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Math Help - proof by induction

  1. #1
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    induction, how do i begin?

    prove by induction that for every natural n and every x>0
    (e^x) > 1 + x/1! + (x^2)/2! + ... + (x^n)/n!

    it says we can assume that we know basic properties of derivatives and integrals, but i just am not sure where to begin.
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  2. #2
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    Quote Originally Posted by cen0te
    prove by induction that for every natural n and every x>0
    (e^x) > 1 + x/1! + (x^2)/2! + ... + (x^n)/n!
    In duction follows a pattern:

    1. First we show that what we wish to prove for general
    n actualy holds for the first n which is realevant to the
    problem. In this case it's n=0, and we need to show that
    for every x>0 that:

    e^x\ >\ 1,
    which I presume you know is true.

    2. Second we show that if what we have to prove is true
    when n=k, then it is also true for n=k+1. So assume it
    true for n=y, then we have

     e^x\ >\ 1 + x/1! + x^2/2! + ... + x^k/k!<br />
,

    for all x\ >\ 0.

    Now consider:

    \int_0^y\ e^x\ dx\ =\ [e^x]_0^y\ =\ e^y-1\ \ \ \ \ ...[1]

    also:

    \int_0^y\ 1 + x/1! + x^2/2! + ... + x^k/k!\ dx\ =
    \ [x + x^2/2! + x^3/3! + ... + x^{(k+1)}/(k+1)!]_0^y\ =

    y + y^2/2! + y^3/3! + ... + y^{(k+1)}/(k+1)!\ \ \ \ \ ...[2]

    Now the integrand in [1] is strictly greater than the integrand in
    [2] at every point over which the integral is taken. So
    \int_0^y\ e^x\ dx\ >\ \int_0^y\ 1 + x/1! + x^2/2! + ... + x^k/k!\ dx
    which can be rewritten from what we have found above as:
    <br />
e^y-1\ >\ y + y^2/2! + y^3/3! + ... + y^{(k+1)}/(k+1)!<br />
    So rearranging and replacing y by x we have:
    <br />
e^x\ >\ 1\ +\ x + x^2/2! + x^3/3! + ... + x^{(k+1)}/(k+1)!<br />

    3. That is we have proven that if what we have to prove for all n \geq\ 0,
    is true for n=k, then it is true for n=k+1, also we have proven that
    it is true for n=0, so by the principle of induction we have proven
    it true for all n \geq\ 0

    RonL
    Last edited by CaptainBlack; November 20th 2005 at 09:33 PM.
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  3. #3
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    Ok, I think I understand your proof. I understand induction, it is calculus that I am horribly rusty on. I have not taken any in over 5 years, and I do not even remember the notation.

    Many thanks for your help. As an aside, could the statement be reworked to support x<0?
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  4. #4
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    You can also use Taylor's Theorem, but that too requires Calculus.

    P(x)=\sum_{n=0}^{\infty}\frac{f^n(a)(x-a)!}{n!}
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  5. #5
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    Quote Originally Posted by cen0te
    Ok, I think I understand your proof. I understand induction, it is calculus that I am horribly rusty on. I have not taken any in over 5 years, and I do not even remember the notation.

    Many thanks for your help. As an aside, could the statement be reworked to support x<0?
    It's not true for x<0, since the series has alternating signs and
    so is alternatly greater and less than e^x as n
    increases.

    RonL
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  6. #6
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    Quote Originally Posted by cen0te
    Ok, I think I understand your proof. I understand induction, it is calculus that I am horribly rusty on. I have not taken any in over 5 years, and I do not even remember the notation.
    The problem is that the instructions given with the problem
    indicate that you are expected to use calculus in the proof.

    In fact I was a bit worried about this problem when I first
    saw it because induction is not the natural way to prove it.

    If you know the series expansion for e^x the
    result is hovering around the obvious pile.

    RonL
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  7. #7
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    There is a second part to the homework problem which says

    "State and prove the corresponding theorem for x<0."

    So I have tried, but I am not sure what is being sought. I assume the the statement needs to be reworked, but again...I am at a loss.
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  8. #8
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    Quote Originally Posted by cen0te
    There is a second part to the homework problem which says

    "State and prove the corresponding theorem for x<0."

    So I have tried, but I am not sure what is being sought. I assume the the statement needs to be reworked, but again...I am at a loss.
    I don't know if this is what they are looking for but: suppose x<0,
    and let y=-x, then y>0, so for all n\ \epsilon\ \mathbb{N} (the set of natural
    numbers 0,1, .. or 1,2, .. depending on your preference for how
    they are defined).

    <br />
e^y\ >\ 1 + y/1! + y^2/2! + ... + y^n/n!<br />
    now substitute -x for y to get:
    <br />
e^{-x}\ >\ 1 - x/1! + x^2/2! - ... + (-1)^n\ x^n/n!<br />

    or:

    <br />
e^x\ <\ \frac{1}{1 - x/1! + x^2/2! - ... + (-1)^n\ x^n/n!}<br />


    RonL
    Last edited by CaptainBlack; November 21st 2005 at 10:54 PM.
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  9. #9
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    Hmm, I actually thought of that, but for some reason I thought that was basically just cheating. I had assumed that it would have something to do with how e^x with x<0 would just keep approaching 0. However, I can't come up with anything that satisfies me, so I will go with your more experienced suggestion. Thanks again.
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  10. #10
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    Quote Originally Posted by cen0te
    There is a second part to the homework problem which says

    "State and prove the corresponding theorem for x<0."

    So I have tried, but I am not sure what is being sought. I assume the the statement needs to be reworked, but again...I am at a loss.
    For all x < 0 and all positive integers n,
    <br />
(-1)^n e^x\  <\   (-1)^n(1 + x/1! + x^2/2! + ... + x^k/k!)<br />
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