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Condensing the series gives

f(θ) = = Cosine (θ), Maclaurin Series.

Let θ = 0.1

f (0.1) =

Then from excel

Formulas

Solution

f (0.1) = Cosine (0.1) = 0.9950.

b. For θ = 0.25

f(0.25) =

Excel Solution

f (0.25) = Cosine (0.25) = 0.9689

And for θ = 0.4

f(0.4) =

Excel Solution

f (0.4) = Cosine (0.4) = 0.9211

c. The series converges to 1 as the value of θ approaches 0 or 1

Task 2

a. Let the original length be 4 metres

The series becomes

= 2 + 1 + length of string put in the box.

b. The series is a geometric with common ratio r =

Let S = 2 + 1 + Implying S = 1 + + …

Now, S - S = 2

S = 2, S = 4. The series converges to 4.

c. Let the original length be 20 metres

The series becomes

= 10 + 5 + length of string put in the box.

d. The series is a geometric with common ratio r =

Let S = 10 + 5 + Implying S = 5 +

Now, S - S = 10

S = 10, S = 20. The series converges to 20.

e. The total length of the pieces going to the box will not exceed the original length of the string. Therefore, the results will be consistence regardless of the original length of the string.

f. The result is not unexpected since a geometric series with r < 1 converges to

where a is the first term. The condition is satisfied in all the examples above.

Task 3

a. Given and , let k = 6

implying n = 1.

Now,

= 3.5000

For implying n = 2.

Now,

= 2.6071

For implying n = 3.

Now,

= 2.4543

b. From excel

2.44949436673213

2.44948974278754

2.44948974278318

2.44948974278318

Therefore, the series seems to be converging to 2

c. Given and , let k = 20

implying n = 1.

Now,

= 4.50000000000000

For implying n = 2.

Now,

= 4.47222222222222

For implying n = 3.

Now,

= 4.47213595583161

From excel

4.47213595499958

4.47213595499958

4.47213595499958

4.47213595499958

Therefore, the series seems to be converging to

d. Given and , let k = 10

implying n = 1.

Now,

= 2.333333333

For implying n = 2.

Now,

= 2.206349206

For implying n = 3.

Now,

= 2.246241102

From excel

2.232707652

2.237191458

2.235693860

2.236192725

Therefore, the series seems to be converging to

e. The series will converge to

Let , let k = 20

implying n = 1.

Now,

= 3.000000000

For implying n = 2.

Now,

= 2.416666667

For implying n = 3.

Now,

= 2.673132184

From excel

= 2.538747975

2.604161731

2.571044072

2.587496254

Therefore, the series seems to be converging to

Summary

In task 1, I learnt that Maclaurin series are representation of trigonometric ratios. Therefore, they converge to the value of a particular trigonometric ratio. Besides, in task 2 geometric series can be used to model real life situation and they converge to the sum of the series. For instance, they will converge to (a/(1-r)) for common ratio r < 1 and (a/(r-1)) for r > 1. Finally, in task 3 approximation of a series through iteration outcome depend on the starting value for the iteration.

September 25, 2023

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