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I am having trouble understanding the contradiction established in the proof of Lindemann-Weierstrass Theorem. It somehow contradicts $$\vert \int_0^t e^{t-x}f(x) dx\vert\le\int_0^t\vert e^{t-x}f(x)\vert dx\le\vert t\vert e^{\vert t \vert}F(\vert t\vert)$$ Where $f(x)=\sum a_ix^i$, and $F(x)=\sum \vert a_i \vert x^i$. The first part of the inequality statement is pretty obviously true and proved in any intro to real analysis class. How do you derive/prove the second part of the inequality? Also how is it conrtradicted?

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    $\begingroup$ Can you show the proof in question? I would hate to do a Google search and not find the one you're thinking about. $\endgroup$ – Sean Roberson Aug 4 '16 at 23:59
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    $\begingroup$ What's $I$? What's $f$? What's $F$? $\endgroup$ – Erick Wong Aug 5 '16 at 0:03
  • $\begingroup$ Just edited it. Hope that clarifies my question some. $\endgroup$ – Drunk Deriving Aug 5 '16 at 1:19
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The second inequality is the observation that for $\int_{0}^{t}g.h(x) \ dx \leq (t - 0)g(0)H(|t|)$ for positive valued decreasing function $g$ because maximum value that a polynomial $h(x) \in \mathbb{R}[x]$ can take between 0 and $t$ is $H(|t|)$ where $H$ is formed by making all coefficients of $h$ positive.

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