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Someone help me understand how e is irrational proof

  1. Nov 25, 2007 #1


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    Well there is the proof i am reading and trying to understand...

    can someone tell me how they knew that [itex]0<R_n<\frac{3}{(n+1)!}[/itex]
  2. jcsd
  3. Nov 25, 2007 #2


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    Taylor's theorem.
  4. Nov 25, 2007 #3


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    Well I didn't really learn Taylor's theorem in much depth..so I don't really understand how they got the remainder to be less than or equal to that
  5. Nov 26, 2007 #4

    Gib Z

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    Taylor's theorem states that any function that is n-times differentiable by a Taylor series, plus a remainder term, which with some functions goes to zero as the number of terms of the series increase. We can find a taylor series around a certain value of x, but 0 is the easiest and what is required here.

    So, basically the theorem states an n-times differentiable function can be expressed as such:

    [tex]f(x) = f(0) + f'(0)x + f''(0) \frac{x^2}{2} + f'''(0) \frac{x^3}{6} ..... + R_n = \sum_{k=0}^n f^k (0) \frac{x^k}{k!} + R_n[/tex]

    There are several, and actually equivalent ways of expressing the remainder term (for those with watchful eyes, will regard it as a form of the mean value theorem) but for this one, it looks like you'll need the Cauchy form, search "Cauchy Remainder term" in google.

    Basically that form states that [tex]R_n = \frac{(x - t)^n}{n!} x f^{n+1} (t) [/tex] where t is some number in the closed interval [0, x]. Note this is for around the point zero again.

    Hopefully you can construct an inequality with the Remainder term?

    EDIT: O I do forget to state, in this notation f^k denotes the k-th derivative of f, not an exponent. f^0 should be interpreted as f.
  6. Nov 26, 2007 #5


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    Hi rockfreak. Remember that in this problem all we're trying to do is to show that b*R_n is between zero and one (for a given positive integer b and for suitably large n) and therefore that b*R_n cannot be an integer. The exact bound we find is not all that important so long as we can make b*R_n less than one. What I'm saying is that the bound doesn't necessarily have to be lowest one we can find and it doesn't even have to be a good approximation to the actual series remainder, it just has to be larger than the series remainder while still being able to make b*R_n less than one.

    Sometimes just making a simple comparison with the series under question and another known series is a very easy way to get a bound. In this case R_n is :

    [tex]R_n = 1/(n+1)! + 1/(n+2)! + 1/(n+3)! + ...[/tex]

    [tex]R_n = \frac{1}{(n+1)!}\, \{\, 1 + \frac{1}{(n+2)} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + \, \ldots \}[/tex]

    [tex] R_n < \frac{1}{(n+1)!} \, \{\, 1 + 1/2 + 1/4 +1/8 + \, \ldots\}[/tex]


    [tex]R_n < \frac{2}{(n+1)!}[/tex]

    Can you see how I put put the expression in curley brackets {} in comparision with a simple geometric series to find an appropriate bound.
    Last edited: Nov 26, 2007
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