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Math Help - Prove

  1. #1
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    Prove

    Prove that e^x \geqslant x+1

    using the taylor series for all x > 0  i can use the taylor series for e^x its obvious that the inequality is true for positive values.

    then using the fact that e^x > 0 works for all x less than -1.

    but i am jut a bit stuck on proving the inequality for values of x -1 \leqslant x \leqslant 0
    Last edited by bobak; January 1st 2008 at 10:35 AM.
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  2. #2
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    Do not use Taylor series. Instead define f(x) = e^x - x -1 and show that f'(x) > 0 on (0,\infty) now argue that it is posititive.
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  3. #3
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    thank you TPH I see how that argument is better, but it does not prove the inequality for -1 \leqslant x \leqslant 0

    or did i missed something ?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobak View Post
    thank you TPH I see how that argument is better, but it does not prove the inequality for -1 \leqslant x \leqslant 0

    or did i missed something ?
    TPH's method will work for all x. just show that the derivative is positive for x < 0 as well (try it, it's easier than you think)

    note: we have equality when x = 0
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    TPH's method will work for all x. just show that the derivative is positive for x < 0 as well (try it, it's easier than you think)
    It is not quite that easy. The derivative is of e^x - x -1 is negative if x<0.
    However, that function has an absolute minimum at x=0.
    Hence that function is always non-negative.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    It is not quite that easy. The derivative is of e^x - x -1 is negative if x<0.
    However, that function has an absolute minimum at x=0.
    Hence that function is always non-negative.
    indeed. i had graphed the function and everything and noticed it was non-negative. mixed that up with the derivative being positive somehow.
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  7. #7
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    Its all very clear now, thanks a lot.
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  8. #8
    Super Member PaulRS's Avatar
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    Read G. Pólya's proof of the Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
    it is interesting and involves the inequality e^x\geq{1+x}
    Last edited by PaulRS; January 2nd 2008 at 05:46 AM.
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