0

I've been set a question about Madhava's approximation of pi. The first part of it is to create a vector which contains the first 20 terms in the series. I know I could just input the first 20 terms into a vector, however that seems like a really long winded way of doing things. I was wondering if there is an easier way to create the vector?

Currently I have the vector 

g = c((-3)^(-0)/(2*0+1), (-3)^(-1)/(2*1+1), (-3)^(-2)/(2*2+1), (-3)^(-3)/(2*3+1), (-3)^(-4)/(2*4+1), (-3)^(-5)/(2*5+1), (-3)^(-6)/(2*6+1), (-3)^(-7)/(2*7+1), (-3)^(-8)/(2*8+1), (-3)^(-9)/(2*9+1), (-3)^(-10)/(2*10+1), (-3)^(-11)/(2*11+1), (-3)^(-12)/(2*12+1), (-3)^(-13)/(2*13+1), (-3)^(-14)/(2*14+1), (-3)^(-15)/(2*15+1), (-3)^(-16)/(2*16+1), (-3)^(-17)/(2*17+1), (-3)^(-18)/(2*18+1), (-3)^(-19)/(2*19+1), (-3)^(-20)/(2*20+1))

And 

  h = sqrt(12)

So I have done g*h to get the approximation of pi. Surely there's an easier way of doing this? Apologies if this is relatively basic, I am very new to R and still learning how to properly use stack overflow. Thanks.

F. Wotton
  • 45
  • 1
  • 6
  • 1
    Are you actually asking how to do sums in R, as suggested by the comment "I have no idea how to do sums in R"? If so, look at `help(sum)`. I suspect what you really want is someone to help you do all the question parts, including creating the series vector. If so, please make your question more clear and indicate what you've tried in terms of solving your homework and where specifically you're stuck or confused rather than just asking to have the question solved for you. You will get higher quality help that way. – duckmayr Oct 20 '17 at 11:22
  • Hi, thanks duckmayr, I realise that it was completely stupid of me the way I asked the initial question. I had a look over the sum function and it makes sense how to do part b, I'm just really struggling finding a code which will help create the vector in part a. Thanks for the advice. – F. Wotton Oct 21 '17 at 10:49
  • No problem -- your edits have really helped zero in on where you need help. I can post an answer shortly that should help you along and more importantly have a little explanation to help you with similar issues in the future. – duckmayr Oct 21 '17 at 11:02

1 Answers1

0

One of the best features of R is that it is vectorised. This means that we can do operations element-wise on entire vectors rather than having to type out the operation for each element. For example, if you wanted to find the square of the first five natural numbers (starting at one), we can do this:

(1:5)^2

which results in the output

[1]  1  4  9 16 25

instead of having to do this:

c(1^2, 2^2, 3^2, 4^2, 5^2)

which gives the same output.

Applying this amazing property of R to your situation, instead of having to manually construct the whole vector, we can just do this:

series <- sqrt(12) * c(1, -1) / 3^(0:19) / seq(from=1, by=2, length.out=20)
sum(series)

which gives the following output:

[1] 3.141593

and we can see more decimal places by doing this:

sprintf("%0.20f", sum(series))
[1] "3.14159265357140338182"

To explain a little further what I did in that line of code to generate the series:

  • We want to multiply the entire thing by the square root of 12, hence the sqrt(12), which will be applied to every element of the resulting vector
  • We need the signs of the series to alternate, which is accomplished via * c(1, -1); this is because of recycling, where R recycles elements of vectors when doing vector operations. It will multiply the first element by one, the second element by -1, then recycle and multiply the third element by 1, the fourth by -1, etc.
  • We need to divide each element by 1, 3, 9, etc., which is accomplished by / 3^(0:19) which gives / c(3^0, 3^1, ...)
  • Lastly, we also need to divide by 1, 3, 5, 7, etc. which is accomplished by seq(from=1, by=2, length.out=20) (see help(seq))
duckmayr
  • 16,303
  • 3
  • 35
  • 53
  • Ahh that makes a lot of sense, loads easier than manually doing it as well. Thank you! For the last part where you used the `sprintf("0.20f", sum(series))`, can you explain this briefly? As I used the format function to see more decimal places. – F. Wotton Oct 21 '17 at 11:34