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Question Number 8127 by Yozzia last updated on 30/Sep/16

F o r ∣ x ∣< 1, w e h a v e t h a t

{(1 + x)}^{1 / 2} = 1 + \frac{1}{2} x + \frac{(\frac{1}{2}) (\frac{1}{2} - 1)}{2!} x^{2} + \frac{\frac{1}{2} (\frac{1}{2} - 1) (\frac{1}{2} - 2)}{3!} x^{3} + \dots

{(1 + x)}^{1 / 2} = 1 + \underset{r = 1}{\sum^{\infty}} \frac{\underset{k = 0}{\prod^{r - 1}} (0.5 - k)}{r!} x^{r} .

L e t g (r) = \underset{k = 0}{\prod^{r - 1}} (0.5 - k) .

I s i t t r u e t h a t f o r x = \frac{1}{2} i \Rightarrow∣ x ∣= 0.5 < 1

{(1 + \frac{1}{2} i)}^{1 / 2} = 1 + \underset{r = 1}{\sum^{\infty}} \frac{g (r)}{r!} \times \frac{1}{2^{r}} i^{r} ?

(i = \sqrt{- 1})

Commented by prakash jain last updated on 01/Oct/16

It is a Taylor series expansion so

it is true for any f (z) .

f (z - a) = \underset{k = 1}{\sum^{\infty}} \frac{f^{(k)} (a)}{k!} {(z - a)}^{k}

If series on RHS converges .

Commented by prakash jain last updated on 01/Oct/16

f (z) = \underset{n = 0}{\sum^{\infty}} c_{n} {(z - a)}^{n}

f r o m r a t i o t e s t

\lim_{n \to \infty} ∣ \frac{c_{n + 1} (z - a)}{c_{n}} ∣< 1

or ∣ z - a ∣< \lim_{n \to \infty} ∣ \frac{c_{n}}{c_{n + 1}} ∣

for {(1 + z)}^{y} (a = 0)

c_{n} = \frac{y (y - 1) \dots (y - n + 1)}{n!}

c_{n + 1} = \frac{y (y - 1) \dots (y - n + 1) (y - n)}{(n + 1)!}

\frac{c_{n}}{c_{n + 1}} = \frac{n + 1}{y - n}

\lim_{n \to \infty} ∣ \frac{c_{n}}{c_{n + 1}} ∣= 1

s o t h e s e r i e s c o n v e r g e s i f ∣ z ∣< 1

so the series expansion of {(1 + \frac{i}{2})}^{1 / 2}

is correct .

Commented by Yozzias last updated on 01/Oct/16

T h a n k s! W h e n w o u l d c o n v e r g e n c e

o c c u r ?

Answered by prakash jain last updated on 02/Oct/16

answer in comments