• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# The Koch Snowflake

Extracts from this document...

Introduction Math Portfolio

The Koch snowflake

Introduction: The Koch snowflake fractal is built by starting with an equilateral triangle, and removing the inner third of each side, building another equilateral triangle at the location where the side was removed.  When n = 0, 1, 2and 3 each value of Nn, ln, Pn and An can be shown as the following table:

 n 0 1 2 3 Nn 3 12 48 192 ln 1   Pn 3 4  An    And the process I use to get the value are shown below:

The number of the side:

To get the next snowflake, we can found that, each side of the triangle will break into 4 new sides, hence the number of each side will always 4 times than previous one. Thus we get

Nn=4Nn-1

Thus we assume that Nn is the single of the number of sides and get the process below: Shape1: we get a triangle. N1=3

Shape2: We break into 4 new sides. So the N2=3×4

Shape 3: Each side then brake into 4 new sides, So N3=3×4 ×4

Shape 4:We repeat the same step, thus we get N4=3×4×4×4

And then we graph our data:

The number of the side We can found that the shape of the graph is a geometric series. Thus, if we want to get the next shape, we should time the same number.

Middle

×  = P3=3×  ×  ×  = Then we can put our data into the diagram below:

The length of perimeter We found that if we link all point together, we may get an up-sloping curve which is a geometric series. Thus, if we want to get the next shape’s perimeter, we should time the same number. Thus it’s a geometric series which has a first term of 3 and a common ratio of  , obviously. So we can get the geometric formula:

Pn=3  To verify the formula, we put n=0, 1,2and 3 into the formula. So we get:

P0= =3         P1= =4         P2= = P3= = The results are already proved above.

The area of the snowflake:

From the observation, we can found a rule between the number of the side and the number of new triangle appeared. Firstly, Nn=3

Secondly, Nn=12

The number of new triangle appeared=3

Thirdly, Nn=48

The number of new triangle appeared=12

Fourthly, Nn=192

The number of new triangle appeared=12

We can found that, the number of new equilateral triangle appeared in the shape is equal to the number of the side in the previous triangle (Nn-1).

And then we show the area of the each addition one.

Conclusion

 n An Nn ln Pn 0 0.433013 3 1 3 1 0.57735 12 1 4 2 0.6415 48 0.333333 5.333333 3 0.670011 192 0.111111 7.111111 4 0.682683 768 0.037037 9.481481 5 0.688315 3072 0.012346 12.64198 6 0.690818 12288 0.004115 16.85597 7 0.69193 49152 0.001372 22.47462 8 0.692425 196608 0.000457 29.96616 9 0.692645 786432 0.000152 39.95488 10 0.692742 3145728 5.08E-05 53.27318 11 0.692786 12582912 1.69E-05 71.03091 12 0.692805 50331648 5.65E-06 94.70788 13 0.692813 201326592 1.88E-06 126.2772 14 0.692817 805306368 6.27E-07 168.3696 15 0.692819 3221225472 2.09E-07 224.4927 16 0.69282 12884901888 6.97E-08 299.3237 17 0.69282 51539607552 2.32E-08 399.0982 18 0.69282 2.06158E+11 7.74E-09 532.1309

To find out the value for n where An+1 equals An to six places of decimals. We get the help from the Excel software to work out the first 18 term to check if we get the number we need.

From the result we can see that the An for n=16 is equal to for n=17 to six decimal places. So the value of n is 16 where An is equal to An+1 to six decimal places.

So when n = 16:

N16 = 12884901888   l16 = 2.323057 10-8

P16 = 299.323656    A16 = 0.692820

Then we consider the expression for Pn which:

Pn= Pn-1

We can see that Pn is always increasing in the same ratio of . So the value of Pn increase to positive infinity as n→∞.

For the expression of An which is:

An= An-1+ According to the formula of sum of geometric series to infinity:

sn  If  Thus,  Thus we know: in this case as n→∞ we can have an infinity length of perimeter in a limited area.

we use the math induction to prove this conclusion:

As we know P(1) is true since Assume that n=m is true for integer t≥1. Consider n=m+1

Am+1 =  Prove    Thus P(m+1) is true when P(m) is true. Therefore P(n) is true for all integers m≥1.

Richard ouyang

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

-ï¿½+ï¿½5ï¿½nï¿½Fq3ï¿½x"ï¿½ï¿½ï¿½7ï¿½{ï¿½#ï¿½ï¿½|#3/4?ï¿½ï¿½Å¯"Mï¿½.ï¿½?!(tm)p"PKï¿½"<"ï¿½ï¿½iiï¿½hthï¿½hBhï¿½hï¿½h.ï¿½ ï¿½1/4ï¿½ï¿½J\$ï¿½D{"ï¿½ï¿½I,#^" ï¿½ï¿½hÉ´'ï¿½Fï¿½ï¿½ï¿½ï¿½ï¿½ï¿½i{hï¿½ï¿½~%'Hï¿½\$}' )ï¿½[email protected] ï¿½&1/2&ï¿½ cï¿½"ï¿½ï¿½ ï¿½ï¿½ "ï¿½kï¿½ï¿½ï¿½BO ï¿½7 ï¿½Cï¿½B_Bï¿½ï¿½ï¿½ï¿½A"ï¿½ï¿½ï¿½Âï¿½Pï¿½ï¿½ï¿½ï¿½a...'ï¿½Qï¿½qc8ï¿½aï¿½Fï¿½{ï¿½sdY"lBï¿½'ï¿½kÈ·ï¿½ï¿½L(&!&#&?ï¿½}Lï¿½LL3ï¿½Xf1f ï¿½ï¿½|ï¿½fï¿½aï¿½E2ï¿½2ï¿½ KK%ï¿½M-)V"("kï¿½Qï¿½+ï¿½ï¿½?ï¿½xï¿½ ï¿½ï¿½ï¿½ï¿½.ï¿½ï¿½ï¿½ï¿½ï¿½sï¿½ï¿½ ï¿½boaï¿½"ï¿½ï¿½ï¿½"#"ï¿½8G;ï¿½+N4ï¿½\$ï¿½=g"ï¿½)ï¿½ï¿½.f.-.?(r)C\Wï¿½^pï¿½ï¿½'ï¿½ ï¿½(c)ï¿½5ï¿½Cï¿½+<1/4<f<Q<ï¿½<ï¿½yxYyï¿½yCxï¿½y"yï¿½ï¿½tï¿½ï¿½|ï¿½|ï¿½ï¿½>ï¿½ï¿½ï¿½ï¿½-ï¿½ï¿½/ p ï¿½ ï¿½ ï¿½Xtï¿½l|%D#ï¿½.(T,ï¿½'ï¿½(ï¿½'l#1/4Wï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½Hï¿½H(c)ï¿½]'UQ1QWï¿½ï¿½ï¿½sbï¿½bb)bï¿½^ï¿½"ï¿½Ä£ï¿½"ÅH`%ï¿½%B%ï¿½\$F\$aIï¿½`ï¿½Jï¿½GRï¿½"ï¿½Uï¿½Jï¿½4FZC:Bï¿½Zï¿½(tm) ï¿½ï¿½ï¿½Lï¿½ï¿½(tm)ï¿½ï¿½ï¿½ï¿½Ö²Ù²ï¿½_ï¿½"ï¿½<ï¿½ï¿½Ýï¿½'W'"ï¿½ï¿½ï¿½T +X*d+t),+J*ï¿½)V*>Q")(tm)*e(u(-)K) (ï¿½Rï¿½PaRï¿½Q9 Ò§ï¿½(c)ï¿½ï¿½ï¿½zQu^MXï¿½Gï¿½ï¿½3ufu;ï¿½ï¿½ï¿½ Cï¿½ ï¿½kï¿½ï¿½ï¿½qï¿½W4ï¿½ï¿½'ï¿½ ï¿½jÔï¿½ï¿½ ï¿½(r)Õï¿½Ô¡ï¿½ï¿½(tm)ï¿½ï¿½ï¿½ï¿½=ï¿½;ï¿½' Gï¿½"ï¿½{ï¿½/ï¿½ï¿½_ï¿½?k abï¿½dï¿½ï¿½Pï¿½0Æ°ï¿½pï¿½Hï¿½(Í¨ï¿½elf|ï¿½xï¿½"lï¿½lRaï¿½ï¿½Tï¿½4ï¿½ï¿½ï¿½(tm)ï¿½Yï¿½Yï¿½9ï¿½ï¿½ï¿½ï¿½ï¿½3 - ?ï¿½ï¿½EK5ï¿½4ï¿½~+Z+G" "wï¿½'ï¿½1ï¿½]6ï¿½ï¿½ï¿½M'ï¿½K[ï¿½ï¿½]`-ï¿½(r)ï¿½]ï¿½ï¿½ï¿½ï¿½ï¿½(r)ï¿½cï¿½ï¿½ï¿½+ï¿½?8(8ï¿½uï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Ó¤ï¿½ï¿½s1/4sï¿½ 1/2ï¿½ï¿½Kï¿½Ëª"ï¿½kï¿½ï¿½"ï¿½ï¿½[ï¿½ï¿½wNwï¿½{ï¿½ ï¿½ï¿½Å£ï¿½ceï¿½ï¿½ï¿½"g<U<s=ï¿½1/2ï¿½1/4'1/4ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½71/27ï¿½ï¿½ï¿½ï¿½Õ§ï¿½gï¿½ï¿½ï¿½RMYï¿½ï¿½=ï¿½"ï¿½gï¿½Wï¿½ï¿½_ß¿ï¿½>@' 0`6P'ï¿½0p.H'ï¿½(h>X/ï¿½\$xï¿½jDï¿½ ....ï¿½ï¿½ï¿½ Y ï¿½z>t;ï¿½5ï¿½%ï¿½ï¿½Aï¿½ï¿½ï¿½ï¿½Lï¿½|%ï¿½5ï¿½}"z1ï¿½*ï¿½.ï¿½ï¿½ï¿½ï¿½cF'áxï¿½ï¿½ot*~\$ï¿½\$^MbLï¿½HJ-Lï¿½Kï¿½M1M9-ï¿½Nï¿½Kï¿½ï¿½+ï¿½7kï¿½ï¿½4ï¿½ï¿½ï¿½ï¿½Pï¿½oz_ï¿½PFNï¿½Lï¿½Yf}MVhï¿½ï¿½lï¿½ï¿½ï¿½ï¿½oï¿½\ï¿½uï¿½ï¿½ï¿½dï¿½Lï¿½7ï¿½!-.7&ï¿½ï¿½ï¿½ï¿½ï¿½R ï¿½)ï¿½ï¿½mï¿½?t?_>ï¿½\$ï¿½ï¿½ï¿½Gï¿½"ï¿½.,>ï¿½zï¿½ï¿½1ì±cï¿½ï¿½ï¿½ï¿½2ï¿½Nï¿½ï¿½ï¿½*ï¿½vï¿½ï¿½1/2ï¿½'Ó¥4ï¿½ï¿½Seï¿½eï¿½ï¿½ï¿½ï¿½ï¿½7*ï¿½+ï¿½VVï¿½ï¿½ï¿½>(tm)wrï¿½Ê¿jï¿½"ï¿½(c)ï¿½ï¿½yNï¿½yï¿½zfï¿½ï¿½Ù¶jï¿½ï¿½'lMBÍZ-ï¿½"ï¿½ï¿½ï¿½5ï¿½qï¿½ï¿½ï¿½mï¿½ï¿½8?Uï¿½Pï¿½ï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½xï¿½|!ï¿½ï¿½|"gï¿½Hï¿½qsï¿½E(tm)ï¿½g[X[ï¿½/ï¿½Kï¿½->^ï¿½ï¿½<~ï¿½ï¿½Jï¿½Uï¿½"ï¿½ï¿½\;ï¿½ï¿½ï¿½zï¿½ jKn[lnï¿½ï¿½pï¿½xï¿½iï¿½ï¿½×¥ï¿½ï¿½z]ï¿½ï¿½ï¿½7*oï¿½ï¿½<ï¿½MÓï¿½1/2}+ï¿½ï¿½JOTï¿½BoPï¿½tï¿½wï¿½ï¿½mï¿½ï¿½Oï¿½ï¿½ ï¿½ï¿½Þ¹}ï¿½ï¿½ï¿½Aï¿½ï¿½ï¿½4ï¿½uï¿½Wï¿½ï¿½@ï¿½AÛï¿½Pï¿½Cï¿½ï¿½ï¿½êï¿½mï¿½ï¿½-uï¿½hï¿½t=ï¿½~ï¿½=ï¿½7ï¿½;f<vï¿½ï¿½" Omï¿½>-w-ï¿½xï¿½ljï¿½bï¿½yï¿½ï¿½ /ï¿½'3_b^-zï¿½ï¿½ï¿½5ï¿½ï¿½ï¿½7oZï¿½Tï¿½n3/45~;ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½7fr>ï¿½>"ï¿½ï¿½ï¿½6ï¿½)ï¿½Ý7ï¿½ï¿½ï¿½ï¿½ï¿½Ì§ï¿½Oï¿½ ï¿½ï¿½?ï¿½ï¿½"ï¿½ï¿½ï¿½_ï¿½ -ï¿½-ï¿½,ï¿½,m/ï¿½ï¿½ï¿½7ï¿½o}+v+ï¿½ï¿½ï¿½__=ï¿½ï¿½Gï¿½ï¿½ï¿½ï¿½Ý(r)?gï¿½7peï¿½ï¿½][V[/ï¿½÷ï¿½ï¿½(1"_ï¿½ iï¿½@-ï¿½#5";R;ï¿½@Cï¿½"ï¿½ï¿½Å"+Â` !ï¿½-6'ï¿½ï¿½aï¿½ ï¿½Aï¿½0

2. ## Koch Snowflake

The general formula for a geometric progression is as follows - Un = U0*Rn Where R = ratio between two successive terms, U0 = 1st term, and Un = nth term of the sequence. Substituting Un with Nn can give us the general formula for the number of sides of

1. ## The Koch Snowflake

Nn = 3 x 4n and Ln= 1 x (1/3)n for the reasons previously explained, therefore the formula for the perimeter is 3 x (4/3)n . In graph 4, which shows the area of each shape, we can see that as we progress in stage the area increases.

2. ## MAths HL portfolio Type 1 - Koch snowflake

Pn Scale X-axis - 1 step = 0.5 units Y-axis - 1 step = 1 unit Relationship between n and Perimeter of the shape: Pn Perimeter of the shape n (stage)

1. ## El copo de nieve de Koch

con los valores 0,1,2 y 3 respectivamente al n�mero de fase deseada. Al comparar estos resultados con los de la Tabla 1 (Valores para el n�mero de lados, longitud de un lado, per�metro y �rea para n=0, 1, 2 y 3"), vemos que son los mismos valores.

2. ## Investigating the Koch Snowflake

and then add the area of all of the new smaller triangles added. And the general expression can be derived from the values of An from the zero stage till the third stage A0 = (V3)/4, A1= (V3)/3, A1= [(V3)/4] + (1/3)

1. ## Math Portfolio - The Koch snowflake investigation.

Area scale factor Extra triangles 0 3 - - 1 12 3 2 48 12 3 192 48 A0: ? 12 = : A0 (1 + new triangles ? ) : (1 + 3 ? ) A1: (1 +) Stage 2 has 12 new triangles and also the scale factor becomes square of the previous scale factor.

2. ## The Koch Snowflake

Since when finding the perimeter the length of all sides are added together, I was able to realize that if I multiplied the answers I would get my answer for perimeter. In order to prove my idea I estimated my answer for n=2 which was,16-3.. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 