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  1. #1
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    Real Analysis

    If x>0, show that 1 + x + x^2/2 < e^x < 1 + x + x^2/2 e^x
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  2. #2
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    Quote Originally Posted by dhhnerd View Post
    If x>0, show that 1 + x + x^2/2 < e^x
    I do the easier inequality, hopefully, the second one is similar.

    First we will prove that $\displaystyle 1+x < e^x$ for $\displaystyle x>0$.
    Define $\displaystyle f(x) = e^x - x - 1$.
    Then $\displaystyle f(0) = 0$ but $\displaystyle f'(x) = e^x - 1 > 0$ for $\displaystyle x>0$.
    Therefore, $\displaystyle f$ is increasing for $\displaystyle x>0$.
    But since we have $\displaystyle f(0)=0$ it must mean that $\displaystyle f(x) > 0$ for $\displaystyle x>0$.
    Therefore, $\displaystyle e^x - x - 1 > 0 \implies e^x > 1+x$.

    Second define $\displaystyle g(x) = e^x - \tfrac{1}{2}x^2 - x - 1$.
    Then $\displaystyle g(0)=0$ but $\displaystyle g'(x) = e^x - x - 1>0$ for $\displaystyle x>0$ by above.
    Therefore, $\displaystyle g$ is increasing for $\displaystyle x>0$.
    But since we have $\displaystyle g(0)=0$ it must mean that $\displaystyle g(x)>0$ for $\displaystyle x>0$.
    Therefore, $\displaystyle e^x > \tfrac{1}{2}x^2 + x + 1$
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dhhnerd View Post
    If x>0, show that 1 + x + x^2/2 < e^x < 1 + x + x^2/2 e^x
    What is the second inequality? $\displaystyle e^x<1+x+\frac{x^2}{2}+e^x$?

    Or $\displaystyle e^x<1+x+\frac{x^2}{2}\cdot{e^x}$

    Both are very simple
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