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Calculate e

  1. May 14, 2005 #1
    Would you please tell me how to calculate natural number e ? How that number came into being ?
  2. jcsd
  3. May 14, 2005 #2
  4. May 14, 2005 #3
    So I just realized that all this doesnt get me to the numerical value of e, just one of the nice properties of it. So if anyone wants the limit derivation for e^x, let this be your guide:

    [tex]\lim_{h\rightarrow 0} \frac{n^{x+h} - n^x}{h} \Rightarrow \frac{n^xn^h - n^x}{h} \Rightarrow \frac{n^x(n^h-1)}{h}[/tex]

    The limit of the product is the product of the limits:

    [tex] \lim_{h\rightarrow 0} \frac{n^h}{h} \lim_{h\rightarrow 0} n^x [/tex]

    L'hospitals simplifies the left to [itex] n^hln(n)[/itex]

    [tex] \lim_{h\rightarrow 0} n^hln(n) \lim_{h\rightarrow 0} n^x = n^0ln(n)n^x[/tex]

    Result: [tex] n^x' = n^x(ln(n))[/tex]

    For some n, n^x' = n^x, and that is only when n = e.
  5. May 14, 2005 #4


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    The definition is with the limit of that sequence,but the calculation is done with the series of [itex] e^{x} [/itex] for x=1.

  6. May 14, 2005 #5
    Or you could use the limit [tex]e = \lim_{x\rightarrow\infty} (1 + \frac{1}{x})^{x}[/tex]

  7. May 14, 2005 #6


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    As to "How that number came into being ?": it has the very nice property that the rate of change of the function ex is just ex itself- no other function has that property. That's why it was recognized as being important. I wouldn't begin to speculate as to how any number "came into being"!
  8. May 14, 2005 #7
    yet another way would be to integrate the hyperbola, and then take the inverse of that function:

    [tex]\int \frac{1}{x} dx = ln(x)[/tex]

    [tex] ln^{-1}(x) = e^x [/tex]

    this integration could be done numerically, and a table of [itex]ln[/itex] values be built (ala Napier and friends) and then a column could be made of [itex]e^x[/itex] by indexing backward into the table.

    tables of logs were actually very popular at one time (and led to the slide rule) because multiplication and division could be reduced to addition and subtraction (think algebra of exponents). this greatly benefited fields like artillery and naval navigation since complicated calculations could be done quickly by indexing into a precomputed table (same for trig functions).
    Last edited: May 14, 2005
  9. May 14, 2005 #8


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    of course all functions of form ce^x have that property as well, and these are the only ones.
  10. May 14, 2005 #9
    The way I was introduced to e was to try to find the exponentional which was its own derivative. So say we have a^x, we know that a^0 is 1, so try to find the value of a such that the gradient of a^x is 1 at the point (0,1). From this you get the definition of a, or as we know it e: Lim (1+(1/n))^n as n tends to infinity. I think this is how it was approached, I think Eli Maor has a book on the history of e if you're interested, it's on Amazon. Cheers, Joe
  11. May 14, 2005 #10

    James R

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    You could always use:

    e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...

    which is from the Taylor series for [itex]e^x[/itex].
  12. May 15, 2005 #11
    James R: You could always use:

    e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...

    which is from the Taylor series for e^x.

    This is a really effective way, that is if the student seeks a good approximation. As my professor once said, "e gets to its limit very fast."

    After n terms, starting with 0, the error is less than 1/(n!) for n =1 or greater. This can be shown. Consider:

    1/n! >1/(n+1)! + 1/(n+2)! +1/{(n+3)! ++++

    Since 1>1/(n+1) +1/{(n+1)(n+2)} + 1/{(n+1)(n+2)(n+3)} ++++

    Since for n=1, we have 1>=1/2 + 1/(2*3) + 1/(3*4) +1/(4*5) +1/(5*6) ++++++ =(number of terms)/(number of terms +1)> the series above.

    So that even at n=1, we can conclude 2<e<3. At n=3 we find that 2+2/3<e<2+5/6. At n=7, we have 2+3620/5041<e<2+3621/5041.
    Last edited: May 15, 2005
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