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    • CommentRowNumber1.
    • CommentAuthorjcmckeown
    • CommentTimeMar 19th 2014

    I only wanted to say, on the one hand, bravo on the sine page in your private nlab web; on the other hand, the main inspiration for the goofy inequality was that I wanted a similar presentation to that of the natural exponential function which, as everyone knows who knows it, satisfies

    • 1+xe x 1 + x \leq e^x
    • e 2x=(e x) 2 e^{2x} = (e^x)^2

    and furthermore that this pins it down exactly.


    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeMar 19th 2014

    I see; thanks for the explanation! And sorry for the rudeness (“goofy”); I’ll get rid of it.

    So this characterization of the sine is due to you? Very interesting; I’d never seen that before. You must have devised your own proof; I’d be interested in hearing it!

    • CommentRowNumber3.
    • CommentAuthorjcmckeown
    • CommentTimeMar 19th 2014

    I don’t mind “goofy”.

    • CommentRowNumber4.
    • CommentAuthorjcmckeown
    • CommentTimeApr 6th 2014

    Goodness, I just lost a lot of editing, and I’ve got the last assignment marking of the year still to do…

    The Tricky Part of the argument is to consider that the function we want is of the form g(x)=xh(x)g(x) = x h(x), and then construct an equivalent functional equation for hh:

    h(x)=h(x/3)427x 2h(x/3) 3 h(x) = h(x/3) - \frac{4}{27} x^2 h(x/3)^3

    iff g(x)=3g(x/3)4g(x/3) 3g (x) = 3 g(x/3) - 4 g(x/3)^3 .

    In terms of the (nonlinear nonlocal) transformation TT

    T:fxf(x/3)427x 2f(x/3) 3 T : f \mapsto x \mapsto f(x/3) - \frac{4}{27} x^2 f(x/3)^3

    one calculates

    (TFTf)(x)=(Ff)(x/3)(1427x 2(FF+fF+ff)(x/3)) (T F - T f)(x) = (F - f)(x/3) \left( 1 - \frac{4}{27} x^2 (F F + f F + f f) (x/3) \right)

    which shows

    1. TT preserves the ordering of (small) FF and ff, on small intervals [δ,δ][-\delta,\delta], and
    2. in the same circumstances also that the supremum distance between TFT F and TfT f on [δ,δ][-\delta,\delta] is at most the supremum distance between FF and ff on [δ/3,δ/3][-\delta/3,\delta/3].

    Now consider the particular bounds F 0=1F_0 = 1 and f 0(x)=1x 2f_0 (x) = 1-x^2. An otherwise uninteresting calculation gives

    TF 0(x)=1427x 2 T F_0 (x) = 1 - \frac{4}{27} x^2 Tf 0(x)=11627x 2+x 4P(x) T f_0 (x) = 1 - \frac{ 16 }{ 27 } x^2 + x^4 P(x)

    for an explicit polynomial PP; in brief,

    f 0Tf 0TF 0F 0 f_0 \le T f_0 \le T F_0 \le F_0

    on some interval [r,r][-r,r], which need not be bigger than [1,1][-1,1]. This is the start of an induction argument that

    f 0T nf 0T n+1f 0T n+1F 0T nF 0F 0 f_0 \le T^n f_0 \le T^{n+1} f_0 \le T^{n+1} F_0 \le T^n F_0 \le F_0

    while at the same time (induction via item 2) we have the bounds

    |T nF 0(x)T nf 0(x)|x 29 n. | T^n F_0 (x) - T^n f_0 (x) | \leq \frac{x^2}{9^n} .

    It follows that TT has a unique fixedpoint within the specified bounds, over the interval [r,r][-r,r], and hence a unique fixedpoint over the whole real line. So that’s existence and uniqueness. Since, obviously, the functional equation and the bounds are concocted to hold for sine, we might be happy with that.

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 6th 2014

    Thanks very much, Jesse – I think I get the overall idea; I can run over this with a fine-toothed comb maybe a little later. (That 4/27 looks weirdly suggestive…)

    But where on earth does all this come from? Did you find this characterization in a book somewhere, or what? It looks just a bit off the beaten track, shall we say, at least to my eyes.

    • CommentRowNumber6.
    • CommentAuthorjcmckeown
    • CommentTimeApr 6th 2014

    I wanted to impress on some calculus students just how much easier everything is with the right tools; so, here is a complete characterization of a familiar-ish thing, but what on earth can you do with it? But after developing some calculus, e.g. Taylor series, one can start crunching digits of things like sin(1)sin(1), or prove that it’s irrational, and so forth.