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Jul
23
comment Fundamental Theorem of Calculus and the left endpoint of the interval.
Well, that makes sense. Thank you
Jul
23
comment Fundamental Theorem of Calculus and the left endpoint of the interval.
He does mention that the value of $b$ can be negative, but that is mentioned as notation for the integral itself, and here in the definitions we have actual intervals where it makes no sense to have $b<a$ (or at least it wasn't defined)
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