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I have a little problem with the following question:

Given that $e^x$ is approximately equal to $1+x+\frac{x^2}{2}$. Show that $e^x + e^{-x} = 2 + x^2$

I can do this by simply long dividing $$\frac{(1+x+\frac{x^2}{2})^2 +1}{1+x+\frac{x^2}{2}}$$

But that takes some time, especially squaring out the quadratic.

I assume there is a faster way because another question later on asks you to apply whatever method you used to another question where it is not $e^x$ but $e^\frac{1}{79}$

I would appreciate any help.

Thanks

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If $e^x \approx 1 + x + \frac{x^2}{2}$, then $e^{-x} \approx 1 + (-x) + \frac{(-x)^2}{2}$. Now add the two equations.

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    $\begingroup$ I could not see the obvious, thank you $\endgroup$ – user Mar 18 '14 at 23:29
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    $\begingroup$ No problem! :-) $\endgroup$ – wckronholm Mar 18 '14 at 23:31
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If $e^x\approx 1+x+\frac{x^2}{2}$, then $$e^x+e^{-x}\approx (1+x+\frac{x^2}{2})+(1-x+\frac{(-x)^2}{2})=2+x^2. $$

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