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Math Help - tricky question, prove for all positive integers n

  1. #1
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    tricky question, prove for all positive integers n

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

    e^x > 1 + x + x^2/2! + x^3/3!+ ... + x^n/n!


    Hint:
    e^x = 1 + \int _0\,^x\!e^tdt > 1 + \int _0\,^x\!dt = 1+ x


    e^x = 1 + \int _0\,^x\!e^tdt > 1 + \int _0\,^x\!(1+t)dt = 1+ x + x^2/2, and so on.

    Prove by induction on n >= 1, must jutify the hint
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by wopashui View Post
    Prove that for all x > 0 and all positive integers n

    e^x > 1 + x + x^2/2! + x^3/3!+ ... + x^n/n!


    Hint:
    e^x = 1 + \int _0\,^x\!e^tdt > 1 + \int _0\,^x\!dt = 1+ x


    e^x = 1 + \int _0\,^x\!e^tdt > 1 + \int _0\,^x\!(1+t)dt = 1+ x + x^2/2, and so on.

    Prove by induction on n >= 1, must jutify the hint
    Excause me,
    e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \;\ \forall \;\ x \in R
    Its the well-known mac. series for e^x !
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  3. #3
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    Quote Originally Posted by General View Post
    Excause me,
    e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \;\ \forall \;\ x \in R
    Its the well-known mac. series for e^x !
    Exactly! He's simply claiming that e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!} > \sum_{k=0}^{n} \frac{x^k}{k!}...
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  4. #4
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    Quote Originally Posted by Defunkt View Post
    Exactly! He's simply claiming that e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!} > \sum_{k=0}^{n} \frac{x^k}{k!}...

    Sorry, I don't understand how you guys got that from? explain more please!
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