<s>[INST] Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
 [/INST] I want to point out that $\frac{13}{32}$ is the value of the Pade $(2,1)$ approximation of $\ln (1+x)$ at $x=\frac{1}{2}$.
In detail, the Pade $(2, 1)$ approximation of $\ln (1+x)$ is $g(x) = \frac{x^2+6x}{6+4x}$.
It is easy to prove that $\frac{x^2+6x}{6+4x} > \ln (1+x)$ for $x > 0$.
Indeed, let $f(x) = \frac{x^2+6x}{6+4x} - \ln(1+x)$. We have
$f'(x) = \frac{x^3}{(3+2x)^2(1+x)} > 0$ for $x > 0$. Also, $f(0) = 0$. The desired result follows.
We have $g(\frac{1}{2}) = \frac{13}{32} > \ln \frac{3}{2}$.
We may find the Pade approximation by hand. We may first try the Pade $(1, 1)$ approximation, second try the Pade $(2, 1)$ approximation,
third try the Pade $(1, 2)$ approximation, and so on, until we find enough approximation.
More details about the Pade $(2, 1)$ approximation:
The Taylor expansion of $\ln(1+x)$ is $x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots$
Let $g(x) = \frac{a_0 + a_1x+ a_2x^2}{1 + b_1x}$. Comparing the coefficients of $x^k$ for $k=0, 1, 2, 3$ of
$$(x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots)(1 + b_1x)
= a_0 + a_1x+ a_2x^2, $$
we obtain $a_0 = 0, a_1 = 1, a_2 = \frac{1}{6}, b_1 = \frac{2}{3}$. Then, $g(x) = \frac{x^2+6x}{6+4x}$.
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