# E^x expansion

## 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!

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djeitnstine
Gold Member
$$e^x = 1 +(-0.1) + \frac{(-0.1)^2}{2}$$ Substituting you get $$2.7 (1- \left( 1 +(-0.1) + \frac{(-0.1)^2}{2} \right))$$ Watch your minus signs.

jbunniii
Homework Helper
Gold Member

## 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

$$I(t) = 2.7(1 - e^{-0.1t})$$? (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,

$$(-0.1)^2 = 0.1^2$$, not $$-0.1^2$$

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 $$0.1^3$$ term as well.

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:
$$I = 2.7 (1- \left( 1 +(-0.1) + \frac{(-0.1)^2}{2} + \frac{(-0.1)^3}{6}\right))$$

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

$$2.7(-0.1) + \frac{2.7(-0.1)^2}{2} + \frac{2.7(-0.1)^3}{6}$$

Pull out the fractions to please the prof:

$$2.7(-0.1) + \frac{1}{2}2.7(-0.1)^2+ \frac{1}{6}2.7(-0.1)^3$$

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

$$I(t) = -0.27t + 0.0135t^2 - 0.00045t^3$$ 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.

jbunniii