The line is the shortest path between two points

Let p, q be two points of \mathbb{R}^n , u be a unit vector and let \gamma be a curve passing through those points. (that is there exist a, b such that \gamma(a)=p , \; \gamma(b)=q.) Prove that the shortest path between these two points is the line.

Solution

The following facts hold:

(1)   \begin{equation*} \gamma'(t)\cdot u \leqslant \left \| \gamma'(t) \right \| \end{equation*}

(2)   \begin{equation*} \overrightarrow{q}-\overrightarrow{p}=\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,{\rm d} t}\end{equation*}

Integrating (1) we get that:

(3)   \begin{equation*} \int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \leqslant \int_{a}^{b} {\| \overrightarrow{\gamma}\,'(t) \|\,dt}=\mathcal{L}(\overrightarrow{\gamma})\end{equation*}

where \mathcal{L}(\overrightarrow{\gamma}) is the length of the curve between the points p, q. Taking \overrightarrow{ u}=\frac{\overrightarrow{q}-\overrightarrow{p}}{ \|\overrightarrow{q}-\overrightarrow{p} \|} we have

    \begin{align*} \|{\overrightarrow{q}-\overrightarrow{p}}\|&= \frac{\|{\overrightarrow{q}-\overrightarrow{p}}\|^2}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\ &=\bigl({\overrightarrow{q}-\overrightarrow{p}}\bigr)\cdot\frac{\overrightarrow{q}-\overrightarrow{p}}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\ &= \bigl({\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)}\bigr)\cdot\overrightarrow{ u}\\ &\stackrel{(2)}{=}\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,dt}\cdot\overrightarrow{u}\\ &=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \\ &\stackrel{(1)}{\leqslant} \int_{a}^{b} { \| \overrightarrow{\gamma}\,'(t) \|\,dt}\\ &\stackrel{(3)}{=}\mathcal{L}(\overrightarrow{\gamma})\,. \end{align*}

 

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Order of a group’s element

Let \mathcal{G} be a group and a, b \in \mathcal{G} such that a^5=e and

(1)   \begin{equation*} aba^{-1}=b^2  \end{equation*}

where e is the identity element of the group. Find the order of b.

Solution

We will begin stating a lemma:

Lemma: If aba^{-1}=b^r then a^n b a^{-n}=b^{r^n}.

Proof:


First we multiply with a and a^{-1} from right and left respectively. Thus one can see that

(2)   \begin{equation*}a^2ba^{−2}=ab^ra^{−1}  \end{equation*}

Thus

(3)   \begin{equation*} ab^ra^{−1}=(aba^{−1})^r \end{equation*}

and

(4)   \begin{equation*} a^2ba^{−2}=(aba^{−1})^r \end{equation*}

Now, we use the main relation and so

a^2ba^{−2}=b^{r^2}

By repeating the previous procedure, one can prove the result.

Using the lemma we see that a^nba^{-n}=b^{2^n} and thus b^{31}=e. Since 31 is prime the order of b will be either 31 or 1.

 

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A contour integral

Define

\displaystyle f(z) = \frac{1}{z} \cdot \frac{1-2z}{z-2} \cdots \frac{1-10z}{z-10}

Evaluate the contour integral \displaystyle \ointctrclockwise \limits_{|z|=100} f(z) \, {\rm d}z.

Solution

We are applying the substitution u=\frac{1}{z} thus:

    \begin{align*} \ointctrclockwise \limits_{\left | z \right |=100} f(z) \, {\rm d}z &=- \ointclockwise \limits_{\left | w \right |=1/100} f \left ( \frac{1}{w} \right ) \frac{{\rm d}w}{w^2} \\ &=\ointctrclockwise \limits_{\left | w \right |=1/100} \frac{1}{w} \prod_{n=1}^{5} \frac{w-2n}{1-2nw} \, {\rm d}w \\ &=- 2\pi i 3840 \end{align*}

since the function \displaystyle g(w) = \frac{1}{w} \prod_{n=1}^{5} \frac{w-2n}{1-2nw} has only one pole in the specific contour , namely w=0.

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Finite matrix group

Let \mathcal{G} be a finite subgroup of {\rm GL}_n(\mathbb{C})  this is the group of the  n \times n invertible matrices over \mathbb{C}). If \sum \limits_{g \in \mathcal{G}} {\rm Tr}(g)=0 then prove that \sum \limits_{g \in \mathcal{G}} g =0.

Solution

Let us suppose that |\mathcal{G}| = \kappa and x= \frac{1}{\kappa} \sum \limits_{g \in \mathcal{G}} g . We note that for every h \in \mathcal{G} the depiction \varphi: \mathcal{G} \rightarrow \mathcal{G} such that \varphi(g)=h g is 1-1 and onto. Thus:

    \begin{align*} x^2 &=\left ( \frac{1}{\kappa} \sum_{g \in \mathcal{G}} g \right )^2 \\ &= \frac{1}{\kappa^2} \sum_{g \in \mathcal{G}} \sum_{h \in \mathcal{G}} gh\\ &= \frac{1}{\kappa^2} \sum_{g \in \mathcal{G}} \sum_{h \in \mathcal{G}} g\\ &= \frac{1}{\kappa} \sum_{h \in \mathcal{G}} \left (\frac{1}{\kappa} \sum_{g \in \mathcal{G}} g \right ) \\ &= \frac{1}{\kappa} \sum_{h \in \mathcal{G}} x\\ &= \frac{1}{\kappa} \kappa x \\ &=x \end{align*}

Thus the matrix x is idempotent. thus its trace equals to its class. (since we are over \mathbb{C} which is a field of zero characteristic.) Hence

\displaystyle {\rm rank} \;(x) = {\rm trace} \;(x) = \frac{1}{\kappa} \sum_{g \in \mathcal{G}} {\rm trace} \; (g) =0

This implies that x=0 hence \sum \limits_{g \in \mathcal{G}} g =0.

The exercise can also be found at mathematica.gr

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The volumes are equal

Prove that for every constant c>0 the set

\mathcal{B}_{f, g} = \{ (x, y, z) \in \mathbb{R}^3 : (x-f(z))^2 + (y-g(z))^2 \leq c, \quad z \in [a, b] \}

has the same volume for all continuous functions f, g: [a, b] \rightarrow \mathbb{R}.

Solution

For every z_0 \in [a, b] on the plane z=z_0 the set

\{{(x,y,z_0)\in\mathbb{R}^3\,|\,(x-f(z_0))^2+(y-g(z_0))^2\leqslant c} \}\,, \quad c>0

is a disk of constant radius \sqrt{c}. Thus the set

\mathcal{B}_{f,g}=\{{(x,y,z)\in\mathbb{R}^3\,|\,(x-f(z))^2+(y-g(z))^2\leqslant c\,,\;a\leqslant z\leqslant b}\}

is a “cylinder” which axis is the curve

 \big\{{(f(z),g(z),z)\in\mathbb{R}\;|\;z\in[a,b]}\big\}

and its radius is \sqrt{c}.More specifically , the set \mathcal{B}_{f, g} is bounded by the planes z=a and z=b and for every t \in [a, b] the intersection of \mathcal{B}_{f, g} with the plane z=t is the disk

\mathcal{D}_t=\big\{{(x,y,z)\in\mathbb{R}^3\,|\,(x-f(t))^2+(y-g(t))^2\leqslant c\,,\;z=t}\big\}

The area of this disk is the same with the disk

\mathcal{D}=\big\{{(x,y,z)\in\mathbb{R}^3\,|\,x^2+y^2\leqslant c\,,\;z=0}\big\}

The latter one has an area of

    \begin{align*} \oiint \limits_{\mathcal{D}} \left \| \bar{N} \right \| \, {\rm d} \sigma &= \oiint \limits_{\mathcal{D}} \left \| \bar{e}_3 \right \|\, {\rm d}\sigma \\ &= \int_{-\sqrt{c}}^{\sqrt{c}} \int_{-\sqrt{c-x^2}}^{\sqrt{c-x^2}} \, {\rm d} (y, x)\\ &=\pi c \end{align*}

It follows from Cavalieri’s Principal that \mathcal{B}_{f, g} has the same volume and that is equal to

\displaystyle \mathcal{V}\left ( \mathcal{B}_{f, g} \right ) = \int_{a}^{b} \pi c \, {\rm d}t = \pi c (b-a)

which is the same for all continuous functions f, g:[a, b] \rightarrow \mathbb{R}.

A somewhat visualization would be the following:

Rendered by QuickLaTeX.com

This was an exam’s question somewhere in Greece. The answer was migrated from mathematica.gr .

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