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Math question (calculation of e)

Posted by: SuperShuki - Wed Jun 12, 2013 6:05 pm
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Math question (calculation of e) 
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Post Math question (calculation of e)   Posted on: Wed Jun 12, 2013 6:05 pm
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

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Post Re: Math question (calculation of e)   Posted on: Wed Jun 12, 2013 9:43 pm
I like π. :lol:

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.


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Post Re: Math question (calculation of e)   Posted on: Wed Jun 12, 2013 11:05 pm
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

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Last edited by SuperShuki on Wed Jun 12, 2013 11:11 pm, edited 1 time in total.



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Post Re: Math question (calculation of e)   Posted on: Wed Jun 12, 2013 11:10 pm
JamesG wrote:
I like π. :lol:



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

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Post Re: Math question (calculation of e)   Posted on: Thu Jun 13, 2013 12:00 pm
"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 ;-).
Code:
.    x
.  de         (  (x+h)    x )
. ---- = lim  ( e      - e  )
.  dx    h->0 (-------------)
.             (      h      )

(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
Code:
.   (0+x)    0      x
.  e      - e      e - 1
. ------------- = -------
.      x             x

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.

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Post Re: Math question (calculation of e)   Posted on: Thu Jun 13, 2013 3:04 pm
you're reading too much into a typo!

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

:D

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Post Re: Math question (calculation of e)   Posted on: Thu Jun 13, 2013 7:13 pm
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 ;-).

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Post Re: Math question (calculation of e)   Posted on: Thu Jul 25, 2013 12:11 pm
Sorry dude no idea. I have no idea about this.


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Post Another e math question   Posted on: Sat Sep 21, 2013 6:33 pm
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?

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Post Re: Math question (calculation of e)   Posted on: Sun Sep 22, 2013 1:50 pm
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 :).

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Last edited by Lourens on Sun Sep 22, 2013 10:14 pm, edited 1 time in total.



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Post Re: Math question (calculation of e)   Posted on: Sun Sep 22, 2013 8:07 pm
Thanks, laurens!

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Post More help - parametric equation   Posted on: Mon Nov 18, 2013 10:05 pm
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

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Post Re: Math question (calculation of e)   Posted on: Sun Dec 01, 2013 2:45 pm
There has to be a typo somewhere, because what you posted reduces to x = y, not x^2 - y^2 = 4.

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Post Re: Math question (calculation of e)   Posted on: Sun Dec 01, 2013 9:47 pm
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!

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Post Re: Math question (calculation of e)   Posted on: Sat Dec 07, 2013 12:56 pm
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

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