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I was asked by my teacher to prove that $7<e^2<8$ using only algebraic methods and knowing that $2<e<3$.

I don't know how to do this, where to start from, but I guess that I would need some kind of a function where: $f'(\xi )=\frac{f(b)-f(a)}{b-a}$ so as to exploit monotonicity of $f'$. Any hint?

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    $\begingroup$ The first sentence doesn't really give much leeway and asks for something impossible. $\endgroup$ Commented Feb 4, 2016 at 6:11
  • $\begingroup$ I think you should be more specific as of which methods are deemed "algebraic": there are several trascendent numbers in the interval $(2\sqrt{2},3)$, and I don't really see how to "algebraically" prove that they are not $e$. $\endgroup$
    – user228113
    Commented Feb 4, 2016 at 6:19
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    $\begingroup$ can we use $e^x = 1 + x + \frac{x^2}{2!} ..$ ? $\endgroup$
    – Max Payne
    Commented Feb 4, 2016 at 6:19
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    $\begingroup$ How do you define e $\endgroup$ Commented Feb 4, 2016 at 6:20
  • $\begingroup$ @Samuel By algebraic method, I mean a mathematic proof not based on calculation of $e^2$ using a computer or calculator. $\endgroup$
    – user171110
    Commented Feb 4, 2016 at 7:31

2 Answers 2

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Use the (convergent) series $$e^x=1+ x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$$

For $x=2$, all the terms are positive and to get the lower bound take the partial sum using first 5 terms, $1+ 2+ \frac42 + \frac86 +\frac{16}{24}=7$. So $e^2>7$.

Now look at the omitted (infinite) tail, starting from $\frac{32}{120}$. It is term-wise bounded above by the geometric series with ratio $\frac12$, and first term $\frac{32}{120}$.

Using the formula $\frac{a}{1-r}$ for the sum of an infinite geometric series with first term $a$ and common ratio $r$ we get the tail sum to be bounded above by $\frac{32}{120}\big/\frac12= \frac{64}{120}<1$.

So to the full series for $e^2$, split as the sum of head and tail is $e^2< 7+1$.

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It's not possible.

Since you aren't allowed to use any special property of e, for all practical purposes you can replace it with the variable x in the inequality and treat the inequality like 2

This means that x=2.1 is a valid case. But $x^2=4.41$ which is less than 7.

So, given only this property about e, we cannot prove 7

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  • $\begingroup$ I didn't say that you should use only the inequality but any property of $e^x$ can be used. $\endgroup$
    – user171110
    Commented Feb 4, 2016 at 7:29

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