Let's Make Robots!

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



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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Don't know if you've managed to crack this yet, there is a lot of posts in this thread, and I cannot read them all! 


If I understand correctly, you're looking to find out for a value of y, with what size of x you can add x to itself y times to sum to 1?

So something along the lines of: 1 = (2^y)*x

Which if you know y, you can solve for x as: x = 1/(2^y) 

So for y=4 for example that gives you x=0.0625 

which would work out as 

0.0625 + 0.0625 = 0.125

0.125 + 0.125 = 0.25

0.25 + 0.25 = 0.5

0.5 + 0.5 = 1


And for y=214.7787009 x=2.214E-65

But then I might be wrong about what it is you are after! 




I still not understand why the term "y times" should represent an exponential function rather than a simple multiplication. In the link you will find the text "3 multiplied by 4 (often said as "3 times 4")".

But let's say, the term "y times" should represent, the addition (x+x=2x) shall be exponential repeated. Then we have to follow the exponential law:

This would lead to following euqation:

Solving for x

This means this equation has only one solution if y0. But that's not what was originally intended :)

I AM to late to the party, sadness

cause if it is only for addition/multiplication, then it is only limited to what the given X is

so.....you are looking for the formulae for x? hmmm...how about:

y=any number but not = to zero, since having zero calculations gives nothing :P

since y is the times of the number that is going to be calculated......y can be the addition of a number that ends up into 1

ie: A=0.25, Y=4, X=1, X=A x Y


or 0.25+0.25+0.25+0.25=1

X=gravity, since as you've said, yer gravity=1 and x=1...so it is fixed

so if you are looking for the calculation of the gravity, then. A is the object(or mass...i think) and Y is the distance

lol, x+x=x? like 2+2=2 if x=2? hmm...should be undefined if the formula was that :P

if that was a typo...do you mean y=x+x? cause if it is, then if y=2, 2=1+1, x=1 :P

if you are asking about addition...then to know what x is, with a given y, then just devide the y into 2 and there you have the x :P





formulae: y/2=x

sol'n: 214.7787009 / 2=107.3893505

checking: 107.3893505 x 2 or 107.3893505+107.3893505= 214.7787009


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?

Don't calculate it, use a good scientific calculator :P Good ones have limits, differnetiation and integration in it as well!!

Or if you are a guy who likes it the old school way, I believe I can send you a copy of my class 12th Maths Book (if it still exists btw. Dunno, have to check).

Shame on you, vishu, if you, who has this education, would need a scientific calculator to do such simple limit calculations. My most beloved Indian mathematician Srinivasa Ramanujan would be turning in his grave!!

Just that it can be done that way as well...

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.