Join over 1.2 million students every month
Accelerate your learning by 29%
Unlimited access from just £6.99 per month

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

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.

Found what you're looking for?

Start learning 29% faster today
150,000+ documents available
Just £6.99 a month

Join over 1.2 million students every month
Accelerate your learning by 29%
Unlimited access from just £6.99 per month

Related International Baccalaureate Maths essays

Extended Essay- Math

ï¿½ï¿½[ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½}ï¿½ï¿½ï¿½ï¿½Ëï¿½ $ï¿½ï¿½FYï¿½"8oï¿½Tï¿½>ï¿½-ï¿½yï¿½"oï¿½ï¿½ï¿½]Gï¿½ï¿½ï¿½ï¿½>ï¿½?ï¿½ï¿½ï¿½...'Ð£,ï¿½ï¿½7ï¿½*ï¿½S?ï¿½-ï¿½<ï¿½ï¿½7ï¿½ï¿½(r)ï¿½ï¿½fï¿½?ï¿½gï¿½rï¿½ï¿½ï¿½ï¿½ï¿½?ï¿½Q-ï¿½ ï¿½ï¿½ ï¿½)ï¿½"ï¿½K~-ï¿½ï¿½ï¿½WQï¿½ï¿½qï¿½ï¿½ï¿½ï¿½9c?ï¿½padï¿½ï¿½(ï¿½?ï¿½ ï¿½Êï¿½ï¿½ï¿½ï¿½%ï¿½?ï¿½ï¿½+ï¿½ï¿½Yï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½?ï¿½ï¿½Oï¿½eï¿½ï¿½Cï¿½ï¿½ï¿½Aï¿½ cï¿½ ï¿½ßï¿½ï¿½Fï¿½{ï¿½ï¿½ï¿½ï¿½ï¿½Gï¿½Cï¿½ï¿½ Xï¿½\ï¿½Y'ï¿½ 2ï¿½!ï¿½ï¿½ ï¿½...1ï¿½oï¿½ï¿½#|=ï¿½ï¿½ï¿½?ï¿½n#ï¿½ï¿½ï¿½qï¿½ï¿½,gï¿½.ï¿½,"ï¿½...gï¿½ï¿½PÂï¿½Aï¿½D*ï¿½ï¿½3/4-ï¿½ï¿½uï¿½7ï¿½ï¿½8ï¿½ï¿½-3ï¿½- ï¿½Iï¿½Bï¿½ï¿½ï¿½pï¿½ï¿½ï¿½e?ï¿½ï¿½^ï¿½ï¿½ï¿½ï¿½z-ï¿½ï¿½ï¿½-Qï¿½ï¿½ï¿½...ï¿½ï¿½ï¿½ï¿½'L1/4I(c)ï¿½$iï¿½ï¿½ï¿½mï¿½ï¿½ï¿½Mï¿½Xkï¿½ï¿½ï¿½ï¿½/ï¿½ï¿½s] (ï¿½ï¿½ï¿½ï¿½1ï¿½9ï¿½ï¿½tï¿½T"\ï¿½wï¿½ï¿½uwg(c)ï¿½+ï¿½pï¿½l._ï¿½ï¿½ï¿½Iï¿½Uï¿½...ï¿½Jï¿½ï¿½3/4(ï¿½ï¿½ï¿½ï¿½ï¿½^ï¿½(c)ï¿½Wï¿½ï¿½ï¿½:ï¿½ï¿½ï¿½ï¿½ï¿½l?ï¿½7ï¿½-ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½ ï¿½ ( ï¿½ ( ï¿½ ( ï¿½ ( ï¿½ ( ï¿½ ( ï¿½ ( ï¿½ ( ï¿½ kHï¿½Èï¿½!aï¿½ï¿½Tn`ï¿½NXï¿½ï¿½Íp9<Pï¿½(r)ï¿½ÑÕU"ï¿½V-ï¿½ï¿½ ï¿½phï¿½ï¿½ï¿½A9-A ï¿½=Aï¿½@ @hï¿½{ï¿½ï¿½1/2ï¿½ï¿½ -,ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Mrï¿½?ï¿½.ï¿½?ï¿½ï¿½ "ï¿½"aï¿½ï¿½ï¿½ï¿½1/4ï¿½5ï¿½ï¿½}ï¿½P ï¿½ @[email protected]@[email protected]@[email protected]@[email protected]@P Âï¿½ï¿½ï¿½ï¿½;ï¿½ï¿½3/4ï¿½ï¿½ï¿½[ï¿½Fï¿½ï¿½_ï¿½Vï¿½Cï¿½ ;Aï¿½|aï¿½ï¿½dï¿½"|Uï¿½1/2ï¿½1jï¿½7ï¿½lï¿½ï¿½Wï¿½^ï¿½ï¿½ï¿½ #ï¿½Iï¿½2Ö³Cï¿½ï¿½_ ï¿½Oï¿½ï¿½ï¿½<#ï¿½l3/4*ï¿½"ï¿½ï¿½~ï¿½S^}ï¿½ï¿½- ï¿½4 Vï¿½_ï¿½*|xï¿½ï¿½ï¿½ï¿½kj:Ö£ï¿½h~ï¿½C"ï¿½ H'ï¿½ï¿½ï¿½Qï¿½-ï¿½0kï¿½X~ï¿½ï¿½mï¿½ï¿½Oï¿½zï¿½ï¿½ h>1ï¿½ï¿½}3Eï¿½ï¿½ï¿½Å¦iï¿½ï¿½'ï¿½ï¿½$ï¿½ï¿½CgBï¿½:ñ|_ï¿½4ï¿½nï¿½ï¿½ï¿½i(c)...ï¿½P "ï¿½/ï¿½ï¿½ï¿½ï¿½9h|e x"ï¿½ï¿½?| 'ï¿½Oï¿½-"ï¿½> ï¿½ ï¿½ï¿½hï¿½ï¿½ï¿½5ï¿½^ï¿½ï¿½(r)ï¿½ï¿½Zï¿½I'ï¿½uA4V(r)ï¿½ï¿½e""Wï¿½ï¿½+"p5ï¿½ï¿½ï¿½+ï¿½cswï¿½ï¿½NYVSï¿½ï¿½ï¿½ln2Uï¿½ï¿½ï¿½&ï¿½ZxjPJNï¿½ï¿½W'%cï¿½ï¿½<Aï¿½ï¿½...ï¿½dXï¿½>ï¿½Uï¿½ï¿½cï¿½AVï¿½ï¿½ï¿½ï¿½ï¿½ï¿½'pï¿½j'n<ï¿½Ò"ï¿½ï¿½ï¿½ï¿½Jï¿½xï¿½ï¿½3/44ï¿½ï¿½ï¿½ï¿½-ï¿½ï¿½...jï¿½ï¿½ï¿½ï¿½Uï¿½Kï¿½ï¿½oï¿½h6ï¿½ï¿½ï¿½"<ï¿½ï¿½mï¿½(r).-ï¿½ï¿½Gko-ï¿½ï¿½h"-~ï¿½ï¿½ï¿½ ,ï¿½ï¿½' ï¿½*ï¿½0YM wdY...L&ï¿½hï¿½Xï¿½U\|ï¿½+E^sï¿½Pï¿½(c);ï¿½q"\ï¿½woï¿½1/4-ï¿½cqY>pï¿½ï¿½ï¿½F:(r)ï¿½3,+ï¿½jJï¿½GJï¿½,*ï¿½1/4ï¿½ï¿½Ü¥ï¿½F.Rï¿½Jï¿½%ï¿½ ï¿½/ï¿½ï¿½i/Ú£ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½_?fï¿½Ù¶ï¿½ï¿½ZGï¿½ _Cï¿½-ï¿½ï¿½|Bï¿½ï¿½ï¿½;ï¿½Z <]%ï¿½ï¿½ï¿½ï¿½Z.ï¿½ï¿½7)ï¿½\ï¿½ï¿½ï¿½[eï¿½Apï¿½} ï¿½'ï¿½pgpï¿½Sï¿½|Qï¿½|eW0ï¿½e8|N(tm)arï¿½ï¿½/(c)

Math Studies I.A

$19,100 (2007 est.) $18,500 (2006 est.) note: data are in 2008 US dollars Belarus $11,800 (2008 est.) $10,700 (2007 est.) $9,900 (2006 est.) note: data are in 2008 US dollars Belgium $37,500 (2008 est.) $37,100 (2007 est.) $36,200 (2006 est.)

Math IA - Logan's Logo

have: a=-2.2 b= c=3.4 d=-1.3 y=-3.15sin(1.01x+3.0)+0.35 Combining the determined values of a, b, c, and d, we now have a complete sine function to fit the data with: This then gives us the following graph: However, we can see, in the blue circled area, places where the curve doesn't fit the data well.

Math Studies - IA

These two numbers can then be compared. Whenever the calculated value is below 1, US wins, and when it is above 1, Europe wins. A value of 1 represents a tie because for example a Ryder Cup final score of 14 - 14 is ()

Infinite Summation Internal Assessment ...

Here is another example of the sequence taking the same concept where , a x n t Sn 2 7 0 1.000000 1.000000 1 4.852030 5.852030 2 11.771099 17.623129 3 19.037909 36.661038 4 23.093128 59.754166 5 22.409711 82.163878 6 18.122099 100.285977 7 12.561282 112.847259 8 7.618465 120.465724 9 4.107225 124.572949

SL Type 1 PF - Infinite Summation - A general statement has been reached, ...

Figure 4: Values of when Along with the data values of , the exponential graph of is plotted for comparative analysis. Observing the values suggests that the exponential graph of matches the values of , which allows a generalized statement: In order to reach a conclusion, values of for , , and will be analyzed.

This assignments purpose is to investigate how translation and enlargement of data affects statistical ...

Height (cm) Frequency Cumulative Frequency 130-139 17 17 140-149 14 31 150-159 5 36 160-169 9 45 170-179 15 60 The graph below would represents

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

Now a general sequence where x=1, will be considered. Taking all data collected into account a relation between a and ? has shown. So as the two sets of results show a=2 or a=3, the total of sum won?t exceed 2 or 3, when x=1. To prof this suggestion right, two more examples will be made, again when x=1

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).

Found what you're looking for?

Related International Baccalaureate Maths essays

Found what you're looking for?

Looking for expert help with your Maths work?

Check out our FREE Study Guides:

Or get inspiration from these FREE essays:

pH titration curves Lab Report. How does the use of a strong acid ...

China One Child Policy