Question

What are the first four terms in the expansion of (2+x)^(1/2)? In what interval is the expansion valid?

Answer 1

I don't understand why the interval is -2<x<2. Could someone explain that?

Answer 2

(2+x)^1/2
means
\[\sqrt (2+x)\]
If anything is square rooted then it must be greater then or equal to 0 because if it is less than that it would be complex!
SO
\[2+x \ge 0\]

Answer 3

Then x>=-2 why -2<x<2?

Answer 4

You are right! :D
The answer key is mistaken!

Answer 5

Oh... Thanks!!!

Answer 6

:)

Answer 7

\[\left( 2+x \right)^{\frac{ 1 }{ 2 }}=2^{\frac{ 1 }{ 2 }}\left( 1+\frac{ x }{ 2 } \right)^{\frac{ 1 }{ 2 } }\]
The expansion is valid if
\[\left| \frac{ x }{ 2} \right|<1, or-1<\frac{ x }{2 }<1, or -2<x <2\]

Answer 8

Thanks, surjithayer.

Answer 9

if you want i can expand.

Answer 10

i got
sqrt(2) (1+x/4-x^2/8+3x^3/16-...)

the answer given by my book is different...
sqrt(2) (1+x/4-x^2/32+x^3/128-...)

Answer 11

@Surjith Can you expand?

Answer 12

I've figured that out. But I couldn't solve another problem... Could you help?
Find the coefficient of x^3 in x^(-1/2)/[2+sqrt(x)]^3
x^(-1/2)/[2+sqrt(x)]^3
= x^(-1/2) * 2^(-3) * (1+sqrt(x)/2)^(-3)
(1+sqrt(x)/2)^(-3) = (1-3x^(1/2)/2+3x^(3/2)/2-5x^(5/2)/4+15x^(7/2)/16+...) (Is this right?)
Coefficient of x^3:
15/16 * 2^(-3) = 15/128

The answer should be -9/256.
Thanks a lot!!!

Answer 13

\[\left( 1+x \right)^{n}=1+nx+\frac{ n \left( n-1 \right) }{ 2! }x ^{2}+\frac{ n \left( n-1 \right)\left( n-2 \right) }{3! }x ^{3 }+...\]

Answer 14

\[n =\frac{ 1 }{ 2 }, 2!=2*1=2,3!=3*2*1=6,x=\frac{ x }{2 }\]
plug the values and multiply by sqrt{2} to get the solution.

Answer 15

if still problem persists i will solve

Answer 16

solved. thanks :)