#how to do this?

57 messages · Page 1 of 1 (latest)

sage mauve
cedar nimbusBOT
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silver scaffoldBOT
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Graph = G for Garbage

wise tiger
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@sage mauve

sage mauve
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Dot product 0

wise tiger
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so <R,R'>=0

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what does this tell us

sage mauve
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<> iska mean?

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if you asking me what r.r' then it will be 0

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dot product

wise tiger
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yes

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what's R and R'?

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A+B and A-B correct?

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$(A+B)\cdot(A-B)=\norm{A}^2-\norm{B}^2$

silver scaffoldBOT
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Graph = G for Garbage

wise tiger
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can you verify this?

sage mauve
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@wise tiger

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A=B

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1:1

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B

wise tiger
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llAll=llBll

sage mauve
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Yeah so??@wise tiger

sage mauve
sage mauve
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Bro

wise tiger
sage mauve
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Mod sign

wise tiger
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${\mathbf{x}\in\bR^d:\norm{\mathbf{x}}_2=1}$

silver scaffoldBOT
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Graph = G for Garbage

sage mauve
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Bro you are making it more complicated to read

wise tiger
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the 1-sphere

wise tiger
sage mauve
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Yeah so?

wise tiger
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$f(\theta)=\left(\cos\theta,\sin\theta\right)$

silver scaffoldBOT
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Graph = G for Garbage

wise tiger
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$\forall\theta\in\bR,\norm{f(\theta)}_2=1$

silver scaffoldBOT
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Graph = G for Garbage

wise tiger
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you get what i mean?

sage mauve
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So??

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What it mean here?

wise tiger
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$\norm{\mathbf{x}}=\norm{\mathbf{y}}$ doesn't imply $\mathbf{x}=\pm\mathbf{y}$ here

silver scaffoldBOT
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Graph = G for Garbage

wise tiger
# sage mauve

they are vectors but the vectorspace is not mentioned

sage mauve
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|x|=|y|

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Represents lines

wise tiger
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x and y are vectors here

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not real numbers

wise tiger
silver scaffoldBOT
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Graph = G for Garbage

wise tiger
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but not in $\bR^d$ for $d>1$

silver scaffoldBOT
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Graph = G for Garbage

wise tiger
sage mauve
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Can you tell me what i have to do with this next ... I'm not understanding the sphere theory here sorry

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+close