Does anyone know a simple way to manipulate x(1-x) to get a double

>>8705165
>Does anyone know a simple way to manipulate x(1-x) to get a double humped function such that f(x) still equals 0 at x=0 and x=1?
How about multiplying x(1-x) by a function that doesn't diverge at 0 or 1 or adding a function that also vanishes at 0 or 1?

[math] (c\,x^n)^2 \cdot \left( 1 - x^{2n} \right)^k [/math]

You are suggesting you want symmetry but havent explicitly said you want -1 as a root. It isnt going to work with something like x^m*(1-x)^n, because it will have terms with an odd power and so wont have symmetry. As long as you delete the +0.2 from the function in the picture it seems to have everything you want.

>>8705186
I would not bet on saying this has 2 maxima most of the time.

>>8705183
The simplest examples of continuous and smooth functions that do this are polynomials themselves so this is fairly pointless

>>8705165
Seriously? Just take off the +0.2

File: bumps.gif (560KB, 599x1896px) Image search: [Google]

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>>8705186
Plot[(c x^n)^2 (1 - x^(2 n))^k, {x, -1.3, 1.2}, PlotRange -> {0, 1}]

Thanks for the replies. I wasn't clear in the OP. I want the double maxima (they don't even have to be symmetrical) within the domain (0,1).

So in other words I want the original picture squeezed such that both the maxima fall between 0 and 1.

>>8705235
sound like the second interation of the logistis equation

>>8705256
*iteration, *logistic

>>8705235
They already do that. Just take the 0.2 off.

>>8705235
Just replace x by (2x-1) in the formula

File: wp_ss_20170226_0004.png (156KB, 768x1280px) Image search: [Google]

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>>8705263
The +0.2 just shifts it up. If I remove it, then there will be a minimum at 0, but there will still be only one maxima between 0 and 1. I want two maxima between 0 and 1. Basically, I want a curve to look like an M, with the two maxima between 0 and 1, and the intercepts at exactly 0 and 1.

>>8705283
3.56x(1-x)*(1-3.57x(1-x))

>>8705292
Thank you!

>>8705165
(coshx+1)(x^2-1)

>>8705804
(-coshx+1)(x^2-1)
*

File: asda.png (23KB, 661x535px) Image search: [Google]

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-5/2(2x -1)^2((2x - 1)^4 - 1)

you can just translate the graph up and down (f(x) + c or f(x) - c)
left and right (substitute x for x + c or x - c)

and

stretch/compress the function larger (|k| > 1 for k * f(x)) or smaller (|k| < 1 for k * f(x))
or substitute x for kx to stretch it left and right untill you get the function you want (provided it looks something like what you're starting with)