Question

EXPAND sqrt{/frac{1-x}{1+x}}

Answer 1

\[\Huge \sqrt{\frac{1-x}{1+x}}\]

Answer 2

up to power x^2

Answer 3

i get \[1-x+\frac{x^2}{2} - \frac{x^3}{8} - \frac{3x^4}{64}\]

Answer 4

@saifoo.khan

Answer 5

wolfram got \[-\frac{3 x^5}{8}+\frac{3 x^4}{8}-\frac{x^3}{2}+\frac{x^2}{2}-x+1\]

Answer 6

Unfortunately i forgot the formula! D:

Answer 7

Is it a maclaurin series?

Answer 8

you can try that

Answer 9

I want to know what is meant by Expand, since that's the only thing that comes to mind.

Answer 10

you can use this
\[(x+a)^n=\sum ^{n}_{ k=0} \left(\begin{matrix}n \\ k\end{matrix}\right) x^k a^{(n-k)}\]

Answer 11

or
\[(1+x)^n = 1+nx+\frac{(n)(n-1)}{2!}(x^2)+...\]

Answer 12

That works for \(n\in \mathbb{N}\).

Answer 13

In this case you have a square root...

Answer 14

still can, watch

Answer 15

\[\sqrt{\frac{1-x}{1+x}} \\ \\ \frac{\sqrt{1-x}}{\sqrt{1+x}} \\ \\ (1-x)^\frac{1}{2}(1+x)^{-\frac{1}{2}}\]

Answer 16

expand separately

Answer 17

mix them by multiplying together

Answer 18

Yeah, my problem is with \(1/2 \notin \mathbb{N}\)

Answer 19

I don't think it should be natural number

Answer 20

i mean real number will do

Answer 21

How can you have summations or binomial coefficients with rational numbers?

Answer 22

interval of x is ?

Answer 23

the formula list says can use

Answer 24

we use the second one (1+x)^n

Answer 25

http://en.wikipedia.org/wiki/Binomial_series
look at the History section

Answer 26

or maybe its \[-\frac{3 x^5}{8}+\frac{3 x^4}{8}-\frac{x^3}{2}+\frac{x^2}{2}-x+1\]

Answer 27

hmm

Answer 28

based from the formula as long as its in the from ("1"+x)^n should be fine to expand, i tried it

Answer 29

from \[(1-x)^{1/2} \\ \\ n=-1/2, x=-x(for formula) \\ \\ [1-\frac{x}{2}+\frac{(\frac{1}{2})(\frac{1}{2}-1)}{2}(x^2)] \\ \\ 1-\frac{x}{2}-\frac{x^2}{8}\]