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## New answers tagged constants

0

I would use Newton's method on the function $f$ defined on $(1,\infty)$ by: $$f(x)=\log\left(\frac{1}{\log(x)}\right)-1$$ $$f'(x)=\frac{-1}{x\log(x)}$$ Newton's method converges quadratically. The computation of the logarithm could use the Arithmetic-Geometric Mean that also converges quadratically. (See this article) It means that asymptotically, to ...

1

Hint: Let $x=\sin y$ then $\sqrt{1-x^2}=\dots?$ And $\sin2y=\dots?$

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The antiderivative should include the absolute values, so is $g(x)=2x+\ln |x+3| +\ln |x+2|$ to start. The expression so far is defined at all $x$ other than $-3,-2$. Removing these two points from the reals, there remain three sections, each of which can have an independent constant of integration added to the "antiderivative" $g(x).$ So in this sense the ...

1

Antonio Vargas's observation means that $1$ starts closer and closer to the fixpoint, so that maybe there is less and less difference between $C_k$ and the first term in the sequence defining it ; and maybe that first term converges to $\log 2$. Let $f_k(x) = \sqrt[k]{1+x}$ for $x \ge 0$ and $k > 1$. Let $\alpha_k$ the unique positive fixpoint of $f_k$ ...

0

Hint : $\displaystyle x(n)=\underset{k=0}{\overset\infty\Xi}\left(a,b\,;\tfrac1n\right)\iff x^n=a+bx\iff n(x)=\frac{\ln(a+bx)}{\ln x}\iff n(1)=\infty$ , $n(\infty)=1$ . Now show, using l'Hopital, that $\displaystyle\lim_{x\to1}\Big[n(x)\cdot(x-1)\Big]=\ln2$.

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