Results 1 to 10 of 10

Math Help - e question

  1. #1
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60

    e question

    How are \lim_{x \to \infty}(1+\frac{1}{x})^x and 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}  ... similar or related to each other, other than their results being e?
    Last edited by Jameson; April 26th 2006 at 11:30 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    It should be \lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right  )^x and I don't really know how to answer your question. Is this out of a book, looking for something specific or just you pondering?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by cinder
    How are \lim_{x \to \infty}(1+\frac{1}{x})^x and 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}  ... similar or related to each other, other than their results being e?
    Consider instead:

    <br />
e_1(x)=\lim_{k \to \infty}e_1(x,k)<br />

    where:

    <br />
e_1(x,k)=\left(1+\frac{x}{k}\right)^k<br />

    and:

    <br />
e_2(x)=1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}  ...<br />

    expand e_1(x,k) and look at the behaviour of the coefficients of
    powers of x that occur in both, and consider what happens when
    k\to \infty.

    Maybe?

    RonL
    Last edited by CaptainBlack; April 26th 2006 at 01:01 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by cinder
    How are \lim_{x \to \infty}(1+\frac{1}{x})^x and 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}  ... similar or related to each other, other than their results being e?
    Well, we definie
    \ln x=\int^x_1 \frac{dx}{x} for x>0.
    And define, e such as \ln e=1 (it does exists and well-defined.)
    Then we can prove that,
    e^x=1+x+\frac{x^2}{2!}+...
    Where e^x is the inverse function of \ln x because this function is bijective.
    Thus if, x=1 then,
    e=1+1+\frac{1}{2!}+\frac{1}{3!}+...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    You don't have to define exp(x) as the inverse of log(x) which you defined through an integral. That's certainly possible (and is often done) but is not of any use here since we can just as well start by defining exp(x) (and after that define log(x) as its inverse, though that's not necessary here).

    What the topic starter was asking, I think, is to prove the equivalence between these two possible definitions. Of course, this follows trivially from the equivalence of the analogous definition for exp(x), but proving that equivalence would be a bit 'overkill' to prove this one.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by TD!
    You don't have to define exp(x) as the inverse of log(x) which you defined through an integral. That's certainly possible (and is often done) but is not of any use here since we can just as well start by defining exp(x) (and after that define log(x) as its inverse, though that's not necessary here).

    What the topic starter was asking, I think, is to prove the equivalence between these two possible definitions. Of course, this follows trivially from the equivalence of the analogous definition for exp(x), but proving that equivalence would be a bit 'overkill' to prove this one.
    I never met two mathematicians from Analysis that define some function the same way. Every person has his own way, everything else is biconditional. I just happen to like the the integral defintion, because it demonstrates the importance of the fact that countinous functions are Riemann integrable.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    That's a bit strange, there aren't too many ways on how log(x) and exp(x) are conventionally defined. I know at least 3 books (and thus authors, i.c. professors) who start by defining log(x) as that integral; then defining exp(x) as its inverse (which is why I remarked that it is often done ).
    One of them being "my professor" (the one I had for analysis); on the other hand you have for example Rudin (from 'Principles of Mathematical Analysis') who defines exp(x) first, as the series.

    There are indeed different possibilities, but fundamentally there aren't too many different definitions in use (log through the integral, the series, the limit, the initial value probem y' = y with y(0) = 1).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    No I was implying that about functions not log and exp.
    For example, you can define arctan as an integral, or inverse of tan function. Also you can define it intuitively as a trigometric interprestation. This happens with a lot of functions.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    I see, that's true of course although I still think you'll find that many functions are often defined in a similar way - within the rather limited number of fundamentally different ways of defining them of course.

    Apart from that, I just read the original question carefully. What are you exactly looking for? As said before, both expressions are completely equivalent which means, in this case, that they're the same - being e as you said already. You can prove this, but is that what you're looking for?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60
    I couldn't tell you exactly what I'm looking for. My calculus teacher mentioned it might be part of a bonus question on a test, but wanted us to do figure it out. After looking but not really finding anything, I figured I'd ask here.

    Thanks for your replies!
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum