Expansion of $(\frac{x+2}{x})^{-\frac{1}{2}}$ and then find an approximate value...

Question

First, obtain the four terms in the expansion of $ (\frac{x+2}{x})^{-\frac{1}{2}}$ then let x = 100 and use the result to find an approximate value of $(\frac{450}{51})^{\frac{1}{2}}$.

I am stuck with this problem and would be very happy if someone could help me how to move on with it...

Here is what I've attempted so far:

Factoring out $\sqrt{\frac{x}{2}}$ from the original equation. I ended up with $\sqrt{\frac{x}{2}}(1+\frac{x}{2})^{-\frac{1}{2}}$.

I expanded it using this formula $ (1+x)^{n}=1+\frac{nx}{1!}+\frac{n(n-1)x^{2}}{2!}+\ldots $ and simplified the resulting expansion.

My solution was $ \sqrt{\frac{x}{2}}-\frac{x^{\frac{3}{2}}}{4\sqrt{2}}+\frac{3x^{\frac{5}{2}}}{32\sqrt{2}}+\ldots $ which is incorrect.

I feel there must be something simpler that I'm missing.

Answer 1

I wouldn't factor the way you did. The binomial expansion as a radius of convergence of 1, but with your factorisation, you have $\frac{x}{2}$, wich is clearly $>1$ when $x=100$.

Try to just rewrite $\frac{x+2}{x}$ as $1+\frac{2}{x}$, and then you can use the binomial expansion (because $|\frac{2}{x}|<1$ if x=100)

Answer 2

First, the binomial expansion: $$(\frac{x+2} x)^{-1/2}=(1+\frac 2 x)^{-1/2}$$ $$=1+\frac2 x(\frac{-1} 2)+(\frac 2 x)^2+(\frac{-1} 2)(\frac{-3} 2)+(\frac 2 x)^3(\frac{-1} 2)(\frac{-3} 2)(\frac{-5} 2)+(\frac 2 x)^4(\frac{-1} 2)(\frac{-3} 2)(\frac{-5} 2)(\frac{-7} 2)+...$$ $$=1-\frac 1 x+\frac 3 {x^2}-\frac 5 {2x^3}+\frac {35} {8x^4}-\frac {63} {8x^5}+...$$ Now, for x=100, we would simply have, $$=1-\frac 1 {100}+\frac 3 {100^2}-\frac 5 {2.100^3}+\frac {35} {8.100^4}-\frac {63} {8.100^5}+...$$

Consider $(\frac{450}{51})^{1/2}=(\frac{9\times 50}{51})^{1/2}=\sqrt 9(\frac 1 {1+1/50})^{1/2}=3(1+\frac 2 {100})^{-1/2}$, which can be calculated from the above expression using the first four terms after the initial "1" as 2.970442631

The actual value of $(\frac{450}{51})^{1/2}=2.9704426289...$