## philosophy of mathematics

use alone*elementary methods*Let's prove the fundamental theorem of algebra. Along the way, we'll make friends with some key ideas from elemental analysis and take a peek into the beautiful world of complex numbers.

Regardless, it's been a great quarantine project for me :)

The aim of theory is, in large part, to organize past experience in a systematic way so that the next generation, our students and their students, etc., can absorb the essence of it as painlessly as possible. possible, and only then can each scientific activity continue to accumulate without reaching a dead end.

Michael Atiyah, Fields-Medaillenge Award recipient, on "How Research Is Conducted"

The language of elementary analysis is that of continuity, limits, open and closed sets, and calculus in one variable. In hindsight, we can see that these concepts and definitions reduce impossible problems to very viable problems. (Or, I found, *doable* with sporadic attempts over several months and many failures.)

In trying to prove this theorem, I used this language of elementary analysis. First, you will be introduced to the meaning of some of these (at first scary) concepts. So let's use them to prove something that otherwise seems impossible.

I've tried to include the bare minimum of what's needed. If you are familiar with some of them, you can skip the sections you are familiar with. You can also get a feel for the test and go back to the prerequisites if needed. If this is your first time encountering much of this material, don't necessarily expect it to make a lot of sense; The goal is to get the gist of the argument, not all the details.

Suppose we have a polynomial over the complex numbers and we want to know if a complex number exists where the polynomial evaluates to 0. (See below for an introduction to complex numbers if needed.)

The fundamental theorem of algebra states that every polynomial over complex numbers has at least one root. This is in stark contrast to real numbers, where many polynomials have no roots, such as B. x² + 1. For complex numbers, z² + 1 has two roots: +i and -i. i²=-1, so they both evaluate to -1+1 = 0.

not how,**Polynomials over real numbers**Sohn**NO**Guaranteed to have at least one root. For example, the polynomial below has no roots, as can be seen from its diagram.

It is very difficult to give a general outline of what will be a very technical test. Don't skip this section, it's important to get an overview of what we're about to do!

So what I'm doing here is testing it out for a very simple function. I prove for P(z) = 2z and show that P(z) = 2z has a root. Well obviously yes, say just plug in P(0+0i) = 2(0+0i) = 0. That's it.

But I'm going to pretend I didn't notice and use a method that is*generalize*very well for more polynomials.

First, we take a point w. We can see where P sends w: P(w) = 2w. But what about a small region around w? On the complex plane, draw a small circle around w and see where all the points on it stop. It turns out that if we shaded all the spots where they landed, another circle would appear. [Spoiler: when we represent complex numbers, we do it on a 2D plane. This is because z = x+iy can be plotted as (x,y) coordinates on a graph]

Now let's assume that for*any complex number w***except**For a finite number, say 4 points, we could draw a shaded circle around w, then see where it ends, and finally we could write a circle on the "picture" where that circle ends. A diagram shows the process of inscribing a circle into the "picture" where our old circle ended.

Our function P(z) = 2z was good because*Exactly*reassigned the circle to another circle. But we could have a more horrible polynomial, maybe z⁷-3z² +1. So a natural question is: if we can see where a circle goes, can we*enroll*a circle? It doesn't matter if the circle is very, very small. In the diagram above, the circle corresponds to an odd shape, but it doesn't matter because we can draw this little green circle with a red border that fits the slightly odd shape perfectly.

Why is it useful? Well, it turns out that we can inscribe P(w) as a small circle, as long as the derivative at w is not 0. We will also note that the derivative can only be 0 at a finite number of points.

We used to see where a circle maps, but now let's see where the entire plane maps. In particular, if we don't have a point corresponding to the origin, then for each line that passes through the origin in the greenish patch below, we can look for the point on the line that is closest to the origin.

Let me take some time to explain the diagram below. Look around first. The gray dot represents the complex plane. It continues endlessly in all directions. We then place a green dot at each point where a gray dot falls under the P-map. The P-map is like a projectile that shoots from one plane to another. We don't know much about this green form; even the diagram itself assumes some things tacitly, hence the need for formal proofs and pretty pictures! - but if there is no point to send to 0 then the origin is not shaded. It turns out that there must be an area around 0+0i where nothing is sent.

We then draw a purple line from the origin at the bottom of the graph. We identify the point on that line that is closest to the origin and figure out which point was sent there.

Here's the trick. Suppose that the point with the cross in the gray diagram is called w and that in the green diagram is P(w). Now suppose the gradient at w is non-zero. So we can draw a little circle around w and see where it lands

In particular, the small gray area can contain a small circle based on our previous discussion. However, this is clearly a contradiction: the gray area, along with all the other circles centered on P(w), clearly (visually) contain points on the purple line closest to the origin. But remember that of all the possible complex numbers z that mapped to end up on the purple line, w was the closest. Therefore, finding those points on the purple line closest to the origin that are on the P-map image is a contradiction.

We can apply this to any of the lines passing through the origin, of which there are infinitely many. This argument is not valid only if the derivative is 0, which is at most for a finite number of points. This still leaves infinite lines where we can apply our argument and still find our contradiction.

But for now we keep our horses. That argument was good, and the graphics were good (well, sort of; I cobbled them together in Microsoft Word!). They give a good intuition for the approach, but that's not it.*a test*. For that we need definitions that capture these geometric intuitions and allow us to apply them rigorously to any polynomial.

There are several ways to approach a complex number. The "rigorous" definition is probably only useful for people who already know what a complex number is. Then comes some visual understanding and intuition. If you are already familiar with complex numbers, skip this section.

Visually, a complex number z can be written as its "real" and imaginary parts. The real part is written on the x-axis of a graph and the imaginary part is written on the y-axis. So z = x + iy.

The addition is done as expected: you add the "real parts" and you add the "imaginary parts".

Multiplication can be viewed in a slightly different way: it corresponds to a*rotation*it is a*to change*on the radio. Here, the square root of negative 1 makes a lot of sense, as we've expanded the definition of multiplication. Point i has an angle of 90 degrees and a length of 1. So if we multiply a point z by i, we rotate it 90 degrees and scale the radius by 1. Of course, if we scale a radius by 1, it stays the same . If we multiply z by 2i, it will be rotated 90 degrees and stretched by a factor of 2.

The complex derivative of the polynomial

It is given by

Although it is difficult to show that the polynomial has a root,*demarcate*the number of roots is less difficult. A polynomial of order n, that is, whose power term is z^n, has at most n roots. This is because we can factor the roots of the polynomial individually.

After factoring the first n roots, it becomes clear that all other numbers do not add up to zero because not all terms of the multiplication are zero.

Polynomials are "continuous" functions. Since its derivative is a polynomial, its derivative is also continuous. But I'll go into more detail about what exactly continuity means and what complex derivation means in the next section.

## 3.1 What is a limit? Sequences and Convergence

Let's understand what sequences and convergence are.

A sequence of complex numbers is simply a*List*of complex numbers, where for every integer n you can get the number n from the list.

An example of a sequence of complex numbers is:

0,9 + 0,9i, 0,99 + 0,99i, 0,999 + 0,999i, 0,9999 + 0,9999i, 0,99999 + 0,99999i, …

where the number N in the list has nines after zero for both the real and complex parts

We say that this sequence "converges" to 1 + 1i. Why? The first term is just 0.1+0.1i of 1+1i, the second term is just 0.01+0.01i of 1+1i, and the nth term is 0.00…01+0.00…1i away from 1+1i. where N zeros precede 1. If we use*distance between manhattan*then the distances are (0.2, 0.02, 0.002, 0.0002, ...), which clearly tends to 0. If we use Euclidean distance, we find that the distances (sqrt(0.02), sqrt(0.002) , sqrt(0.0002), …) also tend to zero. In general, when a sequence converges to the Euclidean distance, it converges to the Manhattan distance at the same point, and vice versa.

So a good definition of convergence for 1+1i is: for the maximum target distance you give, I can find an N such that all terms after the sequence N are within that 1+1i distance.

This definition makes sense. Setting the target to be 0.00234 away in the real and complex parts, we see that 0.001 is less than 0.00234 starting with the third term:

- 1–0,999=0,001≤0,00234
- 1–0,9999=0,0001≤0,00234
- 1–0,99999=0,00001≤0,00234
- etc.

Of course, regardless of the maximum allowable distance of one, we can eventually find a 0.000000000000000000…00000001 less than that number, using the same argument as above.

## 3.2 Continuous Functions

A continuous function maintains limits. In the example above, a continuous function f would ensure that the sequence (f(0.9 + 0.9i), f(0.99 + 0.99i), f(0.999 + 0.999i), ...) converges to f (1+1i). An alternative but mathematically equivalent definition is that if you set a target to proximity, either 0.1 or 0.0000001, then whatever it is, you can find a distance d within which all points are within that proximity.

This is best illustrated with an example. Suppose you set the target to 0.001 at a z point. Then I find a distance that can be 0.0142, where if |w-z|<0.0142, that is, if w is contained in the circle centered on z and radius 0.0142, then |f(w)-f(z) |<0.001 .

The latter definition is captured in the famous epsilon-delta definition of continuity at a point z.

Or for functions on real variables it is intuitively understood (though with somenotable exceptions*) by a 'line that can be drawn without lifting the pen from the paper'.

*These types of functions are often referred to as pathological by mathematicians. They usually appear when a definition captures the essence of something, like B. our intuitive notion of continuity, but it also happens that that definition contains several functions that are not*NO*agree with our intuitions.

## 3.3 Open and Closed Balls

An important concept is the open and closed ball. The open sphere of radius r centered on z is made by drawing a circle around z with a "dotted line". This symbolizes that it contains all the points inside the circle, but none on the boundary. The icon we use to represent this is below

The closed sphere is like the open sphere, but it encloses the boundary of the circle. The symbol we use for the closed sphere of radius r centered on a point z_0 is below. (Z_0 is used because we need to find names for several complex numbers at once, and it's easier to put small numbers at the end than to find a new alphabet.)

We say there's a point to it*Within*a set if we can find an open or closed sphere completely enclosing the point contained in a set. then a*closed sphere*it has some points that are not inside, i.e. the points on its boundary. Any circle drawn around a point on the boundary will contain some points that are not inside the circle. This can be seen below.

On the other hand, all points on the open sphere are inside, as the open sphere does not contain any of the boundary points. No matter how close we get to the boundary, if we get close enough there will be enough room to fit in a small circle around our point.

general one*closed sentence*is the one that, like the closed sphere, contains its boundary (technically it contains all of its boundary points). An open set, like an open sphere, contains no boundary, and you can always find a small open sphere around a point where the open sphere is entirely contained within the set.

For example, a rectangle with a "dotted line" (a rectangle that does not contain its boundary) is an open set, while a rectangle with a hard line (which contains its boundary) is a closed set. If you place a circle on the boundary of the closed rectangle, the circle will always contain points outside the set.

## 3.4 Image of a function

The image of a function is an array containing all the points to which the function can send something. For example, if a company has a function that takes a flight number as input and a destination country as output, the image is the set of all countries the company flies to. In our case they are all points v, so there is a z with P(z) = v. (all countries v then there is a flight number such that P(flight number) = v)

We can look at the image of a smaller subset of the complex plane. For example, say we want to consider all points v such that there is a z on circle B such that P(z) = v (wow, that's awesome). We would then call this the image of B under P.

A worked example. Consider the function P(z) = 2z. The image of P is then the entire complex plane: for any point v there is a point z - in this case v/2 itself - such that P(z) = v. P(v/2) = 2v/2 = v.If B were all points with radius less than or equal to 1, that is, H. Points on a closed sphere of radius 1 centered at the origin, then the image of B under P would be the circle of points of radius 2. For example, P(1+0i) = 2+0i.

There are several ways to represent a complex continuous and differentiable function.

Often we just write f(z). For example, f(z) = z + z³ + 1+2i is a complex function. We can also divide it into real and imaginary parts. In this case we write f(z) = f(x+iy) = u(x,y) + iv(x,y). It turns out that a complex function is continuous only if u(x,y) and v(x,y) are continuous.

Another slightly weird thing about complex numbers is how to tell them apart. We want the following expression to converge as h tends to 0

However, since we are dealing with complex functions (that is, a function that takes a complex number as input and a complex number as output), the restrictions are a little tighter than for real functions. That's because complex numbers are a plane: a two-dimensional object. So our edge can come from any direction: left, right, up, down, spiral, whatever. Whereas in normal analysis we can only approximate a number, say 3, in one dimension (i.e. from the left as 2.9, 2.99, 2.999, 2.9999 and from the right as 3.1, 3.01, 3 ,0001, 3,00001, …).

The conditions for the differentiability of a complex function can be summarized in two very simple but fundamental equations.

These two equations show that we want the derivative to be the same in whatever direction we approach it.

There are many wonderful theorems that follow from this simple assumption that actually offer another way to prove the Fundamental Theorem of Algebra. Our proof is more elementary: it is not based on the most powerful theorems of complex analysis, but it is more complicated.

## Part I: At most N-1 points of a polynomial are not in the interior.

In the first part of the proof, we notice that all but a finite number of points of the polynomial are not in the interior.

## AI part.

We learned that the number of roots of the derivative is at most N-1. The idea is to explore this by considering closed and open balls where the derivative is different from zero for any point on the ball.

Exercise: Draw the 2D plane. Draw three dots. Can you draw a circle around a fourth point that doesn't contain those three points?Likewise, given a finite number of points where the derivative is zero and all other points, we can draw a circle small enough not to contain any points where the derivative is zero.

We take a point z_0. We want to show that P(z_0) is inside the range of P (where P is our polynomial function). This means that we want to draw a small ball around where z_0 landed after being evaluated by P and for every point on that ball there is a point z such that P(z) is evaluated up to this point.

`An example where this is not possible is in the function T(z), which sends all points in the complex plane to 1+0i. Therefore, every circle that contains 1 also contains points that are not 1. If the circle's radius were 0.05, it would contain 0.99 + 0i. But there is no point z with T(z) = 0.99, because T sets all points to 1.`

[This property of being "inside" may be a little confusing to examine at first. Will pay dividends]

## Part IB. Interior points are preserved during multiplication.

Suppose we are at a point w, given the polynomial P, and we have a derivative P'(w) = a+ib at that point. (Note the apostrophe to indicate the derivative.) If the derivative is not zero, we can consider another polynomial, P(z)/(a+ib). This new polynomial behaves very similarly to the previous one, but its derivative at w is 1+0i. (That's because when we scale a function by a constant, the derivative scales by the same constant.)

This will be a very useful simplification later in the demo. But we have to prove that w is inside P(z) if and only if w is inside P(z)/(a+ib). One way to visualize this is that complex multiplication scales the radius and rotates the angle. So we multiply the "ball" around P(w)/(a+ib) by (a+ib), which rotates and stretches each point of the ball, resulting in a new ball that encompasses the surroundings of P(w) .

## CI part. A useful inequality when P'(w) = 1+0i

Here we show a useful inequality that we can use when the derivative of P at w is 1+0i, that is, H. P'(w) = 1+0i. Let's prove that in a circle containing w, if w and v are some distance L on that circle, then P(w) and P(v) are separated by at least 0.8 L when the distance is measured with Manhattan distance.

For this inequality, we will represent P(z) as u(x,y) + iv(x,y), which is one of the methods of expressing a complex function discussed earlier.

First we use the triangular inequality

Second, remember the Cauchy-Riemann equations.

Also, both partial derivatives are continuous.

`(Note to math geeks) It's okay to assume this because for a polynomial both u(x,y) and v(x,y) are just polynomials in two variables, so the apparatus is usually used to calculate the It is not necessary to prove the continuity of partial derivatives. In general, proving that partial derivatives are continuous is not so easy.`

In this case we assume that P'(w) = 1+0i. Therefore, we update our equations to include this information.

Using the continuity of these derivatives, we can find a circle where they lie within these limits. Remember this inequality for later!!!

`It is worth noting that the numbers 0.1, -0.1, 0.9, 1.1 are nothing special. There are many numbers you could have chosen that form an equally useful inequality.`

We used our calculus knowledge on one of the following variables to make it useful here.

`For example, if x = 0.2, then u(0,2, y) is a function in one variable and the normal one-variable calculus rules apply. Basically, we found a way to transfer our calculus knowledge into this new environment.`

We use a similar idea again:

If we put these two equations together, we get:

The integral over x is always greater than 0.9|x_1 — x_0| because its length |x_1 — x_0| and the magnitude of the function is always greater than 0.9. [Remember the disparity I asked you to remember later?]. The integral over y is never more negative than -0.1|y_1 — y_0| because its height is never less than -0.1 and its width |y_1 — y_0| It is

In general, in mathematics, you cannot abuse a good idea. Using the same logic as before, we get a similar equation for v(x,y)

With reasoning almost identical to that of u(x,y), we obtain

If we combine the two, we can infer that

## party identification

*Some parts of this section are moderately technical. Try to understand the arguments without necessarily getting bogged down in them. Most of this is the technical implementation of the overview in section '0. game schedule'*

We now use proof by contradiction to complete Part I of the proof. Remember that our point w has the property that P'(w) is not 0+0i, and we want to show that it is inside. Also remember that it suffices to show that P'(w) is inside when P'(w) = 1 + 0i to prove the more general case when P'(w) is anything but can be 0 + 0i

Since the derivative is nonzero and continuous, we choose a small circle around z_0 that is small enough that the derivative is bounded from 0. We call this circle B.

Suppose our point P(z_0) is not inside the image of B under function P. So every time we draw a circle around P(z_0) there is some point*NO*in the photo of P more*es*in this circle.

`note. If you are not comfortable with this terminology, see the introductory section on function screenshots.`

This can be seen visually below. If the circle around the point is not entirely contained in the image of P, we can find at least one point that is both on the circle and*NO*in the image of P. We can do this for a circle of arbitrarily small radius, which allows us to construct a*Series*of points outside the image of P that are 'arbitrarily' close to P(z_0).

If we decrease the radius of the circle that we draw successively, we conclude that there is a sequence of points that converge to P(z_0) but are not in the image of P, where the domain is a small circle around z_0.

These red dots (see photo) cannot exist in isolation. The image of B is a "closed set", which means it contains its boundary points. Therefore, if the points in the image of B were arbitrarily close to the red point, the red point would be a boundary point of B and therefore included in B. Therefore, the red point is surrounded by an open sphere that contains no points. in B's photo

Behind the scenes we have to play with the fact that the domain is a compact set - closed and bounded - the function is continuous and so we can guarantee that the image of B is compact.In the following argument, the existence of some of the points also requires the use of compactness and continuity arguments.

For each of the red dots, we can look at the point in the image of B that is closest to it. Let's call this P(z_1). Using our inequality Ic, as long as P(z_1) is close enough to P(w), we know that z_1 is not on the boundary of B, ie H. on the edge of the circle. In particular, z_1 is inside B.

`Intuitively, we show that if P(z) is 'close' to P(w), then z is 'close' to w. As long as P(z) is "close enough" to P(w) in terms of the distance between the bat and B's boundary, we can be sure that z is not at B's boundary.`

Well written in LaTeX:

## Part 1E:

*This section is also quite technical.*

`Some intuition. Since z_1 is inside Ball B, we can take a small step in either direction and stay on Ball B. We can use the derivative to approximate P with a linear function. With a small error term, we can roughly calculate where this step will take us. By choosing our correct direction and our small enough step, the error term is not large enough to prevent us from approaching the red dot.`

Take our point z_1. It was the point on the figure of B closest to one of the red dots. We want to prove that there is a closer point and thus arrive at a contradiction.

We define z_2 as shown below.

Above, h is a small value that we haven't specified yet, and the second expression is a linear approximation. This linear approximation is very good locally because it is defined by the derivative P'(z_1). In particular, Er is the error term of the linear approximation. The error term decreases very quickly:

`The above is almost a definition of the derivative: we can do a linear approximation and the linear approximation error disappears faster than our z_1 distance.`

Now let's do a very simple but complicated algebra (just manipulating the definitions above). Trust me if you are a little frazzled right now :)

`Repeating the previous intuition, since z_1 is on Ball B, we can take a small step in any direction and stay on Ball B. Thanks to the derivative as an approximation, we can calculate approximately where this step will be. land us with a little notion of error. By choosing our correct direction and our small enough step, the error term is not large enough to prevent us from approaching the red dot.`

## Part II: Conclusion of the argument

The idea for completing the proof is that, for any straight line that passes through the origin, we can examine the point on that line that is closest to 0+0i.* Everyone except N-1 knows that they are in the condition. Since there are an infinite number of straight lines passing through the origin, even dropping those N-1 points leaves us with an infinite number of points * which are (1) the point on that line in the image of P, which is closest to the origin , and (2) inside the image of P. Inside means we can draw a small circle around it that is included in the image of P. But this small circle contains points on the line that are closer to the origin - a contradiction.

From this we can conclude that our polynomial has a root.

*Testing this 'rigorously' requires a bit more complexity, which we leave here because it's more boring than insightful.

## one last word

*If you made it this far, thank you and well done :) I hope you enjoyed it. The argument I have presented here is not all the detail, but it is most of the detail, and I think more would be illegible (or more illegible!). You can follow me on twitter wherever I am**ethan_the_mathmo**, Shortcut**Here**. Let me know below any comments/thoughts/suggestions/requests/whatever you want.*

Thanks to Bartek Pierzchała for reading the proof and helping with readability.