Limits and While Loops

Question

Question: Consider the following program. Does $f(1)=\infty$?

\begin{align*} f(i):=&|\text{while } \frac{1}{i}>0\\ &||i\leftarrow i+1\\ &|i \end{align*}

I would say that $f(1)=\infty$ is a true statement. The program does not terminate, but one could consider the sequence of points $\{x_i\}_{i\in \mathbb{N}}$ given by $x_i:=\frac{1}{i}$, so that $\{x_i\}_{i\in \mathbb{N}}$ converges to the limit $0$ which makes $\underbrace{\frac{1}{i}}_{=0}>0$ false which means $f(1)=\infty$. However, I could be wrong.

Answer 1

The final result of the function would depend on the method of division you'd end up using, at least if this function is executed by a computer.

One thing we can be sure of, however, is that the result of f(1) will under no circumstance be 0.

After all, the input of the function is 1 and the loops keeps increasing the value of i.

For example, if the function is executed using integer division, (1 / 2) > 0 would be false, and the function would end up yielding a result.

However, the result would be 2, not 0.

Therefore, the function either runs out of resources, or returns something larger than 0 - but never 0.

---

Although I'm answering this question as a programmer, rather than as mathematician, the logic remains true whether it be mathematical or programmatic: Even if we assume the loop can run to infinity, and we assume 1 divided by infinity as greater than 0, the function would produce infinity, not 0.

Pure logic dictates that f(1) != 0, whichever notation is preferred :)

---

Since I'm writing this as a programmer on the internet anyway, I might as well provide proof along with this answer.

I wrote the following snippet:

<?php

function f($i) {
    while ((1 / $i) > 0) {
        $i = increase($i);
    }
    return $i;
}

function increase($n) {
    if ($n < 100) {
        return $n + 1;
    }
    return INF;
}

(I would have used if ($n < INT_MAX) but that results in a 3+ seconds runtime)

Long story short, assert(f(1) === INF); passes, and

dump(f(1));

Is indeed float(INF).

Run this code

Answer 2

It's a bit of a funny question. Looks like the best answer is "no." The number $1$ is not in the domain of $f$, because the program doesn't terminate.

It's true that as a human we can see that the in-memory value of $1/i$ is headed toward $0$, but $f$ was never going to return $1/i$, it was going to return $i$, which is headed off to $\infty$. Even if it were going to return $1/i$, the answer still shouldn't be $0$, because the program doesn't terminate and there's no notion of a limiting operation made explicit here.

Answer 3

The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.

We cannot say that $f(1)=0$.

Answer 4

The function is undefined at $i=1$ since $\not \exists b$ such that $b=f(1)$.

For a formal proof, indicating with $i_k$ the value assumed by $i$ at the $k^{th}$ loop, we can show by induction that $\forall k$

• $i_k\ge 1 \implies \frac 1{i_k}>0$

therefore the loop will never end.

More in general $f(i)$ is undefined for any $i>0$ and $f(i)=i$ for any $i< 0$.

For $i=0$ it depends on how the "while" command returns to the operation $1/0$.

Answer 5

Instead of a while-loop, a better programming analogy of limits is the lazy list.

In Haskell:

naturals = [1..]
my_limit = [1/n n <- naturals]
take 100 my_limit

Outputs:

[1.0,0.5,0.3333333333333333,0.25,0.2,0.16666666666666666,0.14285714285714285,0.125,0.1111111111111111,0.1,9.090909090909091e-2,8.333333333333333e-2,7.692307692307693e-2,7.142857142857142e-2,6.666666666666667e-2,6.25e-2,5.8823529411764705e-2,5.555555555555555e-2,5.263157894736842e-2,5.0e-2,4.7619047619047616e-2,4.5454545454545456e-2,4.3478260869565216e-2,4.1666666666666664e-2,4.0e-2,3.8461538461538464e-2,3.7037037037037035e-2,3.571428571428571e-2,3.4482758620689655e-2,3.333333333333333e-2,3.225806451612903e-2,3.125e-2,3.0303030303030304e-2,2.9411764705882353e-2,2.857142857142857e-2,2.7777777777777776e-2,2.702702702702703e-2,2.631578947368421e-2,2.564102564102564e-2,2.5e-2,2.4390243902439025e-2,2.3809523809523808e-2,2.3255813953488372e-2,2.2727272727272728e-2,2.2222222222222223e-2,2.1739130434782608e-2,2.127659574468085e-2,2.0833333333333332e-2,2.040816326530612e-2,2.0e-2,1.96078431372549e-2,1.9230769230769232e-2,1.8867924528301886e-2,1.8518518518518517e-2,1.818181818181818e-2,1.7857142857142856e-2,1.7543859649122806e-2,1.7241379310344827e-2,1.694915254237288e-2,1.6666666666666666e-2,1.639344262295082e-2,1.6129032258064516e-2,1.5873015873015872e-2,1.5625e-2,1.5384615384615385e-2,1.5151515151515152e-2,1.4925373134328358e-2,1.4705882352941176e-2,1.4492753623188406e-2,1.4285714285714285e-2,1.4084507042253521e-2,1.3888888888888888e-2,1.36986301369863e-2,1.3513513513513514e-2,1.3333333333333334e-2,1.3157894736842105e-2,1.2987012987012988e-2,1.282051282051282e-2,1.2658227848101266e-2,1.25e-2,1.2345679012345678e-2,1.2195121951219513e-2,1.2048192771084338e-2,1.1904761904761904e-2,1.1764705882352941e-2,1.1627906976744186e-2,1.1494252873563218e-2,1.1363636363636364e-2,1.1235955056179775e-2,1.1111111111111112e-2,1.098901098901099e-2,1.0869565217391304e-2,1.0752688172043012e-2,1.0638297872340425e-2,1.0526315789473684e-2,1.0416666666666666e-2,1.0309278350515464e-2,1.020408163265306e-2,1.0101010101010102e-2,1.0e-2]

In Python:

import itertools
naturals = itertools.count(1)
my_limit = (1/n for n in naturals)
list(itertools.islice(my_limit, 100))

Outputs:

[1.0, 0.5, 0.3333333333333333, 0.25, 0.2, 0.16666666666666666, 0.14285714285714285, 0.125, 0.1111111111111111, 0.1, 0.09090909090909091, 0.08333333333333333, 0.07692307692307693, 0.07142857142857142, 0.06666666666666667, 0.0625, 0.058823529411764705, 0.05555555555555555, 0.05263157894736842, 0.05, 0.047619047619047616, 0.045454545454545456, 0.043478260869565216, 0.041666666666666664, 0.04, 0.038461538461538464, 0.037037037037037035, 0.03571428571428571, 0.034482758620689655, 0.03333333333333333, 0.03225806451612903, 0.03125, 0.030303030303030304, 0.029411764705882353, 0.02857142857142857, 0.027777777777777776, 0.02702702702702703, 0.02631578947368421, 0.02564102564102564, 0.025, 0.024390243902439025, 0.023809523809523808, 0.023255813953488372, 0.022727272727272728, 0.022222222222222223, 0.021739130434782608, 0.02127659574468085, 0.020833333333333332, 0.02040816326530612, 0.02, 0.0196078431372549, 0.019230769230769232, 0.018867924528301886, 0.018518518518518517, 0.01818181818181818, 0.017857142857142856, 0.017543859649122806, 0.017241379310344827, 0.01694915254237288, 0.016666666666666666, 0.01639344262295082, 0.016129032258064516, 0.015873015873015872, 0.015625, 0.015384615384615385, 0.015151515151515152, 0.014925373134328358, 0.014705882352941176, 0.014492753623188406, 0.014285714285714285, 0.014084507042253521, 0.013888888888888888, 0.0136986301369863, 0.013513513513513514, 0.013333333333333334, 0.013157894736842105, 0.012987012987012988, 0.01282051282051282, 0.012658227848101266, 0.0125, 0.012345679012345678, 0.012195121951219513, 0.012048192771084338, 0.011904761904761904, 0.011764705882352941, 0.011627906976744186, 0.011494252873563218, 0.011363636363636364, 0.011235955056179775, 0.011111111111111112, 0.01098901098901099, 0.010869565217391304, 0.010752688172043012, 0.010638297872340425, 0.010526315789473684, 0.010416666666666666, 0.010309278350515464, 0.01020408163265306, 0.010101010101010102, 0.01]

Your question might be interpreted to ask what the last element of such a lazy list is.