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May
11
comment Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
I added an inductive argument
May
11
revised Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
added 324 characters in body
May
11
comment Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
I wasn't sure if you were being asked to use induction or you were just asked the question and you were attempting an inductive proof. At any rate, I still think this is worth pointing out.
May
11
comment Decay of Fourier Transform
The decreasing condition on the Fourier transform's absolute value is incompatible with the conclusion also.
May
11
answered Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
May
11
comment Decay of Fourier Transform
That really doesn't change the issue.
May
11
comment Decay of Fourier Transform
Since $|\hat{f}(y)|^2$ is decreasing, you can't have $|\hat{f}(0)| = 1$ and $|\hat{f}(y)| < 1$ for $y < 0$.
May
7
comment How to determine the limit of $f(\mathbf{x}) = \frac{2x^6y}{x^8+y^4+5x^4y^2}$ as $\mathbf{x} \rightarrow \mathbf{0}$?
The denominator is $(x^2)^4 + (y)^4 + 5 (x^2)^2(y)^2$. If you switch the roles of $x^2$ and $y$ this expression doesn't change. And the limit doesn't exist, as I explained in the answer.
May
7
comment How to determine the limit of $f(\mathbf{x}) = \frac{2x^6y}{x^8+y^4+5x^4y^2}$ as $\mathbf{x} \rightarrow \mathbf{0}$?
No he means as the vector ${\bf x} = (x,y)$ approaches $(0,0)$
May
7
answered How to determine the limit of $f(\mathbf{x}) = \frac{2x^6y}{x^8+y^4+5x^4y^2}$ as $\mathbf{x} \rightarrow \mathbf{0}$?
May
6
answered Tricks to solve inequalities
May
2
comment $L^{p}$ Boundedness of Fourier Multiplier without Littlewood-Paley
$L^p$ boundness results for singular integrals or multiplier operators are rarely very simple when $p \neq 2$.. as far as I know even $L^p$ boundedness for the Hilbert transform in one dimension doesn't have that easy a proof.
May
2
revised $L^{p}$ Boundedness of Fourier Multiplier without Littlewood-Paley
added 5 characters in body
May
2
answered $L^{p}$ Boundedness of Fourier Multiplier without Littlewood-Paley
May
1
comment Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible
You can't immediately get $d = b$ from the $x$ term since $a$ and $c$ might both be zero.
May
1
answered Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible
Apr
25
answered Given $r>0$, find $k>0$ such that $\sqrt{(x-2)^2+(y-1)^2}<k$ implies $|xy-2|<r $
Apr
25
answered Find the exact value of $\cos(11\pi/12)$.
Apr
25
comment $y_{2n}, y_{2n+1}$ and $y_{3n}$ all converge. What can we say about the sequence $ y_n$?
No, I just called them $L, M,$ and $N$.. there's no relation to the $2M$ and $2N$ you wrote there. So yes you would call your index $N'$ or something to distinguish.
Apr
25
answered $y_{2n}, y_{2n+1}$ and $y_{3n}$ all converge. What can we say about the sequence $ y_n$?