Triangular and Stellar Numbers
Math I.B Internal Assessment: SL Type 1 Stellar Numbers 6/26/2011 St. Dominics International School Raj Devraj TRIANGULAR NUMBERS TRIANGULAR NUMBERS WITH THREE MORE TERMS GENERAL STATEMENT: NTH TRIANGULAR NUMBERS IN TERMS OF N. The differences between the sequences of terms: X Y Y= number of dots on triangle X= number of dots on 1 side of triangle 0 0 2 3 3 5 According to Finite Differences if the 3rd difference of a pattern is 1, then the general term is a quadratic equation: Thus to find the general term we must first find the value of 'c': In order to find the values of 'a' and 'b' we must solve a quadratic using simultaneous equations, thus: Substitute the value of 'a' of one equation: To find the General statement that represents the nth triangular number in terms of 'n', we substitute the value of 'y' by Un and the value of 'x' by n, thus: STELLAR NUMBERS NUMBER OF DOTS TO S6 STAGE S1 S2 S3 S4 S5 S6 3 37 73 21 81 Thus, using finite difference: The most obvious pattern is that the 1st row all numbers and odd and the second row all are even. Also all these numbers are some multiples of 12 + 1, for example: 12 also turns out to be the half of 6. 6 STELLAR NUMBER AT STAGE S7 + 1 (12) + 2 (12) + 3 (12) + 4(12) + 5(12) + 6(12) = 253 GENERAL STATEMENT FOR 6 STELLAR NUMBER AT STAGE SN IN TERMS OF N If you notice the multiples are
Music and Maths Investigation. Sine waves and harmony on the piano.
Transfer-Encoding: chunked Page EXPLORATION OF THE RELATIONSHIP BETWEEN MATHEMATICS AND MUSIC ________________ INTRODUCTION I have chosen to investigate the mathematical properties of music; to be more precise, the piano. Since an early age I had a great interest and appreciation for music, in addition, I had decided to play the piano and now I have been playing it for 11 years. I enjoy listening to various types of composers and musical artists varying from classical pieces to modern pieces. However, during my last year in my piano classes, I’ve got a spark of interest in figuring out how composers manage to compose pieces that are appealing to me and the public, especially, when some composers had health problems, which made the journey of creation quite hard for them, in this case, Beethoven created wonderful musical pieces while being deaf. After some personal research, I’ve found out that mathematics has a huge role in determining how it appeals to the public, as mathematics is related to the production of harmony. Thus, the aim of my exploration is to explore and understand how mathematics are related to harmony of music. All of the aspects of mathematics in relation to harmony will be focused on my main instrument – the classical piano. Although music is made out of all of the aspects such as melody, harmony and rhythm, however, harmony is very intricate
Crows Dropping Nuts
Min Hua Ma 9-18-09 IB SL MATHEMATICS Internal Assessment Type II IB SL Type II Internal Assessment: Crows Dropping Nuts This assignment is an investigation to find a function that models a given set of data. By using various methods, such as matrixes, different types of regressions, and technology, it allows the investigator/student to create various equations to model the data. This assessment is about birds dropping different sized nuts on a hard surface in a range of heights in order to break open the shells. There are three variables in this investigation: the size of nuts, the heights of drops, and last but not least, the number of drops. The first set of data is on crows dropping large nuts: Height of drop .7 2 2.9 4.1 5.6 6.3 7 8 0 3.9 Number of drops 42 21 0.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2 To begin this investigation, I began plotting the given points on a scatter plot: Then, I decided to begin with using matrixes to formulate an equation. I wanted to do a matrix using all the points to create this polynomial: ax9+bx8+ cx7+dx6+ex5+fx4+gx3+hx2+ix+j I put all the y values in matrix [A] and all the x values in matrix [B].Then I took the inverse of matrix [B], multiplied by matrix [A], in order to find the values of each letter in matrix [C]: [B] x [C] = [A] [C] = [B]-1 x [A] But, this method was unable to solve for what [C] was
Quadratic Polynomials. Real and Imaginary components
MATHS IA Alex Chen ________________ PART A (Quadratic Polynomials) The investigation is to find out if the zeros and to determine the real and imaginary components of the complex zeros of . From the function given, The coordinates of the vertex is by using the Quadratic equation: where Hence, has zeros , and By subbing in different numbers of into the equation: For: , it is given that , which is equal to : : Value of a value of b value of y1 1 1 2 2 3 3 4 4 Value of a value of b value of y1 1 1 2 2 3 3 4 4 For: : : Value of a value of b value of y1 1 1 2 2 3 3 4 4 Value of a value of b value of y1 1 1 2 2 2 3 3 4 4 After subbing different values for and From the above results, by comparing with , it can be seen that their values are opposite, have negative results, ’s results are always a positive number or bigger than 0. A graph of y1 and y2 is shown below when a= 3 and b=5, We know that has zeros , while has opposite concavity to ,which is in the form . From the graph, it can be seen that, is a reflection of , Therefore, the equation of the quadratic is : . When
Math SL Fish Production IA
003717-011 Fish Production I-Shou International School Mathematics Standard Level Internal Assessment Type 2 Candidate Name: Hung Li Chu Candidate Number: 003717-011 Word Count: 2886 Date of Submission: 26th October 2012 The aim of this internal assessment is to consider commercial fishing in a particular country in two different environments- the sea and fish farms (aquaculture). The table of values below is taken from the UN Statistics Division Common Database. Table 1: This shows the total mass of fish caught in the sea between 1980 and 2006 in thousands of tonnes. (1 tonne = 1000 kilograms) Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 Total Mass (tonnes) 426.8 470.2 503.4 557.3 564.7 575.4 579.8 624.7 669.9 Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 Total Mass (tonnes) 450.5 370.0 356.9 447.5 548.8 589.8 634.0 527.8 459.1 Year 1998 1999 2000 2001 2002 2003 2004 2005 2006 Total Mass (tonnes) 487.2 573.8 503.3 527.7 566.7 507.8 550.5 426.5 533.0 Graph 1: This shows the total mass of fish caught between the years 1980 and 2006. Based on the graph, the x-axis represent the years between 1980-2006. The y-axis represents the total mass of fish caught in tonnes per year. The graph illustrates that in some years the total mass of fish caught increases
Maths SL, Type 1 Portfolio - triangular numbers
Maths Practice Portfolio Maths Portfolio Type I ________________ Special numbers go back in history and there is a great relation between the theorists and the maths they discovered. They are numbers with unique properties, making them different to other ordinary numbers. ‘The origins of the concept of the shape of number’ is a topic which can be directly related to this fact. The idea behind this is that there are many origins of the concept of the shape of number. In the following task we were investigating patterns in geometric shapes that will lead to the formation of special numbers. More specifically, we will look at triangular patterns that will enable us to discover a pattern of special numbers. The first part of the investigation we looked at a triangular pattern formed with dots in the shape of triangles to and calculates the nth term for this pattern. Original Sequence Counting the number of dots in each of the triangles, we can see that there is a pattern. The numbers of dots increase by (n+1) adding 2, 3, 4 and 5. Therefore, this hints that the next three terms will be as we will be adding 6, 7, 8. Next Three Terms From the above triangular pattern, we can deduce a general statement which can represent the nth triangular number in terms of n. Butin order to do this, a table of n, Tn,1st difference and 2nd difference should be drawn: n 1 2 3 4 5
IB Math Methods SL: Internal Assessment on Gold Medal Heights
International Baccalaureate IB Mathematics SL Y2 Internal Assessment Task: Gold Medal Heights (Type 2) Introduction The aim of this investigation is to consider the winning height for the men’s high jump in the Olympic Games. Firstly, we are given a table that lists the record height achieved by gold medalists in each competition from 1932 onto 1980. Given Information Year 1932 1936 1948 1952 1956 1960 1964 1968 1972 1976 1980 Height (cm) 197 203 198 204 212 216 218 224 223 225 236 N.B. The Olympic games were not held in 1940 and 1944. Let us first plot the points on a graphing application. After plotting the above points in the graphing programme (see Figure 1 in Appendix), we arrive at the graph below. Graph 1: Height achieved by gold medalists in various Olympic Games Domain: 1920 t 1990 Range: 195 t 240 This graph demonstrates the record height changing over time through every Olympic game. The h-axis represents the height of the record jump (in centimeters). The t-axis represents the years in which the jumps took place. It should be noted that there are certain limitations to this data set; primarily that there was no men’s high jump taking place in 1940 and 1944; none in 1940 due to the revocation of Tokyo as the host venue for the Games due to the Sino-Japanese war; and none in 1944 due to the outbreak of the Second
I am going to go through some logarithm bases, by continuing some sequences and finding an equation to find the nth terms in the sequences
Rachel Brustad Logarithm Bases In the next few examples I am going to go through some logarithm bases, by continuing some sequences and finding an equation to find the nth terms in the sequences in terms offor the first part. Then for the second part, find an equation to figure out the third term in terms of from the given logarithms, after I will check my equation with several examples created by myself and check the validity of the equation, and the scope and limitations of a, b, and x. PART 1: With the four sequences given I will find the next two terms in each sequence then change the first three sequences into the form of the forth sequence which is necessary to find the equation to find the nth term. Finally for all four sequences I will check the validity by using my calculator and finding the 10th term and proving that my equation works. . Log28, log48, log88, log168, log328, log648, log1288 2. Log381, log981, log2781, lg8181, log24381, log72981 3. Log525, log2525, log12525, log52525, log312525, log1562525 4. , , , , , By changing the first three sequences to the form of the forth sequence it changes the subscripts. Each subscript will have a raised power and that is the same as the number in the sequence it is or the nth terms. By setting one of the logarithms in the sequence to y with a raised power of n subscript you can change the equation to
Math IA type I. Here is Lacsaps Fractions (the symmetrical pattern given) from n=1 until n=5, again with red numbers representing n:
Lacsap’s Fractions A Math Internal Assessment By Kelsey Kennelly Lacsap’s Fractions…Ha Clever, IB. You took the word “Pascal” and spelled it backwards to make “Lacsap”. I thought this was simply a Math Assessment but I guess it’s also a word scramble. Anyways, with that being said, here is Pascal’s Triangle starting at n=0 until n=9, with red numbers representing n: . 1 2. 1 1 3. 1 2 1 4. 1 3 3 1 5. 1 4 6 4 1 6. 1 5 10 10 5 1 7. 1 6 15 20 15 6 1 8. 1 7 21 35 35 21 7 1 9. 1 8 28 46 70 46 28 8 1 0. 1 9 36 84 126 126 84 36 9 1 ________________ *Column 0 1 2 3 4 5 6 7 8 9 *Columns rise diagonally from left to right. Column numbers will be represented by “c” followed by a subscript number. Here is “Lacsap’s” Fractions (the symmetrical pattern given) from n=1 until n=5, again with red numbers representing n: . 1 1 2. 1 1 3. 1 1 4. 1 1 5. 1
Graph Theory review notes.
Graph theory review Graph types Simple: no loops or multiple edges Complete: has every pair of vertices adjacent. One move can get you from one vertex to any other vertex. (n vertices--->Kn) Connected: has every vertex accessible from every other vertex, but not necessarily directly. Bipartite: has vertices in two sets and each edge joins one vertex from each set. Vertices within each set are not joined. Complete Bipartite: has every vertex in one set joined to every other vertex in the other set (m vertices in one set, n vertices in the other--->Km,n) Wheel: has every vertex connected to a central hub. (Wn) Two graphs are isomorphic: Equal size + Equal order + Same Degrees + Connectivity is preserved The complement of a graph will have all of the same vertices but its edges will be all of the possible edges that the original graph does NOT have. Handshaking Lemma: For any graph G, the sum of degrees of the vertices in G is twice the size of G. Pigeonhole Principle: n pigeons in m holes, and n>m, then there must be at least one hole containing more than one pigeon. Planar graph: a graph without any edges crossing each other For graph, G, to be planar: If G is a simple graph: If G is a bipartite graph: Any subgraph that contains K5 or K3,3 will not be planar A connected graph has an Eulerian circuit if and only if ALL of its vertices are even. A