6 ms·

The Fundamental Theorem of Algebra: A Visual Approach [pdf]

122 points by loop22 10y ago

stablemap 10y ago
Apart from the typesetting, is this different from this submission from earlier in the day? https://news.ycombinator.com/item?id=13686519 https://news.ycombinator.com/item?id=13686519
jgamman 10y ago
ironically, linked article has no visuals...
- rjtobin 10y ago
  I thought so too, but they are after the references. Still I was disappointed that it wasn't a "proof without words" article.
  - Smaug123 10y ago
    I'd be very impressed to see a proof without words of the FTA :)
- stablemap 10y ago
  Look at the end, past the bibliography.
- kmill 10y ago
  If you want some visuals, here is something I made a few months ago: https://math.berkeley.edu/~kmill/math54fa16/roots/roots.html https://math.berkeley.edu/~kmill/math54fa16/roots/roots.html (it accepts polynomials in any reasonable format, factored or not, and gives both the domain coloring and the circle image). I thought I had implemented a visualization for Gauss's original proof, but I only did that for the reverse problem of giving the roots (and poles) and seeing the zero curves for the real and imaginary parts: http://www.kylem.net/stuff/homsim/homsim2.html http://www.kylem.net/stuff/homsim/homsim2.html (hit tab a few times). For dealing with non-polynomials, I made https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html shortly after. It handles most complex functions, like polynomials, quotients, exponentiation, the exponential function, logarithms, and trigonometric functions (though possibly not inverse trigonometric functions). This gives the domain coloring along with the Riemann sphere. One day I'll add the ability to define functions so that one may, for instance, explore Moebius transformations. This second program works by compiling the input complex function into a shader program (which is output to the Javascript console, and it might be more interesting than the complex analysis itself to some people here).
  - tempodox 10y ago
    I love it, especially that the natural logarithm is acually named `ln'. I wrote myself a mapper of the complex plane for fractal generation that compiles expressions to in-memory closures to compute iterations and coloring. If I set the iteration count to 1, it produces similar images. But it's a shell command only, eating an ASCII file and producing PNG. No web iterface (yet).
    - kmill 10y ago
      I just added an iteration operator to https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html Example: z^2-0.4+0.57i+0.03*e^(i t) iterate 150 Use shift+left and shift+right to adjust the rate at which t varies. It starts at constant 0. It also now has a composition operator: "z^2 then e^z then 1/z" computes 1/(e^(z^2)). The precedence rule is that "f1(z) iterate 3 then f2(z) iterate 3" is "(f1(z) iterate 3) then (f2(z) iterate 3)", which is f2(f2(f2(f1(f1(f1(z)))))).
      - tempodox 10y ago
        Neat :)
- pjdorrell 10y ago
  Also: http://thinkinghard.com/domkol/main.html http://thinkinghard.com/domkol/main.html This interactive visualization lets you see that the number of times the polynomial function "winds" around the circumference of the circle is equal to the number of zeroes inside the same circle. (Caveat: the interaction to move the zeroes around requires a mouse.)
raister 10y ago
Typo: Guass --> Gauss
cousin_it 10y ago
I think the simplest explanation is by analogy to the one-dimensional case: 1) Let f be a polynomial of odd degree over the reals. Take a large "circle", i.e. a pair of points (x, -x) where x is some large number. Since f behaves like its highest degree term, the circle's image under f lies on different sides of 0. If you gradually diminish the circle to nothing, its image must cross 0 at some point. So the polynomial has a root. 2) Let f be a polynomial of positive degree over the complex numbers. Take a large circle, i.e. a set of points z such that |z| is some large number. Since f behaves like its highest degree term, the circle's image under f winds n times around 0. If you gradually diminish the circle to nothing, its image must cross 0 at some point. So the polynomial has a root. (It crosses 0 each time the winding number decreases by 1, which is why you have n roots.)
- ice109 10y ago
  that escalated quickly lol.

122 points by loop22 10y ago

Apart from the typesetting, is this different from this submission from earlier in the day? https://news.ycombinator.com/item?id=13686519 https://news.ycombinator.com/item?id=13686519

jgamman 10y ago

ironically, linked article has no visuals...

rjtobin 10y ago
I thought so too, but they are after the references. Still I was disappointed that it wasn't a "proof without words" article.
- Smaug123 10y ago
  I'd be very impressed to see a proof without words of the FTA :)
stablemap 10y ago
Look at the end, past the bibliography.
kmill 10y ago
If you want some visuals, here is something I made a few months ago: https://math.berkeley.edu/~kmill/math54fa16/roots/roots.html https://math.berkeley.edu/~kmill/math54fa16/roots/roots.html (it accepts polynomials in any reasonable format, factored or not, and gives both the domain coloring and the circle image). I thought I had implemented a visualization for Gauss's original proof, but I only did that for the reverse problem of giving the roots (and poles) and seeing the zero curves for the real and imaginary parts: http://www.kylem.net/stuff/homsim/homsim2.html http://www.kylem.net/stuff/homsim/homsim2.html (hit tab a few times). For dealing with non-polynomials, I made https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html shortly after. It handles most complex functions, like polynomials, quotients, exponentiation, the exponential function, logarithms, and trigonometric functions (though possibly not inverse trigonometric functions). This gives the domain coloring along with the Riemann sphere. One day I'll add the ability to define functions so that one may, for instance, explore Moebius transformations. This second program works by compiling the input complex function into a shader program (which is output to the Javascript console, and it might be more interesting than the complex analysis itself to some people here).
- tempodox 10y ago
  I love it, especially that the natural logarithm is acually named `ln'. I wrote myself a mapper of the complex plane for fractal generation that compiles expressions to in-memory closures to compute iterations and coloring. If I set the iteration count to 1, it produces similar images. But it's a shell command only, eating an ASCII file and producing PNG. No web iterface (yet).
  - kmill 10y ago
    I just added an iteration operator to https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html https://math.berkeley.edu/~kmill/tools/zgraph/zgraph.html Example: z^2-0.4+0.57i+0.03*e^(i t) iterate 150 Use shift+left and shift+right to adjust the rate at which t varies. It starts at constant 0. It also now has a composition operator: "z^2 then e^z then 1/z" computes 1/(e^(z^2)). The precedence rule is that "f1(z) iterate 3 then f2(z) iterate 3" is "(f1(z) iterate 3) then (f2(z) iterate 3)", which is f2(f2(f2(f1(f1(f1(z)))))).
    - tempodox 10y ago
      Neat :)
pjdorrell 10y ago
Also: http://thinkinghard.com/domkol/main.html http://thinkinghard.com/domkol/main.html This interactive visualization lets you see that the number of times the polynomial function "winds" around the circumference of the circle is equal to the number of zeroes inside the same circle. (Caveat: the interaction to move the zeroes around requires a mouse.)

raister 10y ago

Typo: Guass --> Gauss

cousin_it 10y ago

I think the simplest explanation is by analogy to the one-dimensional case: 1) Let f be a polynomial of odd degree over the reals. Take a large "circle", i.e. a pair of points (x, -x) where x is some large number. Since f behaves like its highest degree term, the circle's image under f lies on different sides of 0. If you gradually diminish the circle to nothing, its image must cross 0 at some point. So the polynomial has a root. 2) Let f be a polynomial of positive degree over the complex numbers. Take a large circle, i.e. a set of points z such that |z| is some large number. Since f behaves like its highest degree term, the circle's image under f winds n times around 0. If you gradually diminish the circle to nothing, its image must cross 0 at some point. So the polynomial has a root. (It crosses 0 each time the winding number decreases by 1, which is why you have n roots.)

ice109 10y ago
that escalated quickly lol.