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E in compound interest

  1. Dec 29, 2015 #1
    Originally (as far as I know) the number e (ie. 2.71828...) came up in compound interest calculations.

    For example, if you have 1 dollar, and a compound interest of 100% per year, and the interest is continuously calculated, after one year you'll have exactly e dollars.

    The generic formula for this is (1+1/n)n, where n is the amount of times compound interest is calculated during the year. As n approaches infinity (which means it's continuously calculated), that formula approaches e.

    Is the fact that ex is its own derivative just a coincidence, or is there a correlation with the above?
     
  2. jcsd
  3. Dec 29, 2015 #2

    BvU

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    The way we were first acquainted with the existence of ##e## was: We are looking for the derivative of ##f(x) = a^x##. From visual inspection there must be a real number ##a## such that this derivative is equal to 1 in ##(0,1)##.
     
  4. Dec 29, 2015 #3
    That's not what I heard. But either way, it's not really what I asked...
     
  5. Dec 29, 2015 #4

    BvU

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    You are right. I got carried away by memories ...:rolleyes:

    You could investigate the Taylor series of your expression and see that it yields something looking suspiciously like ##f'= f##

    [edit] that doesn't make much sense. Dug up a dog-eared yellowish textbook, copyright 1969 :smile:, on infinitesimals and found: we call this number ##e##.
    So the derivative definition gives us ##\displaystyle f'(0) = \lim_{h\rightarrow 0}{e^h-1\over h} = 1## hence ##\displaystyle f'(x) = \lim_{h\rightarrow 0}{e^{x+h}-e^x\over h} = e^x \lim_{h\rightarrow 0}{e^h-1\over h}= e^x##.
    Not interesting for you, but: a few pages further, on limits:
    ##f(x)=\log(1+x)\Rightarrow f'(x) = (1+x)^{-1} \Rightarrow \displaystyle f'(0) = \lim_{x\rightarrow 0}{\log(x+1)\over x} = 1## and now we do have an answer to your question !
     
    Last edited: Dec 29, 2015
  6. Dec 29, 2015 #5

    SteamKing

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    Although e is now called Euler's number, it was actually discovered by Jakob Bernoulli (among other mathematicians) before Euler was born when he, Bernoulli, was investigating compound interest:

    https://en.wikipedia.org/wiki/E_(mathematical_constant)

    Euler's singular contribution was his formula which relates e, i, π, 0, and 1.
     
  7. Dec 29, 2015 #6

    Ssnow

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    Considering the derivative of ## f(x)=e^{x}##:

    ##f'(x)= \lim_{h\rightarrow 0} \frac{e^{x+h}-e^{x}}{h}=\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}\cdot e^{x},##

    you can see that the derivative of ##e^{x}## differs from ##e^{x}## by the factor ##\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}##. If we write:

    ## \lim_{h\rightarrow 0}\frac{\lim_{n\rightarrow +\infty}\left(1+\frac{1}{n}\right)^{nh}-1}{h},##

    we can examine the sequence ##\frac{\left(1+\frac{1}{n}\right)^{nh}-1}{h}##. If ##h=1## then for ##n\rightarrow \infty## we have ##e-1## as limit, because it is a variation of the neper sequence you cited before. The fact that ##h\rightarrow 0## ''deforms'' the result of this sequence that under the limit for ##h\rightarrow 0## tends to ##1##. This permit to have the equality between ##f'## and ##f## ... this is the way how the growth of neper sequence is correlated with the derivative of the function ##e^{x}##.
     
  8. Dec 29, 2015 #7
    There is a correlation - this is a direct consequence of the fact that e is its own integral.
     
  9. Dec 29, 2015 #8
    Could you explain in vernacular how?
     
  10. Dec 29, 2015 #9

    BvU

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    Did you read #4 or #6 ?
     
  11. Dec 31, 2015 #10

    Mark44

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    Or, rather, that ##\int e^x dx = e^x## plus an arbitrary constant.
     
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