GCSE Maths Coursework: Round and Round

Categories: CourseworkMath

I have been asked to investigate the equation –

n–>(+1) –>(?2) –>

* ?

(a) (b)

I will do this by first of all, changing the first number (a) to find out if that it has any relevance to the answer that comes out of the equation at the end.

Then I am going to change the (b) number to find out weather that has anything to do with the outcome of the final number, I will also be looking for patterns and sequences in the answers.

Investigation 1

6–>(+1)–>(?2)–> = 3.5

2.25

1.625

1.3125

1.15625

1.078125

1.0390625

1.01953125

1.009765625

1.004882813

1.002441406

1.001220703

1.000610352

1.000305176

As you can see, when the sum is entered in to a graphical calculator, the numbers that come out are as above. The numbers are decending from 3.5 to 1.000305176, that is the most that I have done down to so far. I predict that the numbers will eventually stop ay the number 1 as near the end there is three 0’s and before that there were two 0’s and before that there was one 0, so I estimate that there will eventually be nine 0’s and the number will be finally 1.

000000000.

Investigation 1:carried on

1.000152588

1.000076294

1.000038147

1.000019074

1.00009537

1.000047685

1.000023843

1.000011922

1.000005961

1.000029805

1.000014903 here an extra “0” is added

1.000007452

1.000003726

1.000001863

1.000000932

1.000000466

1.000000233

1.000000177

1.000000089 here one extra “0” is added.

1.000000045

1.000000023

1.000000012

1.000000006 here one more “0” is added

1.000000003

1.000000002

1.000000001

1.000000001

1.000000001

At this point, it is the furthest that the number is been to the number one, I was almost right in my prediction.

I will now try a change in the sequence, I am going do the sequence as followed;

6–>(+2)–>(?2)–> = 4

3

2.5

2.25

2.125

2.0625

2.03125

2.016625

2.0078125

2.00390625

2.001953125

2.000976563

2.000488281

Now that I have changed the “+1” to a “+2” I can see that the answer goes up.

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When the (a) number is “+1” the number eventually rounds down, close to 1, when the (a) number is “+2” the number will eventually also round down to the (a) number which is now “2”.

I can now see quite clearly, that if the (a) number in the sequence is for example, “20” the number, when divided by two and after been passed through the sequence, the answer will be “20”.

I am now going to put my method to practice:

6 –>(+20)–>(?2)–> = 13

16.5

18.25

19.125

19.5625

19.78125

19.890623

19.9453125

19.97263625

19.98632813

19.99316406 19.99658203

19.99829102

19.99914551

Investigation 2

I think that I have found a pattern in the numbering, I think that the higher the “a” number, the longer it takes to get to the final answer.eg. When the “a”number is 6, it takes only a short time to get to the final answer, but with the “a” number as 20, it takes a lot longer to get the final answer.

I will show this now:

High number

6 –>(+20)–>(?2)–> = 13 19.99998665

16.5 19.99999333

18.25 19.77777667

19.125 19.88888834

19.126 19.94444417

19.5625 19.97222209

19.78125 19.98611105

19.890623 19.99305553

19.9453125 19.99652777

19.97263625 19.99826389

19.98632813 19.99913195

19.99316406 19.99956598 19.99658203 19.99978299

19.99829102 19.9998915

19.99914551 19.9994575

19.99957276 19.99972875

19.99978638 19.99986438

19.99989319 19.99993219

19.9999466 19.9999661

19.9999733 19.99998305

Lower number

6–>(+1)–>(?2)–> = 3.5

2.25

1.625

1.3125

1.15625

1.078125

1.0390625

1.01953125

1.009765625

1.004882813

1.002441406

1.001220703

1.000610352

1.000305176

1.000152588

1.000076294

1.000038147

1.000019074

1.00009537

1.000047685

1.000023843

1.000011922

1.000005961

1.000029805

1.000014903

1.000007452

1.000003726

1.000001863

1.000000932

1.000000466

1.000000233

1.000000177

1.000000089.

1.000000045

1.000000023

1.000000012

1.000000006

1.000000003

1.000000002

1.000000001

1.000000001

1.000000001

As you can see, when the “a” number is high, it takes a longer time to get to the final answer then when the number is lower.

When the “a”number was +1, it took only 42 attempts to get to the closest that it would get to 1, but when the “a”number was +20, after 40 attempts, it was still not even close to becoming 20. However, my analysis that the “a” number decides how long the answer takes to find could be wrong, so I am going to carry out one more investigation to certify my prediction and give me a clearer view.

Investigation 3

6–> (+10)–>(?2)–> = 89 9.999969483

9.5 9.999984742

9.75 9.999992371

9.875 9.999996186

9.9375 9.999998093

9.96875 9.999999047

9.984375 9.999999524

9.9921875 9.999999762

9.99609375 9.999999881

9.998046875 9.999999941

9.999023438 9.999999971

9.999511719 9.999999986

9.99975586 9.999999995

9.99987793 9.999999997

9.999938965 9.999999999

10

It took 31 attempt to get to the final number, 10, but this shows me that the size of the number, does not decide how long it takes to find the final answer.

Investigation 4

I am now going to change some more numbers in the sequence, I will change the “n” number in the sequence to see if that has any relevance to the final number given.

18–>(+1)–>(?2)–> = 9.5

5.25

3.125

2.0625

1.53125

1.265625

1.1328125

1.06640625

1.033203125

1.016601563

1.008300782

I now believe that it is not the “a” number that decides how long it takes to find the answer, but the “n” number.

I will now see how long it takes to get to the final answer when the “n “number is 1.

1–>(+1)–>(?2)–> = 1

1

1

it takes only one attempt to find the answer in this sequence, but I will now try with 2 as the “n” number.

2–>(+1)–>(?2)–> = 1.5

1.25

1.125

1.0625

1.03125

1.015625

1.0078125

1.00390625

1.001953125

1.000976563

1.000488282

1.000244141

1.000122171

1.000061086

1.000030543

1.000015272

1.000007636

1.000003818

1.000001909

1.000000955

1.000000478

1.000000239

1.000000165

1.000000083

1.000000042

1.000000021

1.000000011

1.000000006

1.000000003

1.000000002

1.000000001

1.000000001 this is the lowest that the number will go.

It took 32 attempts to get the final answer and so this shows me that not the “n” number or the “a” number have anything to do with how long the equation takes to solve.

Now I will investigate what the “b” number has to do with the equation. I will do this by changing the number, higher and lower, looking for patterns and rhythms.

Investigation 5: Changing the “b” value

6–>(+1)–>(?3)–> = 2.333333333

1.111111111

0.703703703

0.567901234

0.522633744

0.507544581

0.50251486

0.75125743

0.583752476

0.791876238

0.895938119

0.631979373

0.543993124

0.514664374

As you can see, when the “b” number has been changed, it alters the whole answer completely. There are no patterns and the number is not heading in a pattern that I can see.

Now that I have done all my working, I think that I have found a solution.

To see if this solution works, I am going to predict something:

I predict that 6 ? (+1)? (?5) = 1/4

With my equation: +x ?y ? x

?

y-1

I predict that the number will eventually round down to 0.25 (1/4).

1.4 1.2 0.44 0.288 0.2576 0.25152 0.250304 0.2500608 0.25001216 0.250002432 0.25000486 0.250000486 0.250000097 0.250000019 0.250000003 0.250000000 = 1/4

My prediction was correct. I will now see if it works on other sequences.

I predict that

6 ?(+3) ? (?2) = 3

will eventually end up at the number three:

4.5

3.75

3.375

3.1875

3.09375

3.046875

3.0224375

3.01171875

3.005835375

3.002917688

3.001458844

3.000729422

3.000364711

3.000182356

3.000091178

3.000045589

I was correct, it does eventually end up at 3.

This means that my equation works.

X? (a) ?(?b) = a

?

(b-1)

e.g. 1?(+2)?(?2)=

2

?

2-1 =

2

?

1 =

2

So, I have worked out that the final rule is +x ?y = x

?

y-1

Cite this page

GCSE Maths Coursework: Round and Round. (2020, Jun 02). Retrieved from http://studymoose.com/gcse-maths-coursework-round-round-new-essay

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