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E^x expansion

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A resistor and inductor are connected in series to a battery. The current in the circuit ( in A ) is given by I = 2.7(1-e-0.1), where t is the time since the circuit was closed. By using the series for ex, find the first three terms of the expansion of this function.


    2. Relevant equations
    ex = 1 + x + x2/2! + ... + xn/n!


    3. The attempt at a solution
    Would I substitute -0.1 for x in this expansion like so: ex = 1 - 0.1 - 0.12/2! ? And then place this expansion into the original function 2.7(1-(1 - 0.1 - 0.12/2!)) ?? And finally, multiply the 2.7 back in?

    This isn't coming out to one of my answer choices. Where have I gone wrong in my setup?

    Thanks!
     
  2. jcsd
  3. May 26, 2009 #2

    djeitnstine

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    [tex]e^x = 1 +(-0.1) + \frac{(-0.1)^2}{2}[/tex] Substituting you get [tex]2.7 (1- \left( 1 +(-0.1) + \frac{(-0.1)^2}{2} \right))[/tex] Watch your minus signs.
     
  4. May 26, 2009 #3

    jbunniii

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    Do you mean

    [tex]I(t) = 2.7(1 - e^{-0.1t})[/tex]? (There's no "t" in your formula.)

    I see a couple of problems. First,

    [tex](-0.1)^2 = 0.1^2[/tex], not [tex]-0.1^2[/tex]

    Second, notice that the "1" terms cancel, leaving you with only two terms. The problem asks for three, so you probably need to include the [tex]0.1^3[/tex] term as well.
     
  5. May 26, 2009 #4
    It kills me when I try to get clever and my signs get skewed.

    So, now I've added a couple steps to insure I'm sign-correct and also added the third term to account for the 1's canceling.

    Here is the setup:
    [tex]I = 2.7 (1- \left( 1 +(-0.1) + \frac{(-0.1)^2}{2} + \frac{(-0.1)^3}{6}\right))[/tex]

    Clear the 1's and multiply by 2.7...

    [tex] 2.7(-0.1) + \frac{2.7(-0.1)^2}{2} + \frac{2.7(-0.1)^3}{6}[/tex]

    Pull out the fractions to please the prof:

    [tex] 2.7(-0.1) + \frac{1}{2}2.7(-0.1)^2+ \frac{1}{6}2.7(-0.1)^3[/tex]

    The numbers come out almost right. And the missing 't' from the problem shows in the answer key so I've arbitrarily added the [tex]t,t^2,[/tex] and [tex]t^3[/tex]

    The answer I got is:
    [tex]I(t) = -0.27t + 0.0135t^2 - 0.00045t^3[/tex] but to match the answer, I'd have to multiply by -1 to get the signs right. I'm close but there must be another tidbit I've overlooked. Or there's a typo in the answer key...

    I wonder if this is like a falling object where the negative sign is removed(or multiplied by -1) in relation to time.
     
  6. May 26, 2009 #5

    jbunniii

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    You only applied the - sign to the first term inside the inner parentheses. It should have been applied to all of them! (Distributive law.)
     
  7. May 26, 2009 #6
    Ahhhhh... it's not 1-(...) it's 1-1(....) Dangit! I'm so caught up in the new stuff I forgot the details.

    Thanks so much!!!
     
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