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Math question (calculation of e)
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Space Station Commander ![]() ![]()
Joined: Mon May 31, 2004 9:47 pm
Posts: 827 Location: Yerushalayim (Jerusalem) - capital of Israel! ![]() |
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 _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Moon Mission Member ![]() ![]()
Joined: Mon Dec 18, 2006 11:15 pm
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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. |
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Space Station Commander ![]() ![]()
Joined: Mon May 31, 2004 9:47 pm
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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 _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous Last edited by SuperShuki on Wed Jun 12, 2013 11:11 pm, edited 1 time in total. |
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Space Station Commander ![]() ![]()
Joined: Mon May 31, 2004 9:47 pm
Posts: 827 Location: Yerushalayim (Jerusalem) - capital of Israel! ![]() |
JamesG wrote: I like π. ![]() My father was in a Jewish fraternity in college. It was called iata kosher pie. _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Space Station Commander ![]() ![]()
Joined: Thu Oct 27, 2005 7:44 am
Posts: 707 Location: Haarlem, The Netherlands ![]() |
"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. _________________ Say, can you feel the thunder in the air? Just like the moment ’fore it hits – then it’s everywhere What is this spell we’re under, do you care? The might to rise above it is now within your sphere Machinae Supremacy – Sid Icarus |
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Space Station Commander ![]() ![]()
Joined: Mon May 31, 2004 9:47 pm
Posts: 827 Location: Yerushalayim (Jerusalem) - capital of Israel! ![]() |
you're reading too much into a typo!
OK, thanks Tovarich Laurens! I guess commies are good for something after all! ![]() _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Space Station Commander ![]() ![]()
Joined: Thu Oct 27, 2005 7:44 am
Posts: 707 Location: Haarlem, The Netherlands ![]() |
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
![]() _________________ Say, can you feel the thunder in the air? Just like the moment ’fore it hits – then it’s everywhere What is this spell we’re under, do you care? The might to rise above it is now within your sphere Machinae Supremacy – Sid Icarus |
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Joined: Thu Jul 25, 2013 10:53 am
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Sorry dude no idea. I have no idea about this.
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Joined: Mon May 31, 2004 9:47 pm
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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? _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Space Station Commander ![]() ![]()
Joined: Thu Oct 27, 2005 7:44 am
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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 ![]() _________________ Say, can you feel the thunder in the air? Just like the moment ’fore it hits – then it’s everywhere What is this spell we’re under, do you care? The might to rise above it is now within your sphere Machinae Supremacy – Sid Icarus Last edited by Lourens on Sun Sep 22, 2013 10:14 pm, edited 1 time in total. |
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Thanks, laurens!
_________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Joined: Mon May 31, 2004 9:47 pm
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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 _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Space Station Commander ![]() ![]()
Joined: Thu Oct 27, 2005 7:44 am
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There has to be a typo somewhere, because what you posted reduces to x = y, not x^2 - y^2 = 4.
_________________ Say, can you feel the thunder in the air? Just like the moment ’fore it hits – then it’s everywhere What is this spell we’re under, do you care? The might to rise above it is now within your sphere Machinae Supremacy – Sid Icarus |
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Space Station Commander ![]() ![]()
Joined: Mon May 31, 2004 9:47 pm
Posts: 827 Location: Yerushalayim (Jerusalem) - capital of Israel! ![]() |
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! _________________ “Once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return.” -Anonymous |
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Space Station Commander ![]() ![]()
Joined: Thu Oct 27, 2005 7:44 am
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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 _________________ Say, can you feel the thunder in the air? Just like the moment ’fore it hits – then it’s everywhere What is this spell we’re under, do you care? The might to rise above it is now within your sphere Machinae Supremacy – Sid Icarus |
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