# Is infinitely small error the same as $0$ error?

I am really confused over the idea.

If I have a circle I can approximate its area by using a regular polygon inside of it, with $n$ sides, and I can just split that polygon into triangles and compute the area. If I want better accuracy then I can use a polygon with more sides. The accuracy becomes better since the area between the polygon and circle (the "error") decreases as $n$ increases.

So it stands to reason that if we could keep adding sides our answer would keep getting better since the error would get closer to $0$.

I don't know what it means to "add infinitely many sides" because to me the idea doesn't make sense. No matter how many sides there are we can always add one more. But it does make sense to ask what's the value we never actually reach but get closer and closer to? For that we use the concept of a limit.

The error becomes "infinitely small" but I don't understand what this really means. Is "infinitely small" the same as $0$? Because by definition it's never quite getting to $0$. But conceptually then how can we say something with nonzero error gives us an exact answer as if it had $0$ error?

• Everything you have written makes perfect sense. It's great that you don't understand infinitely small because there is no need to understand it. The idea that error can be made smaller than any margin we choose by having suitable number of sides of polygon is all that matters here. – Paramanand Singh Jan 27 '18 at 13:12
• You tagged this question "limits", so I assume you're familiar with the concept? Yes, we can never have an infinite sided regular polygon, but as ${n \to \infty}$, the error get's closer and closer to $0$ without ever reaching it. – TreFox Jan 27 '18 at 13:12
• @user525456 Exactly! So what's the question? – TreFox Jan 27 '18 at 13:16
• @ParamanandSingh But that sort of gets to my question. The error can be made smaller by any margin we pick but it's never exactly $0$. – user525456 Jan 27 '18 at 13:17
• "infinitely small" is a confusing formulation. "arbitarily small" woule be much better. – Peter Jan 27 '18 at 13:21

Infinitely small error and $0$ error surely aren't the same thing.

Even though it seems like you already grasp the concept, let me reword part of the train of thought exposed:

Draw a circle of radius $1$ and fix a number $\epsilon > 0$. Make it really as small or as big as you want, and now ask me to create a regular polygon whose area misses the exact area of the circle by less than $\epsilon$. If you make $\epsilon$ really big, then I won't need to draw a polygon with many sides but if you make $\epsilon$ really, really small, then I may need to draw a polygon with a very big number of sides! Either way, we can agree that whatever $\epsilon > 0$ you choose, my task isn't impossible. Right?

Thus you probably can't say that my best approximation to the area of a circle has error $\delta$ with $\delta > 0$. Why not? Just set $\epsilon = \delta$ and build a polygon with approximation error inferior to $\epsilon$. So if you really insisted on finding a number that you could say "the approximations by polygons have error blah", then no positive number would do... so it is just like $0$ is the first number that isn't just wrong... or it is like $0$ is the less wrong answer, in a sense...

That is the way I would explain this relation, but certainly infinitely small and $0$ aren't the same thing.

In a not-so-precise sentence, infinitely small is not $0$, but it is as if it were for all practical purposes.

• Would it make sense to say that we use the definition of a limit to give us the area if we were to remove/chop off/truncate/ignore the arbitrarily small error? – user525456 Jan 27 '18 at 13:36
• I believe it would... talking about these sort of things there is a clear problem: when we just want to get an intuitive idea, there's some things that we like to say because they help us grasp some concept, but if we want to be completely precise, then we can't say them. That is why, for example, I ended my answer with "In a not-so-precise sentence, infinitely small is not 0, but it is as if it were for all practical purposes." – RGS Jan 27 '18 at 13:46
• I believe both me and the OP would benefit immensely from knowing why my answer was downvoted! – RGS Jan 28 '18 at 20:16

Maybe it helps to look at the problem from another side:

You want to determine the area of the circle (well, to start with, you assume that the circle actually has an area, which strictly speaking isn't guaranteed anyway: There do exist sets which don't have an area). But you only know how to determine the area of polygons.

Now you observe that if a polygon is inscribed in the circle, it certainly cannot have a larger area than the circle. This is true for every such polygon, therefore you know that the area of the circle, if it exists, is an upper bound to the set of areas of inscribed polygons. Therefore it certainly cannot be less than the least upper bound to those areas, that is, their supremum.

Now it turns out that if you calculate the area of an inscribed $n$-gon, it is monotonously growing with $n$, therefore their supremum equals the limit for $n\to\infty$.

Note however that at this point, all you know is that the limit gives a lower bound to the area of the circle. For all you know, the circle might be larger than that.

However you can now do a second observation, namely that the area of a circumscribed polygon is always larger than the area of the circle, and therefore their greatest lower bound (the infimum) is an upper bound of the circle's area.

Now the area of the circumscribed polygons decreases with $n$, therefore now the infimum of those equals their limit for $n\to\infty$.

So now you have a lower bound to the area of the circle, which is given by the limit of the areas of the inscribed polygons, and an upper bound to that area, which is given by the limit of the areas of the circumscribed polygons. And now it turns out that both bounds are the same. Now if the lower bound to the circle's area equals the upper bound, then obviously the area of the circle must equal that. Note that it also proves that area of the circle actually exists.

• I agree that if the lower and upper bounds were equal that gives the area, but it doesn't touch on my question, which is that those two bounds will never actually meet. You can get closer and closer but never quite there, there's always an error term under this definition of limits. – user525456 Jan 27 '18 at 17:11
• @user525456: it can be proved that those two bounds are equal without much difficulty. As explained in my answer only when these bounds are equal the region is said to have a well defined area. – Paramanand Singh Jan 27 '18 at 17:16
• @user525456: Read more carefully: The two bounds are equal. They don't meet, because they don't move. Note that those two bounds are not the areas of the polygons, but the supremum of the areas of the inscribed/infimum of the areas of the of the circumscribed polygons. – celtschk Jan 28 '18 at 7:16

Responding to your last sentence, the answer is not the area of polygons, but rather the area of circle. We have techniques developed to find the limit given the function and point under consideration. By the way the example of circle is bad one to explain the way the answer is obtained.

The simpler option is to consider a triangle with vertices $A(0,0),B(1,0),C(1,1)$ and consider a set of rectangles $R_i$ with vertices $(i/n, 0),((i+1)/n,0),(i/n,i/n),((i+1)/n,i/n)$ for $i=1,2,\dots n-1$. The area of each rectangle $R_i$ is clearly $i/n^2$ and the total area of all such rectangles is $(1+2+\cdots +(n-1))/n^2=(n-1)/2n$ and the desired area of triangle $ABC$ is the limit of this expression which is clearly $1/2$.

From your comments it appears that you think that area of plane regions is a pre-defined concept and we are just trying to approximate it by various methods. No!!

The area of a plane region which is a rectangle is defined as product of its width and height/length. For plane regions which are not rectangles the area is defined by a somewhat lengthy procedure. Essentially first we find a large rectangle say $R$ which contains region of interest say $I$. Next we divide the rectangle $R$ into multiple rectangles by drawing lines parallel to its edges. Suppose the width is divided into $m$ parts (not necessarily equal) and height is divided into $n$ parts (again not necessarily equal. Count all the small rectangles which are contained in $I$ and add their areas to get $s$. Further count all the small rectangles which contain some point of $I$. Add their areas to get another number $S$.

These numbers $s, S$ depend on the region of interest $I$ as well as mode of division of $R$ into $mn$ small rectangles. Consider the infimum $a$ of all numbers like $S$ and supremum $A$ of all numbers like $s$. If $a=A$ then we say that the region $I$ has area $A=a$ otherwise the region can not be assigned any area. Using some more effort we can show that the area $A$ can be obtained as a limit of such approximations like $s, S$.

• My question would still be the same there, the rectangles approximate the triangle, but the error is always $>0$. – user525456 Jan 27 '18 at 13:30
• I am not saying that the area of rectangles give you the area of triangle. Area of rectangles is an approximation. But from that approximation one can obtain the area of triangle by using the methods of evaluation of a limit. To put in more formally the area of triangle is defined as the limit of such approximations. Area of a plane region is defined using a limit procedure and its not a pre-existing concept. – Paramanand Singh Jan 27 '18 at 13:31
• But the definition of a limit only mentions the idea of closeness, not exactness, it tells us what we get closer to but never quite reaching, the difference between the two also being a type of "error". $(n-1)/(2n) = (1 - 1/n)/2 = (1/2) - 1/(2n)$. So as $n$ increases we get closer to $1/2$ and we have an error term of size $1/(2n)$ no matter how large $n$ is. – user525456 Jan 27 '18 at 13:32
• @user525456: if you understand the difference then there should be no issue. Area of a circle turns out to be the limit of area of polygons and hence it need not necessarily be equal to area of those polygons. – Paramanand Singh Jan 27 '18 at 17:20
• @user525456 : the definition of area as explained in my answer is such that it matches the limit of such approximations. This is a thing which needs to be proved. The error is in the approximation and not limit of the approximation. This is how an approximation differs from its limit. – Paramanand Singh Jan 27 '18 at 17:36

First there is a technical problem that should be addressed.

The accuracy becomes better since the area between the polygon and circle
(the "error") decreases as n increases.

Actually it only looks like the "error" decreases as n increases unless you can prove it. This is done by considering polygons that are inscribed in and circumscribed about the circle. In which case the area of the circle must be between those two areas. Then you show that the difference of the areas of the two polygons goes to $0$ as n increases.

Adding "infinitely many sides" is nonsense. It is really just an impractical (colorful ?) way of saying let n approach infinity.

However, there is a field called nonstandard analysis in which the real numbers are augmented with numbers that are not equal to $0$ but whose magnitudes are smaller than any positive real number. It would be too much work to describe how that works but you might want to check it out.

• Nonstandard analysis is all well and good but I don't see how it helps you understand the limit of the area of a regular polygon as it approaches a circle. – Rahul Jan 27 '18 at 13:49

When you are dealing with a framework including rigorous infinitesimals, then infinitely small error is indeed the same as zero error in the following sense. Consider for instance the problem of differentiating $y=f(x)=x^2$. One chooses an infinitesimal $x$-increment $\Delta x$ and computes the corresponding $y$-increment $\Delta y=f(x+\Delta x)-f(x)$, obtaining $2x+dx$.

The latter expression already contains all the information about the slope at the point, meaning that the derivative can now be computed by applying the standard part: $f'(x)=\textbf{st}(2x+dx)=2x$ where $\textbf{st}$ is the standard part function. In this sense there is no error involved once you know the answer up to an infinitesimal "error".