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Math Help - Deriving the number e

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    Deriving the number e

    Is it possible to derive the number e, starting with the assumption that there exists a base number for an exponential function that is equal to its own derivative?

    I've seen proofs that [e^x]' = e^x, using the definition of e. What I'm looking for is a way to start with the assumption that the above is true, and get to the limit definition of e.

    I searched around a bit, without any luck, so I thought I'd ask here.
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    Re: Deriving the number e

    Quote Originally Posted by gralla55 View Post
    Is it possible to derive the number e, starting with the assumption that there exists a base number for an exponential function that is equal to its own derivative? What I'm looking for is a way to start with the assumption that the above is true, and get to the limit definition of e.
    To be quite honest, I do not follow your question?
    Have you read this thread? Is that close to what you are asking?
    If not, please try to explain exactly what you mean.
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    Re: Deriving the number e

    I've attached a picture which should clarfy. What I mean is if there is a way to start with the assumption that there exisits a number^x which is its own derivative, and solve for that number to get to the last expression in my picture.
    Attached Thumbnails Attached Thumbnails Deriving the number e-e.jpg  
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    MHF Contributor Siron's Avatar
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    Re: Deriving the number e

    So you're looking for a proof that:
    \lim_{x \to 0} (1+x)^{\frac{1}{x}}=e
    across the derivative of e^x=e^x
    ...?
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    Re: Deriving the number e

    Yes.

    I've seen proofs that [e^x]' = e^x, using the definition of e. What I'm looking for is how to do it the other way around. Starting with [e^x]' = e^x, and getting to the definition of e.
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    MHF Contributor chisigma's Avatar
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    Re: Deriving the number e

    Quote Originally Posted by gralla55 View Post
    I've attached a picture which should clarfy. What I mean is if there is a way to start with the assumption that there exisits a number^x which is its own derivative, and solve for that number to get to the last expression in my picture.
    If Your task is to find...

    \lim_{ x \rightarrow 0} (1+x)^{\frac{1}{x}} (1)

    ... then the most confortable way is to use the identity...

    (1+x)^{\frac{1}{x}}= e^{\frac{\ln (1+x)}{x}} (2)

    ... and compute the limit of the exponent in (2) using series expansion or l'Hopital rule...

    Kind regards

    \chi \sigma
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