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Calculate what number x needs to be, to make y 0 ?

  1. Oct 1, 2004 #1
    #m = 0g

    Hi all,
    Is it possible to calculate what number x needs to be, to make y 0 ?
    An experiment shows that:

    x = y
    0 = +0.16
    1 = -0.38
    2 = -0.43

    Can I work out from that what x needs to be (it can be as many decimals as you want :tongue2:) for y to equal 0?

    Thanks alot,
    [r.D]
     
  2. jcsd
  3. Oct 1, 2004 #2

    arildno

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    What do you mean by "="??
    It most certainly is not "equality" since 1 is NOT equal to -0.38
     
  4. Oct 1, 2004 #3

    Alkatran

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    :rolleyes:
    I think he means:
    (x, y)
    (0, .16)
    (1, -.38)
    (2, -.43)

    find f() where f(x) = y then find the x that makes y = 0
     
  5. Oct 2, 2004 #4

    arildno

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    DrWarezz:
    Note that there's a lot of f()'s which satisfy your demand.
     
  6. Oct 6, 2004 #5
    Yeah, sorry. I mean:
    When x = 0, y = +0.16
    When x = 1, y = -0.38
    When x = 2, y = -0.43

    And the question is, what must x be, so that y = 0?

    Sorry for not being to specific at the start..

    Can this be worked out?

    Sorry, when you say:
    What do you mean, by the f() function? :eek:\

    Thanks alot, and sorry for the delay :)
    [r.D]
     
  7. Oct 6, 2004 #6

    Gokul43201

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    It can be worked out only if the relationship between x and y is understood.

    What experiment is this, and what do x and y refer to in the experiment ?

    Without this information, it is impossible to determine what you want.
     
  8. Oct 6, 2004 #7

    Integral

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    So happens that this little data set is a prefect fit to some function. It is so perfect, that I cannot believe that it is truly data.

    What have you done? Have you plotted the numbers? Remember that a picture is worth a 1000 words.
     
  9. Oct 6, 2004 #8

    Gokul43201

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    What did you fit it to integral ?

    Looking at the numbers, I'd guess an exponential decay with a negative floor.

    Of course, you can always fit it to a polynomial of second or higher degree....but I guess that's not what you mean.
     
    Last edited: Oct 6, 2004
  10. Oct 6, 2004 #9

    Integral

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    I tried a 2nd order poly. , got R=1. It is a perfect fit,

    I did this at home, I am now at work I think the equation was


    [tex] .245 x^2 -.785x + .16 [/tex]

    Here is an interesting way of finding the root in this data set is to iteratively evaluate the expression

    [tex] x_{n+1} = \frac {.245 x^2 + .16} {.785 x_n} [/tex]

    edit:
    Let x0 = .1
     
    Last edited: Oct 6, 2004
  11. Oct 6, 2004 #10

    Gokul43201

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    Shouldn't you always be able to exactly fit a second order polynomial to a set of three points ?
     
  12. Oct 7, 2004 #11

    Integral

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    Of Course, you are correct. With the amount of information given that is as good a solution as any.
     
  13. Oct 7, 2004 #12
    :eek:\ Thanks everyone! lol
    I'm kind of confused. I'm just a young student, and I got those numbers from a preliminary experiment.
    I figured that by using those numbers, could give a pretty accurate result, without even doing the final.

    It appears that you lot have figured it out? Could you simplify it a bit, I don't understand half of the jargon you used :eek:\ lol.

    Thanks alot for the help :)
    [r.D]
     
  14. Oct 7, 2004 #13
    btw; this is an experiment of the concentration of potato chips.. based around osmosis (how boring eh :P).
     
  15. Oct 7, 2004 #14

    matt grime

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    So, you've got some data points, 3 in fact. You could try and plot these three points on a graph. The next thing is then to 'guess' the graph of the curve that describes the relationship. You can do this in many ways, and the exact method you choose depends on the situation you're in.

    The most common is called 'least squares regression' and this is where you try and draw a straight line that fits 'best' where best is something you can make precise.

    There is another method though. Given n points there is a polynomial relation, that is one that looks like:

    [tex]y=a_0 + a_1 x +a_2x^2 + \ldots a_{n-1}x^{n-1}[/tex]

    that fits exactly through those points.

    There are merits for each obviously. Suppose you'd got the data

    (-1,1) (0,0.99) (1,1)

    then least squares is going to give you something very close to the horizontal straight line y=1 the polynomial will give a parabola that goes off to infinity.

    Of course this is an extreme example: in the interval -1 <=x<=1 they're very similar. No one would be silly enough to use such data to predict what the behaviour is like at x = 1000.

    Well, I say that but people are that silly.

    Just recently someone at Oxford did some studies of data collected over the last 40 years or so about sprinters and extrapolates to predict the woman will win the 100m in less time than the man at the olympics in 150 years time. Very dody science.

    More seriously of course was the challenger shuttle disaster when scientists looked at the behaviour of a plastic ring at relatively warm (20deg C) when it behaved perfectly and ignored the fact that at temperatures on the launch pad, near to zero, it became brittle. Feynman I think demonstrated this with a simple experiment on the investigating committee when no one could believe this would be allowed to happen.
     
  16. Oct 7, 2004 #15

    Gokul43201

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    I think the o-rings were made of rubber, which has very poor resilience at freezing temperatures.

    Anyways, DrWarezz, the important thing to take away from matt's post is that you can only guess what the number would be at y=0, if you know what the relationship between x and y is expected to be.

    If you thing y should decrease linearly with increasing x, you can fit a straight line to the data (as matt suggested - using a least squares fit). On the other hand, you might think that it would make more sense for y to decay exponentially, in which case, you would draw a different curve.

    And we can't make a guess for what behavior to expect, because we don't know what x and y are.
     
  17. Oct 7, 2004 #16
    Okay, thanks alot everyone.
     
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