Investigating a sequence of numbers

Type 1: Investigating a sequence of Numbers This is an investigation about series and sequences involving permutations. From a given series, I find the pattern of numbers that result from different values and use graphs to conjecture an expression from the series. By using mathematical induction and direct proof, I prove the general terms that I derived for the series. Part 1: The sequence of numbers is defined by , , , ... From the pattern of different values of n in above, I conclude that! Part 2: Let If n=1 = were ! ! If n=2 were !, ! + (2 If n=3 were !, !,! + (2 Part 3: From Part 2, I know that: ! To conjecture an expression of , I first organize the results that are derived in Part 2 to discover a pattern in the value of as n increases. Table 1.1: n 2 4 5 3 8 23 - - - - - - n ! ? The same results of from Table 1.1 can be represented as follows: Table 1.2 : n 2-1= (1+1)!=1 2 6-1= (2+1)! =5 3 24-1= (3+1)! =23 - - - - n From the patterns exhibited in Table 1.2, I notice that which is further illustrated in Graph 1.1. In Graph 1.1, I plotted the graph of for the first three values (represented by green dots) and I assumed that (n+1)! will lead to a conjecture forand plotted its values for n=1,2,3 (represented by red dots) . From the two graphs, I notice that (n+1)! is exactly 1 unit above for all three points

  • Word count: 846
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Math Investigation - Sine Law

Part 1 Look at the graphs of y = sin x Compare the graphs of: y = sin x y = 2sin x y = 1/3sin x y = 5sin x The difference between all these graphs is a variable known as A, or amplitude of wave. When A > 1, the graph stretches vertically. When 0 < A < 1, the graph compresses vertically. Also, A is the number that manipulates how far the graph compresses or stretches to. For example, in the graph of y = 2sin x, the graph stretches out to +2 and down to -2. The characteristics of the waveform are altered because of this. The range of the graph is increased or decreased in conclusion. The domain and range of the graphs are: y = sin x D: {x| x?R} R: {y| -1 < y < 1, y?R} y = 2sin x D: {x| x?R} R: {y| -2 < y < 2, y?R} y = 1/3sin x D: {x| x?R} R: {y| -1/3 < y < 1/3, y?R} y = 5sin x D: {x| x?R} R: {y| -5 < y < 5, y?R} Also, if A was negative in the initial equation, it would flip the graph around like a mirror as so: y = 2sin x y = -2sin x Part 2 Investigate the family of curves y = sin (x + C), where 0o ? C ? 360o. How does the value of C transform the standard curve y = sin x? The C variable in y = sin (x + C) is a variable that creates a horizontal translation. In example, if C is substituted with 90, the equation becomes y = sin (x + 90): y = sin x y = sin (x+90) It is evident that the entire graph shifted to the left by 90 unit. This can

  • Word count: 840
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Portfolio SL - Matrix

Math Portfolio SL MATRIX BINOMIALS NONAME STUDENT 3/17/2009 MATRIX BINOMIALS In this portfolio I will try to find a general statement and patterns for given matrix binomials exercises. For data processing I will use TI-83 Plus (Sliver Edition) graphing calculator. I will use my knowledge from patterns and matrix in order to find suitable formulas. X=, Y=. Find X2, X3, X4; Y2, Y3, Y4. X2== X3= X4= Y2= Y3= Y4= Find expressions for Xn , Yn, (X+Y)n * The entries double for every higher power of X, i.e.: X2= 2X= 21X X3= 4X= 22X X4= 8X = 23X follows Xn=2n-1X I will test this formula with a random number: X10= X10= 210-1= X19= 219-1= Xn=2n-1X formula valid for all natural numbers N; = {1,2,3,...} as it is shown in examples. * The entries double for every higher power of Y, i.e.: Y2= 2Y Y3= 4Y= 22Y Y4= 8Y = 23Y follows Yn=2n-1Y Again I will test is with random numbers using GDC: Y8= Y8=28-1= Y16= Y16=216-1 Yn=2n-1Y formula valid for all natural numbers N; = {1,2,3,...} as it is shown in examples. * In the case of (X+Y)= 2I (X+Y)= = =2I (X+Y)n= n = n = = 2n In this case we have two slightly different formulas, the first one includes the sum of (X+Y) and the second one includes the identify matrix, i.e. I

  • Word count: 840
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Math IA Logarithm bases

Logarithm Bases Consider the following sequences. Write down the next two terms of each sequence. Log 2 8 , Log 4 8 , Log 8 8 , Log 16 8 , Log 32 8 , Log 64 8 , Log 128 8 Log 3 81 , Log 9 81 , Log 27 81 , Log 81 81 , Log 243 81 , Log 729 81 Log 5 25 , Log 25 25, Log 125 25 , Log 625 25, Log 3125 25 , Log 15625 25 Find an expression for the nth term of each sequence. Write your expressions in the form P/Q. Log 2 8 , Log 4 8 , Log 8 8 , Log 16 8 , Log 32 8 , Log 64 8 , Log 128 8 This sequence can be expressed 3/N. N being the nth term of the sequence. 1st Log 2 8 = 3 3/1 = 3 2nd Log 4 8 = 1.5 3/2 = 1.5 3rd Log 8 8 = 1 3/3 = 1 4th Log 16 8 = .75 3/4 = .75 5th Log 32 8 = .6 3/5 = .6 6th Log 64 8 = .5 3/6 = .5 7th Log 128 8 = .43 3/7 = .43 Find an expression for the nth term of each sequence. Write your expressions in the form P/Q. Log 3 81 , Log 9 81 , Log 27 81 , Log 81 81 , Log 243 81 , Log 729 81 This sequence can be expressed 4/N. N being the nth term of the sequence. 1st Log 3 81 = 4 4/1 = 1 2nd Log 9 81 = 2 4/2 = 2 3rd Log 27 81 = 1.33 4/3 = 1.33 4th Log 81 81 = 1 4/4 = 1 5th Log 243 81 = .8 4/5 = .8 6th Log 729 81 = .66 4/6 = .66 Find an expression for the nth term of each sequence. Write your expressions in the form P/Q. Log 5 25 , Log 25 25, Log 125 25 , Log 625 25, Log 3125 25 , Log 15625 25 This

  • Word count: 835
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Logarithm Bases Math IA

Logarithmic Sequences IA Andrew Cherny Math SL Scott Learned /26/11 In math class, I was given the assignment to evaluate multiple logarithmic sequences to see if any patterns were evident within these sequences. Using logarithmic rules that I previously learned in math class, I was able to discover multiple patterns within each logarithmic expression. To begin, I was given the general logarithmic expression: ,... In order to establish patterns within this general logarithmic expression, I will use multiple examples to help establish a common pattern between all the examples. The first sequence is as followed: ,... Next, the same logarithmic sequence will be evaluated but more in depth, to try and find a common pattern. Note: a pattern is already starting to show. While the numerator stays the same throughout the logarithmic sequence, the denominator increases linearly by one. The pattern is also evident within these next two examples of logarithmic sequences. ) 2) Cleary, the pattern is noticeable as the sequence goes on. Therefore, I was able to find the term for each sequence, writing it in the form where . For the general logarithmic sequence, the term is the following. ..., That being the case, the nth term for the three examples of logarithmic sequences are the following: 1) 2) 3) To justify my answer using technology, I used excel to

  • Word count: 828
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Volumes of Cones

IB Mathematics HL Portfolio (Type III) Volumes of Cones By Diana Herwono IB Candidate No: D 0861 006 May 2003 In this assignment, I will use WinPlot, a graphing display program. . Find an expression for the volume of the cone in terms of r and ?. The formula for the volume of a cone is: V = 1/3 x height x base area To find the base area, we must find the radius of the base. The circumference of the base of the cone = the length of arc ABC Therefore: 2? ? rbase = r? rbase = r? / (2?). The area of the base can therefore be calculated: Abase = ? ? rbase 2 = ? ? ( r? / (2?))2 Next, we must find the height of the cone, h. Notice that for the cone, h rbase r is a right angled triangle, with r as the hypotenuse. Therefore, using the Pythagorean Theorem, we can find h. r2 = h2 + R2 r2 = h2 + (r?/2?)2 h2 = r2 - (r?/2?)2 h = ? [r2 - (r2?2/4?2)] Therefore, substituting the values for rbase and h, we can find the volume of the cone. V = 1/3 x height x base area V = 1/3 ? ? [r2 - (r2?2/4?2)] ? ? (r2?2/4?2) 2. By using the substitution x = ?/2? , express the volume as a function of x. x = ?/2? ? = x2? V = 1/3 ? ? [r2 - (r2?2/4?2)] ? ? (r2?2/4?2) V = 1/3 ? ? [r2 - (r2(x2?)2/4?2)] ? ? (r2(x2?)2/4?2) = 1/3 ? ? [r2 - (r24x2?2/4?2)] ? ? (r24x2?2/4?2) = 1/3 ? ? [r2 - (r2x2)] ? ? (r2x2) 3. Draw the graph of this function using the calculator. Hence find the

  • Word count: 823
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Ib math SL portfolio parabola investigation

___________ IBHL 2 Math Parabola Investigation In this investigation, relationships between the points of intersection of a parabola and two different lines were examined.[tt1] First, the parabola and the lines and were used as an example. As seen on the adjacent graph, the points of intersection were labeled left to right as x1, x2, x3, and x4. Using a graphing calculator, these values were found: x1=1.764, x2=2.382, x3=4.618, and x4=6.234. At this point, x1 was subtracted from x2, and x3 from x4, and the resulting numbers were labeled SL and SR respectively: After this, a value D was found: To further investigate this relationship, I followed this same process with many different parabolas and lines. These lines and their values are shown in the table below (Figure 2.1). A graph of the displayed functions is below this (Figure 2.2). An explanation for each significant situation is provided below Figure 2.2. Figure 2.1 Situation # 2 3 4 4 2 5 6 7 8 9 0 [tt2] Situation 3: This situation caused me to hypothesize the conjecture that, where A is the A value from the equation. In addition, g(x) was tangent to f(x) at the point (10,10). This meant that x2 and x3 were the same (10), and showed that the conjecture held true for tangents. Situation 6: A concave down parabola in the 1st quadrant still holds the conjecture, provided that the

  • Word count: 815
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Crows Dropping Nuts Math Portfolio

In this project, I will attempt to model the function of a group of crows dropping various size nuts from varying heights. The model will help to predict the number of drops it takes to break open nuts from even more heights. The following table shows the average number of drops it takes to break open a large nut from varying heights. Large Nuts Height of drop (m) .7 2.0 2.9 4.1 5.6 6.3 7.0 8.0 0.0 3.9 Number of drops 42.0 21.0 0.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2 When graphed in Excel, the points form this graph: To model this graph a power function can be used. There is an x variable and a y variable. For my model of the graph, I used a function with the following parameters: (a(x-b) c) + d where a controls the curve of the model, b controls the shifts of the model, c is the rate of fall, and d is the horizontal asymptote. My function was f(x) = (47(x-1)-1.5) + 1. I chose this function because it seems to fit closely with the data provided. I used the parameters stated above to create the equation in my Graphic Display Calculator and adjusted based on how well it fit the data provided. This is how my model appears by itself: And with the data points included: When graphed with Excel, a function of y = 46.096x-1.154 is found to be the best model of the data provided. Here it is graphed with comparison to my model: And with the data points: As you

  • Word count: 810
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

mathsbreaking

Mathematics Course Work Dominique Albert-Weiss "Stopping disntances" Table 1 Speed v (km/h) Thinking distance (m) Braking distance (m) 32 6 6 48 9 4 64 2 24 80 5 38 96 8 55 12 21 75 .) Use a GDC or graphing software to create two data plots: speed versus thinking distance and speed versus braking distance. Describe your results. Graph 1 The given values in Table 1 of speed (km/h) and thinking distance (m) were plotted against each other in this graph. With close observation, it is noticeable that the speed is proportional to the thinking distance. In other words: the points construct a straight line that is going through the origin. In this example, we are able to assume that the more the driver increases his/her speed, the more time it takes him/her to apply the brakes. Graph 2 In graph 2, the values of Table 1 of speed (km/h) and braking distance (m) were plotted against each other in this graph. Unfortunately, no straight line can be drawn as the thinking distance is increasing more than proportional. From the 1st point to the 2nd point it is increasing by 8, from the 2nd to the 3rd by 10, from the 3rd to the 4th by 14....and so on. The conclusion of this graph is that the faster the car is moving, the longer it takes the car to brake. 2.) Using your knowledge of functions, develop functions that model the behaviours noted in step 1.

  • Word count: 801
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay

Binomios Matriciales-Portafolio

Binomios Matriciales En este trabajo se va a tratar con matrices. Una matriz es un ordenamiento o base de datos de elementos específicos. Este orden tiene forma rectangular, es decir que se distribuyen en filas y columnas. El Objetivo del trabajo es llograr crear una proposición general de un binomio matricial, a partir de distintas fórmulas con incógnitas. Para poder hacer esto, se resolverán diversas operaciones con el fin de llegar a estas mismas fórmulas, luego se comprobarán las fórmulas encontradas mediante ejemplos. Finalmente luego de hallar la proposición general, esta será comprobada del mismo modo, hallando además los límites que esta presenta, es decir hasta que punto la fórmula hallada puede ser empleada. A partir de las matrices X= e Y= encontré, con el uso de la calculadora los valores de: X2, X3, X4; Y2, Y3, Y4. Inmediatamente pude encontrar una relación entre las respuestas de ambas matrices, con lo que hallé una fórmula para las expresiones Xn e Yn. A continuación muestro el proceso que llevé a cabo para hacerlo: Una vez hecho esto, hallé los valores de (X+Y)1, (X+Y)2, (X+Y)3 y (X+Y)4, para encontrar la fórmula de la expresión (X+Y)n y así luego poder relacionar las respuestas de la misma forma como lo hice en el proceso anterior. Luego, siendo las matrices A=aX y B=bY, tomé distintos valores constantes para a y b,

  • Word count: 801
  • Level: International Baccalaureate
  • Subject: Maths
Access this essay