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OH MY GOD, I proved a theorum that I made up. PLEASE READ

  1. Jan 4, 2005 #1
    Please find any errors in this theorum. I am 16 yearsold, and after thinking abuot how to find the position in change of time in supply and demand curves, I came to this conculsion and developed a theorum, (please excuse my lack of using Latex):

    [tex]\lim n->0 = sigma n=1 to n=x [ 1 * 2 = 2(2) = 4 ... (2n - 2) ] \int n to x plusorminus \partial f(x,n)/\partial f(t) dn dt dx.[/tex]

    Where dn = some instant rate in time,
    [tex]dt[/tex] = the position of time at some point
    dx = some number at change in dn/dt.

    I will call the formula the fundamental theorum of economics. When supply equals demand..


    Think I should publish it?
     
    Last edited: Jan 4, 2005
  2. jcsd
  3. Jan 4, 2005 #2

    Gokul43201

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    Please, don't try and publish it just yet.

    Right now, it just looks like a lot of gibberish. Please don't feel insulted by this, but there's a lot of fixing and explaining that needs to be done first. And even after that, it's likely to be nothing new or morelikely, it's wrong. So I suggest you don't keep your hopes high - the number of real scientific discoveries made by people in their teens is negligible.

    First, let's try and understand what you've written, as this is incredibly hard to decipher.

    [tex] lim_{n \rightarrow 0} \sum_{n=0}^x [1 \cdot 2 = 2 \cdot 2 = 4...(2n-2)] = \int n^x \frac{\partial f(x,n)}{\partial f(t) }~dndtdx [/tex]

    This is an attempted first correction, but clearly there are problems with this, especially the terms in the summation. Clearly, it makes no sense to have "equal signs" in the sum.Also, it makes no sense to take the limit of the summing parameter. Please fix these. There are other problems too, but let's do things one at a time.


    Please click quote, and correct the TeX to make it mean what you have in mind. Also, what is the function [itex]f [/itex] ?
     
  4. Jan 4, 2005 #3

    [tex] lim_{n \rightarrow x} \sum_{n=0}^{x} [1 \cdot 2 .. 2 \cdot 2 .. 4...(2x-2)] = \int_{n}{x} \frac{\partial f(x,n)}{\partial f(t) }~dndtdx [/tex]


    In order to find the position of the function (to find the difference between the change in demand and supply), should I square it so that:
    [tex] lim_{n \rightarrow x} \sum_{n=0}^{x} [1 \cdot 2 .. 2 \cdot 2 .. 4...(2x-2)] = \int_{n}{x} \frac{\partial^2 f(x,n)}{\partial f(t) }~dndtdx [/tex]
     
    Last edited: Jan 4, 2005
  5. Jan 4, 2005 #4
    To know what f is you must know all the varibles of f.

    [tex]f(n)dn, f(t)dt, f(x)dx.[/tex]

    If n is some point, then [tex]{\partial dn^2/dt[/tex] is the position in time of a point, and what that point is at that value.

    dn = some instant rate in time.
    dt = the value of that point in time
    dx = some number at (delta)dn/(delta)dt.
     
    Last edited: Jan 4, 2005
  6. Jan 4, 2005 #5

    Gokul43201

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    I'm sorry QuantumTheory,

    Your expression is riddled with so many mathematical mistakes that it would serve no good to debug it. It looks like you have not learnt very much yet of calculus or series summation. I strongly suggest you be patient and learn your basics first. This will require having to complete school and college, so please don't be impulsive. You have a lot to learn, so keep at it.
     
  7. Jan 4, 2005 #6
    Heh, I do understand summation as well as all calculus about vector calculus. However, I don't do the problems, (I read the books for pleasure;not for work.) So when I try to make one of my own, it is full of errors because I'm not used to writing those types of equations. I DO UNDERSTAND it though. And I tried and made a few mistakes because what I was THINKING didn't get put in equations exactly the right way I intended it to.

    I do know the basics...And I've read several books. I have not been in a calculus class, however, since my school will not allow that until I've taken algebra, geometry, trigonmetry, and pre calculus. I'm in my junior year, and if each one of those classes is 1 year long, obviously I will not have enough time.

    Plus, I go to a charter school. The highest they offer is .. Geometry :(

    Anyway, it's very likely I expressed what I was thinking in an equation wrong. How I thought of the equation was like this:

    I was in Economics class and was reading and learned about a theory by a person someone developed it in the 1800's (Please remember the teacher does not speak during this. It is all computerized). It was that as population increases, food supply decreases. And over time, we will not be able to feed everyone.

    So I decided to make a limit problem with this information, thinking about how geometrically this would happen. I came to the conclusion people would grow at a rate of (considering the rate is constant) (2n - 2).

    The growth formula I planned this out of goes as follows:
    1 x 2 = 2 . 2 x 2 = 4. 4 x 2 = 8. 8 x 2 = 16. 16 x 2 = 32 ... (2n) <-- (correct?)

    I saw how the equation was going, and tried to find a formula if you didn't know the next number. I thought it was (2n - 2) but I think I was wrong even on this.
     
  8. Jan 4, 2005 #7
    Personally, I don't think anybody can really learn calculus without doing problems. I think you can read all the calculus books you want, but if you don't do any problems, you'll never really understand calculus. I bet there are others out there who will also echo this opinion.

    that's what solving problems will do for you. it gets you used to writing out equations, understanding the notation and teaches you where things go and how things are done.

    it's good that you're reading about calculus and learning about it, but don't forget about the problems. you shouldn't skip them just because you think you understand calculus by reading a few books. there's a reason why those problems are there and why you will get seemingly endless lists of problems to solve from your instructors. It's not for punishment despite what some people might think.

    If you're going to go through the trouble of reading the books, do the problems. it will benefit you in the long run.
     
  9. Jan 4, 2005 #8

    Gokul43201

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    Yes, even though this is a very basic kind of growth known as exponential growth.

    [tex] p(n) = p_0 \cdot 2^n [/tex]

    where n = number of years, p(n) is the population after n years and po is the initial population = 1, say. Then, by the formula, you'll see that :

    [tex] p(1) = 2, ~p(2) = 4,~ p(3) = 8, ~etc. [/tex]

    That you are having trouble with this simple (for someone in college, at least) concept means that you must just be patient and go through the process of learning things rigorously.

    You can not learn math by reading.

    And as far as the economics goes, supply and demand schedules are thoroughly understood, and are typically described from statistical data rather than theoretical models, mostly because of the number of factors that influence the equilibrium, and because of the uncertainty of human behavior.
     
  10. Jan 5, 2005 #9

    selfAdjoint

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    Gokul, just a comment on economics. There is actually a large and active field of theoretical ecionomics, with postulates and theorems and all. In the last few years it has been extended to cover (simple, but often counterintuitive) human behavior. Check out the topics which received the Nobel Memorial prize in Economics.
     
  11. Jan 6, 2005 #10

    russ_watters

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    You will never be able to convince a math teacher or technical journal publisher that you understand math unless you can demonstrate it. That should be self-evident. It is one thing to look at equations in a book and think to yourself "hmm, that makes sense," but quite another to reproduce the calculations or derivations yourself.
    May I ask what level math you are taking/have taken? I thought it was required to take more than that to graduate high school (and I'm certain colleges require more). Most people who intend to go to college take algebra 1 and geometry with trig in junior high. Please don't take this the wrong way, but it seems you may need some remedial math classes to get you back up to speed.
     
  12. Jan 6, 2005 #11

    Yes, in fact, I do need to know how to factor. I'm having problems with advanced factoring techiniques. You gotta walk befoer you can run, ya know :)

    I reproduced an equation using partial derivatives yesterday. I had f(x,y) = x^2 + 2y. I differentated f(x) while holding f(y) constant. I got 2x. I then differentated 2y, I got 2. My answer from [tex]f(x,y) = \partial x/\partial y
    = 2x/2 = x[/tex]

    The math I need to know to graduate from high school is at least up to geometry.

    I've taken the following in school: Pre-algebra, part of Algebra I, Geometry. . The rest I've learned on my own accord.
     
    Last edited: Jan 6, 2005
  13. Jan 9, 2005 #12
    Learn to run before learning how to walk, that's what I can say. There's a difference between knowing a topic and being competent in it. Any violinist can play a note, but a good violinist is one who plays that note, and owns it.
     
  14. Jan 9, 2005 #13

    Curious3141

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    [tex]f(x,y) = x^2 + 2y[/tex]
    [tex]\frac{\partial f}{\partial x} = 2x[/tex]
    [tex]\frac{\partial f}{\partial y} = 2[/tex]
    [tex]df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = 2xdx + 2dy[/tex]

    As others have said please learn to crawl before you walk. I say this with absolutely no malice, but at the moment you really have no idea what you're writing with all those math symbols. Please take the time and effort to learn existing theory with rigour and dedication before trying to come up with new theory.
     
  15. Jan 10, 2005 #14
    Please understand that I do understand what these symbols mean. I am dyslexic, and tend to write things backwards.

    What you wrote is what I meant. The symbol [tex]\frac{\partial f}{\partial x}[/tex] means differentating f with respect to x.

    Differentating [tex]\frac{\partial f}{\partial y}[/tex] means taking the derivitive of y with respect to f.

    Therefore, if a forumla has two functions, let's say f(x,y), then the answer could be expressed in 2 parts (In real life examples.), or one.

    So, yes, I do know what I am talking about. I just don't know how to express it, and that is what I need to pratice. I can pratice this by doing the problems on paper.

    Edited: I'd appreciate it if you could give me a link to some of these theories you say of. Thank you.
     
    Last edited: Jan 10, 2005
  16. Jan 10, 2005 #15
    First sentence is correct.

    Second sentence is not.

    Third sentence : f(x,y) is not a "formula with two functions". It is a "function of two variables" and nothing else.

    It is crucial to use proper terminology.
     
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