<< We have cheerfully assumed that the sum of two unimodal functions is necessarily at worst bimodal. This may be false in general, even when they are symmetric, and shifts of one another. So why exactly is it true in this case? >> Writing down the second difference again, but solving \Delta^2 f(i) = 0 for i in terms of n, k instead, yields a quartic in i with at most 4 real roots; so f has at most 3 maxima and minima, and is at worst bimodal. These chimaerical arguments --- mixing discrete differences and continuous notions like inflection --- do strike me as rather unsatisfactory --- not to mention liable to confusion if not outright error. But if the alternative is involving higher transcendental functions ... Turning to the more general question of summing a pair of unimodal functions: here's an example of a trimodal sum of a symmetric unimodal with its own shift: [ 1 3 5 7 9 11 12 15 16 19 19 16 15 12 11 9 7 5 3 1 0 0 0 0 0 ] + [ 0 0 0 0 0 1 3 5 7 9 11 12 15 16 19 19 16 15 12 11 9 7 5 3 1 ] = [ 1 3 5 7 9 12 15 20 23 28 30 28 30 28 30 28 23 20 15 12 9 7 5 3 1 ] (shame about the spacing). The summand is however not concave on its support. Is there an example which remedies this defect? Fred Lunnon On 6/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Writing f(i) = n_C_i + n_C_j where i+j = n+k , I asked for the boundary n = n_0(k) between f(i) bimodal and unimodal.
What I mistermed "binomial distribution" actually just meant regarding f as a function of a continuous variable: solving for the second differential vanishing appeared to give n_0 = k^2 - 4/3 - O(1/k^2) for no dip in the middle of the smooth curve at all.
What I mistermed the "discrete" (sic) bound was actually chimaerical, taking the first difference of f then treating it as a function of a continuous variable, which appeared to give n_0 = k^2 - 5/3 - O(1/k^2) for maybe a very small dip.
Gareth's "really discrete" argument simply divides out common factorials from the second difference of f(i) in the centre of the combined support, which vanishes at n_0 = k^2 - 6/3 permitting wobbles between several successive integers.
So yup, it really is obvious after all, once I can kick my addiction to floating-point numerical computation.
But 'ang abaht there --- we have cheerfully assumed that the sum of two unimodal functions is necessarily at worst bimodal. This may be false in general, even when they are symmetric, and shifts of one another.
So why exactly is it true in this case? Back to the drawing board! (But I can't work up much enthusiasm for tangling with digamma ...)
Fred Lunnon
On 6/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
My thanks to Warren, Gareth, Gene --- I will digest their replies.
In the meantime, I realised that I conflated discrete (Pascal triangle) and continuous (binomial distribution) problems: the answers are not quite the same. For the continuous case instead I find experimentally
n_0 = k^2 - 4/3 - (32/45)/k^2 - (4/9)/k^4 + O(1/k^6) .
WFL
On 6/28/14, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
In the Gaussian approximation, normalized to mean 0 and variance 1, we have
exp(-(x-s/2)^2/2)) + exp(-(x+s/2)^2/2)) = 2 exp(-(x^2+s^2)/2) cosh sx.
This has extrema where the derivative of the log is 0.
s tanh sx = x.
When s < 1, there is a single maximum at 0. When s > 1, there is a minimum at 0 and 2 maxima. The critical point is s = 1, when there is a triple root. That is, the critical shift s is 1 standard deviation, which for the binomial(n,k) is sqrt(n)/2.
-- Gene