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Binomial expansion problem

  • #1

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 [Broken]. 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.
 
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Answers and Replies

  • #2
eumyang
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You'll need to find x such that
[tex]\frac{1+x}{1-x} = 11[/tex]
 
  • #3
You'll need to find x such that
[tex]\frac{1+x}{1-x} = 11[/tex]
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 [Broken], i get (11)1/2=157/72 which is incorrect... what answer did you get..? ><
 
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  • #4
vela
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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:

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

You want to find x such that (1+x)/(1-x) = 11/9.
 
  • #5
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:

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

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?
 
  • #6
vela
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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.
 

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