It's often been brought to my attention, recently, that a graduate student --- indeed, any honest research professional --- spends most of his time being confused. Sometimes this confusion is consciously felt; sometimes it isn't. When it isn't consciously felt, it can lead to people declaring things like "Of Course the Riemann Hypothesis Ought to be True" (this is something most number theorists expect, and can't prove yet) or "Of Course You Can Square the Circle" (this is something actually true, in the right context, but not the context most people claiming it intend) or "Of Course All Widgets are Thingumy" (may or may not be true, depending on W and T, but usually not).
This happens to (honest) research professionals all the time, as well as to the more-normal core of Humanity; but there is a key difference between (honest) r.p.s and a particular subgrouping, not really quite fitting in the latter; which is that an h.r.p. is likely not to mind having been wrong, and will happily acknowledge it and receive correction. The weirdos are convinced that they Are Not Wrong, and no ammount of argument will convince them --- because the wrong conclusion derives from honest unknown confusion. Exposed to truth, honest unknown confusion learns only deceptive felt confusion, and prefers what it doesn't find confusing (even unknown confusion). I trust I make myself clear?
Anyways. I get confused a lot, and sometimes it shows. It's actually kind-of fun! Have a good day, everyone.
Monday, April 23, 2012
Monday, April 9, 2012
Monday Musings
Dear Herecomeseverybody,
If I may be allowed to imagine, I should imagine that today would be the liturgical anniversary of "Twin" Thomas' assertion "nisi videro ... non credam". Certainly, it wasn't yesterday, and it will certainly come before the Octave, when we shall hear "beati qui non viderunt et crediderunt".
At other times I've wondered why might Thomas not have been there that Easter Sunday; today, being in a mood for punning, it occurs to me that, in Vulgate and Douay, the Low Sunday Beatitude is, as all beatitudes, written in the perfect tense: "blessed are they who have not seen, and have believed". But it raises the curious question of whether this particular beatitude applies in that moment to anyone at all. The only candidates that spring to mind are Mary the Mother of Our Lord, Peter, and the disciple whom Jesus loved --- but that disciple ("he that saw these things... his testimony is true, and he knows his testimony is true...") was careful to write, earlier
Nor is Our Lord's visible revelation of his resurrection made to make up for what is wanting in his disciples' faith, but to make up for what is wanting in ours. It's one thing for the Eleven survivors among the Twelve to learn of the empty tomb and believe, but another for them to say so to those same priests and scribes and lawyers who, mere pages earlier were crying "surely thou hast a devil". That is, not only has He given them faith in His rising, he has also given them the power to say and we have seen Him. I have not seen, but I believe because Peter saw.
Let us think on these things with humble joy, and take every help to live as people new-raised from the death that is sin.
Happy Easter,
a simple one among everybody
If I may be allowed to imagine, I should imagine that today would be the liturgical anniversary of "Twin" Thomas' assertion "nisi videro ... non credam". Certainly, it wasn't yesterday, and it will certainly come before the Octave, when we shall hear "beati qui non viderunt et crediderunt".
At other times I've wondered why might Thomas not have been there that Easter Sunday; today, being in a mood for punning, it occurs to me that, in Vulgate and Douay, the Low Sunday Beatitude is, as all beatitudes, written in the perfect tense: "blessed are they who have not seen, and have believed". But it raises the curious question of whether this particular beatitude applies in that moment to anyone at all. The only candidates that spring to mind are Mary the Mother of Our Lord, Peter, and the disciple whom Jesus loved --- but that disciple ("he that saw these things... his testimony is true, and he knows his testimony is true...") was careful to write, earlier
4 ... they both ran together, and that other disciple did outrun Peter, and came first to the sepulchre.Nonetheless, let us not think on these things into thinking of ourselves as being better than Peter or John were in those moments of great amaze. They had to live through the Gospel before telling it to anyone, whereas we grow up hearing about Easter as a thing accomplished, every year about this time --- we know the ending of this story long before we know what the story is even about, or where it really started.
5 And when he stooped down, he saw the linen cloths lying; but yet he went not in.
6 Then cometh Simon Peter, following him, and went into the sepulchre, and saw the linen cloths lying,
7 And the napkin that had been about his head, not lying with the linen cloths, but apart, wrapped up into one place.
8 Then that other disciple also went in, who came first to the sepulchre: and he saw, and believed.
Nor is Our Lord's visible revelation of his resurrection made to make up for what is wanting in his disciples' faith, but to make up for what is wanting in ours. It's one thing for the Eleven survivors among the Twelve to learn of the empty tomb and believe, but another for them to say so to those same priests and scribes and lawyers who, mere pages earlier were crying "surely thou hast a devil". That is, not only has He given them faith in His rising, he has also given them the power to say and we have seen Him. I have not seen, but I believe because Peter saw.
Let us think on these things with humble joy, and take every help to live as people new-raised from the death that is sin.
Happy Easter,
a simple one among everybody
Saturday, April 7, 2012
Tomorrow, we'll sing...
echo << eof >> /dev/null
eof
II 1 So the heavens and the earth were finished, and all the furniture of them. 2 And on the seventh day God ended his work which he had made: and he rested on the seventh day from all his work which he had done.
eof
Wednesday, March 28, 2012
Thursday, March 22, 2012
More Pythagoras than Pythagoras
So, a nifty theorem: that in a tetrahedron of which the three angles at one vertex are right angles, the square of the opposite face is the sum of the squares on the three adjacent faces. By homogeneity, it follows that the square of any plane area in 3-space is the sum of the squared areas of its three orthogonal shadows on any three orthogonal planes. Here, by homogeneity I mean that orthogonal shadows of parallel plane figures are proportional to eachother as those plane figures are.
A sketch of the hypothesis, as stated:
The "opposite" face is left implicit; the adjacent faces coloured blue, red, and grey.
There are, of course, several approaches to this theorem.
One might calculate relations among the sides and areas, and proceed algebraically to conclude that the two quantities described have equivalent expressions (e.g., in terms of the three adjecent edge lengths).
Alternatively, one might take as prior what we have stated as corollary, and argue (by suitable means) that the sum-of-squared-shadows is independent of which three orthogonal planes are chosen for catching the shadows; then, since the equation obviously holds when our plane figure lies in one of the three planes --- for then it is its own shadow on its own plane, and its other two shadows have no area, and the full proposition follows.
I rather like circles, lately, so I'll do something between the two: a special plane shape (indeed a circular disc) has three orthogonal shadows on three orthogonal planes, and their squares sum to the square on the circle; one returns to general plane figures by homogeneity again.
The niftyness of this arrangement begins with the observation that the orthogonal shadow of a circular disc is an ellipse; and moreover that the major axis of this ellipse equals the diameter of the circle. Indeed, the shadow doesn't depend on where the disc lies, only on its relative attitude. Supposing the disc's center were on the shadow plane, then the shadow and disc meet in a line, precisely this major axis. Since the area of an ellipse is universally proportional to the product of its principal axes, this reduces the problem to arguing that the squares on the minor axes of the three elliptical shadows sum to the square of the circle's diameter.
To get there, we must be sneaky: consider a line segment perpendicular to our circular disc, and equal to a diameter. With some thought, one can see (this means "prove", but it's not too hard) that the minor axis and the shadow of this perpendicular lie on the same line; but they are shadows of equal and perpendicular line segments, so their squares must sum to the square on the diameter --- once in each ellipse.
At the same time, the three orthogonal shadows of the perpendicular segment squared sum to twice the square on the diameter (this is a fun exercise) adding up six squared line shaddows in three ellipses should give three times the squared diameter (since each ellipse gives one), so the three squared minor axes sum to a single squared diameter, as promised.
A sketch of the hypothesis, as stated:
The "opposite" face is left implicit; the adjacent faces coloured blue, red, and grey.
There are, of course, several approaches to this theorem.
One might calculate relations among the sides and areas, and proceed algebraically to conclude that the two quantities described have equivalent expressions (e.g., in terms of the three adjecent edge lengths).
Alternatively, one might take as prior what we have stated as corollary, and argue (by suitable means) that the sum-of-squared-shadows is independent of which three orthogonal planes are chosen for catching the shadows; then, since the equation obviously holds when our plane figure lies in one of the three planes --- for then it is its own shadow on its own plane, and its other two shadows have no area, and the full proposition follows.
I rather like circles, lately, so I'll do something between the two: a special plane shape (indeed a circular disc) has three orthogonal shadows on three orthogonal planes, and their squares sum to the square on the circle; one returns to general plane figures by homogeneity again.
The niftyness of this arrangement begins with the observation that the orthogonal shadow of a circular disc is an ellipse; and moreover that the major axis of this ellipse equals the diameter of the circle. Indeed, the shadow doesn't depend on where the disc lies, only on its relative attitude. Supposing the disc's center were on the shadow plane, then the shadow and disc meet in a line, precisely this major axis. Since the area of an ellipse is universally proportional to the product of its principal axes, this reduces the problem to arguing that the squares on the minor axes of the three elliptical shadows sum to the square of the circle's diameter.
To get there, we must be sneaky: consider a line segment perpendicular to our circular disc, and equal to a diameter. With some thought, one can see (this means "prove", but it's not too hard) that the minor axis and the shadow of this perpendicular lie on the same line; but they are shadows of equal and perpendicular line segments, so their squares must sum to the square on the diameter --- once in each ellipse.
At the same time, the three orthogonal shadows of the perpendicular segment squared sum to twice the square on the diameter (this is a fun exercise) adding up six squared line shaddows in three ellipses should give three times the squared diameter (since each ellipse gives one), so the three squared minor axes sum to a single squared diameter, as promised.
Monday, March 19, 2012
Since we're re-writing actual real positable Law like nobody's business...
Dear Sir Isaac,
I just had a bit of brie fall, if you will believe it, to the flooring of my computer's tower case (er... I keep it open because, being an eleven-year-old obsolescence it overheats otherwise... Joseph Fourier and William Thomson, you're next on my list!)
And so I'm writing to protest, sir, this so-called "law" of so-called "universal" gravitation. Obviously we don't want cheese in our computers (any more than we can avoid it) so obviously, it's time we extended to dairy curd the same exceptions that Helium and small birds seem to have been enjoying for centuries!
Yours, most gravely,
a physical progressive
I just had a bit of brie fall, if you will believe it, to the flooring of my computer's tower case (er... I keep it open because, being an eleven-year-old obsolescence it overheats otherwise... Joseph Fourier and William Thomson, you're next on my list!)
And so I'm writing to protest, sir, this so-called "law" of so-called "universal" gravitation. Obviously we don't want cheese in our computers (any more than we can avoid it) so obviously, it's time we extended to dairy curd the same exceptions that Helium and small birds seem to have been enjoying for centuries!
Yours, most gravely,
a physical progressive
Sunday, March 18, 2012
A little more, about quadrilaterals
Before getting quite lost in why the isoperimetric problem should have exactly one solution (the answer, in brief, is relative convexity, but that's a mouthful, before we know how to spell it), a little bit about Steiner's proof, adapted slightly.
Now, last time it was pointed out that rotating a figure preserves both its area and its perimeter; and this in fact remains true of separate pieces of the figure; so, for instance, pertaining to the area within the black curve below:
one may rotate the mauve or blueish pieces, preserving their area as well as the lengths of both black and red curve or line segments. This is clearly a nifty thing to consider, as it means also that if there's a quadrilateral with the same sides as the red quadrilateral, but with greater area, then aranging the mauve and green bits around that gives a new shape with the same black perimeter, and greater area (the mauve and bluey bits will be the same, and the transparent bit in the middle can be made greater).
It just so happens that the best arrangement of vertices for the red quadrilateral is (surprise!) as a cyclic quadrilateral, aka "chord" quadrilateral, which is to say, such that the vertices are all on a circle. Perhaps the other surprise is that this is always doable. There is a bit of annoyance about proving the optimality of a cyclic arrangement, in that it works out to be more natural to maximize the square of the area --- in a calculation attributed to one Bretschneider, this is a sum of two terms, one of which is determined entirely by the four side lengths, the other proportional to their product and otherwise depending only the sum of opposite vertex angles... you can see how this is getting messy, yes? The calculus is quite straight-forward, but the geometry gets to be rather icky. I think there ought to be a 4-dimensional scissors congruence proof of this fact, considering how similar it is to Heron's formula. But we mustn't jump to conclusions! (Also, such things are difficult to read).
Anyways. Suppose, then, that the curve we had wasn't a circle. Then there must be some four points of our curve that weren't cyclic. (Three points determine a circle, a fourth is either on that circle or off it). And this in turn means there was then another curve with the same perimeter, and containing greater area.
Steiner's original proof relied on a simpler case of this quadrilateral argument, specifically that of parallelograms with fixed sides. It's easier to see that the greatest such area is a rectangle; but reducing to this case uses slightly more surgical trickery beforehand. I'm undecided, just now, which approach is tidier.
Now, last time it was pointed out that rotating a figure preserves both its area and its perimeter; and this in fact remains true of separate pieces of the figure; so, for instance, pertaining to the area within the black curve below:
It just so happens that the best arrangement of vertices for the red quadrilateral is (surprise!) as a cyclic quadrilateral, aka "chord" quadrilateral, which is to say, such that the vertices are all on a circle. Perhaps the other surprise is that this is always doable. There is a bit of annoyance about proving the optimality of a cyclic arrangement, in that it works out to be more natural to maximize the square of the area --- in a calculation attributed to one Bretschneider, this is a sum of two terms, one of which is determined entirely by the four side lengths, the other proportional to their product and otherwise depending only the sum of opposite vertex angles... you can see how this is getting messy, yes? The calculus is quite straight-forward, but the geometry gets to be rather icky. I think there ought to be a 4-dimensional scissors congruence proof of this fact, considering how similar it is to Heron's formula. But we mustn't jump to conclusions! (Also, such things are difficult to read).
Anyways. Suppose, then, that the curve we had wasn't a circle. Then there must be some four points of our curve that weren't cyclic. (Three points determine a circle, a fourth is either on that circle or off it). And this in turn means there was then another curve with the same perimeter, and containing greater area.
Steiner's original proof relied on a simpler case of this quadrilateral argument, specifically that of parallelograms with fixed sides. It's easier to see that the greatest such area is a rectangle; but reducing to this case uses slightly more surgical trickery beforehand. I'm undecided, just now, which approach is tidier.
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