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I think I made a mathematical discovery! Check it out!

  1. Sep 16, 2006 #1
    Don't laugh if this basic stuff and has already been discovered centuries ago but I am just in basic Calculus. Here is what I have came up w/.

    In y=x^odd number(just pick any) graph, the greater the coeffecient, the closer it is to the graph x=0. But if we were to imagine what the graph of y=infinityx^odd number(just pick any) look like, I believe it will look like a vertical line at x=0.(you can check this theory by finding greater and greater secant line from the origin of this graph to a farther and farther point, the slope keeps increasing.)

    Same thing happens w/ the graph of y=x^infinity. Just at an exponentially faster rate but as long as you have a vertical line, I don't think slower and faster matters.

    So I think that y=infinityx^odd number(just pick any) is equivalent to y=x^infinity which is equivalent to the graph x=0.

    Now, y=smallestpositive#x^2 is equal to y=0 graph. And same w/ exponent but I think we all know that already; y=x^0. Which equals to y=1 graph.

    I am like an infinitologist, always trying to understand infinity in other ways, I have read so many books and articles over infinity that you won't even believe it. I think if we can understand infinity, we well get answers to all our questions.:smile:

    What do you think?
    Last edited: Sep 16, 2006
  2. jcsd
  3. Sep 16, 2006 #2
    I think your function would look something like this:

    y(x) = 0,\,\,\,\,-1<x<1,
    y(x) = \pm 1, \,\,\,\, x=\pm 1,
    y(x) = -\infty, \,\,\,\, x<-1,
    y(x)= \infty, \,\,\,\, x>1.

    If you take a number less than one, and multiply it by itself, you get a number small than the one you started with. If you then mulitply the result by the same number you get an even smaller number. So in the limit of doing this infinitely many times you will get 0. Similarly, if you multiply a number greater than 1 (or less than -1) by itself inifinitely many times the result will diverge. If you multiply 1 by itself infinitely many times the result will always be 1.
    (Sorry for the messy tex but I can't get arrays working).
    Last edited: Sep 16, 2006
  4. Sep 16, 2006 #3
    So is this something new or common sense?(I am a senior in high school)
  5. Sep 16, 2006 #4

    matt grime

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    What makes you think that 'infinity', whatever you may mean by it, is not understood mathematically? We seem to have a prefect grasp of it. Writing things l ike y=infinityx^odd power doesn't mean anything. Just put limits in and it all drops out properly, which is after all what infinity in the sense you seem to want to use it means.
  6. Sep 16, 2006 #5

    matt grime

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    what coefficient?

    Just take time to write things out properly and remember to treat things like infinityx as inherently meaningless at the moment (It is unless you chose to adopt some meaning for it).
  7. Sep 16, 2006 #6
    By the way that was for the function [tex]y(x)=\lim_{n\to\infty}x^{2n+1}[/tex]. This kind of thing isn't new I suppose. In physics you sometimes use something called the Fermi distribution function, which is a smooth function of x say, except when you take one of it's parameters to 0 then it becomes step-like, in much the same way as your function does. For [tex]0\leq x < x^{\prime}[/tex] the Fermi function is [tex]1[/tex], at [tex]x=x^{\prime}[/tex] it is [tex]\frac{1}{2}[/tex], and for [tex]x>x^{\prime}[/tex] it is 0.

    I may aswell write it here:

    f(E) = \frac{1}{\exp((E-E_f)/kT)+1}

    When you take the limit of [tex]T\to 0[/tex] for this function you get the above step like behaviour.
  8. Sep 16, 2006 #7
    I think he means the exponent
  9. Sep 16, 2006 #8

    I think it'll probably lose its enigma when you learn something about the various notions of infinity in mathematics. They're pretty well-defined :tongue2:
  10. Sep 16, 2006 #9
    since this topic has pretty much ended, I was just wondering, if I take the antiderivitive of a position graph, what do I get? Anything useful?
  11. Sep 19, 2006 #10
    Nvm that, btw, what does an increasing curve(like a half of the positive quadratic) in an acceleration graph indicate?(I know technically but hoping you guys could give me a real life example.

    Also, there is acceleration at linear rate, geometric rate, exponential rate, what comes after that?

    One more thing, for 1,2,3.... there is first, primary, single, and I can't remember, there is one more, but do you guys know more of these? I would really appreciate if you guys could tell me.(don't need to list all of them, just tell me the starting point.)
    Last edited: Sep 19, 2006
  12. Sep 19, 2006 #11
    You mean in an acceleration-time graph? Its just one of those things which are harder to visualize because its hard to distinguish constant acceleration from a change in acceleration just through experience. (Even when you are falling you are just going through a phase of constant acceleration).

    Let me give it a try. Say you are in an intergallactic spaceship which is taveling at a constant velocity (i.e. a horizontal line [y=0] on your acceleration-time graph). How do you know this? Because you are experiencing no force. Now say the ship started accelerating at a constant rate (i.e. a flat horizontal line [y=constant] in the accleration-time graph). How do you detect this? Because you are feeling a constant force (maybe a throbbing pain in your chest). Now say the ship is going through a change in acceleration (i.e. a curve in the acceleration time graph). How do you know? Because each instant of time, you would be experiencing more and more force! (i.e. the pain in your chest would throbb more and more every second!).

    That's basically how you can visualize the three aspects of acceleration: 1) no acceleration; 2) constant acceleration and 3) increasing acceleration.
  13. Sep 19, 2006 #12


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    A time changing acceleration is called JERK. You have experienced this in every day life. A good example is when shifting gears with a manual transmission. Very frequently when shifting to a different gear your head will jerk. That is because your rate of acceleration has changed. This will happen when going from constant acceleration to constant velocity or from constant velocity to an acceleration.
  14. Sep 19, 2006 #13
    Here is what I knew,
    1) no accel, empty graph
    2) constant accel., horizontal line
    3) increasing acceleration, a line w/ positive slope
    4) But what about curved line in an accel?

    You see, when you were explaning, you skipped 3). Increasing accel(linear rate) has a line w/ a positive slope, not a curved line. I think curve line is where accel is inc. at a geometric rate. Then what kind of line would indicate an accel. increasing at an exponential rate?

    Btw, I know about jerk too, that wasn't my question.
  15. Sep 19, 2006 #14


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    1) is actually a horizontal line at 0. You have 2) and 3) correct. A curve (that isn't a line) in the acceleration-time graph simply represents a non-constant derivative of acceleration wrt time. Any non-zero jolt* will cause the graph to be 'curved'. If jolt is constant, acceleration is linear; if jolt is linear, acceleration is quadratic. But you know this since it's just basic calculus (calc I or pre-calc, I can't remember).

    Again, with basic calculus, you know that if any of position, velocity, acceleration, jolt, 'snap', ... are exponential they all are, and you know from high school algebra how to graph exponentials.

    * Oddly, I have always known of the derivative of acceleration wrt time as jolt rather than jerk.
  16. Sep 19, 2006 #15


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    I don't see the connection between these. They are, generically:


    [tex]y=a\cdot b^x[/tex]

    [tex]y=a\cdot b^x[/tex]

    If you're loking for a function that increases faster than a geometric/exponential function as it tends toward infinity, you could consider factorials, double exponentials y=e^(e^x), tetration, or some version of the Ackerman function.
  17. Sep 19, 2006 #16
    I have to say this: a lot of graphs can look in a lot of ways if you stretch the windows in ways to fit those...

    I understand what you're saying...[tex]y=x^{10^{9999}-1}[/tex] does look almoust like y=0 but...I don't understand why that's in any way a 'discovery".

    I mean if you're trying to graph a vertical asymptote as a function...just like squaring a circle, yuo'll never get it perfectly...am I right?
  18. Sep 22, 2006 #17
    I apologize for not making my question clear b/c this isn't even my question. I was wondering if there is a curve on an acceleration graph, can you give me a real life situation that would represent that? I mean none of my teachers could come up w/ one.
  19. Sep 22, 2006 #18

    matt grime

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    Perhaps no examples were forthcoming because the phrase 'a curve on an acceleration graph' doesn't mean much. UNdoubtedly there is some idea you have, but I don't think you're communicating it at all clearly.
  20. Sep 22, 2006 #19
    All I am saying is that you have to imagine a quadratic graph on the acceleration graph.(but only the 1st quadrant of it) Now just like peddling down the accelerator in a car can be described by a line w/ some slope on the acceleration graph, how would you describe a real situation which matches w/ my description of the acceleration graph?
  21. Sep 22, 2006 #20
    You experience what would correspond to a curved acceleration graph all the time.

    Suppose you're driving in a car and you floor the accelerator. At first, your car accelerates very quickly. After a while, friction and air resistance start limiting your acceleration, and it decreases, eventually getting close to zero, as the forces balance.

    So your acceleration is decreasing. But it's not a linear decrease: the rate of decrease depends on how fast you're already accelerating.

    Once the forces have "balanced," you'll still be experiencing a curved acceleration graph: wind will change, maybe the surface of the road will change, and you'll slow down and speed up in an unpredictable pattern (keeping the pedal floored).
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