# Simple math problem.. I think.. If I was good at math :)

I am going nuts over this (possibly) simple equation that I need for a fun Processing project.

I need to have the "calculation" or "equation" for X

Y is the number of times we go (X=X+X), and the end result should always be 1

EXAMPLE:

IF Y is 2 then X = 0.25, because

0.25+0.25=0.5 (Y1)

0.5+0.5=1 (Y2)

HOW on earth does one calculate X? (if Y is for example 214.7787009)?

I imagine it'd be something like X=1/SQR Fish Y Something Hotdog?!?

There's SO much respect to the person cracking this! :) (Or I'll delete the post and pretend this never happened, if the answer is really simple :)

Thanks :) (and respect)

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Domain of function is x IR and not x=0.

That's true.

Where did I say that x=0?

The command x=x+x is valid for most microcontroller, but the function is completely different then the operation in math.

True, that's what I was trying to point out. When Frits used x=x+x he meant it as a programming instruction and that's how I interpreted it, not as an equation.

When I said:

Also, x=x+x is a perfectly valid ecuation, the only solution for x is 0 :)

I was just trying to reply to:

First of all, in math x=x+x makes no sense, because x=x+x <=> x=2x <=> 1=2 (contradiction).

I was trying to point out that x=x+x can be interpreted as an ecuation and if you do, the solution is x=0.

You have even a problem with x=0, because then you have a possible 0 through 0 divison which is not defined in some domains, but anyhow, that's another topic.

So let's simply say, z=2x, to avoid all the problems.

Frits now defies the operation z should be performed y-times and the result of the multiplication should always be 1, which simply means yz=1 <=> 2xy=1.

Can you try to derivate your formula from Frits definition in a common mathematical way that I can see how you came to an exponential equation?

I guess Frits wanted to compute the gravitational force between his dick and Mars but failed to explain it in a proper way :P

Frits now defines the operation z should be performed y-times and the result of the multiplication should always be 1, which simply means yz=1 <=> 2xy=1.

That's the problem right there. The operation z (z=2x) performed y-times does not mean multiply z by y. On each step the x changes and the next step will double the new value.

so if y is 3 you get something like: (2*(2*(2*x)))

for y=4 you get (2*(2*(2*(2*x))))

for y=5 you get (2*(2*(2*(2*(2*x)))))

-> since we don't need the parens we can write this as 2*2*2*2*2*x or (2^5)*x

y-times is defined as a simple multiplication according to the propositional logic, otherwise you would define the variable not as y, but let Frits simply do a lookup table with some x/y values. Then we are more clear what he wants to compute. Topic closed for me as long as the definitions and notations are ambiguous which is detrimental to any serious discussion.

Would I be correct in guessing you are using this to identify which samples to examine from a universe of samples? to achieve a certain reliability or satisfaction?

I knew it will happen again. Just commented here http://letsmakerobots.com/node/34018#comment-91690 about "I am the smartest guy in the room" but have to face that I am not the one :-) Seems Markus is beating all of us also when birdmun and TH did a great job trying to explain higher maths :-)

For me it's just another unknown miracle how people can calculate such things.  Ok, I am not the worst in math as I instantly saw the funny thing in Markus's example picture but if it's going to be complicated then I am trying to seek for other solutions.

What do you want to achieve with that formula Frits?

To be fair, I have a mathematician degree (B.Sc.) :P

If you want to take a look, Lumi, this is what I am doing beside robotics. Working on the Collatz conjecture.

I know that you are a mathematician Markus :-) (but had to google B.Sc.just now)

What I don't like is something endless to the top....I like defined areas/ranges/numbers with an END.

So I want to know what Frits is planning with this calculation.