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Isn't this strange

  1. Sep 8, 2007 #1
    Isn't this strange!!!

    Hello,

    Sometime ago I realized sth strange for most of the polynomials;

    For example, [tex]y=x^2[/itex]

    if we take f(x+d) as a multiple of f(x), I mean: f(2)/f(1) = 4 hence 4*f(1) = f(2) so 4 is a multiple of f(1). Now;

    [tex](x+d)^2/x^2[/itex]​

    as x approaches infinity the multiple approaches 1 !
    doesn't this means that when y=x^2 and x is an infinitly large number, y reaches a constant term, ie never approaches ininity?

    I'll be thankfull for your help. I hope I've been able to express what I really mean.
    Thankyou.
     
  2. jcsd
  3. Sep 8, 2007 #2
    Firstly, f(x+d) is not generically a multiple of f(x). (I'm not sure if you were claiming it was). Secondly, whether or not f(x+d) is a multiple of f(x) has nothing to do with anything here, as you can sesnibly talk about f(x+d)/(f(x) anywhere where f(x) is not zero.

    Finally, it is obvious that (x+d)^2/x^2 tends to 1, but this does not suggest that x^2 appraches a constant, merely that x^2 "tends to" (x+d)^2 in some sense, which is intuitive and does not in anyway suggest that either function is bounded above.
     
  4. Sep 8, 2007 #3
    Well, sorry for my wrong words, but what I really mean is different, note the following for y = x^2:

    f(2)/f(1) = 4

    f(3)/f(2) = 2.25

    f(4)/f(3) = 1.7778

    and so on... as we divide these consecutive terms the result approaches 1 and at some point f(x+1) = f(x) where x is extreamly large...

    which means y approaches a constant term when x approaches infinity...

    This is impossible, so I'm puzzeled... :)
     
  5. Sep 8, 2007 #4

    Gib Z

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    Homework Helper

    No. There is no point on f(x) = x^2 where f(x+1) = f(x).

    As as a limit where x approaches infinity then they are equal, but for any Real value, they are not. Expressing f(x+1) = f(x) differently, you are claiming x^2 + 2x + 1 = x^2 is true for large values of x. Indeed, if it was true, then for large values of x, 2x+1 = 0. Quite false. Just because in the limiting sense the 2x+1 becomes negligible compared to the x^2 does not mean it is not there.
     
  6. Sep 9, 2007 #5
    mubashirmansoor:

    You can see that
    [tex]\mathop {\lim }\limits_{x \to \infty } \frac{{x^2 + d}}{{x^2 }} = \mathop {\lim }\limits_{x \to \infty } 1 + d/x^2 = 1[/tex]

    Clearly,
    [tex]\forall \varepsilon > 0,\;\exists x_0 > 0:\forall x > x_0 ,\;\left| {\frac{{x^2 + d}}{{x^2 }} - 1} \right| < \varepsilon [/tex]
    Simply choose [tex]x_0 = \sqrt {\left| d \right|/\varepsilon } [/tex]
     
  7. Sep 9, 2007 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What it means is that f(x)= x2 and g(x)= (x+d)2, for fixed d, are of the same "order": x2= o((x+d)2) which was clear to begin with since they are of the same degree.
     
  8. Sep 9, 2007 #7
    The ratio of F(n) over F(n+1) is going to be smaller as n increases. This is to be expected. So what? It has no bearing on the nature of the infinitesimal.

    After all, just n/(n+1) tends to 1, so does the square. We have [tex]lim\frac{n^2}{(n+1)^2}=lim\frac{n}{n+1}*lim\frac{n}{n+1}\rightarrow1*1=1[/tex].

    Geometrically, from the standpoint of the derivative we are talking about a tangent between two points that tends to a ultimate of a single point.

    From your standpoint, we are looking at [tex] \frac{F(n+1)-F(n)}{1}=2n+1\rightarrow\infty[/tex]

    I take this as a case of wholesale confusion between a constant increase and infinitesimal.
     
    Last edited: Sep 9, 2007
  9. Sep 9, 2007 #8
    ohgodicanthelpmyself

    let x = -0.5

    but I see what you were going for.
     
    Last edited: Sep 9, 2007
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