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If $x > 0$ and $t \leq x $

Prove:

$$ e^{-t}-\left(1-\dfrac{t}{x}\right)^x \geq 0 \>. $$

  • 0
    Do you know the Taylor expansion for $\log(1-z)$?2012-02-07
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    @Thomas Andrews ,I know you will use the method in complex variable, but it is a problem in a calculus book.2012-02-07
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    @gingerjin Then use the binomial theorem and the series expansion for $\exp(t)$ to how the inequality.2012-02-07
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    Hint: first prove $e^{-a} \ge 1-a$.2012-02-07
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    You can start by: $${\left( {1 - \frac{t}{x}} \right)^x} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{x \choose k}\frac{{{t^k}}}{{{x^k}}}} $$ $${\left( {1 - \frac{t}{x}} \right)^x} = 1 - t + \frac{{x\left( {x - 1} \right)}}{{2!}}\frac{{{t^2}}}{{{x^2}}} - \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{{3!}}\frac{{{t^3}}}{{{x^3}}} + \cdots $$ You should focus on the coefficients $${x \choose k}\frac{1}{x^k}$$ to show you inequality adn the fact that $t \leq x \Rightarrow \displaystyle \frac{t}{x} \leq 1$2012-02-07
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    @Peter It's not clear how you reconcile the alternating sign part of the the infinite series. Yes $|{x \choose k}\frac{1}{x^k}|<\frac{1}{k!}$, but putting them in an alternating sum is much harder, isn't it? Especially since $x\choose k$ can be negative?2012-02-07
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    Another approach is to use that $lim_{z\to\infty} (1-\frac{1}{z})^z = e^{-1}$.2012-02-07

1 Answers 1

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This is equivalent to $\mathrm e^{-t}\geqslant\left(1-t/x\right)^x$, which is equivalent to $\mathrm e^{-u}\geqslant1-u$ for $u=t/x$.

Now, $u\mapsto1-u$ is the tangent to the graph of the function $u\mapsto\mathrm e^{-u}$ at $u=0$. This function is convex hence its graph lies above any of its tangents and you are done.

Or, consider the function $b:u\mapsto\mathrm e^{-u}-1+u$ and note that $b(0)=0$ and $b'(u)=1-\mathrm e^{-u}$. Hence the function $b$ is decreasing on $u\leqslant0$ and increasing on $u\geqslant0$. This proves that $b\geqslant0$ everywhere.