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

[markusb]

Sorry, no.Domain of function

Sorry, no.

Domain of function is x∈ IR and not x=0. If x can only be 0, your equation 2^yx=1 makes no sense anyhow, because then 2^y0=1 <=> 0=1.

The command x=x+x is valid for most microcontroller, but the function is completely different then the operation in math. It multiplies the current value of x by two and store then this calculated value into the variable x, deleting the old one firstly. So everytime you call this command it does nothing more then multiply the current value of x by 2, buffer then this value, delete the old one, and store the new buffered one into the variable x. But you can not use this command as an equation in math

 

[antoniocaciuc]

One more time from the top?

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.

[vishurockssrivastava]

Hey Markus!!

The last bit is stolen from Facebook!!

[vishurockssrivastava]

Hey Markus!!

The last bit is stolen from Facebook!!

[vishurockssrivastava]

To Lumi,

Don’t calculate it, use a good scientific calculator 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).

[markusb]

You have even a problem with

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 ■■■■ and Mars but failed to explain it in a proper way

[tucker98]

I saw that on fb too
I saw that on fb too

[markusb]

Nope. I have it in my

Nope. I have it in my collection a long time as many other funny math comics. Never said though, I have created it.

[markusb]

Shame on you, vishu, if you,

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!!

[vishurockssrivastava]

Dude, I don’t need one…

Just that it can be done that way as well…

[metalmonkeelad]

hmm…not good in math…but…

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

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

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

i.e:

x+x=y

y=214.7787009

x=?

formulae: y/2=x

sol’n: 214.7787009 / 2=107.3893505

checking: 107.3893505 x 2 or 107.3893505+107.3893505= 214.7787009

x=107.3893505

[antoniocaciuc]

We’re getting closer

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*(2x)))

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

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

 -> since we don’t need the parens we can write this as 22222x or (2^5)*x

 

[markusb]

y-times is defined as a

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.

[vishurockssrivastava]

My only “Serious comment”…

…is that you are not able to phrase out your problem to us “O Master of LMR”. Also, this is not a mathematical problem but a physical one. Yes we’ll be using a lot of maths to solve it (atleast till I get a PHD in quantum physics and prove that maths is wrong and insufficient).

What I understand from your last para is this- You want to calculate how much gravity is present at the edge of an object relative to a particular surface you define to have gravity equal to 1. All you know is the distance of the object (but from where? Your reference point or what?). What I’m giving below is a possible solution-

Now, going back to what I’ve studied in my old physics book (Concepts of Physics Part-1 by Dr.H.C. Verma Chapter 6, page 204), here is what Newton wrote when the apple fell on his head-

Force(F)=[Gravitational Constant(G)*Mass of 1st object(M₁)Mass of 2nd object(M₂)]/{[Distance b/w the 2 objects®]^2}

Or- F=GM₁M₂/(r^2).

Now, what it simply means is that each object applies a gravitational force on every other object. Simply said, right now as you read this, Pluto is trying its best to pull you off the ground. However, as the distance r is very large and r^2 is just too large for me to mention is, F tends to 0 (limits yes…).

Now coming to the frame of earth (ie all calculations that will be done only on earth),

g(gravitational acceleration aka gravity you feel)=F/M₂ where M₁ is mass of earth and M₂ is mass of any other object.

which implies, g=GM₁/(r^2) where r^2 is the distance of center of object 2 from center of the earth.

Now, as you said, any object will have gravity equal to 1, so g₁ felt by that object is = GM₁/(r₁^2) where r₁ is the distance of that object from the center of the earth. Now, if you know the distance of the object 2, ie the object you want to find out about, g₂=GM₁/(r₂^2). So if g₁=1 and g₁/g₂=(r₂^2)/(r₁^2)=(r₂/r₁)^2, g₂=(r₁/r₂)^2.

If you are out of earth frame ie in general space- g₂=(m₂*r₁^2)/(m₂*r₂^2) if g₁=1. It also means that g₂

>=1

if m₂>m₁ and r₂>r₁.

 

What I believe you are not understanding is that gravity is relative and not just that any object has a particular gravity at one point. For the simplest example, consider your weight. If say, its 60kgs on earth, its not kg as in mass but kgwt. On moon, it’ll be 20kgs only (gravity on moon is 1/3 of earth roughly). Your gravity is not just a quantity. Its an attractive force which acts on both bodies in opposite directions. If earth pulls you, you also pull it back with the same force. That’s one of the reason’s why we say there is no gravity in outer space. That’s because there is nothing to pull you back. You need some 3rd body to get an idea of what you are going to define gravity by and then calculate the reltives of the 2 bodies.

I don’t know if I’ve made my point, inform me.

[metalmonkeelad]

hey vishu, don’t say that

hey vishu, don’t say that you can prove that maths is wrong! no!!! maths is a language! and dispoving it is like disproving language, plus, maths is the one who can make it easier to calculate properties and make theories explain and learn via numbers, it’s hard to just explain a theory with just words, you need to use another language, and that language is maths!

[metalmonkeelad]

is it only for addition?

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

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

0.25x4=1

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

[lumi]

I am out now LOL

I am out now LOL

[mfmc]

Don’t know if you’ve managed

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! 

But, 

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! 

 

 

 

[mfmc]

I AM to late to the party,

I AM to late to the party, sadness

[markusb]

I still not understand why

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 y≠0. But that's not what was originally intended :)