Approximating the Square Root of 11 Using Binomial Expansion

[Michael_Light]

Homework Statement

Prove that , if x is so small that terms in x³ and higher powers may be neglected, then http://www.mathhelpforum.com/math-help/attachments/f37/21081d1299568824-binomial-expansion-question-msp520319eed9935g5h93hf000055b8ic10787h09a9.gif . By substituting a suitable value of x in your result, show that (11)^1/2 is approximately equal to 663/200.

Homework Equations

The Attempt at a Solution

I can solve the first part, but i have difficulty in the second part (show that (11)^1/2 is approximately equal to 663/200). Can anyone help me? Thanks.

 

[eumyang]

You'll need to find x such that
\frac{1+x}{1-x} = 11

 

[Michael_Light]

You'll need to find x such that
\frac{1+x}{1-x} = 11

After solving that, i get x=5/6. Substituting x=5/6 into http://www.mathhelpforum.com/math-help/attachments/f37/21081d1299568824-binomial-expansion-question-msp520319eed9935g5h93hf000055b8ic10787h09a9.gif , i get (11)^1/2=157/72 which is incorrect... what answer did you get..? ><

 

[vela]

That doesn't work well because 5/6 isn't very small. You want x<<1 for the approximation to be good. The trick is to pull out a nearby perfect square:

\sqrt{11} = \sqrt{9+2} = \sqrt{9}\sqrt{1+2/9} = 3\sqrt{11/9}

You want to find x such that (1+x)/(1-x) = 11/9.

 

[Michael_Light]

That doesn't work well because 5/6 isn't very small. You want x<<1 for the approximation to be good. The trick is to pull out a nearby perfect square:

\sqrt{11} = \sqrt{9+2} = \sqrt{9}\sqrt{1+2/9} = 3\sqrt{11/9}

You want to find x such that (1+x)/(1-x) = 11/9.

Allow me to ask.. if i just compare (11/9)^1/2 with ((1+x)/(1-x))^1/2, where did the '3' go? I just ignore it?

 

[vela]

No, it's part of the final answer for the square root of 11. You're using the series approximation to find the square root of 11/9, which when multiplied by 3 gives you the square root of 11.