calculusgeometric-topologygeometry
Find the length of a main diagonal of an n-dimensional cube, for example the one from $(0,0,…,0)$ to $(R,R,…,R)$
I tried to use induction to prove that its $\sqrt{n}R$ but I'm stuck on writing the proof that for an n-dimensional cube, the perpendiculars that with that main diagonal compose the right-angled triangle are the main diagonal of the n-1-dimensional cube and another R-length-ed perpendicular
Thanks
The path goes through the middle point of common opposite side considering two squares only.
As per your working, $a^2+b^2 = 4c^2, a = \sqrt{bc}$
$ \implies b^2 + bc = 4c^2$
$ \displaystyle (b + \frac{c}{2})^2 = \frac{17c^2}{4}$
$ \displaystyle \frac{b}{c} = \frac{-1 \pm \sqrt{17}}{2}$
Now if $A$ and $B$ are two adjacent vertices of rhombus and $O$ is the intersection of diagonals, $\triangle OAB$ is a right angled triangle. Cosine of one of the angles of the triangle will be $\displaystyle \frac{b/2}{c}$.
As a diagonal bisects the vertex angles in the rhombus, one of the angles of the rhombus will be,
$\theta = 2 \arccos (\frac{b}{2c})$ and so the other angle will be $(\pi-\theta)$.
As you know $\frac{b}{c}$, you can now find the angle. Note that you will discard $ \displaystyle \frac{b}{c} = \frac{-1 - \sqrt{17}}{2}$ as angle between diagonal and side is less than $90^0$.
Best Answer
Simple derivation:
From my naive perspective, you are looking for a distance between points $(0,0,\dots,0)$ and $(R,R,\dots,R)$. Since you are in $n$-dimensional Euclidean space, their separation is $\sqrt{(R-0)^2 + \dots + (R-0)^2} = \sqrt{n} R$. Would that be sufficient?
Looking at it geometrically, if the length in $(n-1)$ dimensions is $l_{n-1}$, you can use the fact that, since the $n^{th}$ direction is perpendicular to any direction in the $(n-1)$ dimensional subspace, Pythagorean addition of distances holds and $l_n = \sqrt{l_{n-1}^2 + R^2}.$ Starting from $l_1 = R$, you get $l_n = \sqrt{n} R$ by induction.
More detailed derivation using differential geometry:
To make it more explicit, one can use the metric of $n$-dimensional Euclidean space $g_{ab} = \delta_{ab}$ for $a,b \in [1,2,\dots,n]$. The "distance" $s$ is then defined as $$ (\mathrm{d} s)^2 = \sum_{a,b} g_{ab} \mathrm{d}x^a \mathrm{d}x^b \, , $$ in general. Let's have a curve $x^a = x^a(t)$ parametrised by $t$. Then $$ \frac{\mathrm{d} s}{\mathrm{d} t} = \sqrt{\sum_{a,b} g_{ab} \frac{\mathrm{d}x^a}{\mathrm{d}t} \frac{\mathrm{d}x^b}{\mathrm{d}t}} = \sqrt{\sum_a \left ( \frac{\mathrm{d}x^a}{\mathrm{d}t} \right ) ^2 } \, . $$ The diagonal going from $(0,0,\dots,0)$ to $(R,R,\dots,R)$ can be described by the curve $x^a(t) = Rt$ for $t \in [0,1]$. The total length of the curve is $$ s = \int_0^1 \mathrm{d} s = \sqrt{\sum_a \left ( \frac{\mathrm{d}x^a}{\mathrm{d}t} \right ) ^2 } \mathrm{d}t = \int_0^1 \sqrt{\sum_a \left ( R \right ) ^2 } \mathrm{d}t = \int_0^1 \sqrt{n} R \mathrm{d} t = \sqrt{n} R \, . $$ Therefore the length of the diagonal in $n$ dimensions is $\sqrt{n} R$.