2 a_0 = 0$

$\because \nexists x^2$ in $(1 - x)$

$\therefore 2 a_2+a_1^2=0$

or $2 a_2+\frac{1}{4}=0$

$\therefore a_2=-\frac{1}{8}$

$\therefore$ We can write $(1-x)^{\frac{1}{2}} =\sqrt{1-x}$ as $1-\frac{1}{2} x-\frac{1}{8} x^2+\cdots$

If $x$ is small we aften ignore higher powers.

In a similar way we can find $\sqrt{1+x}$.

say $\sqrt{1+x}=a_0+a_1 x+a_2 x^2+\cdots$

$\therefore(a_0+a_1 x+a_2 x^2+\cdots) \times (a_0+a_1 x+a_2 x^2)\cdots = 1 + x$

$a_0^2=1 \text { or } a_0=1$ (Taking +ve value)

$a_0 a_1+a_1 a_0=1$

or $2 a_0 a_1=1 \quad \therefore a_1=\frac{1}{2}$

$a_0 a_2+a_1^2+a_0 a_2=0$

or $2 a_2+a_1^2=0$

$\therefore 2 a_2=-a_1^2=-\frac{1}{4}$

$\therefore a_2=-\frac{1}{8}$

$\therefore$ We get $\sqrt{1+x}$

$=1+\frac{1}{2} x-\frac{1}{8} x^2 \ldots .$

Find $\sqrt{17}$.

Solution:

We can write it as

$(1+16)^{\frac{1}{2}}$ $\larr$ we are making a mistake

In $(1 + x) \larr$ $|x| < 1$

We write in a different way

$16^{\frac{1}{2}}(1+\frac{1}{16})^{1/2}$

$|{\frac{1}{16}}| < 1$

To obtain $\sqrt{17}$

we write it as $16^{\frac{1}{2}}\left(1+\frac{1}{16}\right)^{\frac{1}{2}}$

$=4 \times (1+\frac{1}{16})^{\frac{1}{2}}$

In the expansion of $(1+x)^{\frac{1}{2}}$

$=1+\frac{1}{2} x-\frac{1}{8} x^2$

Putting $x=\frac{1}{16}$ we get

$=1+\frac{1}{32}-\frac{1}{8 \times 16^2}$

$\therefore \sqrt{17}=4{\times}\left(1+\frac{1}{32}-\frac{1}{8 \times 16^2}\right)$

$4 \times 1=4$

$4 \times \frac{1}{32}=\frac{1}{8}=0.125$

$4 \times \biggl( -\frac{1}{8 \times 16^2}\biggl )=-\frac{1}{2 \times 16^2}$

$=-\frac{1}{2 \times 256} =-\frac{1}{512}$

$=-0.0019$

$\therefore \sqrt{17} =4.125-0.0019$

$= 4.1231$

Therefere the final expansion is $(1+x)^{p/q}$

$=1+\frac{p}{q} x+\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)}{2 !} x^2 +\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)\left(\frac{p}{q}-2\right)}{3 !} x^3 +…..$

What we have done so far is not a proof, we have verified certain results.

The basic assumption was:

Weierstrass Approximation Theorem.

Every continous function in a closed interval can be approximated as closely as possible by a polynomial function.

$(1+x)^k \leftarrow$ negative integral or rational.

How to obtain the coefficients of the given function $f(x)$?

it is possible to have such a polynomial expansion about a point $a.$

$f(x)=a_ 0 +a_ 1(x-a) +a_ 2(x-a)^2 +a_3(x-a)^3 +……$

Advantage of a polynomial function is that

If it is of degree $n$, it can be differentiatal $(x + 1)$ times.

& If we take an infinite polynomial, we can differentiate infinite number of times.

$f(x)=a_0+ a_1(x-a)+a_2(x-a)^2 +a_3(x-a)^3+..$

$\therefore f(a)=a_0 \therefore a_0=f(a)$

$f^{(1)}(a)=a_ 1+2 a_ 2(x-a) +3 a_ 3(x-a)^2 +4 a_4(x-a)^3+…..$

$f^{(2)}=2 a_2+3 \cdot 2 a_3(x-a) +4 \cdot 3 \cdot a_4(x-a)^2+…..$

$\therefore f^{(2)}(a)= 2 \cdot a_2$

or $a_2=f^{(2)}(a) / 2$

$f^{(3)}(x)= 3 \cdot 2 \cdot 1 \cdot a_3$

$\quad +4 \cdot 3 \cdot 2(x-a)+…..$

terms with higher powers of $(x - a)$

$\therefore f^{(3)}(a)=3 ! a_3$

$\therefore a_3=\frac{f^{(3)}(a)}{3 !}$

$f^{(4)}(x)=4 \cdot 3 \cdot 2 \cdot 1 a_4$ + power of $(x - a)$

$\therefore a_4=\frac{f(4)(a)}{4 !}$

This expansion is called Taylor Series expansion of $f(x).$

$\therefore$ $f(x)$ can be written as:

$f(a) + f^{(1)}(a)(x-a) +\frac{f^{(2)}(a)}{2 !} (x-a)^2$

$+\frac{f^{(3)}(a)}{3 !} (x-a)^3$

$+\frac{f^{(4)}(a)}{4 !} (x-a)^4 +…….$

Let us consider $(1-x)^{-2} \ \ \therefore f(x) =(1-x)^{-2}$

$f^{(1)}(x) =-2(1-x)^{-3}(-1)=2(1-x)^{-3}$

$f^{(2)}(x) =-3 \cdot 2 \cdot(1-x)^{-4}(-1) =3 !(1-x)^{-4}$

$f^{(3)}(x) =4 !(1-x)^{-5}$

$\therefore f(0) = 1$

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

$f^{(2)}(0)=3!$

$f^{(4)}(0)=4!$

$\therefore$ we get

$f(x) = f(0) + 2(x - 0) +3 ! \frac{(x-0)^2}{2 !} +4 ! \frac{(x-0)^3}{3 !}$

$\therefore$ We can write

$f(x)= f(a)+f^{(1)}(a) \cdot(x-a) +f^{(2)}(a)\frac{(x-a)^2}{2 !} +f^{(3)}(a)\frac{(x-a)^3}{3 !}$

$+f^{(4)}(a)\frac{(x-a)^4}{4 !} +……$

Putting the values:

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

$=f(0)+f^{(1)}(0) x+f^{(2)}(0) \frac{x^2}{2 !} +f^{(3)}(0)\frac{x^3}{3 !} + f^{(4)}(0)\frac{x^4}{4 !}$

$f(0)+2 \cdot x+3 ! \frac{(x-0)^2}{2 !}+4 ! \frac{(x-0)^3}{3 !}+…..$

$=1+2 x+3 x^2+4 x^3+\cdots$

These are the terms that we have already seen.

Looking Beyond Polynomials.

Consider trigonometric Functions:

$\sin x$

$\cos x$

$\tan x$

$0^{\circ}, \pi / 6, \pi / 4, \pi / 3, \pi / 2, \pi$

$15^{\circ}, 18^{\circ} \ \text{etc.}$

$\sin 1^{\circ}, \sin 5^{\circ}$

Consider $\sin x.$

$f(x) = \sin x \ \ \therefore f(0) = 0$

$f^{(1)}(x)=\cos x \ \ \therefore f^{(1)}(0)=1$

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

$f^{(3)}(x)=-\cos x \ \ \therefore f^{(3)}(0)=-1$

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

$f^{(5)}(x)=\cos x \ \ \therefore f^{(5)}(0)=1$

From Taylor’s series

$f(x)= f(0)+f’(0)(x-0)+f^{(3)}(0) \frac{(x-a)^2}{2 !}$

$+f^{(3)}(0) \frac{(x-a)^3}{3 !}+f^{(4)}(0) \frac{(x-a)^4}{4 !} +f^{(5)}(0) \frac{(x-a)^5}{5 !}+\ldots$

$f(0) = 0$

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

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

$f^{(3)}(0)=-1$

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

$f^{(5)}(0)=+1$

Putting these values we get,

$\sin (x) = \sin (0) +1 \cdot(x-0)+0 \cdot \frac{(x-0)^2}{2 !} -1 \frac{(x-0)^3}{3 !}+0 \cdot \frac{(x-0)^4}{4 !} +1 \cdot \frac{(x-0)^5}{5 !}+\cdots$

$= 1 \cdot x-\frac{x^3}{3 !}+\frac{x^5}{5 !}$

$= x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\frac{x^7}{7 !}+\cdots$

Similarly we can work out for $\cos x$.

$\cos x=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !} \cdots$

Find the Taylor series expansion for $\tan (x).$

Again we expand about $0.$

$f(0) = \tan (0) = 0$

$f^{(1)}(x)=\frac{d}{d x} \tan x=\sec ^2 x$

$=1+\tan ^2 x$

$\therefore f^{(1)}(0)= 1.$

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

$=2 \tan x(\sec^2 x)$

$=2 \tan x(1+\tan^2 x)$

$=2 \tan x+2 \tan^3 x$

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

$f^{(3)}(x)=2(1+\tan^2 x)+6 \tan^2 x(1+\tan^2 x)$

$=2+2 \tan^2 x+6 \tan^2(x) +6 \tan^{(4)} x$

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

$f^{(4)}(x)=\frac{d}{d x}(8 \tan^2 x+6 \tan^4 x)$

$=16 \tan x+16 \tan^3 x + 24 \tan^3 x (1+\tan^2 x)$

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

$f^{(5)}(x)=16(1+\tan^2 x)+$ terms with $\tan x$

$\therefore f^{(5)}(0)=16$

$\tan x=0+\frac{1}{1 !} x+0 \frac{x^2}{2 !}$

$+2 \cdot \frac{x^3}{3 !}+0 \cdot \frac{x^4}{4 !} +16 \frac{x^5}{5 !}+\cdots$

$\tan x= 0+x + 0 \cdot x^2 +2 \cdot \frac{x^3}{3 !}+0 x^4 +16 \frac{x^5}{5 !}$

$=x+\frac{x^3}{3}+\frac{2}{15} x^5 + \text{higher powers of} \ x$

Thank you