The author's original inequality is as follows.
Prove that: $$e-\ln(10)>\sqrt 2-1$$
Is there a good approximation for $$e-\ln 10?$$
Actually, I am also wondering that, Where does $\sqrt 2-1$ come from? Maybe, there exist relevant inequality?
My attempt:
$$-\ln 10>\sqrt 2-1-e\\ \ln 10<e+1-\sqrt 2\\ e^{e+1-\sqrt 2}>10$$
So, can we show that
$$e^x>10$$
when $x\ge e+1-\sqrt 2$?
I don't have a good idea.
On
I hope the following will help. $$\ln10=\ln2+\ln5=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+...\right)+$$ $$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+...\right)<$$$$<2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+\frac{\left(\frac{1}{3}\right)^9}{7}+\frac{\left(\frac{1}{3}\right)^{11}}{7}+...\right)+$$ $$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+\frac{\left(\frac{2}{3}\right)^9}{9}+\frac{\left(\frac{2}{3}\right)^{11}}{11}+\frac{\left(\frac{2}{3}\right)^{13}}{13}+\frac{\left(\frac{2}{3}\right)^{15}}{13}+...\right)=$$$$=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}\cdot\frac{1}{1-\left(\frac{1}{3}\right)^2}\right)+$$ $$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+\frac{\left(\frac{2}{3}\right)^9}{9}+\frac{\left(\frac{2}{3}\right)^{11}}{11}+\frac{\left(\frac{2}{3}\right)^{13}}{13}\cdot\frac{1}{1-\left(\frac{2}{3}\right)^2}\right)=$$ $$=\frac{8166549323}{3546482940}.$$
Can you end it now?
On
I think this is numeric coincidence.
Anyway, let $a = e - 1 + \sqrt{2}$, expand $e^x$ to eighth terms: $$ e^a > \sum_{n=0}^{8}\frac{a^n}{n!} > 10.$$ One still needs a calculator but logically this seems right.
On
By Taylor's series at $196$: $$\sqrt{2}=\frac{1}{10}\sqrt{200}<\frac{1}{10}(14+\frac{1}{7}-\frac{2}{14^3}+10^{-5})<1.4143\tag{1}$$ By Maclaurin's series: $$e>1+1+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}>2,718\tag{2}$$ By using $\ln 10=-\ln(0.9)+2\ln3$: $$-\ln(0.9)<\frac{1}{10}+\frac{1}{2\times10^2}+\frac{1}{3\times10^3}+10^{-4}<0.10544\tag{3}$$ and by using the Maclaurin series $\ln(\frac{1+x}{1-x})=2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$, we have $$2\ln 3<4(1+\frac{1}{12}+\frac{1}{80}+\frac{1}{448}+\frac{1}{2304}+10^{-4})<2.1972\tag{4}$$ Hence by $(3)$ and $(4)$, $$\ln 10<0.10544+2.1972=2.30264\tag{5}$$ Finally by $(1)$, $(2)$ and $(5)$, $$e+1-\sqrt{2}-\ln 10>2.718+1-1.4143-2.30264=0.00106>0.$$
On
Firstly, $$e >\dfrac1{0!}+\dfrac1{1!}+\dfrac1{2}+\dfrac1{6}+\dfrac1{24}+\dfrac1{120}+\dfrac1{720}+\dfrac1{5040}+\dfrac1{40320} = \dfrac{109601}{40320}>2.718278,$$ $$e>2.(7182) = 2+\dfrac{7182}{9999}=2+\dfrac{798}{1111}=\dfrac{3020}{1111}.\tag1$$ Then $$12\cdot 3020^3 = 330523296000 > 330490682071 = 241\cdot 1111^3,$$ $$e^3>\dfrac{241}{12}=20\left(1+\dfrac1{240}\right).\tag2$$ From $\;(1)-(2)\;$ should $$e^{76} > 20^{25}\cdot\left(1+\dfrac1{240}\right)^{25}\cdot \dfrac{3020}{1111}> \left(33554432\cdot 10^{25}\right)\cdot \left(1+\dfrac5{48}\right)\cdot \dfrac{3020}{1111}$$ $$> \dfrac{3\cdot 53\cdot 3020}{48}\cdot10^{29}>10^{33},$$ $$\ln 10<\dfrac{76}{33}.\tag3$$ At last, $$(e+1-\ln10)^2 > \left(\dfrac{3020}{1111}+1-\dfrac{76}{33}\right)^2 = \dfrac{4717^2}{3333^2} = \dfrac{22250089}{11108889}>2,$$ $$\color{green}{\mathbf{e-\ln10>\sqrt2-1}}.$$
On
First we show that $\log 10 < 76/33$, since $$ \exp\Big(\frac{76}{33}\Big) > \sum_{n=0}^9 \frac{76^n}{33^n\,n!} = \frac{1316160031686037871}{131576558279536755}>10\,. $$ Now it is readily seen that $$ \frac{76}{33}<\frac{67997}{29520}\,,=1+1+\frac 1 2+\frac 1 6 + \frac 1{24}+\frac 1{120}+\frac 1{720}-\frac{17}{41}\,, $$ hence $$ \log 10<\frac{76}{33} < e - \frac{17}{41} = e+1-\frac{58}{41}\,, $$ and finally (since $58/41 >\sqrt 2$) $$ e-\log 10 > \sqrt 2 - 1\,. $$
Here is a proof that contains no numbers with more than five decimal digits. \begin{gather*} 10 = \frac{65536}{6561}\times\frac{65610}{65536} = \left(\frac43\right)^8\!\!\times\frac{65573 + 37}{65573 - 37} \\ \therefore\ \ln10 = 8\ln\left(\frac{1 + 1/7}{1 - 1/7}\right) + \ln\left(\frac{1 + 37/65573}{1 - 37/65573}\right) \\ < 16\left(\frac17 + \frac{(1/7)^3}{3(1 - (1/7)^2)}\right) + 2\left(y + \frac{y^3}{3(1 - y^2)}\right), \end{gather*} where $y = 37/65573.$
We have $$ 16\left(\frac17 + \frac{(1/7)^3}{3(1 - (1/7)^2)}\right) = \frac{16}7 + \frac1{3\cdot3\cdot7} = \frac{145}{63}, $$ and $y < 1/1000,$ therefore $1 - y^2 > 2/3,$ therefore $$ 2\left(y + \frac{y^3}{3(1 - y^2)}\right) < 2y + y^3 < 2y + 10^{-9}. $$ Therefore: $$ \ln10 < \frac{145}{63} + 2y + 10^{-9}. $$
On the other hand, from e Continued Fraction - from Wolfram MathWorld, we have: $$ e > \frac{106}{39}, $$ and from Square root of 2 - Wikipedia, or by simple calculation: $$ 2 < \frac{9801}{4900} = \left(\frac{100 - 1}{70}\right)^2, \quad \therefore\ \sqrt2 < \frac{99}{70}. $$ Putting all the inequalities together, we get: \begin{align*} e - \sqrt2 + 1 - \ln10 & > \frac{106}{3\cdot13} - \frac{29}{7\cdot10} - \frac{145}{7\cdot9} - 2y - 10^{-9} \\ & = \frac{106\cdot210 - 29\cdot117 - 145\cdot130} {7\cdot9\cdot10\cdot13} - 2y - 10^{-9} \\ & = \frac{17}{8190} - \frac{74}{65573} - 10^{-9} \\ & > \frac{16}{8192} - \frac{74}{65536} - 10^{-9} \\ & = \frac{16}{8192} - \frac{128}{65536} + \frac{54}{65536} - 10^{-9} \\ & = \frac1{512} - \frac1{512} + \frac{27}{32768} - 10^{-9} \\ & > 0. \end{align*}