CODEWITHSUNDEEP
r
Daman deep
Daman deep

Reputation: 631

Is there any other way to solve this Mathematical problem in R?

Calculate the $\sum_{j=1}^{n}r^j$ where r has been assigned the value $1.06$,and compare with \frac{(1-r^{n+1})}{(1-r)} for $n=10,20$.

This is what I have done so far.

j=c(1:10)
r=1.06
A=r^j
A
sum(A)
compare_with=(1-(1.06)^(11))/(1-(1.06))
compare_with

Upvotes: 0

Views: 113

Answers (3)

Rui Barradas
Rui Barradas

Reputation: 76470

I believe this is simpler than what the discussions in the comments are making it look. In R, all arithmetic operations are vectorized, so the code below works for a vector n.

compare_with <- function(n, r) (r - r^(n + 1))/(1 - r)

n <- c(10, 20, 30, 40)
j <- 1:10
r <- 1.06
A <- r^j

sum(A)
#[1] 13.97164

compare_with(n, r)
#[1]  13.97164  38.99273  83.80168 164.04768

If the function needs to also be vectorized over r, it's once again not that complicated.

CompareWith <- Vectorize(compare_with, "r")

r_vec <- c(1.06, 1.08)
CompareWith(n, r_vec)
#          [,1]      [,2]
#[1,]  13.97164  15.64549
#[2,]  38.99273  49.42292
#[3,]  83.80168 122.34587
#[4,] 164.04768 279.78104

Upvotes: 1

Bhas
Bhas

Reputation: 1834

You have made an error in the formula for compare_with. First of all you define r but don't use it in compare_with. It is much better to use variables you define so that when you change a value you don't have to change it all over the place with the risk of forgetting to change some part.

Your compare_with is wrong. It should read

compare_with=(1.06-(1.06)^(11))/(1-(1.06))

You can prove this the standard way for geometric series.

Secondly there is absolutely no need to use c in the definition of j; just j <- 1:n is sufficient. And you should define n before you start.

A nicer way of writing your formulas is

n <- 10
j <- 1:n
r <- 1.06
A <- r^j
A
sum(A)
compare_with=(r-r^(n+1))/(1-r)
compare_with

If you follow @Rui_Barradas's advice the compare_with function should be written as:

compare_with <- function(n) (r - r^(n + 1))/(1 - r)

BTW: In your question you say that r has been assigned the value 1.08. Yourcalculations use the value 1.06. So what is it?

Additional method

To do what you specify in your comment you could do that this way. Define a function A that can take a vector of values for n like this

A <- function(n) { Asum <- function(n) sum(r^(1:n)); sapply(n,Asum) }

Then this will do what you want

n <- c(10,20,30,40)
compare_with(n)
A(n)

Upvotes: 2

beavis11111
beavis11111

Reputation: 574

j <- 1:10
j.compare <- c(10,20)
sum(1.08^j) #sum
(1-1.08^j.compare)*((1-1.08)^-1) #compare with

Upvotes: 2

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