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Series expansions

  1. Sep 22, 2009 #1
    use series expansion to fine the limit as x--->0 of

    1/(sin^2)(x) - 1/x^2






    Sin^2(x)=(X-X^3/3!)^2 ive assumed this gave me
    Sin^2(x)=X^2-2X^4/6+X^6/36
    flip this over is that equivalent to 1/Sin^2(x)?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 22, 2009 #2

    Dick

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    It's not totally equivalent because you omitted higher order terms. Can you show you have enough terms to capture the behavior of 1/sin^2(x)-1/x^2 in the sense that terms that you omitted will go to zero as x->0?
     
  4. Sep 23, 2009 #3
    Im sorry im not quite sure what you mean by that?
     
  5. Sep 23, 2009 #4

    Dick

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    I'm just saying sin(x) isn't equal to x-x^3/3!. You omitted the higher order terms, like x^5/5!. As you work out the limit you'll want to convince yourself that including them doesn't affect the limit as x->0. Do you know how to find the limit of 1/(x-x^3/3!)^2-1/x^2?
     
  6. Sep 23, 2009 #5
    Actually no i dont, maybe you could explain to me how?
     
  7. Sep 23, 2009 #6

    Dick

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    Factor the denominator of x-x^3/6 as x*(1-x^2/6). So you've got (1/x)*(1/(1-x^2/6)). Use that 1/(1-a)=1+a+a^2+a^3+... (the usual geometric series thing) to move the second factor into the numerator.
    Now you've got (1/x)^2*(1+x^2/6+...)^2-1/x^2. Expand it. Now go back and figure out why I didn't need to keep any higher powers of x than I did.
     
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