DescentBB

I've been trying to figure out for an hour an equation that would make a sigmoid shape, and have points at (0,0), (½,½), and (1,1). The range of numbers entered into this formula would only be between 0 and 1, so anything else doesn't matter.

I've toyed with instead converting the number range of 0 to 1 to -1 through 1, and then using x² for numbers above 0 and -(x²) for numbers below 0 and somehow 'resizing' the graph so (-1,-1) would be at (0,0), (0,0) at (½,½), and (1,1) being left alone, but I can't figure out how to do that final step!

Anyone have an idea? I swear I saw this formula/graph before in math class, but I can't figure it out!

Edit: I think I actually need that graph flipped and then rotated 90°; I basically need a function where I enter time in, and it speeds up to 0.5, then slows to 1.

Oh, that's a logistic function. Try looking up logistic functions and see what you can find out.

You can do it with a third order equation I think.

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Snoopy is exactly right.

You know y=x^3 gives you roughly the shape you want in the (-1,1) range, right? Just adjust it a bit to put it exactly where you want it:

y = ((2x-1)^3 + 1) / 2

Easy to verify this hits all the points you want.

Foil wrote:Snoopy is exactly right.

You know y=x^3 gives you roughly the shape you want in the (-1,1) range, right? Just adjust it a bit to put it exactly where you want it:

y = ((2x-1)^3 + 1) / 2

Easy to verify this hits all the points you want.

Thank you so much, that is exactly what I want! I was just going to go the long method but you saved me

No prob.

Oh, I think a logistic function has finite extrema, but that third-order polynomial ranges over all real numbers. It's not easy to tell with that image in your link.