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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 $\frac{1}{2}$
Jun
9
revised Using second derivative to find a bound for the first derivative
deleted 26 characters in body
Jun
9
asked Using second derivative to find a bound for the first derivative
Jun
9
awarded  Suffrage
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
revised Evaluate $\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})$
deleted 83 characters in body
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
9
revised Evaluate $\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})$
added 6 characters in body
Jun
9
answered Evaluate $\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})$
Jun
7
accepted 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 $x>0$
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
7
revised 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 $x>0$
edited body