# $x^3-3x-3=0$, prove that $10^x<127$

$x$ 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 $$10^x>\frac{1}{127}.$$

• @Arjang $x\in (-2,-1)$. – Ma Ming May 7 '13 at 11:41
• @Arjang,maybe you inputed $x^3-5x+8=0$. – Next May 7 '13 at 11:42
• 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 7 '13 at 13:11
• @Gerry Myerson I know it can be proved by computing,but I want to find a better way to prove it,like this problem math.stackexchange.com/questions/380302/… – Next May 7 '13 at 13:26
• $40/237$ 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 7 '13 at 15:30

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.