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E^x expansion

  • #1

Homework Statement


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.


Homework Equations


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


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!
 

Answers and Replies

  • #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.
 
  • #3
jbunniii
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Homework Statement


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.
Do you mean

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

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!)) ??

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

Thanks!
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.
 
  • #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.
 
  • #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.)
 
  • #6
You only applied the - sign to the first term inside the inner parentheses. It should have been applied to all of them! (Distributive law.)
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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