Math.SE report 201507
My overall SE posting volume was down this month, and not only did I
post relatively few interesting items, I've already written a whole
article about the most interesting
one. So this will be a short
report.
I already wrote up Building a box from smaller
boxes on the blog
here. But maybe I have a
couple of extra remarks. First, the other guy's proposed solution
is awful. It's long and complicated, which is forgivable if it had
answered the question, but it doesn't. And the key point is “blah
blah blah therefore code a solver which visits all configurations of
the search space”. Well heck, if this post had just been one
sentence that ended with “code a solver which visits all
configurations of the search space” I would not have any complaints
about that.
As an undergraduate I once gave a talk on this topic. One of my
examples was the problem of packing 31 dominoes into a chessboard
from which two squares have been deleted. There is a simple
combinatorial argument why this is impossible if the two deleted
squares are the same color, say if they are opposite corners: each
domino must cover one square of each color. But if you don't take
time to think about the combinatorial argument you could waste a lot
of time on computer search learning that there is no solution in
that case, and completely miss the deeper understanding that it
brings you. So this has been on my mind for a long time.
I wrote a few posts this month where I thought I gave good hints.
In How to scale an unit vector !!u!! in such way that !!a u\cdot
u=1!! where !!a!! is a
scalar I think I
did a good job identifying the original author's confusion; he was
conflating his original unit vector !!u!! and the scaled, leading
him to write !!au\cdot u=1!!. This is sure to lead to confusion. So
I led him to the point of writing !!a(bv)\cdot(bv)=1!! and let him
take it from there. The other proposed solution is much more rote
and mechanical. (“Divide this by that…”)
In Find numbers !!\overline{abcd}!! so that
!!\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}!!
the OP got stuck partway through and I specifically addressed the
stuckness; other people solved the problem from the beginning. I
think that's the way to go, if the original proposal was never going
to work, especially if you stop and say why it was never going to
work, but this time OP's original suggestion was perfectly good and
she just didn't know how to get to the next step. By the way, the
notation !!\overline{abcd}!! here means the number
!!1000a+100b+10c+d!!.
In Help finding the limit of this series !!\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!! it would
have been really easy to say “use the formula” or to analyze the
series de novo, but I think I almost hit the nail on the head
here: it's just like !!1+\frac12 + \frac{1}{4} + \frac{1}{8} +
\frac{1}{16} + \frac{1}{32} + \cdots!!, which I bet OP already
knows, except a little different. But I pointed out the wrong
difference: I observed that the first sequence is onefourth the
second one (which it is) but it would have been simpler to observe
that it's just the second one without the !!1+\frac12!!. I had to
review it just now to give the simpler explanation, but I sure wish
I'd thought of it at the time. Nobody else pointed it out either.
Best of all, would have been to mention both methods. If you can
notice both of them you can solve the problem without the advance
knowledge of the value of !!1+\frac12+\frac14+\ldots!!, because you
have !!4S = 1+\frac12 + S!! and then solve for !!S!!.
In Visualization of Rhombus made of Radii and
Chords it seemed
that OP just needed to see a diagram (“I really really don't see how
two circles can form a rhombus?”), so I drew one.
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