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Infinite Summation- The Aim of this task is to investigate the sum of infinite sequences tn.

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Introduction

Infinite Summation

Math Portfolio

Introduction:

Aim of this task is to investigate the sum of infinite sequences tn. An infinite sequence is a listing of numbers with no limits, like:

(1;2;3;4;5;…)

Whereas the three dots (…) at the end of the sequence states the infinity. The sum of an infinite sequence can be determined, depending on the sequence. Sequences are categorised in geometric sequences and arithmetic sequences, where in both notation and calculations differ from each other. An arithmetic sequence is a series of numbers, where the common difference (d) between terms is constant; the common difference is added or subtracted to the term before. A geometric sequence on the other hand, has a common ratio (r), where r will be multiplied to the term before, which also is constant. For both types of sequences, it is possible to calculate specific terms and the sum to certain point, whereas the number of terms is given by (n).

In this task the infinite sequence of tn, will be examined, where

t0=1, t1=(xlna)1,t2=(xlna)22×1,t3=(xlna)33×2×1 …,tn=(xlna)nn!…

As the general sequence states n!, which is factorial notation and is used to simplify. Factorial notation (n!), is defined by

n!=nn-1n-2…3×2×1          Note that 0!=1

To observe, how the sum of the sequence changes, it will be assumed that x=1 and a=2,

1, (ln2)1,(ln2)22×1,(ln2)33×2×1…

The sum of the first 11 terms, 0≤n≤10, starting with S0 and ending with S10, will be calculated with the help of the GDC. The t th term will be used to indicate the used term and Sn states the sum of the term given by n.

Middle

n≤10.

For the fourth example, it will follow the same rules as in the two previous examples, but now x=1 and a=e. The Euler value will be considered as the value 2.718281828…

Again, the sum of the first 11 terms, 0≤n≤10, starting with S0 and ending with S10, will be calculated with the help of the GDC. The same method and notation will be used, as in the sequences before.

S0= t0=1

S1= 1+t1=2

S2=S1+t2=2.5

S3= S2+t3=2.666667

S4= S3+t4=2.708333

S5= S4+t5=2.716667

S6= S5+t6=2.718056

S7= S6+t7=2.718254

S8= S7+t8=2.718279

S9= S8+t9=2.718281 527

S10= S9+t10=2.718281 803

 n Sn 0 1 1 2 2 2.5 3 2.666667 4 2.708333 5 2.716667 6 2.718056 7 2.718254 8 2.718279 9 2.718281 10 2.718281

This diagram shows the relation between Sn and n using the gained results. (Diagram Microsoft Excel)

It is possible to see relating on the diagram and the table, that when n increases, the value of Sn increases as well, so for all four examples this statement is true which makes it reliable, for all positive numbers. The complete data collection for the equation, suggests that when n is ∞, when x=1 and a =e, the sum will equal to e. This may suggest that the domain for the function is 1≤Sne. This may suggest that when n approaches ∞, n will approach a, all four data sets from befor support this statement.

From the collected data a general statement can be deduced, which simplifies calculations to the point of ∞. For this the general sequence given in the beginning should be taken into account,

t0=1, t1=(xlna)1,t2=(xlna)22×1,t3=(xlna)33×2×1 …,tn=(xlna)nn!…

Conclusion

e)3/3x2x1 +…

=19.4125

T7(e,4) = 1+(4ln e)/1 + (4 ln e)2/2x1 + (4 ln e)3/3x2x1 +…

=48.555556

T7(e,5) = 1+(5ln e)/1 + (5 ln e)2/2x1 + (5 ln e)3/3x2x1 +…

=113.118056

T7(e,6) = 1+(6ln e)/1 + (6 ln e)2/2x1 + (6 ln e)3/3x2x1 +…

=244.6

T7(e,7) = 1+(7ln e)/1 + (7 ln e)2/2x1 + (7 ln e)3/3x2x1 +…

=493.168056

T7(e,8) = 1+(8ln e)/1 + (8 ln e)2/2x1 + (8 ln e)3/3x2x1 +…

=934.155556

T7(e,9) = 1+(9ln e)/1 + (9 ln e)2/2x1 + (9 ln e)3/3x2x1 +…

=1675.5625

T7(e,10) = 1+(10ln e)/1 + (10 ln e)2/2x1 + (e ln e)3/3x2x1 +…

=2866.555556

 x T7(e,x) 1 2.718056 2 7.355556 3 19.4125 4 48.555556 5 113.118056 6 244.6 7 493.168056 8 934.155556 9 1675.5625 10 2866.555556

This diagram shows the relation between T7(10,x) and x using the gained results. (Diagram Microsoft Excel)

This diagram shows that when x increases the value of T7(e,x) will increase as well.

To prove that this thereoy will respond to a change, the singne will be change to negative, to prove this a few negative values of x will be taken where a is constant.  The negative values of x will be -1,-3,-5 and -7 and a=2 when T9(a,x).

T9(2,-1) = 1+(-1ln 2)/1 + (-1 ln 2)2/2x1 + (-1 ln 2)3/3x2x1 +…

=0.500000

T9(2,-3) = 1+(-3ln 2)/1 + (-3 ln 2)2/2x1 + (-3 ln 2)3/3x2x1 +…

=0.126654

T9(2,-5) = 1+(-5ln 2)/1 + (-5 ln 2)2/2x1 + (-5 ln 2)3/3x2x1 +…

=0.177919

T9(2,-7) = 1+(-7ln 2)/1 + (-7 ln 2)2/2x1 + (-7 ln 2)3/3x2x1 +…

=2.743859

 x T9(2,x) -1 0.500000 -3 0.126654 -5 0.177919 -7 2.743859

This diagram shows the relation between T9(2,x) and x using the gained results. (Diagram Microsoft Excel)

This graph shows none of the previous observations, which states that when this sequence is taken into the negative area, the general statement becomes false.

To sum up the result, it is possible to say that the formula used in the different steps is only valied for the statement Tn(a,x) for positive values.

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

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