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 Jun 30 comment A negative third derivative implies a positive first derivative at a point. @HagenvonEitzen Why must $f'$ average around 0? Jun 30 asked A negative third derivative implies a positive first derivative at a point. Jun 29 comment Calculating $\lim_{x\to\infty} (\sin\frac{1}{x}+\cos\frac{1}{x})^x$ without l'Hopital @anomaly I agree, but I'm preparing for a test, and the instructions say "without l'Hopital's rule" Jun 29 asked Calculating $\lim_{x\to\infty} (\sin\frac{1}{x}+\cos\frac{1}{x})^x$ without l'Hopital Jun 29 comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions @NikolajK Correct me if I'm wrong, but I am supposed to accept an answer once I understood the solution to my question, am I not? I really wouldn't mind leaving questions open for longer but it does seem like I'm supposed to accept it when I solved it to prevent people putting in effort for nothing instead of answering still open questions. Jun 29 comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions That is exactly what I got to at the end, but thanks for the confirmation :) Jun 29 accepted Proving that $\ln ^3|x|=x$ has exactly 3 real solutions Jun 29 comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions Actually just the fact that the function must be negative on $(0,1)$ is the obvious thing I was missing ><. Thank you Jun 29 comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions @ClementC. I actually tried that, from the second derivative I found out the minimum and maximum of $f'$ Now I could go to a third derivative as it is already pretty simple, but it's so far I have no idea how to use it in my original problem Jun 29 asked Proving that $\ln ^3|x|=x$ has exactly 3 real solutions Jun 17 comment For any closed set $A$ of $\mathbb R$ , does there exists a function $f:\mathbb R \to \mathbb R$ such that, $f$ is discontinuous exactly on $A$? That what happens when I don't think... Removed Jun 17 revised For any closed set $A$ of $\mathbb R$ , does there exists a function $f:\mathbb R \to \mathbb R$ such that, $f$ is discontinuous exactly on $A$? deleted 258 characters in body Jun 17 answered For any closed set $A$ of $\mathbb R$ , does there exists a function $f:\mathbb R \to \mathbb R$ such that, $f$ is discontinuous exactly on $A$? Jun 17 comment For any closed set $A$ of $\mathbb R$ , does there exists a function $f:\mathbb R \to \mathbb R$ such that, $f$ is discontinuous exactly on $A$? What about $f(x)=0$ for $x\in\mathbb{R}\setminus A$ and $f(x)=1$ if $x\in\mathbb{Q}\cap{A}$ and $f(x)=2$ if $x\in\mathbb{I}\cap{A}$. It will be continuous at each point $x_0$ outside of $A$ as it will be $0$ at an environment of $x_0$ (As $\mathbb{R}\setminus A$ is open), and obviously not continuous at $A$. Jun 11 comment Confused with $x$ and $a$ of Taylor Series @user247433 $f(x)$ would stay the same, the Taylor polynomial is equal to the function (as long as it converges to it). Yves gave a good answer by showing you the power series of $sin$ at a general $a$. Notice that (alao as Yves stated) the power series converges for any $a\in \mathbb{R}$ thus $sin(x) =T_a(x)$ for any $a$ and any $x$ Jun 11 comment Confused with $x$ and $a$ of Taylor Series You can take any origin you want as long as the the power series around it converges to the function. People usually expand around 0 as it's the easiest to evaluate. If you want to expand around 1, for example, you need to be able to calculate $sin(1)$ and $cos(1)$ Jun 9 accepted Using second derivative to find a bound for the first derivative Jun 9 awarded Autobiographer Jun 9 awarded Vox Populi Jun 9 comment Using second derivative to find a bound for the first derivative @TedShifrin That's a fair point, changed it. About the maximum, the hint I was given was to use the minimum, when I wasn't able to use that I tried thinking of the maximum, but even if I do, everything else is pretty much the same, and in the given expansion I'm not sure how to prove it's smaller than $\frac{1}{2}$