Question

If $a_0+\frac{a_1}{2}+\dots + \frac{a_n}{n+1}=0$ then $a_nx^n+\dots + a_0$ has a root in $[0,1]$

score 145 · Answer 1 · 2016-08-29 02:28:14

145

Views

If $a_0+\frac{a_1}{2}+\dots + \frac{a_n}{n+1}=0$ then $a_nx^n+\dots + a_0$ has a root in $[0,1]$

Published on 29 Aug 2016 - 2:28

score 1.7k · Answer 2 · 2016-08-30 22:03:17

1.7k

Views

Alternative proof : If E and F have positive measures then E+F contains an interval

Published on 30 Aug 2016 - 22:03

score 258 · Answer 3 · 2026-04-02 11:12:06

258

Views

Proof for an inequality problem using the rearrangement inequality

Published on 02 Apr 2026 - 11:12

score 1.4k · Answer 4 · 2026-03-26 21:12:45

1.4k

Views

the converse of Pythagoras Theorem

Published on 26 Mar 2026 - 21:12

score 74 · Answer 5 · 2026-03-30 23:17:59

74

Views

Absolute value inequality - reflection

Published on 30 Mar 2026 - 23:17

score 427 · Answer 6 · 2026-03-25 06:50:39

427

Views

Alternative computation of eigenvalues of this tridiagonal matrix

Published on 25 Mar 2026 - 6:50

score 72 · Answer 7 · 2016-09-04 11:34:38

72

Views

$\lim_{y \to \infty} e^{-y}y^s=0$ and $\Gamma(s)$

Published on 04 Sep 2016 - 11:34

score 416 · Answer 8 · 2026-03-28 11:54:07

416

Views

Can we specifically prove that if $n\in\mathbb{Z^+}$ is composite and square-free then the ring of integers of $\mathbb{Q}(\sqrt{-n})$ is not a UFD?

Published on 28 Mar 2026 - 11:54

score 90 · Answer 9 · 2016-09-05 18:34:25

90

Views

Finding a different proof.

Published on 05 Sep 2016 - 18:34

score 224 · Answer 10 · 2016-09-06 02:16:43

224

Views

Easy proof of $A(x)$ is bounded under some condition.

Published on 06 Sep 2016 - 2:16

score 129 · Answer 11 · 2026-03-26 23:10:45

129

Views

If $f$ and $g$ are isometries, $f(A,B,C) = g(A,B,C)$ then $f=g$.

Published on 26 Mar 2026 - 23:10

score 194 · Answer 12 · 2026-03-25 04:38:09

194

Views

Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_F = 1$

Published on 25 Mar 2026 - 4:38

score 68 · Answer 13 · 2026-03-29 11:00:53

68

Views

Explaination of the Contractive Proof

Published on 29 Mar 2026 - 11:00

score 227 · Answer 14 · 2026-03-25 12:54:02

227

Views

What makes this theorem fundamental, and why is it an "isomorphism theorem"?

Published on 25 Mar 2026 - 12:54

score 77 · Answer 15 · 2016-09-13 15:32:44

77

Views

The map $C : \mathcal{H} \to L^2 (X)$ is bounded whenever it is bounded on a dense subspace $\mathcal{D} \subset \mathcal{H}$.

Published on 13 Sep 2016 - 15:32

List Question

If $a_0+\frac{a_1}{2}+\dots + \frac{a_n}{n+1}=0$ then $a_nx^n+\dots + a_0$ has a root in $[0,1]$

Alternative proof : If E and F have positive measures then E+F contains an interval

Proof for an inequality problem using the rearrangement inequality

the converse of Pythagoras Theorem

Absolute value inequality - reflection

Alternative computation of eigenvalues of this tridiagonal matrix

$\lim_{y \to \infty} e^{-y}y^s=0$ and $\Gamma(s)$

Can we specifically prove that if $n\in\mathbb{Z^+}$ is composite and square-free then the ring of integers of $\mathbb{Q}(\sqrt{-n})$ is not a UFD?

Finding a different proof.

Easy proof of $A(x)$ is bounded under some condition.

If $f$ and $g$ are isometries, $f(A,B,C) = g(A,B,C)$ then $f=g$.

Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_F = 1$

Explaination of the Contractive Proof

What makes this theorem fundamental, and why is it an "isomorphism theorem"?

The map $C : \mathcal{H} \to L^2 (X)$ is bounded whenever it is bounded on a dense subspace $\mathcal{D} \subset \mathcal{H}$.

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