Cuál es mayor$\frac{13}{32}$ o$\ln \left(\frac{3}{2}\right)$

Question

Cuál es mayor $\frac{13}{32}$ o $\ln \left(\frac{3}{2}\right)$

Mi intento:

tenemos $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$ $

Deje $x=1.5$

Ahora considere la función $$f(x)=\frac{1+x^2}{8}-\ln x$ $

$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $ f$ is Decreasing in $ (0,2) $

alguna ayuda aqui?

Answer 1

\begin{align*} \exp\left(\frac{13}{64}\right) & = \exp\left(\frac15\right)\exp\left(\frac1{320}\right) \\ & > \left(1 + \frac15 + \frac1{50} + \frac1{750}\right)\left(1 + \frac1{320}\right) \\ & = \left(1 + \frac{166}{750}\right)\left(1 + \frac1{320}\right) \\ & = \frac{458}{375}\times\frac{321}{320} = \frac{229\times107}{125\times160} \\ & = \frac{24{,}503}{20{,}000} > \frac{24{,}500}{20{,}000} = \frac{49}{40} \\ \therefore\ \exp\left(\frac{13}{32}\right) & > \left(\frac{49}{40}\right)^2 = \frac{2{,}401}{1{,}600} > \frac32. \end{align*}

Answer 2

La diferencia es tan pequeña que no veo otra forma que hacer el cálculo. Nota $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ implies $$e^{13/32} > 1 + \frac{13}{32} + \frac{(13/32)^2}{2!} + \frac{(13/32)^3}{3!} + \frac{(13/32)^4}{4!} = \frac{12591963}{8388608} > \frac{3}{2}.$ $

Answer 3

Creo que podemos probar con $$\log\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\dots\right)$$ and put $ x = 1/5$. The calculations are simple and one can estimate the error easily. Estimation of error gives $$\log\frac{1+x}{1-x}<2x+\frac{2x^3}{3}+\frac{2x^5}{5(1-x^2)}$$ With a little calculation of the easy variety (division by 2,3,5 etc) you can conclude that the right hand side of the above inequality for $ x = 1/5$ is less than $ 13/32 $ .