Function Posted November 13, 2013 Share Posted November 13, 2013 (edited) Hello everyone I've seen 2 definitions for [math]e[/math] (one on the internet, one I have proven myself): [math]e=\lim_{n\to\infty}{\left[1+\frac{1}{n}\right]^n}[/math] and [math]e=\lim_{n\to 0}{\left[n+1\right]^{\frac{1}{n}}}[/math] Now, is there a proof that these two are equal? (I'd first like to know if I can change the first expression to something in the form of [math]n\to 0[/math].) Thanks! Function Edited November 13, 2013 by Function Link to comment Share on other sites More sharing options...
imatfaal Posted November 13, 2013 Share Posted November 13, 2013 According to wiki they are provable equivalent http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Alternative_characterizations Although I haven't looked at their proof Link to comment Share on other sites More sharing options...
studiot Posted November 13, 2013 Share Posted November 13, 2013 Maybe I'm nitpicking but surely both these definitions are flawed? e is the sum of an infinite series, which we can prove by Tannery's theorem converges to the first limit. With the second limit what is the first n in your series ? Link to comment Share on other sites More sharing options...
uncool Posted November 13, 2013 Share Posted November 13, 2013 Hello everyone I've seen 2 definitions for [math]e[/math] (one on the internet, one I have proven myself): [math]e=\lim_{n\to\infty}{\left[1+\frac{1}{n}\right]^n}[/math] and Now, is there a proof that these two are equal? (I'd first like to know if I can change the first expression to something in the form of [math]n\to 0[/math].) Thanks! Function You may want to restrict the second one to the right-handed limit, i.e. [math]e=\lim_{n\to 0^+}{\left[n+1\right]^{\frac{1}{n}}}[/math] Then, we can easily see that the two are equivalent by looking at the first limit and letting [math]n = \frac{1}{n'}[/math]. Link to comment Share on other sites More sharing options...
Function Posted November 13, 2013 Author Share Posted November 13, 2013 You may want to restrict the second one to the right-handed limit, i.e. [math]e=\lim_{n\to 0^+}{\left[n+1\right]^{\frac{1}{n}}}[/math] Then, we can easily see that the two are equivalent by looking at the first limit and letting [math]n = \frac{1}{n'}[/math]. Thanks Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now