The Daddy wrote:

GurtTractor wrote:

The Daddy wrote:

-ln(1-0.5x)+0.5 might be better

These just come up with a syntax error. Thanks though.

Looks like log defaults to natural log in that language. Try y=-log(1-0.5x)+ 0.5

Although maybe something based on tan would be better. Something like:

Y=0.4tan(2x-1)+0.5 looks pretty close to what you posted. Obvs not in the right syntax

The Daddy wrote:

Although maybe something based on tan would be better. Something like: Y=0.4tan(2x-1)+0.5 looks pretty close to what you posted. Obvs not in the right syntax

Yeah -

Got this which looks pretty good, will test it out with the MIDI controller in a min -

The number at the end moves the graph up or down, the number after the x in the bracket moves left or right, the number in front will squash or stretch vertically, the number in front of the x inside the bracket will squash and stretch horizontally. 

Dunno how easy it is to play around with the numbers in your software, using something like Geogebra can be useful to try a few and see how they look:

https://www.geogebra.org/calculator

I- a maths idiot- have a question to ask for a project i'm working on. Maybe for someone here the answer will be obvious. If not, maybe someone can point me in the direction of what concepts I would need to read around to answer it for myself. Apologies in advance as im sure there are definitely better ways of phrasing this all. 

Imagine you receive the jumbled numbers from 1 to 100, and your job is to arrange them into ascending/descending order. However, you do not know how to count, and the only question you can ask to help you sequence the numbers is whether number X is higher or lower than number Y. 
The order you get this list of numbers is not entirely random. In fact, each number's position in the unordered list is a more likely approximation of its final position on the ordered list. We could go with a probability swing of half a percentage point. So, for example, the number 5 is more likely to be in the first 10 numbers (approx 1.5% chance for any given number), less likely to be mid way in the list (around 1% chance for each), and least likely to be at the end (0.5% chance).  

So my question is: what would be the best way of doing this, if the aim is to ask the fewest number of questions possible?

Maybe just bubble sort? Cos swaps will be lower than average?

Merge sort maybe.

So you thought finding that line was tricky, how would you model an ass?

https://mathematica.stackexchange.com/questions/66538/how-do-i-draw-a-pair-of-buttocks

This year’s “it was too hard” gcse maths question:

Got an A at GCSE maths, couldn't do that now but I think that's  a use it or lose it issue.

Yeah it’s not bad at all really. Lots of ways of breaking the shape down too. It’s for the grade 8/9 kids so the ones complaining aren’t who it’s targeted at.

Here's a thing. Just throwing it out there.

According to Turing (I'm reading his biog at the minute - he's one of the very few heroes I have), AI will not be able to solve the halting problem.

In other words, we can't use AI to predict if any program has an infinite loop in it.

Turing's proof uses the absurdity that is created by feeding the definition of a Turing machine that can detect an infinite loop into itself. It creates a paradox.

It throws up rather fundamental questions about Godel's incompleteness proof if it ever does. It would disrupt the underlying mathematics of computing, and the axioms of mathematics itself.

This gets at the heart of the hype around AI and what it can do.  According to the Chinese room thought experiment, AI will never understand anything since all computers are still "just" Turing machines.