都说是偏移的最基础题目 还没开窍

facebooker

1)$x>1$
求证:
\begin{align*}
\ln x>\dfrac{3(x^2-1)}{x^2+4x+1}
\end{align*}
2)求证:
\begin{align*}
\dfrac{a-b}{\ln a-\ln b}<\dfrac{1}{3}(2\sqrt{ab}+\dfrac{a+b}{2}),a,b>0,a \not=b.
\end{align*}

kuing

两个问题是等价的啊
\begin{align*}
(2)&\iff\frac{\frac ab-1}{\ln\frac ab}<\frac13\left( 2\sqrt{\frac ab}+\frac{\frac ab+1}2 \right)\\
&\iff\frac{x^2-1}{2\ln x}<\frac13\left( 2x+\frac{x^2+1}2 \right)\\
&\iff(1)
\end{align*}
PS、单行的行间公式用

1. \[ ... \]

Copy the Code

即可，无需 align* 环境，多行的才需要。

数学的书蠹

当x的1的时候两边相等。求导通分后要么可以因式分解要么恒大与o，这里是前者，找到根之后就简单了，不过我一直很好奇这里东西是怎么被命题者发现的，是研究其他问题恰巧发现还是什么？

hbghlyj

Lambert's Continued Fraction
Natural logarithm: Continued fraction representations
\begin{align*}
\log \left(\frac{z+1}{1-z}\right)&=\frac{2 z}{1-\frac{z^2}{3-\frac{4 z^2}{5-\frac{9 z^2}{7-\frac{16 z^2}{9-\frac{25 z^2}{11-\frac{36 z^2}{13-\ldots }}}}}}}\qquad z\notin(-\infty,-1)\wedge z\notin(1,\infty)\\
\implies\log \left(\frac{z+1}{1-z}\right)&>\frac{2 z}{1-\frac{z^2}{3}}\\
\text{Substitute }z=\frac{x-1}{x+1}&\\
\implies\log x&>\dfrac{3(x^2-1)}{x^2+4x+1}
\end{align*}

hbghlyj

See Jacopo D’Aurizio's answer here:

By the (generalized) Shafer-Fink inequality we may derive arbitrarily accurate uniform approximation for the $\arctan$ and $\arctanh$ functions, and $
\arctanh z=\frac12\log \left(\frac{z+1}{1-z}\right)
$, hence for the logarithm. We have, for instance, $\log(x)\geq \frac{3(x^2-1)}{x^2+4x+1}$ for any $x\in(0,1)$.

战巡

数学的书蠹 发表于 2019-10-5 20:17
当x的1的时候两边相等。求导通分后要么可以因式分解要么恒大与o，这里是前者，找到根之后就简单了，不过我 ...

\[\frac{3(x^2-1)}{x^2+4x+1}\]
就是$\ln(x)$在$x=1$处做$(2,2)$帕德近似的结果

这类东西多了去了，很容易通过帕德近似构造任意一个函数和一个分数式，然后再想办法证明其到底哪个大

hbghlyj

hbghlyj 发表于 2022-8-4 23:35
Lambert's Continued Fraction$$\label1\tag1
\log \left(\cfrac{z+1}{1-z}\right)=\cfrac{2 z}{1-\cfrac{z^2}{3-\cfrac{4 z^2}{5-\cfrac{9 z^2}{7-\cfrac{16 z^2}{9-\cfrac{25 z^2}{11-\cfrac{36 z^2}{13-\ldots }}}}}}}
$$

对于$$
1+\frac12\log \left(\frac{z+1}{1-z}\right)=1 + z + \frac13z^3 + \frac15z^5 + \frac17z^7 + \frac19z^9 + \frac1{11}z^{11} + O(z^{13})
$$有$$(a_0,a_1,a_2,\dots)=\left(1,1,0,\frac13,0,\frac15,\dots\right)$$代入\eqref{2}变成
$$\left\{\begin{array}{l}b_{0}=1 \\ 1=-b_{1} \\0=b_{1}\left(b_{1}+b_{2}\right) \\ \frac13=-b_{1}\left[b_{2} b_{3}+\left(b_{1}+b_{2}\right)^{2}\right] \\ 0=b_{1}\left[b_{2} b_{3}\left(b_{3}+b_{4}\right)+2\left(b_{1}+b_{2}\right) b_{2} b_{3}+\left(b_{1}+b_{2}\right)^{3}\right]\end{array}\right.$$解得\begin{array}{l}b_{0}=1 \\ b_{1}=-1 \\b_2=1 \\ b_{3}=\frac13\\ b_4=-\frac13\end{array}
得到了和\eqref{1}不同的连分式$$1+\frac12\log \left(\frac{z+1}{1-z}\right)=\cfrac{1}{1+\cfrac{-z}{1+\cfrac{z}{1+\cfrac{\frac13 z}{1+\cfrac{-\frac13z}{\ddots}}}}}$$
那么,如何得到\eqref{1}呢?

8.4.2 Continued Fraction Representation of Functions
If a function is represented by a power series about the origin
$$f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$$
we can also write it in a continued-fraction form. The standard approach here is to write
$$f(x)=\frac{b_{0}}{1+} \frac{b_{1} x}{1+} \frac{b_{2} x}{1+} \frac{b_{3} x}{1+} \frac{b_{4} x}{1+\ldots}$$
Evidently, there is a one-to-one correspondance between the Taylor-series coefficients $\{a_n\}$ and the continued-fraction coefficients $\{b_n\}$, which may be determined by expanding the continued fraction in a power series for small $x$. The theory of such a representation is discussed also in Advanced mathematical methods for scientists and engineers (Carl M. Bender, Steven A. Orszag).
Let us consider a function with the property $f (0) = 1$; this is merely a convenient choice of normalization. Then the relation between the continued fraction coefficients and the series coefficients is easily found to be
$$\tag2\label2\begin{array}{l}a_{0}=b_{0}=1 \\ a_{1}=-b_{1} \\ a_{2}=b_{1}\left(b_{1}+b_{2}\right) \\ a_{3}=-b_{1}\left[b_{2} b_{3}+\left(b_{1}+b_{2}\right)^{2}\right] \\ a_{4}=b_{1}\left[b_{2} b_{3}\left(b_{3}+b_{4}\right)+2\left(b_{1}+b_{2}\right) b_{2} b_{3}+\left(b_{1}+b_{2}\right)^{3}\right]\end{array}$$
and so on. This constitutes a nonlinear mapping from the set of numbers $\{b_n\}$ to the set $\{a_n\}$ or vice versa.
This mapping seems to be quite remarkable in that the sequence of $b_n$s is typically much simpler than the sequence of $a_n$s.

hbghlyj

Generalized continued fraction
Euler's continued fraction formula
Proof of \eqref{1} can be found here: Lambert's Original Proof that $π$ is irrational.
We show $$\tan x=\cfrac{x}{1-
\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$
First we treat Lambert's continued fraction expression for the quotient of two power series.
Let $F_0$ and $F_1$ be two power series we may assume without loss of generality that they both have nonzero constant term and that in fact both have constant term of $1$.
Define
$$F_1-F_0=b_1 x F_2$$
$$F_2-F_1=b_2 x F_3$$
$$F_3-F_2=b_3 x F_4$$
$$F_{n+1}-F_n =b_{n+1} x F_{n+2}$$
where at each step we make the assumption that $F_n$ has a constant term of $1$. In general we would simply get higher powers of $x$ in the equation.
Define $$G_n=\frac{F_{n+1}}{F_n}$$ then we have by dividing the last equation by $F_{n+1}$,
$$G_n=\frac{1}{1-b_{n+1}xG_{n+1}}$$
therefore
$$\frac{F_1}{F_0}=G_0=\cfrac{1}{1-
\cfrac{b_1 x}{1-\cfrac{b_ 2 x}{1-\cfrac{b_3 x}{1-\cdots}}}}$$
Now we shall apply this to obtain continued fraction expansions for several functions including $\tan x$.
Let us define the particular power series,
\begin{align*}
F_n&=1+ \frac{x}{1! (\gamma +n)} +\frac{x^2}{2! (\gamma +n)(\gamma +n+1)} +\frac{x^3}{3! (\gamma +n)(\gamma +n+1)(\gamma +n+2)} \cdots \\
&=1+ \sum\limits_{k=1}^{\infty} \frac{x^k}{k! (\gamma +n)\cdots (\gamma +n+k-1)}
\end{align*}
Then we have
$$F_{n+1}-F_n =-\frac{x}{(\gamma +n)(\gamma +n+1)} F_{n+2}.$$
Thus with $b_n =  -\frac{x}{(\gamma +n)(\gamma +n+1)}$ we get
\begin{align*}
\frac{F_1}{F_0}&=\cfrac{1}{1+\cfrac{\frac{x}{(\gamma)(\gamma +1)}}{1+\cfrac{\frac{x}{(\gamma +1)(\gamma +2)}}{1+\cfrac{\frac{x}{(\gamma +2)(\gamma +3)}}{1+\cdots}}}} \\
&=\cfrac{\gamma}{\gamma +\cfrac{x}{(\gamma+1)+\cfrac{x}{(\gamma +2)+\cfrac{x}{(\gamma +3)+\cdots}}}}
\end{align*}
Now let us set $\gamma=\frac{1}{2}$ and instead of $x$ write $-\frac{x^2}{4}$
Then $$F_1= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k+1)!}=\frac{\sin x}{x}$$
and
$$F_0= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}=\cos x.$$
Putting this altogether we get
$$\tan x=\cfrac{x}{1-
\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$

hbghlyj

战巡 发表于 2022-8-5 04:45 \[\frac{3(x^2-1)}{x^2+4x+1}\] 就是$\ln(x)$在$x=1$处做$(2,2)$帕德近似的结果

Mathematica
PadeApproximant[expr,$x$,$x_0$,{$m$,$n$}] gives the Padé approximant to expr about the point $x=x_0$, with numerator order $m$ and denominator order $n$.

Pari/GP Reference Card page 4: Other numerical methods
power series to cont. fraction ($L$ terms) contfracinit(S, {L})
Padé approximant (deg. denom. ≤ $B$) bestapprPade(S, {B})

hbghlyj

hbghlyj 发表于 2022-8-5 15:44
Proof of \eqref{1} can be found here: Lambert's Original Proof that $π$ is irrational.
We show $$\t ...

youtube.com/watch?v=Lk_QF_hcM8A