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# The purpose of this paper is to investigate an infinite summation patter where Ln(a) is a constant and the coefficient of x is an increasing factor to Ln(a).

Extracts from this document...

Introduction

Infinite Summation

Math IA

The purpose of this paper is to investigate an infinite summation patter where Ln(a) is a constant and the coefficient of x is an increasing factor to Ln(a).

Consider the following sequence of terms where x=1 and a=2 under the terms that 0≤n≤10:

tn= n t(n) S(n) 0.000000 1 1 1.000000 0.69314718 1.693147 2.000000 0.24022651 1.933373 3.000000 0.05550411 1.988877 4.000000 0.00961813 1.998495 5.000000 0.00133336 1.999829 6.000000 0.00015404 1.999983 7.000000 1.5253E-05 1.999998 8.000000 1.3215E-06 1.999999 9.000000 1.0178E-07 1.999999 10.000000 7.0549E-09 2 As n  +, Sn  +2

Consider the following sequence of terms where x=1 and a=3:

tn= n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 1.098612 2.098612 2.000000 0.603474 2.702087 3.000000 0.220995 2.923082 4.000000 0.060697 2.983779 5.000000 0.013336 2.997115 6.000000 0.002442 2.999557 7.000000 0.000383 2.999940 8.000000 0.000053 2.999993 9.000000 0.000006 2.999999 10.000000 0.000001 3.000000 As n → +∞, Sn → +3

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive three. Sn approaches a horizontal asymptote when y=3. There is a y-intercept at (0,1). As n → +∞, Sn → +4

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive four. Sn approaches a horizontal asymptote when y=4.

Middle

5.666868

3.000000

0.277521

5.944389

4.000000

0.048091

5.992480

5.000000

0.006667

5.999146

6.000000

0.000770

5.999917

7.000000

0.000076

5.999993

8.000000

0.000007

5.999999

9.000000

0.000001

6.000000 As n → +∞, Sn → +6

Let a=2 and calculate various positive values for x:

 x t(n) S(n) 0.0 1.000000 1.000000 1.0 0.693147 1.693147 2.0 0.480453 2.173600 3.0 0.166512 2.340113 4.0 0.038473 2.378585 5.0 0.006667 2.385252

In the graph, when x is approaching infinite the Sn values are increasing steadily. When various values are used for x then there is an exponential growth.

Let a=3 then calculate for various positive values of x:

 x t(n) S(n) 0.0 1.000000 1.000000 1.0 1.098612 2.098612 2.0 1.206949 3.305561 3.0 0.662984 3.968546 4.0 0.242788 4.211333 5.0 0.066682 4.278016 In the graph, when x is approaching infinite the Sn values are increasing steadily. When various values are used for x then there is an exponential growth.

Evidence:

tn= n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 2.772589 3.772589 2.000000 0.960906 4.733495 3.000000 0.222016 4.955511 4.000000 0.038473 4.993984 5.000000 0.005333 4.999317 6.000000 0.000616 4.999933 7.000000 0.000061 4.999994 8.000000 0.000005 5.000000 9.000000 0.000000 5.000000 As n → +∞, Sn → +6

Sn approaches a horizontal asymptote when y=6. There is a y-intercept at (0,1).

tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 4.158883 5.158883 2.000000 1.441359 6.600242 3.000000 0.333025 6.933267 4.000000 0.057709 6.990976 5.000000 0.008000 6.998976 6.000000 0.000924 6.999900 7.000000 0.000092 6.999991 8.000000 0.000008 6.999999 9.000000 0.000001 7.000000 As n → +∞, Sn → +7

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive seven. Sn approaches a horizontal asymptote when y=7. There is a y-intercept at (0,1).

tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 4.852030 5.852030 2.000000 1.681586 7.533616 3.000000 0.388529 7.922145 4.000000 0.067327 7.989471 5.000000 0.009333 7.998805 6.000000 0.001078 7.999883 7.000000 0.000107 7.999990 8.000000 0.000009 7.999999 9.000000 0.000001 8.000000 As n → +∞, Sn → +8

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive eight. Sn approaches a horizontal asymptote when y=8. There is a y-intercept at (0,1).

tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 5.545177 6.545177 2.000000 1.921812 8.466990 3.000000 0.444033 8.911022 4.000000 0.076945 8.987967 5.000000 0.010667 8.998634 6.000000 0.001232 8.999867 7.000000 0.000122 8.999989 8.000000 0.000011 8.999999 9.000000 0.000001 9.000000 As n → +∞, Sn → + 9

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive nine. Sn approaches a horizontal asymptote when y=9. There is a y-intercept at (0,1).

tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 6.238325 7.238325 2.000000 2.162039 9.400363 3.000000 0.499537 9.899900 4.000000 0.086563 9.986463 5.000000 0.012000 9.998464 6.000000 0.001386 9.999850 7.000000 0.000137 9.999987 8.000000 0.000012 9.999999 9.000000 0.000001 10.000000 10.000000 0.000000 10.000000

Conclusion

tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 5.493061 6.493061 2.000000 3.017372 9.510434 3.000000 1.104974 10.615408 4.000000 0.303485 10.918893 5.000000 0.066682 10.985575 6.000000 0.012210 10.997785 7.000000 0.001916 10.999701 8.000000 0.000263 10.999964 9.000000 0.000032 10.999996 As n → +∞, Sn → + 11

There is a horizontal asymptote as n approaches positive infinite (∞). tn= n t(n) S(n) 0.0 1.000000 1.000000 1.0 7.690286 8.690286 2.0 4.224321 12.914607 3.0 1.546964 14.461571 4.0 0.424878 14.886450 5.0 0.093355 14.979805 6.0 0.017094 14.996898 7.0 0.002683 14.999581 8.0 0.000368 14.999950 9.0 0.000045 14.999995 As n → +∞, Sn → + 15

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive fifteen. Sn approaches a horizontal asymptote when y=15. There is a y-intercept at (0,1).

tn= n t(n) S(n) 0.0 1.000000 1.000000 1.0 8.788898 9.788898 2.0 4.827796 14.616694 3.0 1.767959 16.384653 4.0 0.485575 16.870228 5.0 0.106692 16.976920 6.0 0.019535 16.996455 7.0 0.003066 16.999521 8.0 0.000421 16.999942 9.0 0.000051 16.999994 As n → +∞, Sn → + 17

There is a horizontal asymptote as n approaches positive infinite (∞). Sn approaches a horizontal asymptote when y=17. There is a y-intercept at (0,1).

Checks with various numbers:

tn = n t(n) S(n) 0.0 1.000000 1.000000 1.0 48.283137 49.283137 2.0 38.854356 88.137493 3.0 20.844558 108.982051 4.0 8.387005 117.369057 5.0 2.699673 120.068729 6.0 0.724159 120.792889 7.0 0.166498 120.959387 8.0 0.033496 120.992883 9.0 0.005990 120.998873 10.000000 0.000964 120.999837 tn = n t(n) S(n) 0.0 1.000000 1.000000 1.0 13.961881 14.961881 2.0 32.489020 47.450901 3.0 50.400871 97.851772 4.0 58.640914 156.492686 5.0 54.582497 211.075184 6.0 42.337463 253.412647 7.0 28.148125 281.560772 8.0 16.375032 297.935804 9.0 8.467639 306.403443 10.000000 3.940806 310.344249
 n t(n) S(n) 0.0 1.000000 1.000000 1.0 179.175947 180.175947 2.0 160.520100 340.696047 3.0 95.871136 436.567183 4.0 42.944504 479.511687 5.0 15.389244 494.900931 6.0 4.595637 499.496569 7.0 1.176325 500.672894 8.0 0.263461 500.936356 9.0 0.052451 500.988807 10.000000 0.009398 500.998205 tn = Conclusion:

The limitation to this general statement is that the scope of the evidence is limited to only a few combinations of x and a .

The continuous observation of various graphs with different values used in place of a and x provide evidence for the general statement.

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

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