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- Kochova vločka
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+<!-- Tereza: Děkuji Vám Roberte a Petře za zvážení a zapracování mých připomínek do české verze. Anglickou verzi jsem upravila a měla by odpovídat české. Myslím, že je to pěkný zajímavý příklad. :-) -->
# Fractal geometry
-A fractal is an object whose geometric structure repeats in itself. A
+A fractal is an object whose geometric structure repeats within itself. The
characteristic property of fractals is their self-similarity. Examples
-of fractals in nature can be clouds, trees or a head of
-cauliflower. The word fractal comes from the Latin word fractus,
-translated broken, broken. It was invented by Benoit B. Mandelbrot,
-considered the father of fractal geometry, which he made famous with
-his book The Fractal Geometry of Nature (1982). The word fractal comes
-from the Latin word fractus, translated as broken, shattered. It was
-devised by Benoit B. Mandelbrot, who is considered the father of
-fractal geometry. He made it famous with his book *The Fractal
-Geometry of Nature* (1982).
-
-When studying fractals, the dimension plays an important role. The
-dimension, the so-called topological dimension, which is known from
-classical Euclidean geometry, was not sufficient when describing
-fractals. Therefore it was necessary to introduce another type of
-dimension. Felix Hausdorff introduced the fractal dimension, also
-called the Hausdorff dimension. For simple objects, we can understand
-it as a number
+of fractals in nature include clouds, trees, or a cauliflower head. The word "fractal" comes from the Latin word "fractus", which
+translates broken or shattered. It was coined by Benoit B. Mandelbrot,
+considered the father of fractal geometry, renowned for
+his book The Fractal Geometry of Nature (1982).
+
+In study of fractals, their dimension plays an important role. The topological dimension, known from
+classical Euclidean geometry, proved insufficient in describing
+fractals. Therefore, another type of
+dimension was needed. It was introduced by Felix Hausdorff, known as the Hausdorff dimension. For simple objects, we can understand
+it as the number:
$$d=\frac{\ln N}{\ln\frac{1}{r}}, $$
-where $N$ is the number of parts of which the object is composed and
-which are created using homothety with the coefficient $r$ from the
-original object. For example, it is true for a square that we can
-assemble it from four smaller squares that arise from it using
-homothety with the coefficient $r=\frac{1}{2}$, i.e.
+where $N$ is the number of parts the object is composed of, formed by self-similarity with the coefficient $r$ from the
+original object. For example, for a square is true that it can be
+composed of four smaller squares that arise from it by
+self-similarity with the coefficient $r=\frac{1}{2}$, i.e.,
$$d=\frac{\ln 4}{\ln2}=2.$$
-Thus for a square, its fractal dimension is the same as the normal
-intuitive dimension.
+Thus, for a square, its fractal dimension (Hausdorff dimension) is the same as its normal
+intuitive dimension (topological dimension).
## Koch snowflake
-*Koch snowflake* is created by an iterative process from an
+The *Koch snowflake* is a curve in the plane created by an iterative process from an
equilateral triangle.
At the beginning, there is an equilateral triangle with sides of
length 1. In each subsequent step, the following is performed:
-1. Each segment is divided into thirds.
+
+1. Each line segment is divided into thirds.
2. An equilateral triangle is constructed above the middle third of the segment.
-3. The base of the triangle (the former middle third of the line segment) is removed.
+3. The base of the constructed triangle (formerly the middle third of the line segment) is removed.
-From the figure we can see that in order to determine the length of
-one side of the snowflake in the first iteration, we need 4 sides of
-the triangle that was formed by reducing the side of the triangle in
-the zero step using the identity with the coefficient $r=\frac{1}{3}$,
-i.e.
+From the figure, we can see that to determine the length of
+one side of the snowflake after the first iteration, we need 4 sides of
+the triangle that was formed by reducing the side of the original triangle from the zero step with a self-similarity coefficient $r=\frac{1}{3}$,
+i.e.,
$$
d=\frac{\ln4}{\ln3}\approx 1{,}26.
$$
-Since the Koch flake is a curve, we would expect its dimension to be
-$1$. This discrepancy is due to the fact that the Koch flake is so
-fragmented at the end that the resulting fractal has an infinite
-length, but bounds a finite content structure.
-
-> **Exercise 1.** Calculate the perimeter of the Koch snowflake during the first three iterations.
-
-*Solution.* At the beginning, we have an equilateral triangle with
- perimeter $o_0=3$. In the first iteration, we divide the three
- segments into thirds, remove the middle third of them and replace it
- with two segments of length $\frac{1}{3}$. We extended each side of
- the triangle by $\frac{1}{3}$ and the resulting perimeter is
+Since the Koch snowflake is a curve, we would expect its dimension to be
+$1$. This discrepancy is due to the fact that the Koch snowflake is finally so
+fragmented that the resulting fractal has an infinite
+length but bounds a finite area plane structure.
+
+> **Exercise 1.** Calculate the perimeter of the Koch snowflake after the first, second, and third iterations.
+
+*Solution.* Initially, we have an equilateral triangle with a
+ perimeter $o_0=3$. In the first iteration, each of the three
+ line segments is divided into thirds, and the middle third is replaced by two segments of length $\frac{1}{3}$. Each side of
+ the original triangle is extended by $\frac{1}{3}$, resulting in a perimeter after the first iteration of
$$o_1=3+3\cdot\frac{1}{3}=4.$$
-In the second iteration, on each side of the original triangle, we
-have four line segments of one-third length, which we divide into
-thirds and extend them by $\frac{1}{9}$. This gives us the perimeter
+
+In the second iteration, each side of the original triangle is divided into four line-segments of one-third of the original length, which are further divided into
+thirds and extended by $\frac{1}{9}$. This leads to a perimeter of
$$o_2=3+3\cdot\frac{1}{3}+3\cdot\frac{4}{9}=\frac{16}{3}.$$
-In the third iteration, we will extend 16 line segments on each side
-by $\frac{1}{27}$ and the perimeter will be
+In the third iteration, we extend 16 line-segments on each side
+by $\frac{1}{27}$, resulting in a perimeter of
$$o_3=3+3\cdot\frac{1}{3}+3\cdot\frac{4}{9}+3\cdot\frac{16}{27}=3+1+\frac{4}{3}+\frac{16}{9}=\frac{64}{9}.$$
-> **Exercise 2.** What is the perimeter of the Koch snowflake in the
+> **Exercise 2.** What is the perimeter of the Koch snowflake after the
> $n$-th iteration? Prove that the perimeter of the Koch snowflake
> is infinite.
-*Solution.* From the calculations above, we see that each segment has
-a third length compared to the segment in the previous step, and at
-the same time we extend it by one part, i.e. the line segment is
-extended to $\frac{4}{3}$ of the original length. The perimeter of the
-Koch snowflake in the $n$-th iteration can be expressed using the
-geometric series with quotient $\frac{4}{3}$ for $n\in\mathbb{N}$:
+*Solution.* From the above calculations, we see that each line segment is one-third the length of the line segment from the previous iteration, and at
+the same time each segment is in the next iteration extended by one third, i.e., the segment is
+extended to $\frac{4}{3}$ of its previous length. The perimeter of the
+Koch snowflake after the $n$-th iteration can be expressed using the sum of a
+geometric series with the common ratio of $\frac{4}{3}$ for $n\in\mathbb{N}$:
$$o_n=3+\left(\frac{4}{3}\right)^0+\left(\frac{4}{3}\right)^1+\left(\frac{4}{3}\right)^2+\cdots+\left(\frac{4}{3}\right)^{n-1}=3+\sum_{i=1}^n\left(\frac{4}{3}\right)^{i-1}.$$
-Since the quotient is greater than 1, this is a divergent infinite
-geometric series and the perimeter of the Koch snowflake is infinite.
+If we continued in this way indefinitely, we would obtain an infinite geometric series in the second term of the above sum. Since the ratio of the corresponding geometric sequence is greater than one, the series is divergent, and the perimeter of the Koch snowflake is infinite.
-> **Exercise 3.** Calculate the area of the Koch snowflake for the first two iterations.
+> **Exercise 3.** Calculate the area of the Koch snowflake after the first and second iterations.
-*Solution.* At the beginning, let us realize that the height in an
- equilateral triangle with a side of length $a$ is of length
- $\frac{\sqrt{3}}{2}a$ and the area of the equilateral triangle is
+*Solution.* At the beginning, let us realize that the height of an
+ equilateral triangle with side length $a$ is
+ $\frac{\sqrt{3}}{2}a$, and its area is given by
$$S=\frac{\sqrt{3}}{4}a^2.$$
The area of the initial equilateral triangle is
$S_0=\frac{\sqrt{3}}{4}$. In the first iteration, we divide the three
-line segments into thirds and place an equilateral triangle with a
-side of length $\frac{1}{3}$ on the middle third. The resulting
-content will be
+line segments into thirds and we place a smaller equilateral triangle with
+side length $\frac{1}{3}$ in the middle third. The resulting area after the first iteration is
$$S_1=\frac{\sqrt{3}}{4}+3\cdot\frac{\sqrt{3}}{4}\cdot\left(\frac{1}{3}\right)^2=\frac{\sqrt{3}}{4}\cdot\frac{4}{3}.$$
-In the second iteration, we will have four times the number of
-segments on each side of the original triangle, where we will place
-the triangle with side length $\frac{1}{9}$. The area will increase to
+In the second iteration, each side of the original triangle is divided into four segments, and a smaller equilateral triangle with side length $\frac{1}{9}$ is placed on each segment.
+ The area after the second iteration will increase to
$$S_2=\frac{\sqrt{3}}{4}+3\cdot\frac{\sqrt{3}}{4}\cdot\left(\frac{1}{3}\right)^2+3\cdot4\cdot\frac{\sqrt{3}}{4}\cdot\left(\frac{1}{9}\right)^2=\frac{\sqrt{3}}{4}\left(1+\frac{1}{3}+\frac{1}{3}\cdot\frac{4}{9}\right)=\frac{\sqrt{3}}{4}\cdot\frac{40}{27}.$$
-> **Exercise 4.** What is the area of the Koch snowflake in the $n$-th
+> **Exercise 4.** What is the area of the Koch snowflake after the $n$-th
> iteration? How many times larger is the area of the Koch snowflake
> relative to the original equilateral triangle?
-*Solution.* From the previous considerations, the number of segments
-where we add a new triangle is four times larger in each iteration. At
+*Solution.* From the previous considerations, it follows that the number of segments,
+where we add a new triangle, is four times greater in each iteration. At
the same time, the side of our new triangle shrinks to a third of its
-previous size, so its content shrinks to a ninth. This gives us the
-members of the geometric sequence with the quotient $\frac{4}{9}$ and
-the area of the Koch snowflake after the $n$-th iteration is
+previous size, so its area decreases to one-ninth.
+We obtain terms of a geometric sequence with a ratio of $\frac{4}{9}$, and the area of the Koch snowflake after the $n$-th iteration is formed by the area of the original triangle and the sum of the first $n$ terms of that geometric sequence:
$$S_n=\frac{\sqrt{3}}{4}\left[1+\frac{1}{3}+\frac{1}{3}\cdot\frac{4}{9}+\dots+\frac{1}{3}\left(\frac{4}{9}\right)^{n-1}\right]=\frac{\sqrt{3}}{4}\left[1+\frac{1}{3}\sum_{i=1}^n\left(\frac{4}{9}\right)^{i-1}\right].$$
-In the expression for $S_n$ there is a convergent infinite geometric
-series, therefore the content of the Koch flake is of finite size:
-$$S_n=\frac{\sqrt{3}}{4}\left(1+\frac{1}{3}\cdot\frac{1}{1-\frac{4}{9}}\right)=\frac{\sqrt{3}}{4}\left(1+\frac{1}{3}\cdot\frac{9}{5}\right)=\frac{8}{5}\cdot\frac{\sqrt{3}}{4}.$$
-Koch snowflake has an infinite perimeter bounding a finite area that
-has a area 1,6 times the area of the original equilateral triangle.
+Since the ratio of the geometric sequence is less than one, by continuing to infinity, we obtain a convergent geometric series. Using the formula for its sum, we get the area of the Koch snowflake after an infinite number of iterations.
+$$S=\frac{\sqrt{3}}{4}\left(1+\frac{1}{3}\cdot\frac{1}{1-\frac{4}{9}}\right)=\frac{\sqrt{3}}{4}\left(1+\frac{1}{3}\cdot\frac{9}{5}\right)=\frac{8}{5}\cdot\frac{\sqrt{3}}{4}= 1{,}6 \cdot S_0.$$
+The Koch snowflake has an infinite perimeter enclosing a finite area that is approximately 1.6 times larger than the area of the original equilateral triangle.
## Literature
* MathWorld. *Koch snowflake* [online]. Available from <https://mathworld.wolfram.com/KochSnowflake.html> [cit. 13. 7. 2023].