Proving Boundedness of f(x) = (x+1)/x^2 in Natural Numbers

In summary, we need to prove that the function f(x) = (x+1)/x^2 is bounded in the set of natural numbers (N). To do so, we first prove that for every x,y >= 1, if x<y, then f(x)>f(y). This is shown by manipulating the expression y^2 * (x+1) - x^2 * (y+1) > 0. We then determine that the maximum and supremum of f(N) is 2, and that f(N) has no minimum since it has a limit of 0 at infinity. It could be more organized by first proving that f:R_{\ge 1} \to R is
  • #1
daniel_i_l
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Homework Statement


f(x) = (x+1)/x^2
a)prove that f is bounded in N (N is the set of natural numbers so we have to prove that f(N) is a bounded set)
b)find supf(N) and inff(N).
c) does f have a maximum or minimum in N?


Homework Equations





The Attempt at a Solution


First I proved that for every x,y >= 1, if x<y then f(x)>f(y):
y>x>=1 and so y^2 > x^2 and so y^2 - x^2 > 0
xy=yx and so x^2 * y < y^2 * x and so
y^2 * x - x^2 * y > 0 and together
y^2 * x - x^2 * y + y^2 - x^2 = y^2 * (x+1) - x^2 * (y+1) > 0 and so
(y^2 * (x+1) - x^2 * (y+1)) / (x^2 * y^2) = (x+1)/(x^2) - (y+1)/(y^2) > 0and so (x+1)/(x^2) > (y+1)/(y^2).

Now, f(1) = 2 and so for all x>1 f(x)<2 and so maxf(N) = supf(N) = 2.

Also, for every x>=1 f(x)>0. The limit of f at infinity is 0. So if f(N) has a lower bound c>0 then since f has a limit of zero at infinity we can find some M>0 so that for every x>M (we can find an x in N) |f(x)|<c => f(x)<c which means that c isn't a lower bound so inff(N) = 0 and there's no minimum.

Is that right (especially the proof)? Does it matter that I did a,b and c in the same step?
Thanks.
 
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  • #2
Instead of going 'since xy=yx, x^2y<y^2x' is do this: 'since x<y, multiplying both sides by xy give x^2y<y^2x'. The way you originally put it was a head scratcher.

You haven't really done a,b,c in one step, what you've shown is that the function [itex]f:R_{\ge 1} \to R[/itex] is monotonically decreasing. Then you considered the restriction to natural numbers and then made your arguments based on that.

If you want to break it up into parts and be more organized then you can first do the preliminary work of showing that [itex]f:R_{\ge 1} \to R[/itex] is monotonically decreasing and then do a,b, and c in that order. You haven't explicitly claimed why f(N) is bounded.

The majority of the work is correct. You could be more liberal with your explanations, but not necessary. As an example, the supremum exists because the maximum exists. I don't know if that's what you meant because you just wrote it as maxf(N)=supf(N)=2 when your work only showed that f(N) has a maximum of 2.

Just little things that your professor might pick on.
 
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FAQ: Proving Boundedness of f(x) = (x+1)/x^2 in Natural Numbers

What does it mean for a function to be bounded?

Being bounded means that the values of the function do not exceed a specific range, or that the function does not grow without limit.

How do you prove boundedness of a function?

To prove boundedness, you need to show that the values of the function do not exceed a certain range. This can be done by finding the upper and lower limits of the function and showing that all values fall within those limits.

What are natural numbers?

Natural numbers are the positive integers (whole numbers) that are used for counting and ordering. They do not include fractions, decimals, or negative numbers.

How do you prove boundedness of a function using natural numbers?

To prove boundedness of a function using natural numbers, you would need to show that the function is bounded for all possible natural number inputs. This can be done by using mathematical induction or by analyzing the behavior of the function for different natural number inputs.

Can you give an example of a bounded function in natural numbers?

Yes, f(x) = (x+1)/x^2 in natural numbers is a bounded function. It has an upper limit of 2 and a lower limit of 0, meaning that for all natural number inputs, the values of the function will always be between 0 and 2.

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