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 Mar4 comment What is the solution of cos(x)=x? math.stackexchange.com/questions/227317/explaining-cos-infty/… Mar4 comment Explaining $\cos^\infty$ @GDumphart you can check it. If you eliminate arctan and tan, the result changes, becomes wrong. It you know other ways to simplify it, you are welcome. Mar4 comment Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$ Why the index of the Bessel functions is not equal to the argument in your expression? Mar4 comment Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$ By the way, the formula $$\sum_{n=1}^\infty \frac{2J_n(n)}{n} \sin(\pi n/2)$$ works for Dottier number. Mar4 comment What is the solution of cos(x)=x? How having closed-form expression is related to being transcedential? Mar4 comment What is the solution of cos(x)=x? more likely a link to Omega constant (fixed point of exponent) and an integral form. Mar4 comment What is the solution of cos(x)=x? Without another function? What do u mean? The answer by giorgiomugnaini gives an analytic solution. Mar4 comment What is the solution of cos(x)=x? What's the point of arguing it is trancedential? Fixed point of the exponent is also transcedential, but can be expressed with an integral, for instance. Mar2 comment Does a non-trivial solution exist for $f'(x)=f(f(x))$? @Julien this answer actually better to be on top, because it addresses exactly the issue of solving the equations of the type you proposed and also gives the exact answer in the closed form to your equation. People who follow the link consisting of your caption likely want to see the result. Mar2 comment Does a non-trivial solution exist for $f'(x)=f(f(x))$? @Julien I expected you to accept my answer because it gives you the exact answer. Feb21 comment Why formal power series are not considered a system of hypercomplex numbers? LOL who said you composition of functions should be multiplication? What hypercomplex system is built this way? Feb5 comment Are there divergent series that cannot be summed up by any method? @Jack D'Aurizio are u sure it will not be convergent for any methods listed on the wikipedia's page? Feb5 comment Are there divergent series that cannot be summed up by any method? @Jack D'Aurizio so what's the point? Feb5 comment Divergent Series The series $\sum_{k=1}^\infty \frac{1}{k}$ can be summed up, for instance using Ramanujan's summation or Cauchy principal value. The sum will be $\gamma$, Euler's constant. Jan24 comment Differentiable only at $x=0$ and $f'(0)>0$ The example of David Mitra satisfies the first two conditions. Jan24 comment Differentiable only at $x=0$ and $f'(0)>0$ @user197137 the definition of derivative is $\lim_{h\to 0}\frac{f(x+h)-f(x)}h$. If it is positive, then there are infinitely many such h that $f(-h) < f(0)< f(h)$ Jan24 comment Differentiable only at $x=0$ and $f'(0)>0$ @user197137 for the function to have positive derivative, such h should exist. Jan24 comment Numerical system that includes the limit targets such as $0^+$, $0^-$, $1^+$ etc @ajotatxe limit will be excessive in this case, in this numerical system the function can be evaluated directly: $\sin 0^+=0^+$. You can see it as the simplifyed system of hyperreals, with $0^+$ substituted for any positive infinitesmal. Jan22 comment Why we cannot ascribe values to behavior of functions at poles? I also think whether anythink will break if we remove infinity. That is, prohibit integration with limits at infinity (only allowing integration to $\omega$). Jan17 comment Evaluate the double integral $\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)} \,dx\,dy$ Great! But I wonder how is it related to the Euler's $\gamma$ and $\ln \pi/4$. I thought the answer should be somewhat continuous with them. What if we change the $i$ to somewhat like $(-1)^k$ or $e^{i\pi k}$? How the answer will be k-dependent?