Top Special Offer! Check discount

Get 13% off your first order - useTopStart13discount code now!

# The Convergence of Maclaurin Series

123 views 2 pages ~ 382 words
Get a Custom Essay Writer Just For You!

Experts in this subject field are ready to write an original essay following your instructions to the dot!

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

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.

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
Category:
Subcategory:
Subject area:
Number of pages

2

Number of words

382

35

Rate:

5

Expertise Geometry
Verified writer

LuckyStrike has helped me with my English and grammar as I asked him for editing and proofreading tasks. When I need professional fixing of my papers, I contact my writer. A great writer who will make your writing perfect.

Hire Writer

This sample could have been used by your fellow student... Get your own unique essay on any topic and submit it by the deadline.

Eliminate the stress of Research and Writing!

Hire one of our experts to create a completely original paper even in 3 hours!