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Jun
30
comment A negative third derivative implies a positive first derivative at a point.
@Tryss Well I haven't learned about integration yet, so I don't know what the last part tells me :P
Jun
30
comment A negative third derivative implies a positive first derivative at a point.
@MattSamuel I wrote the last derivative using Lagrange's remainder, which is why it's $c$ instead of $x$. Added clarification. Is it wrong?
Jun
30
comment A negative third derivative implies a positive first derivative at a point.
@HagenvonEitzen Why must $f'$ average around 0?
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
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
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
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
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
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 $\frac{1}{2}$
Jun
9
comment Where did I go wrong in this limit?
I think the first answer here math.stackexchange.com/questions/46065/… is relevant
Jun
9
comment Evaluate $\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})$
Seems like ${\pi}\over{2}$ to me as well
Jun
7
comment Proving the inequality $2^{-1+\frac{1}{n}}\left(x+1\right)\leq\left(x^{n}+1\right)^{\frac{1}{n}}$ for $n\in\mathbb{N} $ and $x>0$
Well that does it...
Jun
7
comment Proving the inequality $2^{-1+\frac{1}{n}}\left(x+1\right)\leq\left(x^{n}+1\right)^{\frac{1}{n}}$ for $n\in\mathbb{N} $ and $x>0$
We haven't really talked about properties of convex function, so I have no idea how to justify what you wrote, any other options?
Jun
7
comment Proving the inequality $2^{-1+\frac{1}{n}}\left(x+1\right)\leq\left(x^{n}+1\right)^{\frac{1}{n}}$ for $n\in\mathbb{N} $ and $x>0$
I tried using Binomial Theorem, It's how I solved the other side of the inequality, but with this side I had pretty much the same thing as with Taylor polinomial, which I had no idea how to proceed with
Jun
7
comment Proving the inequality $2^{-1+\frac{1}{n}}\left(x+1\right)\leq\left(x^{n}+1\right)^{\frac{1}{n}}$ for $n\in\mathbb{N} $ and $x>0$
@Stromael, fixed, thanks...
Jun
6
comment Proving the ranges of repeated iteration of a one-to-one function from $X$ to a subset are disjoint
That is simple enough, thanks!