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Using the series expansion of e^KX, evaluate 2^-3.4 accurate to 3 dp?

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Using the series expansion of e^KX and the fact that a^X=e^Xlna, evaluate 2^-3.4 accurate to 3 dp?

    2. Relevant equations
    3. The attempt at a solution

    So a^X=e^Xlna. Basically we need to expand e^-3.4ln(2).

    e^x=1+X+(X^2)/2!+(X^3)/3!...

    e^-3.4ln(2)= 1-3.4ln(2)+(-3.4ln(2))^2/2+(-3.4ln(2))^3/6… See any errors in the expansion? The answer is supposedly 0.095 Can't seem to get it right.

    Thank you.
     
  2. jcsd
  3. Dec 12, 2011 #2

    vela

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    Looks fine to me. How many terms did you use in the expansion? Because x is relatively large, you'll have to use about 12 terms to get the accuracy you want.
     
  4. Dec 12, 2011 #3
    When I google the expansion it gives me -0.761 when the answer is supposed to be 0.095
     
  5. Dec 12, 2011 #4

    vela

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    You need to use more terms. (I edited my previous post, probably after you read it.)
     
  6. Dec 12, 2011 #5
    Awesome. I am gonna try it in a minute, but how do I know if the X is relatively large or small? Is there any rule of thumb for choosing the number of terms for expansion? I am asking this because all the examples given in the explanation section of my text are easily solved just by expanding e^x up to maximum of 4 terms.
     
  7. Dec 12, 2011 #6

    vela

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    Roughly speaking, the largeness or smallness of the number is in comparison to 1. If |x|<1, then |xn| grows smaller as n increases. Combined with the effect of the denominator in each term, the terms quickly become small, and you need to use only a few terms to get a good approximation. On the other hand, if |x|>1, then |xn| grows larger as n increases, so the terms don't get small until you've gone further in the series.

    In this problem, because x<0, you end up with an alternating series. There's a theorem that says if you sum the first n terms of an alternating series to get an approximation, the error in that approximation is no bigger than magnitude of the next term in the series. (I'm being a bit sloppy here, but that's the essential concept you need.) For example, after you added the first 4 terms, you got an approximation of -0.761. The next term in the series, x4/4!, is equal to 1.285, so the theorem guarantees you that
    [tex]-0.761 - 1.285 < 2^{-3.4} < -0.761 + 1.285.[/tex]To find the answer to three decimal places, you want the error to be less than 0.001, so you want to find n such that |x|n/n! < 0.001. The easiest thing to do here is try different values of n until you find one that works.

    More generally, there's a formula for the remainder if you truncate the Taylor series after n terms. In a similar way, you want to find the value of n that'll make that remainder small enough for your needs. There might be some rule of thumb that lets you estimate the number of terms you need, but I'm not aware of any.
     
  8. Dec 13, 2011 #7
    I appreciate your help. Thank You very much.
     
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