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Proving the inequality

$$e^{\sin(\sqrt{2}/{7})}<11/9$$ without calculator

I tried to prove it by taking some terms of Taylor series to find a value which comparison with the $11/9$ but I countered the following problem:

"How can I use the $\sqrt{2}$ without calculator especially with the odd power"

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    $\begingroup$ Oh, boy... $\endgroup$
    – Lucian
    Commented Jan 12, 2015 at 21:58
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    $\begingroup$ What is the source of this problem? Is it some contest problem or were you just playing around with a calculator? $\endgroup$
    – Aryabhata
    Commented Jan 12, 2015 at 22:07
  • $\begingroup$ Put x= 0.02 in the expansion of $\frac{1}{\sqrt{1-x}} $ to get a good approximation for $\sqrt{2}$ $\endgroup$
    – APGreaves
    Commented Jan 12, 2015 at 22:07
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    $\begingroup$ $11/9 - e^{\sin (\sqrt{2}/7)} = 0.00001435084425...$ So if you came up with it yourself, and need a proof, don't get your hopes up. While there might be a nice proof, it is unlikely to be found... $\endgroup$
    – Aryabhata
    Commented Jan 12, 2015 at 22:09

1 Answer 1

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The inequality is equivalent to $$ \sin\frac{\sqrt2}{7}<\log\frac{11}{9}. $$ We use the facts that the taylor series for $\sin x$ and $\log(1+x)$ are alternating and their terms are decreasing in absolute value.

First step: bound $\sin(\sqrt2/7)$ from above. $$ \sin\frac{\sqrt2}{7}<\frac{\sqrt2}{7}-\frac{1}{3!}\Bigl(\frac{\sqrt2}{7}\Bigr)^3+\frac{1}{5!}\Bigl(\frac{\sqrt2}{7}\Bigr)^5=\sqrt2\,\frac{23847}{168070}<\frac{99}{70}\,\frac{23847}{168070}<0.20067 $$

Second step: bound $\log(11/9)$ from below. $$ \log\Bigl(1+\frac29\Bigr)>\sum_{n=1}^8\frac{(-1)^{n+1}}{n}\Bigl(\frac29\Bigr)^n=\frac{302337356}{1506635235}>0.20067 $$ I used Mathematica to do the computations, but they can be done by hand in a reasonable time.

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  • $\begingroup$ how you know the $\sqrt{2}<99/70$?? $\endgroup$
    – E.H.E
    Commented Jan 12, 2015 at 23:29
  • $\begingroup$ @Ehegh Why don't you square it? $\endgroup$
    – user2345215
    Commented Jan 12, 2015 at 23:32
  • $\begingroup$ ok, you are right $\endgroup$
    – E.H.E
    Commented Jan 12, 2015 at 23:33
  • $\begingroup$ $$2=\frac{100}{50}<\frac{100}{49}=\Bigl(\frac{10}{7}\Bigr)^2.$$ This is not good enough to obtain the desired inequality. So I checked $99/70$ and $98/70$. $\endgroup$
    – Julián Aguirre
    Commented Jan 13, 2015 at 10:12

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