Preprint.

1. 1
PLOS ONE Fractal Manuscript. 2022/11/21 - 2022/12/21. To 2023/01/24. Sharon Xiao Liu.
(This is a typed up version, of the work done in the diary.)
A fresh perspective, on the Fractal Dimension Equation.
Abstract
In this paper, we explore the fractal dimension equation, which was originally written by Niels Fabian Helge von Koch, in
1904. This equation, is D = -(ln N/ln epsilon) . In this manuscript, we have applied it to the snowflake fractal. We find
meanings, for each symbol in the equation. In examining this equation, we find new discoveries. We have found out, a
diagrammatic proof, with regards to epsilon being 3, in the Koch snowflake fractal. We have used Python, to find out a
large negative power, which is currently not possible with just the pow function. We have proved one possible value of a
power base, with python, if y = dy/dx , as defined for e^x . We have found a t value, which tells you when the root of a
number, reaches a value between 1 and 2.
---
(End neat manuscript version, and start of the preprint.)
(Label each day’s work, under its own date.)
1. This paper examines the fractal dimension equation, written by Koch in 1904.
2. The equation is D = - (ln N/ln epsilon) , where D is the fractal dimension, N is the number of self-similar pieces,
and epsilon is the magnification.
3. D relates to the number of fragments, present in the fractal.
4. The bigger epsilon is, the smaller the child pieces.
5. The bigger N is, the greater the number of the child pieces.
6. D relates to how the fractal increases in size.
7. The negative symbol, is about the free space, which decreases as the fractal increases in size.
8. We examine the Koch snowflake, for which N is 4, and epsilon is 3.
9. It is important to examine work related to the weather and climate.
10. Over recent years, the appearance of snow has declined in the UK.
11. Also, it has been predicted, that we are about to experience a further increase in global temperatures.
12. Possibly this year, there has been drought and parched grass. If the grass is affected, so would farmland.
13. Therefore, a reinvestigation into the snowflake, is much needed.
14. In the Koch snowflake fractal, N is 1 + 3 , as ln 1 = 0 , so the value of N must be bigger than 1.
15. N is 1 + 3 = 4 , because there are three smaller pieces, which are added to the larger piece.
16. Diagram of the snowflake fractal.
a. Iteration 1, has an equilateral triangle.
b. Iteration 2, has an upside down equilateral triangle, behind the equilateral triangle facing up.
c. Iteration 3, has smaller upside down equilateral triangles, at each of the six points in the previous
iteration.
17. The D of each fractal iteration, can be calculated.
18. To draw the snowflake fractal iteration, the centre of the equilateral triangle, has to be found.
19. Diagram of the standard 30, 60, 90 degree triangle.
a. The longest side, is 2.
b. The middle length side, is root 3.
c. The shortest side, is 1.
d. The ratio of the sides, is 1, root 3, 2.
20. Draw iteration 2.
21. Diagram of the bisected equilateral triangle.
a. Triangle alpha, is the right-angled triangle, which bisects the equilateral triangle.
b. Triangle alpha, is to the left of the equilateral triangle.
c. The side of the equilateral triangle, is length 1.

2. 2
d. Triangle alpha, has its longest side to the left, the middle length side, through the centre of the
equilateral triangle, and the shortest side, at the bottom left, of the equilateral triangle.
e. For triangle alpha, the ratio of the sides (going clockwise), is 2, root 3, 1.
f. Brought to the scale, of the equilateral triangle, is ratio of the sides (going clockwise), is 1, root 3 divided
by 2, and ½.
22. Diagram of the centre of the equilateral triangle.
a. Triangle beta, is the right-angled triangle, which has a corner at the centre of the equilateral triangle.
b. Triangle beta, has its longest side extending from the centre of the equilateral triangle, the shortest side
measuring the height of the centre of the equilateral triangle, and the middle length side, to the bottom
left, of the equilateral triangle.
c. For triangle beta, the ratio of the sides (going clockwise), is 2, 1, root 3.
d. The base of triangle alpha, is ½.
e. To scale triangle beta, to the equilateral triangle, we need to divide the base of triangle beta by root 3,
and multiply by ½.
f. Brought to scale, triangle beta, the lengths of the sides (going clockwise), is 1 divided by root 3, 1 divided
by (root 3 times ½), and 1.
23. The value of the e number, in exponential calculations, can be found, by looking at the curve of e to the power
of x.
24. As we know, the gradient of e to the power of x, is the same as e to the power of x. Therefore, we take two
points, on the curve, of e to the power of x.
25. The curve of e to the power of x, is shallower, to the left of the y-axis. Therefore, when we take two x points,
which are adjacent integers, the value of e to the power of x, changes very little, and is more accurate.
26. We will use x = -10,000 , and x = -9,999.
27. Diagram of the e to the power of x curve, when x = -10,000 , and x = -9,999 .
a. The curve goes upwards to the right, when compared to the left.
b. There is a right-angled triangle, where the vertical line is to the right.
c. The bottom left, of the right-angled triangle, is x = -10,000 , and y = e^-10,000 .
d. The top right, of the right-angled triangle, is x = -9,999 , and y = e^-9,999 .
e. The gradient, of the right-angled triangle, is the y difference, divided by the x difference.
f. The gradient, is (e^-9,999 – e^-10,000) / (-9,999 - -10,000) = (e^-9,999 – e^-10,000) / (1) = e^-9,999 – e^-
10,000 .
g. We will take e^-10,000 , as the e^x value. This is the y value, which equals the gradient.
h. Then we obtain, e^-10,000 = gradient. This is, e^-10,000 = e^-9,999 – e^-10,000 .
i. Then, we obtain, 2e^-10,000 = e^-9,999.
j. We notice, that e^-10,000 (the left hand side), needs to be the same, as the right hand side, e^-9,999 –
e^-10,000 .
k. Then we know, that e is around 2. We try e values, and if we use x = 10,000 , e is correct to 3 decimal
places. [I think.]
l. e is around 2, because when we take x = 0, and x = 1, we obtain 1 = e – 1. So, e is around 2.
m. The left hand side, e^-10,000, is (by using WolframAlpha)...
n. This argument, is not circular, as you are supposed to try values of e, close to 2, and via iteration, you
get a more accurate value.
o. If you take x = 1, you get e = e – 1. So, this does not, give you 3. [Unlucky.] Wait. Riley says, this is 1
above 2. [What does he mean?]
i. Well, you get 0 = -1. Or, 0 = 1.
ii. Anyway, you know, for x = 10,000 , you do get a value for e, which is above 2.
iii. He means, that you need to take x = 1, and x = 2, and this gives you a value for e, above 2. Is this
right?
iv. That means, e^1 = e^2 – e, which is e = e^2 – e.

3. 3
v. If you try e = 2, 2 = 4-2. Then, e = 2.
vi. That’s not what we want.
vii. Try e = 3. 3 = 4 – 3, 3 = 1. So, it’s lower than 3.
viii. Try e = 1. 1 = 1 –1. So, it’s higher than 1.
ix. Because of the differences, though, between the left and right hand sides, we think this means,
e is between 2 and 3.
x. If you try x = 2, and x = 3, that means, e^2 = e^3 – e^2. We don’t need, to do this now.
p. I thought, it was x = 10,000 . Or, was it x = 100,000 ? It was one, or the other. Try x = 100,000 , and you
get e = 2.718 , for sure.
q. Do we, need to type the JavaScript loop? PLOS ONE, says yes.
28. Loop, to obtain a value for e. See file.
a. Plan the file.
b. Write in JavaScript, and put it on the InfinityFree server.
c. The script, must do the WolframAlpha calculation.
d. This is a^x (initially, anyway). a starts from 1.
e. Math.pow(1,x) .
f. So, what is the loop you require? (The White Johan, suggests you go around in circles. Like, what you do
in your diary, in green.)
g. for(a = 1; a <= 10000; a++)
h. Edge does handle a loop, of this length.
Python IDLE 3.10 code.
def lhs(try_e):
lhs_one_start_power_10t = pow(try_e,10000)
lhs_two_length = len(str(lhs_one_start_power_10t))
lhs_three = (pow(10,3010))/(pow(2,10000))
lhs_four = (pow(10,lhs_two_length-1))/lhs_one_start_power_10t
lhs_five = "times 10, to the power of -3010."
lhs_six = " times 10, to the power of "
lhs_seven = -(lhs_two_length-1)
lhs_eight = str(lhs_four) + lhs_six + str(lhs_seven)
return lhs_eight
"""
Then you need to work out the 10 to the power of something, that is multiplied
to lhs_three .
You could do an array of numbers, for try_e , but it is better, to use a loop.
"""
print ("2 to the power of -10,000 , is:")
print (lhs(2))
def rhs_three(try_e):
rhs_one_start_power = pow(try_e,9999)
rhs_two_length = len(str(rhs_one_start_power))
rhs_three = (pow(10,rhs_two_length-1))/rhs_one_start_power
rhs_three_two = " times 10, to the power of "
rhs_three_three = -(rhs_two_length-1)

4. 4
rhs_three_four = str(try_e) + " to the power of -9,999 is "
rhs_three_five = str(rhs_three) + rhs_three_two + str(rhs_three_three)
rhs_four_second_power = pow(try_e,10000)
rhs_seven_length = len(str(rhs_four_second_power))
rhs_eight = (pow(10,rhs_seven_length-1))/rhs_four_second_power
rhs_nine = -(rhs_seven_length-1)
rhs_ten = str(try_e) + " to the power of -10,000 is "
rhs_eleven = str(rhs_eight) + " times 10, to the power of " + str(rhs_nine)
rhs_twelve = rhs_three_four + rhs_three_five
rhs_thirteen = rhs_twelve + " and " + rhs_ten + rhs_eleven
#The above is an output.
rhs_fourteen = rhs_three_three-rhs_nine
rhs_fifteen = rhs_fourteen*10*rhs_three
rhs_sixteen = rhs_fifteen-rhs_eight
rhs_seventeen = str(rhs_sixteen) + " times 10, to the power of "
rhs_eighteen = str(rhs_nine)
rhs_nineteen = rhs_seventeen + rhs_eighteen
return rhs_nineteen
print ("The difference between the two powers, is:")
print (rhs_three(2))
---
Output of the Python code.
lhs: 2 to the power of -10,000 , is:
0.5012372749206452 times 10, to the power of –3010
rhs_two:
2 to the power of -9,999 is 0.10024745498412904 times 10, to the power of -3009 and 2 to the power of -10,000 is
0.5012372749206452 times 10, to the power of -3010
rhs_three: The difference between the two powers, is:
0.5012372749206452 times 10, to the power of –3010
1. According to this algorithm, e = 2. (This algorithm is included, as it might be important for reaching a further
conclusion.)
2. This is clearly not 2.718...
3. This is because, this algorithm, does not take into account, all of the factors, that determine e^x.
4. In addition to e^x = d(e^x)/dx , dy/dx>0 , and when x is increasing, dy/dx is also increasing.
5. We are interested in the general form C^x.
6. Let us try, 2^x.
a. Use the Newton-Raphson method, to find d(2^x)/dx .
b. x = 0: f(0) = 1. f’(0) = 1 – 0.5 = 0.5 . (From the laws of exponentials.)
c. x = 1: f(1) = 2. f’(x) = f(x) - f(x-1). Therefore, f’(1) = 2 – 1 = 1.
d. x = 2: f(2) = 4. f’(2) = 4 – 2 = 2.
e. x = 3: f(3) = 8. f’(3) = 8 – 4 = 4.
f. Therefore, d(2^x)/dx = 2^(x-1).
g. Therefore, d(2^x)/dx = f(x)/2.
7. Let us try, 3^x.

5. 5
a. x = 0: f(0) = 1. f’(0) = 1 – 1/3 = 2/3.
b. x = 1: f(1) = 3. f’(1) = 2.
c. x = 2: f(2) = 9. f’(2) = 9 – 3 = 6.
d. x = 3: f(3) = 27. f’(3) = 27 – 9 = 18.
e. Therefore, f’(x) = f’(x-1)*3.
f. f'(x-1) = f(x-1) - f(x-2). So, f’(x) = (f(x-1) - f(x-2))*3.
g. d(3^x)/dx = f(x)-f(x-1).
h. d(3^x)/dx = f(x)*2/3.
8. Let us try, 4^x.
a. x = 0: f(0) = 1. f’(0) = 1 – ¼ = ¾.
b. x = 1: f(1) = 4. f’(1) = 4 – 1 = 3.
c. x = 2: f(2) = 16. f'(2) = 16 – 4 = 12.
d. x = 3: f(3) = 40 + 24 = 64. f'(3) = 64 – 16 = 64 – 10 – 6 = 48.
e. x = 4: f(4) = 16*16 = 256. f'(4) = 256 – 64 = 256 – 60 – 4 = 196 – 4 = 192.
f. d(4^x)/dx = f(x)*¾. This does not work.
g. 18*3 = 30 + 24 = 54.
h. d(4^x)/dx = f(x-1)*3. This does not work.
i. d(4^x)/dx = f(x)-f(x-1) = 4^x – 4^(x-1) = 4^x - (4^x)/4.
2023/01/11. Wednesday. To 2023/01/14.
1. I have used, the Newton-Raphson method, and tried to make it more accurate. The calculus way of doing things,
would have been better, but that’s just a set shortcut, with new maths operators, so we’re keeping it simple.
2. I should investigate the calculus shortcut, though. But you see, I might need a new table of values, like what
happened to multiplication, and sin(x).
3. God exists, because everyone wants the food to remain safe. Therefore, we are investigating the snowflake, and
hope that more creators than destroyers, read the work.
4. There are some ways, where the Newton-Raphson method, could be made more accurate.
a. A part of the curve, which is shallow, can be selected.
b. We are trying to find y = dy/dx .
c. Because we are trying to find e correct to 3 decimal places, we would need to find the result of try_e^x ,
when try_e is fractional.
i. This depends on long multiplication, where the answer starts to the left.
ii. Just write out the answer of the long multiplication, complete with zeroes.
iii. 1234 times 1234.
iv. 1000 times 1234, plus 200 times 1234, plus 30 times 1234, plus 4 times 1234.
d. We would also look at the instance, where x is fractional. This is a^(x_d) times a^x , where x_d is a
power of 10 (or, should we say, a root of 10), by which we mean, it is a decimal, that contains only
zeroes and a one.
i. We would look at the case, where 2^0.0001 = rhs.
ii. This is a root, and we could find the answer, by adjusting x (i.e. 0.0001), but, what is 2^0.0001 ,
by hand?
iii. That is, 2^(1/10,000) .
iv. That is, root 10,000 of 2.
v. We could do a Python loop., where a number between 1 and 2, is multiplied by itself 10,000
times.
vi. If root 10,000 of 3 is calculated, the Python loop, would obtain a number between 1 and 2, I
think. wolframalpha.com , agrees with this.
vii. The relative size of 10,000 to 3, determines what the value of the root is.
viii. We define the root extent, as 10,000 . We define the root base, as 3.

6. 6
ix. If a number is multiplied by itself, like 1 times 1, this remains the same. (Type more, faster.) So,
1 times 1, is the same. 1.1 times 1.1, is bigger than 1.1. At what point, does the root extent, get
too big, for the root answer, to be bigger than 1? When the root base, is 4, and the root extent,
is 2, the root answer, is 2. When the root extent, is 3, the root answer is between 1 and 2.
Therefore, we hypothesise, that if the root extent, is bigger than half of the root base, the root
answer is between 1 and 2. This appears to be the case, for a root base, of 4.
x. We will see, whether this is true, for a root base, of 9. When the root extent, is 2, the root
answer is 3. So you see, the root answer hypothesis, is incorrect. The square root, of the root
base, is important. Actually, there appears to be no pattern. How do you find out, when the root
answer, is between 1 and 2? The root answer, which is above 2, is important. Except, there is no
easy way to find this out. As the root base gets bigger, the square root, then the cube root, gives
answers, which decrease rapidly to above a root answer of 1. We need, to bring out a table.
xi. Root base 4: 4 root 2, is 2. 4 root 3, is between 1 and 2.
xii. Root base 9: 9 root 2, is 3. 9 root 3, is between 2 and 3. 9 root 4.
xiii. So, the 2^x table, comes into effect, for determining when the root answer, is between 1 and 2.
For 4, this does fit into the 2^x table. The x value, is 2. For 9, this does fit into the 2^x table. The
x value, is 3. Therefore, when the root extent is below 3, the root answer is between 1 and 2.
(This is a bit of a diversion. You should really get on, with the main part of this manuscript.)
xiv. Is it, 9/(2^x)? Well, that is, when x = 2, 9/4. If this value, t, is less than a half, then root x of 9,
might be between 1 and 2.
xv. What is root 2 of 9? This is 3.
xvi. When x is 3, t = 9/(2^3), which is 9/8. This value, is less than a half. Root 3 of 9, is between 2 and
3.
xvii. This t equation, is proven to be correct, as the remainder is not divisible by 2. This t equation, is
proven to be incorrect.
xviii. Is it, 4/(2^x)? Well, that is, when x is 2, 4/4 = 1.
xix. Is it, when t is greater than 1, when the root value is between 1 and 2? This appears to apply, to
a root base of 9 or 4.
xx. Is it, 16/(2^x)? Well, when x=2, t = 16/4. 16 root 2, is 4. When x = 3, t = 16/8. 16 root 3, is
between 2 and 3. When x = 4, t = 16/16 = 1.
xxi. We are trying to find, a root extent and a root base. In 9/(2^x) , the x is the root extent, and 9 is
the root base.
xxii. We have already looked at 9/(2^x) , and this has been proven to be correct. This t = root base /
(2^root extent) , appears to be correct, for root bases of 9, 4, and 16.
We still need ln(x), epsilon, python code, family of curves, and the calculus way, of obtaining a^x. This can be done, in
your own spare time. We still need the diagrammatic proof of epsilon, so draw a diagram in Scalable Vector Graphics,
after you write about epsilon.
PLOS ONE, said to type to the bottom of this page, and try submitting it. They say, you’ll get comments, to reply to.
2023/01/24: PLOS ONE, says it’s good enough as it is. They command you, to neaten the manuscript.
2023/01/25: Do not delete anything. Write the neat version, at the beginning of the manuscript.
My feeling is, this manuscript does not contain enough content.
Send the entire manuscript, including the working out, to PLOS.
1. Family of curves.
a. I have proven, in symbolic form, that e^x is not the only curve, where y = dy/dx. In fact, any numerical
base, including 10, when raised to the power of x, also satisfies y = dy/dx .
b. Symbols generalise numbers, and we talk about the general case.
c. y = dy/dx . a^x = a^x – a^(x-1) .
d. Or, a^x = a^(x+1) - a^x .
e. y_(0) = 1. y_(1) = a . y_(2) = a^2 . y_(3) = a^3 .

7. 7
f. a^2 = a^3 – a^2 .
g. As x increases, the gap between the y values, gets bigger. However, this is proportional to, the y value.
Therefore, a good approximation of e^x , where y = dy/dx , is a base to the power of x.
i. The gap between the y values, is always a constant multiple.
h. The power base, can be fractional, or irregular. This is because, when you zoom out of a graph, where
the x values are whole numbers, you obtain a fraction. An irregular base, can be adjusted to, with
sequential zooming of the graph, and if with each zoom, y = dy/dx , then the irregular number is likely to
also satisfy y = dy/dx .
2. Looping python code, to prove the family of curves.
a. The Python code, must contain at least one loop.
Python IDLE 3.10
print ("with a loop:")
#Type a loop.
for x in range(1,10):
print(rhs_two(x))
print("and the difference between the two powers is:")
print(rhs_three(x))
---
with a loop:
1 to the power of -9,999 is 1.0 times 10, to the power of 0 and 1 to the power of -10,000 is 1.0 times 10, to the power of
0
and the difference between the two powers is:
-1.0 times 10, to the power of 0
2 to the power of -9,999 is 0.10024745498412904 times 10, to the power of -3009 and 2 to the power of -10,000 is
0.5012372749206452 times 10, to the power of -3010
and the difference between the two powers is:
0.5012372749206452 times 10, to the power of -3010
3 to the power of -9,999 is 0.18389675171857245 times 10, to the power of -4770 and 3 to the power of -10,000 is
0.6129891723952414 times 10, to the power of -4771
and the difference between the two powers is:
1.225978344790483 times 10, to the power of -4771
4 to the power of -9,999 is 0.10049552230794978 times 10, to the power of -6019 and 4 to the power of -10,000 is
0.25123880576987445 times 10, to the power of -6020
and the difference between the two powers is:
0.7537164173096234 times 10, to the power of -6020
5 to the power of -9,999 is 0.9975315584403792 times 10, to the power of -6989 and 5 to the power of -10,000 is
0.19950631168807584 times 10, to the power of -6989
and the difference between the two powers is:
-0.19950631168807584 times 10, to the power of -6989
6 to the power of -9,999 is 0.18435181339635145 times 10, to the power of -7780 and 6 to the power of -10,000 is
0.30725302232725243 times 10, to the power of -7781
and the difference between the two powers is:
1.536265111636262 times 10, to the power of -7781
7 to the power of -9,999 is 0.7323149453141949 times 10, to the power of -8450 and 7 to the power of -10,000 is
0.10461642075917071 times 10, to the power of -8450
and the difference between the two powers is:
-0.10461642075917071 times 10, to the power of -8450

8. 8
8 to the power of -9,999 is 0.10074420348672732 times 10, to the power of -9029 and 8 to the power of -10,000 is
0.12593025435840915 times 10, to the power of -9030
and the difference between the two powers is:
0.8815117805088641 times 10, to the power of -9030
9 to the power of -9,999 is 0.33818015292642273 times 10, to the power of -9541 and 9 to the power of -10,000 is
0.37575572547380304 times 10, to the power of -9542
and the difference between the two powers is:
3.0060458037904243 times 10, to the power of –9542
---
Analyse the results.
1 to the power of -9,999 is 1.0 times 10, to the power of 0 and 1 to the power of -10,000 is 1.0 times 10, to the power of
0
and the difference between the two powers is:
-1.0 times 10, to the power of 0
The calculation is in range. 1 to the power of anything, is still 1. The –10,000th
power, and the difference, is the same. Fix
the script.
3. Calculus way, of obtaining d(a^x)/dx . This has already been done.
a. From Dummies: d(n^x)/dx = n^x times ln(x) .
b. Isn’t it, just n^x? Now that a family of curves, has appeared.
c. Prove this, for specific cases (numbers).
d. The equation, to find d(a^x)/dx, is delta^-1 times (a^(x+delta)-a^x).
i. Which is, delta^-1 times a^x(a^delta – 1).
ii. The smaller delta is, the more accurate the value.
e. Or, do a zoom factor, where 0.1 is zooming in by 10.
i. d(a^x)/dx = (a^f times a^((x+1)*f^-1) - a^f times a^(x*f^-1))/f.
ii. d(a^x)/dx = a^f(a^((x+1)*f^-1) - a^(x*f^-1))/f.
f. Type in general, about the calculus way, of obtaining d(a^x)/dx.
i. As x increases, the gradient increases by a multiple, of everything before. The gradient is the
height, divided by 1 (if x is increasing by 1, each time).
ii. The gradient, could be the value in front of x, and the value behind x, joined together by a
straight line.
iii. There can be no bias, of the forwards or backwards direction, whilst this method is employed.
iv. Therefore, we have established, that n^x, or a^x, is definitely in the derivative.
v. As it increases like this.
vi. Perhaps, the gradient at x=0, needs to be established, and then, this value, needs to be
multiplied by a^x.
vii. The gradient, could be the value in front of x, and the value behind x, but this crosses a^2.
viii. d(a^x)/dx = (a^(x+1)-a^(x-1))/a^2. (?)
ix. This is the specific height differences, divided by how much the height has risen.
x. d(a^x)/dx = (a^(x+1)-a^(x-1))/a^2/2.
xi. The curve is increasing. At x-1, and x+1, it increases by a, and then a again.
xii. If the gradient at x=0, is established, then this, when multiplied by a^x, definitely gives you the
value of dy/dx.
xiii. Is the gradient at x=0, ln(x)?
xiv. If a line is drawn, between x-1 and x+1, the mean gradient is the gradient of the line, divided by
a^(2/2).

9. 9
xv. d(a^x)/dx = (a^(x+1)-a^(x-1))/2/a^(2/2).
xvi. If a line is drawn, between 10^9 and 10^11, the mean gradient is the gradient of the line,
divided by 10.
xvii. This division by 10, draws the gradient back, to between the centre of the two points.
xviii. Now, this method needs to be proven. Type a script. Or, reason, and then type a script later.
xix. d(a^x)/dx = (a^(x+1)-a^(x-1))/2/a.
xx. This could go into the table of standard derivatives.
xxi. It is a lot simpler looking, than n^x times ln(x).
4. ln(x).
a. This is e^x, flipped in y = x .
b. Put y in place of x, in y = e^x , and x in place of y. Then, rearrange to make y the subject. The inverse
function, is y = ln(x) .
5. Epsilon.
a. Diagrammatic proof, in SVG.
b. This is easier, than drawing in draw.io , as the lines are precise, and can be calculated.
c. The hexagon, is split into six equilateral triangles.
d. Verbal diagrammatic proof, is when a set of words, is used to draw an image, in the mind, that proves a
conjecture.
e. As the snowflake’s points, are made up of equilateral triangles of the same size, the side of the
snowflake, is divided into three equal parts. Therefore, epsilon is 3.
2023/01/31. Get a feel, for how much work, is required, for an article to be PLOS ONE worthy. It has to be better, than
this article: https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0280036 . Your work, is already better
than this article. You need to send your work, to PLOS ONE. Print out this page.