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Infinite Summation Portfolio. I will consider the general sequence with constant values

Extracts from this document...

Introduction

Vladislav Tajc

IB Mathematics SL Type 1: Infinite Summation Portfolio

The English School

IB Mathematics SL

Math Portfolio (Type1)

Infinite Summation

Vladislav Tajc Leal

Candidate Code:000185

19th of October 2011

Bogotá, Colombia

Aim: The aim of this portfolio is to research about the summation of infinite series. In first place I will consider the general sequence with constant values for image00.pngimage00.png and variables for image05.pngimage05.png. The image11.pngimage11.png, which is calculate the sum image11.pngimage11.pngof the first image02.pngimage02.pngterms of the above sequence for image01.pngimage01.png will be calculated with the programme of Microsoft Excel and illustrated in graphs with Graphmatica; another computer based program, so that we can create a general statement which can be proven. Furthermore, to expand the investigation I will explore the same general sequence, but with variables in both image05.pngimage05.png and image00.pngimage00.png. I will accomplish this by doing the same method as the first exercises. Finally, I will conclude by showing the scopes/limitations of the general statement.

image18.png

Where,

image19.png , image24.pngimage24.png,  image29.pngimage29.png,  image32.pngimage32.png ,

We must take into account that factorial notation is in succession, this means that the factorial notation shows all the natural numbers from 1 to image02.pngimage02.png, in such way that:

image43.png

Following the first exercise, we will change the variables of terms image45.pngimage45.pngand image05.pngimage05.png, in such way that image58.pngimage58.pngand image46.pngimage46.png. If we replace these values into the original sequence we obtain:

image07.pngimage07.png, image76.pngimage76.png,  image82.pngimage82.png,  image88.pngimage88.png, …

In order to calculate the sum image11.pngimage11.pngof the first image02.pngimage02.pngterms of the above sequence for image01.pngimage01.png we must know that the sum of the first 10 values is in progression. Hence, I will use Microsoft Excel in order to plot results in a suitable table. The first column will contain the different values of image02.pngimage02.png

...read more.

Middle

image06.png, likewise the way image02.pngimage02.png approaches to image03.png.

Next, we will consider the same general sequence where image04.pngimage04.png in such way that:

image20.pngimage20.png ,   image21.pngimage21.png,   image22.pngimage22.png,   image23.pngimage23.png ,

In this sequence we will replace different values for variable image05.pngimage05.png. In my case I will use 3 different values that can replace image05.pngimage05.png: 4, 5 and 6. For each of them I will calculate the sum image11.pngimage11.pngof the first image02.pngimage02.pngterms of the above sequence for image01.pngimage01.png. To do this I will use again the help of Microsoft Excel as done previously in other sequences.

image12.png

Replacing in the form image25.pngimage25.png

image14.png

0

1

1

1

1,38629436

2,38629436

2

0,96090603

3,34720039

3

0,44403287

3,79123326

4

0,15389007

3,94512332

5

0,04266739

3,98779071

6

0,00985826

3,99764897

7

0,00195235

3,99960132

8

0,00033832

3,99993964

9

5,2112E-05

3,99999175

10

7,2242E-06

3,99999897

Table 5: The sum of image16.pngimage16.png for image26.pngimage26.pnggiven that image27.pngimage27.png


image12.png

Replacing in the form image25.pngimage25.png

image14.png

0

1

1

1

1,60943791

2,60943791

2

1,2951452

3,90458311

3

0,69481859

4,5994017

4

0,27956685

4,87896855

5

0,0899891

4,96895765

6

0,02413864

4,99309629

7

0,00554995

4,99864624

8

0,00111654

4,99976278

9

0,00019967

4,99996244

10

3,2135E-05

4,99999458

Table 6: The sum of image16.pngimage16.png for image26.pngimage26.pnggiven that image28.pngimage28.png

image12.png

Replacing in the form image25.pngimage25.png

image14.png

0

1

1

1

1,79175947

2,79175947

2

1,605201

4,39696047

3

0,95871136

5,35567183

4

0,42944504

5,78511687

5

0,15389244

5,93900931

6

0,04595637

5,98496569

7

0,01176325

5,99672894

8

0,00263461

5,99936356

9

0,00052451

5,99988807

10

9,398E-05

5,99998205

Table 7: The sum of image16.pngimage16.png for image26.pnggiven that image30.pngimage30.png

Analysing the results of each of the image11.pngimage11.png values we can notice similar results to previous exercises. To make a deeper analysis of data I will plot the three tables on a single graph, showing only the results of the third column. Each of the final results will be represented by colour: image27.pngimage27.png , image31.pngimage31.png

Graph 3: Shows the relation between image16.pngimage16.pngand image12.pngimage12.png, given image33.pngimage33.pngand image34.pngimage34.png 
image35.png

Looking in analysis, we can see that each of the replaced values of image36.pngimage36.pngincrease exponentially to a certain point where they are stabilized. The point in which they stabilize is the replaced value of image05.pngimage05.png and again, the reason for this is that all of the curves are convergent. So to generalize, we can confirm that the value of image11.pngimage11.png tends to the replaced value of image05.pngimage05.png in such way that the value of image02.pngimage02.png tends to image03.pngimage03.png.

...read more.

Conclusion

image36.pngimage36.pngand image00.pngimage00.png in the sequence, including the result up to 10 and even the infinite, it will always give us the same value that was replaced in the progression, bur this time it will be image75.pngimage75.png. Even though, as the limits fluctuate it is said that it is approximately image75.pngimage75.png, in other words it is precise but not accurate.

To test the validity of my general statement I will take other values of image05.pngimage05.png and image00.pngimage00.png.

I will consider the sequence with the expression of image77.pngimage77.png. So the sequence will look like this. If it is correct the result will be 2401

image19.pngimage19.png , image78.pngimage78.png,  image79.pngimage79.png,  image80.pngimage80.png ,

Hence the table with the summation will look like this:

Table 18: The sum of image16.pngimage16.png for image81.pngimage81.pnggiven that image83.pngimage83.png and image84.pngimage84.png

Hence the table should look like this:

24

0.003941676

2400.998259

25

0.001227224

2400.999487

26

0.000367395

2400.999854

27

0.000105914

2400.99996

28

2.94426E-05

2400.999989

29

7.90245E-06

2400.999997


And therefore the graph should correspond to this:

image85.png

Graph 5: Shows the relation between image86.pngimage86.png vs image00.pngimage00.png

The greatest limitation that it has is the fact that you limit them of the sum not always are the same so the result is not always exact.

The general this general statement was obtained by means of the analysis of the tables and the graphic ones that were used throughout the exercise. When replacing by different values for image05.pngimage05.png and image00.pngimage00.png, the general proposal was fulfilled so that the image87.pngimage87.pngterms gave the result of image75.png. In the first place for the image00.pngimage00.png values when image46.pngimage46.png, and was fulfilled later for the values of image00.pngimage00.png when values of image06.pngimage06.png, which demonstrated that the results could not be the same. After I did a deeper analysis of graphs and tables I was able to confirm the general statement by the importance of the relation that there is between the value of image00.pngimage00.png that is replaced and the respective value for image89.pngimage89.png.

...read more.

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

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