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Math Help - Extremely difficult induction proof problem

  1. #1
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    Extremely difficult induction proof problem

    I dunno who thought up this one, but it is clearly for masochists.

    Prove that for all x > 0 and all positive integers n

    e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^n}{n!}

    n!=n(n-1)(n-2)...3\cdot 2\cdot 1

    This MUST be proven by induction on n\geq 1.

    In addition, these hints were provided for the question. They must be justified.

    e^x=1+\int_0^x e^t dt > 1+\int_0^x dt = 1+x

    e^x=1+\int_0^x e^t dt > 1+\int_0^x (1+t) dt = 1+x+\frac {x^2}{2}

    As I said, whoever came up with this one is a sadist.

    I believe I have correct justifications on the two hints (unless I wrote something wrong), but I can't seem to provide the proof for a n+1 case (for proving by induction). Still, if anyone can help explain this one in detail, I'd be very appreciative.
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  2. #2
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    Quote Originally Posted by Runty View Post
    I dunno who thought up this one, but it is clearly for masochists.

    Prove that for all x > 0 and all positive integers n

    e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^n}{n!}

    n!=n(n-1)(n-2)...3\cdot 2\cdot 1

    This MUST be proven by induction on n\geq 1.

    In addition, these hints were provided for the question. They must be justified.

    e^x=1+\int_0^x e^t dt > 1+\int_0^x dt = 1+x

    e^x=1+\int_0^x e^t dt > 1+\int_0^x (1+t) dt = 1+x+\frac {x^2}{2}

    As I said, whoever came up with this one is a sadist.

    I believe I have correct justifications on the two hints (unless I wrote something wrong), but I can't seem to provide the proof for a n+1 case (for proving by induction). Still, if anyone can help explain this one in detail, I'd be very appreciative.

    Have you already studied Taylor series? If so then you must know that

    e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}\geq 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Have you already studied Taylor series? If so then you must know that

    e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}\geq 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}

    Tonio
    Never heard of Taylor series. And I don't think it would be a valid answer in this circumstance.
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  4. #4
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    Clearly it's true for n = 1. Assume true for n = k so

    <br />
e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0 since x > 0.

    Then

     <br />
\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt<br />

    so

     <br />
\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x<br />

    so

     <br />
e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}<br />

    so

     <br />
e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}<br />

    proving true for n = k + 1. Thus, it's true for all n.
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    Quote Originally Posted by Danny View Post
    Clearly it's true for n = 1. Assume true for n = k so

    <br />
e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0 since x > 0.

    Then

     <br />
\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt<br />


    This step requires, I think, full justification. It isn't true in general, of course, but it is in this case because of continuity.

    Tonio


    so

     <br />
\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x<br />

    so

     <br />
e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}<br />

    so

     <br />
e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}<br />

    proving true for n = k + 1. Thus, it's true for all n.
    .
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  6. #6
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    Quote Originally Posted by tonio View Post
    .
    And your point is . . . ?
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  7. #7
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    OK guys
    be relax
    The life is still countinuing


    and sorry for my bad english
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  8. #8
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    Quote Originally Posted by Danny View Post
    And your point is . . . ?

    Exactly what I wrote: the OP will do wisely to justify that step.

    Tonio
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  9. #9
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    Quote Originally Posted by Danny View Post
    Clearly it's true for n = 1. Assume true for n = k so

    <br />
e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0 since x > 0.

    Then

     <br />
\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt<br />

    so

     <br />
\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x<br />

    so

     <br />
e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}<br />

    so

     <br />
e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}<br />

    proving true for n = k + 1. Thus, it's true for all n.
    This will have to do. Thanks for the help.
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