Question

Legendre polynomials and inner product

score 311 · Answer 1 · 2026-04-12 21:38:10

311

Views

Legendre polynomials and inner product

Published on 12 Apr 2026 - 21:38

score 95 · Answer 2 · 2026-04-08 20:50:23

95

Views

Confusion about inner product between two functions

Published on 08 Apr 2026 - 20:50

score 184 · Answer 3 · 2026-04-13 10:19:02

184

Views

Dot product $u\cdot (v\cdot w)$

Published on 13 Apr 2026 - 10:19

score 127 · Answer 4 · 2026-04-11 09:46:55

127

Views

tensor definition confusion as multilinear map

Published on 11 Apr 2026 - 9:46

score 58 · Answer 5 · 2026-04-16 19:01:03

58

Views

find $f$ such that $f(\frac{1}{2})=\int_{0}^{1}f(t)g(t)dt$

Published on 16 Apr 2026 - 19:01

score 73 · Answer 6 · 2026-04-17 06:13:59

73

Views

Deriving the identity $\langle\phi_1|\phi_1 \rangle + \langle\phi_2|\phi_2 \rangle \geq 2Re(\langle\phi_1|\phi_2 \rangle)$

Published on 17 Apr 2026 - 6:13

score 78 · Answer 7 · 2026-04-17 05:22:02

78

Views

Deriving identity $\|\langle\phi_1|\phi_1 \rangle + \langle\phi_2|\phi_2 \rangle\| \leq \| \space|\phi_1 \rangle \space\| + \| |\phi_2 \rangle \|$

Published on 17 Apr 2026 - 5:22

score 52 · Answer 8 · 2026-04-13 14:51:42

52

Views

Cannot identify mathematical approach

Published on 13 Apr 2026 - 14:51

score 163 · Answer 9 · 2026-04-12 05:19:17

163

Views

If $V=W$ (vector spaces), then $\det A^{\dagger}=\det A$, where $A$ is a map and $A^{\dagger}$ is its adjoint

Published on 12 Apr 2026 - 5:19

score 39 · Answer 10 · 2026-04-10 22:48:50

39

Views

Quick Question: If $\varphi$ is a linear map and inner product $\langle \varphi x, y\rangle = 0$, why is $\varphi \equiv 0$?

Published on 10 Apr 2026 - 22:48

score 84 · Answer 11 · 2026-04-09 12:18:16

84

Views

Proof of Hermitian adjoint

Published on 09 Apr 2026 - 12:18

score 169 · Answer 12 · 2026-04-13 15:00:18

169

Views

Generalizing the geometric interpretation of dot product to simple $k$-vectors

Published on 13 Apr 2026 - 15:00

score 57 · Answer 13 · 2026-04-13 03:44:47

57

Views

Show that $\{ v : \langle v, v\rangle > 0$ is a subspace of $V$

Published on 13 Apr 2026 - 3:44

score 40 · Answer 14 · 2026-04-13 17:05:22

40

Views

Dot product and work

Published on 13 Apr 2026 - 17:05

score 174 · Answer 15 · 2026-04-16 05:53:32

174

Views

Let $A \subset E$ where $E$ is an inner product space. Show that the orthogonal complement $A^{\perp}$ is closed in $E$.

Published on 16 Apr 2026 - 5:53

List Question

Legendre polynomials and inner product

Confusion about inner product between two functions

Dot product $u\cdot (v\cdot w)$

tensor definition confusion as multilinear map

find $f$ such that $f(\frac{1}{2})=\int_{0}^{1}f(t)g(t)dt$

Deriving the identity $\langle\phi_1|\phi_1 \rangle + \langle\phi_2|\phi_2 \rangle \geq 2Re(\langle\phi_1|\phi_2 \rangle)$

Deriving identity $\|\langle\phi_1|\phi_1 \rangle + \langle\phi_2|\phi_2 \rangle\| \leq \| \space|\phi_1 \rangle \space\| + \| |\phi_2 \rangle \|$

Cannot identify mathematical approach

If $V=W$ (vector spaces), then $\det A^{\dagger}=\det A$, where $A$ is a map and $A^{\dagger}$ is its adjoint

Quick Question: If $\varphi$ is a linear map and inner product $\langle \varphi x, y\rangle = 0$, why is $\varphi \equiv 0$?

Proof of Hermitian adjoint

Generalizing the geometric interpretation of dot product to simple $k$-vectors

Show that $\{ v : \langle v, v\rangle > 0$ is a subspace of $V$

Dot product and work

Let $A \subset E$ where $E$ is an inner product space. Show that the orthogonal complement $A^{\perp}$ is closed in $E$.

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