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$x$ is the real root of the equation $$3x^3-5x+8=0,\tag 1$$ prove that $$e^x>\frac{40}{237}.$$ I find this inequality in a very accidental way,I think it's very difficult,because the actual value of $e^x$ is $0.1687763721…$,and the value of $\frac{40}{237}$ is $0.1687763713…$

I have no idea on this problem,I just know that the equation $(1)$ has only one real root.Thank you!

$(Edit)$A similar problem: $x$ is the real root of the equation $$x^3-3x+3=0,\tag 2$$ prove that $$10^x>\frac{1}{127}.$$

lsr314
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  • Wolfram says that x≈ -1.77918 – EricAm May 07 '13 at 11:41
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    @Arjang $x\in (-2,-1)$. – Ma Ming May 07 '13 at 11:41
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    @Arjang,maybe you inputed $x^3-5x+8=0$. – lsr314 May 07 '13 at 11:42
  • @Hecke : yep, that is exactly what I did – jimjim May 07 '13 at 11:44
  • where the heck does this question come from? – Lost1 May 07 '13 at 12:48
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    I don't understand. Evidently, you have proved the inequality, by computing both sides to sufficient accuracy to tell them apart. So what is the problem? – Gerry Myerson May 07 '13 at 13:11
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    @Gerry Myerson I know it can be proved by computing,but I want to find a better way to prove it,like this problem http://math.stackexchange.com/questions/380302/prove-dfrac25-frac25-ln2?rq=1 – lsr314 May 07 '13 at 13:26
  • In what way are the methods given in that other problem better than computing? – Gerry Myerson May 07 '13 at 13:35
  • @Gerry Myerson,maybe when you can do it without a calculator,especially the second problem(it has less digits than the first problem) – lsr314 May 07 '13 at 13:39
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    $40/237$ is what you get from the first five terms of the continued fraction expansion of $e^x$ which is $0+\cfrac{1}{5+\cfrac{1}{1+\cfrac{1}{12+\cfrac{1}{3+\cfrac{1}{\cdots}}}}}$. Since the truncated continued fraction expansion is alternatingly smaller and larger than its limit value, truncating after the $3$ will give you a value that is smaller than $e^x$. You just need to show that $[0,5,1,12,\ldots]$ is indeed the continued fraction expansion of $e^x$ ;-) – Elmar Zander May 07 '13 at 15:30
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    Why is a "proof" necessary for such a fact? One may simply compute the quantities under question to readily obtain the desired results. – ThisIsNotAnId May 07 '13 at 19:33

1 Answers1

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As suggested by the comments, approximations for $x$ may be found in terms of continued fractions. Then compare with continued fraction approximations for $\ln\frac{40}{237}$ (obtained form the Taylor expansion?)

By checking signs at $x=-1$ and $x_=-2$ we see that there is a real root in $]-2,-1[$. Now substitute $x\leftarrow \frac1y-2$ and multiply by $y^3$ to find $$ -6y^3+31y^2-18y+3$$ as new polynomial. Verify by sign changes that there is a root between $y=4$ and $y=5$. Then substitue $y\leftarrow \frac1z+4$ etc.

Unfortunately, the continued fraction of the root is $[-2, 4, 1, 1, 8, 4, 11, 5,\ldots]$ and that of $\ln\frac{40}{237}$ is $[-2, 4, 1, 1, 8, 4, 11, 8,\ldots]$, so there are quite a few laborious steps in front of you.

Hagen von Eitzen
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