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Math Help - Prove

  1. #1
    Member
    Joined
    Dec 2008
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    152

    Prove

    Hi,

    I'm stuck on the following problem:

    Let  h \  \epsilon \ [0,1]
    Show that for every n \ \epsilon \ N

    (1 + h)^n \leq 1 + (2^n -1)h

    is true..


    I tried with math induction:

    (1) \ \ \ n = 1
    1 + h \leq 1 + (2 - 1)h
     1 + h \leq 1 + h   \ \ \  \surd

    (2) \ \ \ n \Rightarrow n + 1

    (1 + h)^{h+1} \leq 1 + (2^{n+1} - 1)h

    (1+h) (1+h)^n \leq 1 + (2 \cdot 2^n - 1)h

    (1+h) (1+h)^n \leq 1 + 2h \cdot 2^n - h

    (1+h) (1+h)^n \leq 1 - h + 2(h \cdot 2^{n-1})

    now i'm lost. don't know how to continue.

    By the way, how do I write the "is element of" symbol? did i use the correct one above?
    and is there any way i could skip the [tex][/ math] in every line?
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  2. #2
    Junior Member utopiaNow's Avatar
    Joined
    Mar 2009
    Posts
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    Thanks
    1
    Quote Originally Posted by metlx View Post
    Hi,

    I'm stuck on the following problem:

    Let  h \  \epsilon \ [0,1]
    Show that for every n \ \epsilon \ N

    (1 + h)^n \leq 1 + (2^n -1)h

    is true..


    I tried with math induction:

    (1) \ \ \ n = 1
    1 + h \leq 1 + (2 - 1)h
     1 + h \leq 1 + h   \ \ \  \surd

    (2) \ \ \ n \Rightarrow n + 1

    (1 + h)^{h+1} \leq 1 + (2^{n+1} - 1)h

    (1+h) (1+h)^n \leq 1 + (2 \cdot 2^n - 1)h

    (1+h) (1+h)^n \leq 1 + 2h \cdot 2^n - h

    (1+h) (1+h)^n \leq 1 - h + 2(h \cdot 2^{n-1})

    now i'm lost. don't know how to continue.

    By the way, how do I write the "is element of" symbol? did i use the correct one above?
    and is there any way i could skip the [tex][/ math] in every line?
    So in the inductive step, we assume (1 + h)^k \leq 1 + (2^k -1)h for some k.

    Then (1 + h)^{k+1} = (1 + h)^k (1+h) \leq (1 + (2^k -1)h)(1+h) and  (1 + (2^k -1)h)(1+h) =  1 + 2^kh + (2^k - 1)h^2

    Now 1 + 2^kh + (2^k - 1)h^2 \leq 1 + 2^kh + (2^k - 1)h = 1 + 2^kh + 2^kh - h = 1 + (2^{k+1 } - 1)h. Where this was true  (2^k - 1)h^2 \leq (2^k - 1)h since h \in [0,1]

    Therefore (1+h)^{k+1} \leq 1 + (2^{k+1 } - 1)h, as required.

    Oh and also, the element of symbol in latex is simply " \in ". So I used that above, you can click and see. And I don't know a shortcut to get around the constant math tags, sorry.
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