infinite summation

8/2/2019 Infinite Summation

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Portfolio Assignment Type 1

Maths Investigation

Infinite Summation


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In this investigation, the sum of infinite sequences tn

Hence the general sequence can be represented as

To understand this series, we will use numerical methods to understand the behaviour for different

values of and . To assist us in our calculations, we will use excel to calculate sequences, and plottheir behaviour. Other technologies used in this investigation are the GDC (TI-84+).

Variation of series as a function ofTo understand this series, as a function of, we will fix to equal 1. The sum of the first terms willbe calculated for 0 10 for different values of.Setting to equal 2For = 2 and = 1, our series becomes:

This calculation is performed in excel to ensure accuracy of results.


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This is done by first, creating our values in column A. This is done by putting a value of 0 into theA2 cell, and then in the A3 cell inserting A2+1, so the computer knows that the A3 cell is +1 of the

previous A2 cell. Therefore in the A3 cell we get 1, and if we copy and paste this formula (A2+1) into

the cells up to A12, the computer lays out the values of up to 10.

Next, we use a similar method to obtain the values of t n. In the B16 and B17 columns, we put thevalues of and to use as a reference for the equation used later so in cell B16 we have 2, and incell B17 we have 1. Because the value of tn is found through the formula

, we need to recreate

this in a format that excel will recognize and apply to our values. We therefore put

(($B$17*LN($B$16))^A2)/(FACT(A2)) in cell B2, where $B$17 represents the value, and $B$18represents the value, and (FACT(A2)) represents the factorial of the value. The dollar signs arecalled Absolute cell references and are inserted in order to keep the values a constant. We then

copy the formula (($B$17*LN($B$16))^A2)/(FACT(A2)) into cells up to B12, and the computer lays

out values of tn up to =10.Finally, in row C we find the values for , by telling the computer to sum up the preceding values oftn. We do this by entering the equation =SUM($B$2:B2) in cell C2, which tells the computer to sum

up the values from B2, up to B2, and keeping the first value of B2 constant by adding the dollar signs.

We again paste the formula (=SUM($B$2:B2) ) into cells up to C12 and get the remaining values, and

therefore in cell C3, for example, we would get the formula =SUM($B$2:B3), where it tells the

computer to sum up the values from B2 up to B3.

By using this method, we can obtain the following values:

0 1 1

1 0.693147 1.693147

2 0.240226 1.933373

3 0.055504 1.988877

4 0.009618 1.998495

5 0.001333 1.999829

6 0.000154 1.999983

7 1.52527E-05 1.999998

8 1.32155E-06 1.999999

9 1.01781E-07 1.999999

10 7.05491E-09 2


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In order to confirm these results, we use the GDC to calculate values.

When we compare the results obtained by the GDC and by excel, we can verify that the two are the

same, and therefore are correct.


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These results were graphed using excel the relation between and were plotted for = 2 and = 1.

From this graph, we can observe that the value of never exceeds 2, and therefore we canconclude that the sum of the first 11 terms, (i.e. up to n = 10) is 2 (correct to six decimal places). We

observe that the sequence converges, as ,

. This is a result of as , so . Additionally as < 1, as . We can therefore say, that for an infinitesummation where the series is summated to infinity:

Setting to equal 3

Next we consider the same series, where remains 1, but this time becomes 3.For = 3 and = 1, our series becomes :

This calculation is again performed in excel to ensure accuracy of results, using the same method.

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10 12

Sn

n

Sn vs n for a=2, x=1


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We obtain:

0 1 1

1 1.098612 2.098612

2 0.603474 2.702086

3 0.220994 2.923081

4 0.060696 2.983778

5 0.013336 2.997114

6 0.002441 2.999556

7 0.000383 2.999940

8 5.26302E-05 2.999992

9 6.42447E-06 2.999999

10 7.058E-07 2.999999

As previously done, we will calculate these results on the GDC as well to ensure the results are

correct.

As the values match from both results, we can conclude that these values are correct.


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Again, we plot the values of against the values of, this time for = 3 and = 1.

Again the series is observed to converge, this time towards a value of 3. We can therefore say that

As with the previous case, of as , so . However as > 1, hence

as . The series converges to a fixed value of 3, as faster than .Hence the long term behaviour of

as .

Next, we will consider a general sequence where = 1, in such a way that

Using this general sequence, I will calculate the sums of the first terms of different values of.

Setting to equal 1000

To check whether

as for large values of , we set = 1000. Hence our seriesbecomes:

The calculations were performed on excel.

0

0.5

1

1.5

2

2.5

3

3.5

0 2 4 6 8 10 12

Sn

n



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0 1 1

1 6.907755 7.907755

2 23.858541 31.766296

3 54.936321 86.702618

4 94.871667 181.574285

5 131.070051 312.644337

6 150.899973 463.544311

7 148.911441 612.455752

8 128.580474 741.036227

9 98.689161 839.725388

10 68.172057 907.897445

11 42.810535 950.707981

12 24.643725 975.351706

13 13.094832 988.44653914 6.461135 994.907674

15 2.975462 997.883137

16 1.284610 999.167748

17 0.521986 999.689734

18 0.200319 999.890054

19 0.072829 999.962884

20 0.025154 999.988038

21 0.008274 999.996313

22 0.002598 999.998911

23 0.000780 999.99969124 0.000224585 999.999915

25 6.20552E-05 999.999977

26 1.6487E-05 999.999994

27 4.21808E-06 999.999998

28 1.04062E-06 999.999999

29 2.47875E-07 999.999999

30 5.70753E-08 1000

The series is not observed to have fully converged for

, so the summation is performed to

.


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We then plot the values of against the values of, this time for = 1000 and = 1.

The series is observed to converge for this large value of, towards a value of 1000. Theconvergence however does take a longer number of terms to do so. We can therefore say that

This calculation indicates that for large values of, where , faster than reaches infinity for as ,. Hence the long term behaviour of as .

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Sn

n



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Setting to equal 0.1

To check whether

as for smaller values of values of that are less than 1we set = 0.1. Hence our series becomes:

The values were calculated and graphed, again, using excel.

0 1 1

1 -2.302585 -1.302585

2 2.650949 1.348363

3 -2.034678 -0.686314

4 1.171255 0.484940

5 -0.539382 -0.054442

6 0.206995 0.152553

7 -0.068089 0.084464

8 0.019597 0.104061

9 -0.005013 0.099047

10 0.001154 0.100202

From the table and the graph, we can observe that the graph fluctuates, and for odd numbers of,we get negative values, but for even numbers of we obtain positive numbers for the first 5 values.Then, the values stabilize and remain positive numbers, and the graph eventually converges and

reaches a value of 0.1. Although the graph is complicated in behaviour, it still follows the general

pattern that as , .Setting to equal a negative number

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8 10 12

Sn

n

Sn vs n for a=0.1, x=1


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According to the natural logarithms rules and properties, is undefined when , therefore itis not possible to calculate for values of that are smaller than or equal to 0.Summary of variation ofwith

We conclude that for the variation of with , for a fixed value of = 1, the series converges to afixed value. The infinite series can hence be described by the general statement:

This behaviour occurs, as in the long term the sequence converges, due to of

as .As this is a numerical analysis, it is possible that the infinite summation does not hold true for very

small values of, (i.e. 1). Naturally as or of anegative number is undefined.

For very small values of a (i.e. 1000), it is unknown whether as , as thisconvergence relies on faster than .Variation of series as a function of and xWe return to our original sequence, to determine the sum of the infinite sequence , where

We will consider the expression (, ), which represents the sum of the first terms, for variousvalues of and .Setting to equal 2Firstly, we will calculate the sum of the first 9 () terms with = 2 ( remains a constant) withdifferent, positive values of, hence (2, ). The different values of that will be used are 0.1, 1, 2,3, and 4.

Using excel, we calculate the values of (2, ) for different values of.


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For = 0.1

0 1 1

1 0.069314 1.069314

2 0.002402 1.071716

3 5.55041E-05 1.071772

4 9.61813E-07 1.071773

5 1.33336E-08 1.071773

6 1.54035E-10 1.071773

7 1.52527E-12 1.071773

8 1.32155E-14 1.071773

For

= 1

0 1 1

1 0.693147 1.693147

2 0.240226 1.933373

3 0.055504 1.988877

4 0.009618 1.998495

5 0.001333 1.999829

6 0.000154 1.999983

7 1.52527E-05 1.999998

8 1.32155E-06 1.999999

For = 2

0 1 1

1 1.386294 2.386294

2 0.960906 3.347200

3 0.444032 3.7912334 0.153890 3.945123

5 0.042667 3.987790

6 0.009858 3.997648

7 0.001952 3.999601

8 0.000338 3.999939


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For = 3

0 1 1

1 2.079442 3.079441

2 2.162039 5.241480

3 1.498611 6.740091

4 0.779068 7.519159

5 0.324005 7.843164

6 0.112292 7.955456

7 0.033358 7.988814

8 0.008671 7.997485

For

= 4

0 1 1

1 2.772589 3.772588

2 3.843624 7.616212

3 3.552263 11.168475

4 2.462241 13.630716

5 1.365356 14.996073

6 0.630929 15.627001

7 0.249901 15.876902

8 0.086609 15.963511

Therefore we can gather the data:

0.1 1.071773

1 1.999999

2 3.999939

3 7.9974854 15.963511

Using excel, we plot the values of against the values of.


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From this plot, we can observe that as increases, as does , therefore the two variables arepositively correlated. We can also notice that the graph behaves exponentially, and resembles an

exponential function, . Interestingly enough, if we compare the values of to thevalues of , we get:

1.071773 1.071773

1.999999 2

3.999939 4

7.997485 8

15.963511 16

From this comparison, we can observe that the values are almost identical (if the sum did not have a

limit ofthen they would be identical), and therefore conclude that and are thesame.

Setting to equal 3Again, we will calculate the sum of the first 9 () terms, but this time with = 3 ( remains aconstant) with different, positive values of, hence (3, ). The different values of that will beused are 0.1, 1, 2, 3, and 4.

Using excel, we calculate the values of (3, ) for different values of.

0

2

4

6

8

10

12

14

16

18

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

T9

(2,

x)

x

T9(2,x) vs x


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For = 0.1

0 1 1

1 0.109861 1.109861

2 0.006034 1.115895

3 0.000220 1.116116

4 6.06969E-06 1.116123

5 1.33365E-07 1.116123

6 2.44194E-09 1.116123

7 3.83249E-11 1.116123

8 5.26302E-13 1.116123

For

= 1

0 1 1

1 1.098612 2.098612

2 0.603474 2.702086

3 0.220994 2.923081

4 0.060696 2.983778

5 0.013336 2.997114

6 0.002441 2.999556

7 0.000383 2.999940

8 5.26302E-05 2.999992

For = 2

0 1 1

1 2.197224 3.197224

2 2.413897 5.611122

3 1.767958 7.3790814 0.971150 8.350231

5 0.426767 8.776998

6 0.156283 8.933282

7 0.049055 8.982338

8 0.013473 8.995811


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For = 3

0 1 1

1 3.295836 4.295836

2 5.431270 9.727107

3 5.966860 15.693967

4 4.916449 20.610417

5 3.240763 23.851180

6 1.780171 25.631351

7 0.838164 26.469516

8 0.345306 26.814822

For

= 4

0 1 1

1 4.394449 5.394449155

2 9.655591 15.05004084

3 14.143668 29.19370975

4 15.538408 44.73211822

5 13.656549 58.38866741

6 10.002168 68.39083592

7 6.279145 74.66998177

8 3.449173 78.11915517

Therefore we can gather the data:

x T9(2,x)

0.1 1.116123

1 2.999992

2 8.995811

3 26.8148224 78.119155

Again, we plot the values of against the values of.


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Once more, the values of and have a positive correlation, and the graph shows anexponential shape. When we compare the values of to the values of , we get:

1.116123 1.116123

2.999992 3

8.995811 9

26.814822 27

78.119155 81

We observe that the values are virtually identical, and therefore we conclude that and are the same.Conclusion

From the calculations of and , we observed that the values we obtained for bothsums were equal to that of a exponential function. Therefore we can conclude that (, ) is equalto .

This can be proven using the Maclaurin series expansion of exponential functions. This states that:

If we look at the original equation for the general sequence,

we notice that the two equations are extremely similar. If we look into it further, we see that theonly difference between the two equations is that in the first equation, is increased to the power

0

10

20

30

40

50

60

70

80

90

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

T9(3,

x)

x

T9(3,x) vs x


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of, while in the second equation, is increased to the power of. Hence we can conclude that is equal to . Therefore in the first equation, we can substitute the value of with , andget . Now, according to the logarithm power rule, the logarithm of raised to the power of is times the logarithm of [ ], so we can change to . As and cancel each other out, we are left with

. Therefore

so as apporaches .We can test the validity of this statement by trying different values of and . For instance, if = 4,and = 2, . We can test this result by using excel once again.

0 1 1

1 2.772588 3.772588

2 3.843624 7.616212

3 3.552262 11.168475

4 2.462241 13.630716

5 1.365356 14.996073

6 0.630928 15.627001

7 0.249900 15.8769028 0.086609 15.963511

9 0.026681 15.990192

10 0.007397 15.997590

As the value of is equal to , we can confirm this statement is correct.This statement only applies for positive integers of, as is undefined when . The value ofmust also be a positive integer, as you cannot have a factorial of a negative number. The value ofcannot equal 0 since anything to the power of 0 is equal to 1. Lastly,

cannot be equal to 1, as 1 to

the power of anything will always equal 1.

infinite summation

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