[guest39538]

And here is the final function which is g and Newtons third law


function3.jpg (33.08 kB . 882x476 - viewed 2858 times)

[guest39538]

Open your mind Mr Chemist


Take the red pill or the blue pill?



nfield.jpg (34.4 kB . 882x476 - viewed 2796 times)

[guest39538]

Enter the Matrix Neo.


nfield1.jpg (43.51 kB . 882x476 - viewed 2903 times)

[guest39538]

1=(a+b)c³    I think maybe. 


second thoughts , scrap the above notion.


Matrix A ,  u= positive


(AT)i,j = Aj,i.

[Bored chemist]

Open your mind Mr Chemist

If you had posted stuff in Chinese and I had said that I don't understand it, would you say that I need to open my mind, or would you accept that you need to post in a different language?

[guest39538]

Open your mind Mr Chemist

If you had posted stuff in Chinese and I had said that I don't understand it, would you say that I need to open my mind, or would you accept that you need to post in a different language?

I would suggest you learn Chinese and the Chinese person should also learn the different language.   

I think if you wanted too, you could create the correct maths I need.   

I am learning by the frequency of about 1 post a day if I am lucky........so I will keep having a stab at the maths until  somebody says I am correct.   I am mostly self learning and abstracting. 

[Bored chemist]

OK, I might learn Chinese in that situation.
Imagine that I learned Chinese and I still couldn't understand what  you are saying because, while you are using legitimate Chinese words, you are putting them in random patterns and claiming that you have invented a new language.
Nobody but you "understands" this "language".

Obviously, I can't understand it.
Is that because I'm not open minded?

[guest39538]

OK, I might learn Chinese in that situation.
Imagine that I learned Chinese and I still couldn't understand what  you are saying because, while you are using legitimate Chinese words, you are putting them in random patterns and claiming that you have invented a new language.
Nobody but you "understands" this "language".

Obviously, I can't understand it.
Is that because I'm not open minded?

Well then if you are open minded , let me teach you  my language while I learn yours.    Let us compare languages. 

This will be productive.

I will start with a few ''words''

1)Δt
2)ΔS
3)ΔV
4)ΔT
5)ΔF

In my words

1)change of time
2)change of entropy
3)change of velocity
4)change of temperature
5)change of force

Your language?

[petelamana]

Oh WOW!

This was one difficult convo to follow.  I would like to say that I understood everything Thebox said, but I can't.  However, I was able to follow about 75% of it...maybe 77%.  I think Bored chemist has a point about the language.  I am not an "abstractian" - coining a new word - but I am one of those guys who knows a little about numbers and those funny squiggly lines, and such.  I do so enjoy Vector analysis, Linear Algebra, and matrix conceptualization.  I think Thebox has some intriguing "stuff" in what he was trying to say, but I had to work exceptionally hard to dig it out.  I would very much like Thebox posting, or emailing me, what he is trying to say, only in a slightly different way.

Again, I enjoyed this convo.  Thank you - both of you.  I bookmarked it, and will be coming back to it as soon as I take some Tylenol.

[guest39538]

Oh WOW!

This was one difficult convo to follow.  I would like to say that I understood everything Thebox said, but I can't.  However, I was able to follow about 75% of it...maybe 77%.  I think Bored chemist has a point about the language.  I am not an "abstractian" - coining a new word - but I am one of those guys who knows a little about numbers and those funny squiggly lines, and such.  I do so enjoy Vector analysis, Linear Algebra, and matrix conceptualization.  I think Thebox has some intriguing "stuff" in what he was trying to say, but I had to work exceptionally hard to dig it out.  I would very much like Thebox posting, or emailing me, what he is trying to say, only in a slightly different way.

Again, I enjoyed this convo.  Thank you - both of you.  I bookmarked it, and will be coming back to it as soon as I take some Tylenol.

Thank you for your kind words , I will try to write a short version and inbox it you on here, I will also post the short version in this thread.

[Bored chemist]

The use of  Δt for a change in time is pretty standard (Though it's not actually something you have said in this thread so defining it is pointless.)

But you must be using a different notation when you say
"Δ0=Δ1/t=1/k "
Because zero doesn't change.

So, rather than pointlessly  explaining fairly standard symbols, why don't you try explaining what you actually said?

[guest39538]

The use of  Δt for a change in time is pretty standard (Though it's not actually something you have said in this thread so defining it is pointless.)

But you must be using a different notation when you say
"Δ0=Δ1/t=1/k "
Because zero doesn't change.

So, rather than pointlessly  explaining fairly standard symbols, why don't you try explaining what you actually said?

I think I know the problem, it is the equals sign, 

Δ0=Δ1/t=1/k

change of 0 dimension is the change of 1 over a period of time which  results in 1 divided/spread by space , missed off the over time at the end.

Δ0=Δ1/t=(1/k)/t

added- Looking at it now I could of left a Delta sign out

Δ0=1/t=(1/k)/t

[petelamana]

Okay, now I think I'm up to about 80%.   

[guest39538]

]Ok, I have had some good advice and help and now I am going to change my approach.  The new discussion will start with :


For the purposes of this discussion consider an array to be a rectangular  arrangement of coordinate values, similar to, but not exactly like the classical "matrix". My first question is , can a mono-pole retain form in a single array?

Now quite clearly all the coordinate points of a mono-pole array,  would be repulsive points to all other points of the same array.

By the laws of Physics and Coulomb's laws , the mono-pole array should always be in a state of expansion.

The array would have no strong nuclear force or gravity.   All the force would be ''centrifugal'' (outwards from a central point) .

I then considered free electrons and free protons, meaning ones that are not paired.   

How could an electron exist when considering the first array?

In a similar fashion to a Matrix, all electron points in the array  of an electron would be repulsive to each other.

I then considered the proton , what  seems true for the electron, must also be true for the proton. 

So in my first array A, I want to describe this part first.   I want to describe that array A cannot retain form and is always in a state of expansion.

[guest39538]

I think I got it.


end of argument.jpg (13.76 kB . 592x335 - viewed 2914 times)

Schematic depiction of the matrix product AB of two matrices A and B.

[guest39538]

Diagram error , sorry my bad , new edit


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[Bored chemist]

change of 0 dimension is the change of 1 over a period of time which  results in 1 divided/spread by space , missed off the over time at the end.

That's exactly the sort of word salad you need to explain.

My first question is , can a mono-pole retain form in a single array?

Ditto

[guest39538]

change of 0 dimension is the change of 1 over a period of time which  results in 1 divided/spread by space , missed off the over time at the end.

That's exactly the sort of word salad you need to explain.

My first question is , can a mono-pole retain form in a single array?

Ditto

Whatever.....who cares

[guest39538]

a + (−a) = 0

b + (-b) = 0

a+b= 1   

proof  if (-a)=(-b) then (-a)+(-b)<(ab)


That must be close?

I think I may be getting the hang of this, in practise

a + (b-b) = a?