Math question (calculation of e)

Hi, folks,
I'm learning calculus off an online course (http://www.math.byu.edu/~smithw/Calculus/), and I have a question.
On page 8 of chapter 4 (http://www.math.byu.edu/~smithw/Calculus/Differential-Chapter4/) theorem D7, it goes into how to calculate e. At one point, it gets to e**(x/e**(x)-1)). Up until that point I'm ok. The next step is, ((e**x)-1)/x =1 as x->0 since (e**x)' as x->0 is (e**0)' = e**0 = 1. But how did they get from one step to the next? e is first being raised to x/((e**(x-1)), and now they write, (e**(x-1))/x? They seem to be using it to prove that (e**(x-1))/x goes to 1 as x->0, but how does that proof work? How do they get from one step to the next?
Anyway, thanks for any help you can give. It may be clearer just to click on the links. I'd cut and paste, but the thing was scanned onto the internet, as opposed to typed.

Thanks again, SuperShuki


**=raised to the power of

I like π.

Ok seriously, there is no "Theorem D7" that I could find. It cuts off at D6 and goes into review exercises and then the "I" series.

Without that reference I'm not sure from your notation.

sorry.

Thanks James
There are page numbers, and on the eighth page (8. in the upper left of the page) It says, "exponential functions, hyperbolic functions". That's where it starts talking about e. At the bottom of the screen, you should see Theorem D7. The second paragraph after that says Theorem: calculation of e. Go past all the exercises and the review questions and the I series and it should be right there.

Thanks again for your time,
SuperShuki

My father was in a Jewish fraternity in college. It was called iata kosher pie.

"iata" being a Greek/Yiddish pun on to steal? Or am I now reading too much into a typo?

As for the proof, I don't think it's supposed to follow from what came before, it follows from the discussion of the derivative above that. Let's see if I can do some old-fashioned ASCII art here .

(Ignore the dots in the first column, the forum software removes leading whitespace, messing up my formatting .)

Replacing x with 0 in the part of the above equation on the right, and then h with x (!), we get

So, as x (used to be h) goes to zero, this converges to the derivative at 0 (since we replaced x with zero), which equals 1. The inverted fraction x / (e^x - 1) must then also go to 1 as x goes to zero, and the rest follows.

The variable substitution is a bit confusing, and it would have been nice if he'd given a warning or a bit more explanation. Anyway, I hope that that clarifies it.

you're reading too much into a typo!

OK, thanks Tovarich Laurens! I guess commies are good for something after all!

Well, you can rest assured that Leonhard Euler died before Karl Marx was born, so the number e is completely free of communist influence and can safely be used by those of the anti-social persuasion .

Sorry dude no idea. I have no idea about this.

Another math question (more of an algebra question, but here goes):
^ = raised to the power of
e^x + e^(-x) = 3

So far, I've got:
e^x = 3 - e^(-x)
ln(e^x) = ln(3 - e^(-x))
x = ln(3 - e^(-x))
now what?

e^x + e^(-x) = 3
e^x + 1 / e^x = 3
(e^x)^2 + 1 = 3e^x
(e^x)^2 - 3e^x + 1 = 0

Note that (e^x)^2 equals e^(2x), but we don't need that here.

Substitute y = e^x to get y^2 - 3y + 1 = 0

D = b^2 - 4ac = 9 - 4 = 5
y = (-b ± sqrt(D)) / 2a
y = (3 ± sqrt(5)) / 2

Substitute back:
e^x = (3 ± sqrt(5)) / 2
x = ln(3/2 ± 1/2*sqrt(5))

Check:
e^x + e^(-x) = 3
e^(ln(3/2 - 1/2*sqrt(5))) + e^(-(ln(3/2 - 1/2*sqrt(5))) = 3
3/2 - 1/2*sqrt(5) + (1 / (3/2 - 1/2*sqrt(5))) = 3

Multiply both sides by (3/2 - 1/2*sqrt(5)):

(3/2 - 1/2*sqrt(5))^2 + 1 = 3 * (3/2 - 1/2*sqrt(5))
(9/4 - 3/2*sqrt(5) + 5/4) + 1 = 9/2 - 3/2*sqrt(5)
18/4 - 3/2*sqrt(5) = 9/2 - 3/2*sqrt(5)

I'll leave the check for x = ln(3/2 + sqrt(5)) as an exercise for the reader .

Thanks, laurens!

Hi,
another question,
they give me a parametric equation
x=e^t+e^(-t), y=e^t-e^(-t) and go from that to x^2-y^2=4 in cartesian coordinates
how did they do that?
thanks
Supershuki
PS
-infinity<t<infinity

There has to be a typo somewhere, because what you posted reduces to x = y, not x^2 - y^2 = 4.

I got the answer from a friend - here it is (sic):


x = e^t+e^(-t)->x^2=(e^t+e^(-t))*( e^t+e^(-t)) ->
x^2= e^2t+1+1+ e^(-2t) ->
x^2= e^2t+ e^(-2t)+2 (rule 1)

y= e^(-t)- e^(-t) ->y^2=( e^t- e^(-t))*( e^t- e^(-t)) ->
y^2= e^2t-1-1+ e^(-2t) ->
y^2= e^2t+e^(-2t)-2 (rule 2)

Using rule 1 and rule 2

x^2- y^2= (e^2t+ e^(-2t)+2) - (e^2t+ e^(-2t)-2) ->
x^2- y^2= e^2t- e^2t+ e^(-2t)- e^(-2t)+2+2 ->
x^2- y^2=4

Thanks anyway, Lourens!

Still doesn't make sense. Perhaps the original was this?

x = e^t + e^(-t)
y = e^t - e^(-t)

then

x^2 = e^2t + e^(-2t) + 2
y^2 = e^2t + e^(-2t) - 2
x^2 - y^2 = 4