If you want to think binomial theorem, you probably should note that this matrix can be written as $A=I+\mu S$ where $S$ is one on the first upper subdiagonal and zero everywhere else. @NaCl
So therefore you'd really interested in $A^k=(1+\mu S)^k$. What's nice here is that $I$ commutes with $S$, so you can expand that using the binomial theorem in the obvious way.
(If you had $A=X+Y$, then $A^2=X^2+XY+YX+Y^2 \neq X^2+2XY+Y^2$. So this really does help.)
What that does leave you with is the task of computing powers of $S$. But that's actually not too bad at all, as you'll see if you compute a few small cases.
Plus, I didn't want to use $i,j$ since those are being used for matrix entries.
And it's not too hard to show that $(S^n)_{ij}=\delta_{j-i-n}$, so that gives $(A^k)_{ij}=\binom{k}{j-i}\mu^{j-i}$. Done.
Actually, I'm being a bit too quick. That only works if $j-i-n=0$ for some $n=0,1,2,\ldots,k$.
So one needs $j\geq i$.
But that's all. :/
This is definitely an example of the method making the problem harder than it really should be. It's basically making you prove the binomial theorem at the same time as you're doing matrix multiplication. @NaCl
You might need to further separate out the case where $j=i+1$, since in that case the second term will have $i+1=j+1$.
If $j>i+1$, though, then that becomes $$(A^{k+1})_{i,j}=\binom{k}{j-i}\mu^{j-i}+\mu \binom{k}{j-i-1}\mu^{j-i-1}=\left[\binom{k}{j-i}+\binom{k}{j-i-1}\right]\mu^{j-i}$$
But that sum of binomials simplifies to $\binom{k+1}{j-i}$, so $(A^{k+1})_{i,j}=\binom{k+1}{j-1}\mu^{j-i}$ if $j>i+1$.
Which matches. Yay.
That leaves the case where $j=i+1$ but that's easy enough.
Beneath the surface, though, this is really just the fact that $(1+\mu x)^{k+1}=(1+\mu x)^k+\mu x (1+\mu x)^k$.
Which expands via the binomial theorem to $\sum_{l=0}^k \binom{k}{l}\mu^l x^l+\sum_{l=0}^k \binom{k}{l}\mu^{l+1} x^{l+1}=\sum_{l=0}^k \binom{k}{l}\mu^l x^l+\sum_{l=1}^{k+1} \binom{k}{l+1}\mu^{l} x^{l}$
(The projections from the Associated Press put them at 321/670, just shy of the 336 needed for the majority.)
They'll still have a plurality, but in order to actually get stuff done they'll need to form a coalition gvmt.
And if they can't, then probably Labour will do so instead.
(Labour had actually been leading in seat counts for the time I've been following it, but the Conservatives are now in the lead and I imagine the remaining seats are in their favor.)
And, well, combinatorial game theory sounds like a pretty good description of what's happening in that collection of seats :P
Thinking about it on my own, i have determined a series of steps to determine the intersection point where the point hits the transition between the two triangles.
Partial algorithm for geodesic motion along a 3D model
The displacement of the point
Find the reflection of the starting point $p$...
and i better get out of here before I get in trouble for posting
[Ordinals] Ok, it seems we can define ordinal left tetration (since all ordinal operations are left associative, (and because of the defining property of epsilon numbers being $\epsilon_{\alpha}=\omega^{\epsilon_{\alpha}}$, i.e. a fixed point wrt right associative (instead of left associative) exponentiation of $\omega$), tetration will normally be defined left associative as well).
Unless you can, er... complete the penteract, then complete the tesseract for the extra terms, then complete the cube and finally complete the square...?
So while finding the range of a quadratic is straightforward (and finding it for a cubic is trivial on a technicality) finding the range of a quartic isn't going to be easy.
Good. I wanted to write an answer here, but kept from doing it because I don't really like that approach (my answer would be equivalent to the existing answer)
currently thinking about how to do it the way I want to
Use the ring isomorphism $\Bbb Z_n\to\Bbb Z_p\times\Bbb Z_q$, if you know how to prove that there are 4 solutions in the (c,n)=1 case almost the same proof works here
I think there's something really strange going on when we build our way from $\omega$ to $\epsilon_0$. Recall that $$\epsilon_0=sup(\{1, \omega,\omega^{\omega},\omega^{\omega^{\omega}},...\})$$
By using what we learnt back when going from $\omega$ to $\omega^{\omega}$ via multiplication, it is clear the exponential tower is building from the bottom up, yet the evaluation of it is top down
but once we reach $\epsilon_0=\omega^{\omega^{\omega^⋰}}$, how on earth are we going to evaluate it from the top down as there is no top?
The fixed points of the lower hyperoperators don't suffer from this problem. For example:
$\omega = 1+1+\cdots$ is perfectly fine evaluating from the left (i.e. is left associative)
Also true for $\omega^2=\omega + \omega + \omega + \cdots$
$\omega^{\omega} = \omega\omega\omega \cdots$
but that works because multiplication and addition are associative, but exponentiation is not
sorry typo, I mean limit ordinals, not fixed points
So that means we can always compute the exponential towers from the top down (like how exponentiation is usually defined) since the first time when that occurs (when we reached $\epsilon_0$) every step in that sequence is not a tower with an infinite ordinal of levels thus the tower can get very big, but always have a top to evaluate it from top down?
I kinda try to reverse engineer the notation from $\omega\omega$ since I knew $\omega^2 = \sup (\{\omega,\omega 2, \omega 3,\cdots\})$ and that $\omega n$ for finite $n$ can be expanded into $\omega +\cdots + \omega$ n times due to left distributivity, (and similar phenomenon holds for $\omega^n$
So I thought intuitively $\omega^2=\omega + \omega + \omega + \cdots$, similar to how the notion is used when talking about $\epsilon_0$
well, the thing about how the exponential tower should be evaluated is now clarified. But I am still trying to understand how the tetration case fails and whether the fix point equation $\alpha=\omega^{\alpha}$ is very different from the fixed point equations that came before it i.e. $\omega\alpha = \alpha$, $\omega + \alpha = \alpha$
ordinal tetration (base case) $\;\; \sideset{_{}^0}{}\alpha = 1$ (successor case) $\;\; \sideset{_{}^{\beta + 1}}{}\alpha \; = \; \left(\sideset{_{}^{\beta}}{}\alpha \right)^{\alpha}\;$ if $\beta \geq 1$ (limit case) $\;\; \sideset{_{}^{\lambda}}{}\alpha \; = \; \sup \left\{\sideset{_{}^{\beta}}{}{\alpha}: \; \beta < \lambda \right\}\;$ if $\lambda$ is a nonzero limit ordinal Reader Exercise $\;\; \sideset{_{}^{\epsilon_0}}{}{\epsilon_0} \; = \; {\omega}^{{\omega}^{{\omega}^{{\epsilon}_0 \cdot 2}}} $ (Not the same as usual tetration when finite ordinals are used, however.) — Dave L. RenfroMay 13 '13 at 19:51
the idea is to figure out if I continue the pattern on $\omega$ in the tetration case like that in the previous cases (addition, multiplication and exponentiation), where I will be relative to the epsilon numbers
Thinking about it on my own, i have determined a series of steps to determine the intersection point where the point hits the transition between the two triangles.
Partial algorithm for geodesic motion along a 3D model
The displacement of the point
Find the reflection of the starting point $p$...
