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 Spivak Calculus on Manifolds Exercises
09-07-2016, 08:01 PM
Post: #1
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Spivak Calculus on Manifolds Exercises
1-1. Prove that $|\mathbf{x}|\le{\small\displaystyle{\sum_{i=1}^n}|x_i|.\quad((|a_1|+\cdots+|a_n​|)^2=|a_1|^2+\cdots+|a_n|^2+2\sum_{1\le i< j\le n}|a_i||a_j|)}$
1-2. When does equality hold in Th 1-1(3)?
Sol. $\small|\mathbf{x+y}|=|\mathbf{x}|+|\mathbf{y}|\iff \langle\mathbf{x,\,y}\rangle = |\mathbf{x}||\mathbf{y}|\iff \exists \lambda,\eta \ge 0\;(\lambda\mathbf{x}=\eta\mathbf{y})\wedge(\lambda^2+\eta^2>0).\;\underset{\,}{\;}\square$
1-3. Prove that $|\mathbf{x-y}|\le |\mathbf{x}|+|\mathbf{y}|\;\;\small(|\mathbf{x-y}|\overset{\text{Th.1-1(3)}}\le |\mathbf{x}|+|-\mathbf{y}|)$
$\qquad |\mathbf{x-y}|= |\mathbf{x}|+|\mathbf{y}|\overset{\text{Ex.1-2}}{\iff}\small\exists \lambda,\eta\ge 0\,(\lambda\mathbf{x}+\eta\mathbf{y}=\mathbf{0})\wedge(\lambda^2+\eta^2>0).\underset{\,}{\quad\square}$
1-4. Prove that $\big||\mathbf{x}|-|\mathbf{y}|\big|\le |\mathbf{x}-\mathbf{y}|$
Proof $\;\because\;|\mathbf{x}|=|\mathbf{y}+(\mathbf{x}-\mathbf{y})|\le|\mathbf{y}|+|\mathbf{x}-\mathbf{y}|,\quad\therefore |\mathbf{x}|-|\mathbf{y}|\le |\mathbf{x}-\mathbf{y}|\underset{\,}{\,}$ and so
$\qquad -(|\mathbf{x}|-|\mathbf{y}|)=|\mathbf{y}|-|\mathbf{x}|\le |\mathbf{y}-\mathbf{x}|=|\mathbf{x}-\mathbf{y}|.\underset{\,}{\;}$Therefore $\big||\mathbf{x}|-|\mathbf{y}|\big|\le |\mathbf{x}-\mathbf{y}|.\;\square$
1-5. Let the distance $d(\mathbf{x},\mathbf{y})$ between $\mathbf{x},\,\mathbf{y}$ be $|\mathbf{y}-\mathbf{x}|$. Show that $d(\mathbf{x},\mathbf{z})\le d(\mathbf{x},\mathbf{y})+d(\mathbf{y},\mathbf{z})\underset{\,}{.}$
Proof $|\mathbf{z}-\mathbf{x}|=|\mathbf{z}-\mathbf{y}+(\mathbf{y}-\mathbf{x})|\underset{\text{Th.1-1(3)}}{\le} |\mathbf{z}-\mathbf{y}|+|\mathbf{y}-\mathbf{x}|.\quad\square\quad$(triangle inequality)
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