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Use Taylor's Theorem with $n=2$ to prove that the inequality $1+x<e^x$ is valid for all $x\in \mathbb{R}$ except $x=0$.

Taylor's Theorem: $$ f(x)=\sum_{k=0}^n{1\over k!}f^{(k)}(c)(x-c)^k+E_n(x) \\ \text{where}\quad E_n(x)={1\over (n+1)!}f^{n+1}(\xi)(x-c)^{n+1} $$

I'm not sure how to do this. Any solutions or hints are greatly appreciated. I'm not sure if I'm supposed to even use this error term $E_n(x)$ since when $n=2$, $f(x)=e^x$, and $c=0$ we obtain something like $e^x=1+x+{1\over 2}x^2$ and since ${1\over 2}x^2>0$ for all reals except $x=0$ we see that $1+x<e^x$. I'm just confused about how to approach this problem.

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  • $\begingroup$ In most statement of Taylor's Theorem, "$n$" refers to the highest power $(x-c)^n$ used in the approximating Taylor polynomial. With that notation, it is much better to use $n=1$. Using $n=2$ makes for trouble in dealing with negative $x$. $\endgroup$ Commented Aug 28, 2015 at 15:13

2 Answers 2

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I suffices $n=1$ and $c=0$. In this way you get $e^x=1+x+E_1(x)=1+x+\frac{1}{2}e^\xi x^2$. Note that $\frac{1}{2}e^\xi x^2>0$ for all $\xi$ provided $x\neq 0$

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  • $\begingroup$ Is the error term independent of the sum in Taylor's Theorem? $\endgroup$
    – 1233dfv
    Commented Aug 28, 2015 at 15:39
  • $\begingroup$ What do you mean by independent? Sorry, I don't understand $\endgroup$
    – user178826
    Commented Aug 28, 2015 at 15:40
  • $\begingroup$ nevermind, I answered my own question $\endgroup$
    – 1233dfv
    Commented Aug 28, 2015 at 15:43
  • $\begingroup$ If $n=2$ would I get $e^x=1+x+{1\over 2}x^2+{1\over 3!}e^{\xi}x^3$? $\endgroup$
    – 1233dfv
    Commented Aug 28, 2015 at 15:45
  • $\begingroup$ That is correct $\endgroup$
    – user178826
    Commented Aug 28, 2015 at 15:48
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The Taylor's expansion for $e^x$ at $x_0=0$ (or Maclurent sereis) is $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$ So for $x\geq 0$ there is no problem. For negetivs the sign of removed terms is agree with the first removed term,then $\frac{x^2}{2!}+\frac{x^3}{3!}+\dots>0$ so $1+x<e^x$.

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