# Isn't this strange

1. Sep 8, 2007

### mubashirmansoor

Isn't this strange!!!

Hello,

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

For example, $$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. Sep 8, 2007 ### DeadWolfe 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. 3. Sep 8, 2007 ### mubashirmansoor 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... :) 4. Sep 8, 2007 ### Gib Z 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. 5. Sep 9, 2007 ### bomba923 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$$

Clearly,
$$\forall \varepsilon > 0,\;\exists x_0 > 0:\forall x > x_0 ,\;\left| {\frac{{x^2 + d}}{{x^2 }} - 1} \right| < \varepsilon$$
Simply choose $$x_0 = \sqrt {\left| d \right|/\varepsilon }$$

6. Sep 9, 2007

### HallsofIvy

Staff Emeritus
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.

7. Sep 9, 2007

### robert Ihnot

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 $$lim\frac{n^2}{(n+1)^2}=lim\frac{n}{n+1}*lim\frac{n}{n+1}\rightarrow1*1=1$$.

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 $$\frac{F(n+1)-F(n)}{1}=2n+1\rightarrow\infty$$

I take this as a case of wholesale confusion between a constant increase and infinitesimal.

Last edited: Sep 9, 2007
8. Sep 9, 2007

### Pathway

ohgodicanthelpmyself

let x = -0.5

but I see what you were going for.

Last edited: Sep 9, 2007