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MathPrinceps's profile
Laurens Gunnarsen
Laurens Gunnarsen
Laurens Gunnarsen
@MathPrinceps

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Laurens Gunnarsen

@MathPrinceps

Mathematical physicist and mentor to mathematically talented youth. Talent is that which bridges the gap between what can be taught and what must be learned.

Joined June 2012

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    1. Timothy Gowers‏ @wtgowers Mar 19
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      #MyQuarantineInSixWords "Daddy this maths lesson was fun." Thanks to @mikeandallie for pointing out this great idea of @JDHamkins. (See in particular this blog post: http://jdh.hamkins.org/draw-an-infinite-chessboard-in-perspective/ ….) Left pics by my 9yo daughter and right ones (accidentally 8x9) by my 12yo son.pic.twitter.com/iblycnzHP6

      9 replies 37 retweets 232 likes
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    2. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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      Replying to @wtgowers @mikeandallie @JDHamkins

      It may be good to relate this project to the solution of quadratic equations. If ax^2 + 2bx + c = 0, then (ax + b)x + (bx + c) = 0 and x = -(bx + c)/(ax + b). That is, a solution, x, is a fixed point of the mapping y --> -(by + c)/(ay + b).

      1 reply 0 retweets 2 likes
    3. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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      Replying to @MathPrinceps @wtgowers and

      And this mapping has an extraordinary property: it "squares" to the identity. That is, if we do it twice, we get back out just exactly what we started with! The best way to understand this is to think of it not as a mapping of y, but as a mapping of z, where z = y + b/a.

      1 reply 0 retweets 1 like
    4. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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      Replying to @MathPrinceps @wtgowers and

      For then we find that the mapping y --> -(by + c)/(ax + b) becomes the mapping z --> D/z, where D = (b^2 - ca)/a^2. And of course the fixed points of that mapping are obviously just +/- sqrt(D). So the roots of the original quadratic are x = -b/a +/- sqrt(D).

      1 reply 0 retweets 0 likes
    5. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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      Replying to @MathPrinceps @wtgowers and

      Now, what does all this have to do with drawing perspective chessboards? Well, the first point to note is that D is the square of the distance between the two roots of the original quadratic: if ax^2 + 2bx + c = a(x - r)(x - s), then (r - s)^2 = D. So z --> [(r - s)^2]/z.

      1 reply 0 retweets 0 likes
    6. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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      Replying to @MathPrinceps @wtgowers and

      Now, the next thing to do is to draw a number line, and to mark on it the two roots, r and s. From these two roots, we're going to see how to recover the mapping y --> -(by + c)/(ay + b) in a purely graphical way, using only a straightedge. Here come the chessboards!

      1 reply 0 retweets 0 likes
      Laurens Gunnarsen‏ @MathPrinceps Mar 19
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      Replying to @MathPrinceps @wtgowers and

      So we add to the number line another mark, which corresponds to the number y. We now want to specify where the "image" number, -(by + c)/(ay + b), Is on the number line -- again, using only a straightedge. Let's put y somewhere between r and s, just to keep things simple.

      9:29 AM - 19 Mar 2020
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        2. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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          Replying to @MathPrinceps @wtgowers and

          The next step is to draw another line through y (i.e., in addition to the original number line.) (I've found that the picture tends to come out nicest if you make this new line a little short of perpendicular to the original number line. But this is certainly not necessary.)

          1 reply 0 retweets 0 likes
        3. Laurens Gunnarsen‏ @MathPrinceps Mar 19
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          Replying to @MathPrinceps @wtgowers and

          Then draw two more lines, this time through r (say), so that each line intersects the line through y in a convenient (i.e., nearby) location. Let's call these two intersection points P and R. And now for the penultimate step: connect P and R to s with two more lines.

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