I need to take a limit of a function $\frac{x^n-1}{x-1}$ in R but I want the answer in terms of n. I tried defining n as a symbol but it did not work. I am new to R so I would appreciate any assistance.
Assuming that you want to take the limit of this function for x->1, you can obtain the result by using the package Ryacas in the following way:
require(Ryacas)
x <- Sym("x")
n <- Sym("n")
Limit((x^n-1)/(x-1),x,1)
which yields the answer:
expression(n)
Related
I am new to R (also not too good at math) and I am trying to calculate this equation in R with some difficulties:
X is some integer data I have, with 550 samples.
Any help is appreciated since I am unsure how to do this. I think I have to use a for loop and the sum() function but other than that I don;t know.
R supports vectorisation, which means you very rarely need to implement for loops.
For example, you can solve your equation like so:
## I'm just making up a long numerical vector for x - obviously you can use anything
x <- 1:1000
solution <- sum(20/x)^0.5
Unless the brackets denote the integral, rather than the sum? In which case:
solution <- sum( (20/x)^0.5 )
I was trying to solve a basic matrix problem. .
I used :
A<- matrix(c(2,7,5,7), 2,2)
b<- c(8,12)
solve(A,b, fractions = TRUE)
However, my result only gives me results in decimal places. How can get fractions results?
I also want to plot this equation above.
I used:
plotEqn(A,b)
However, it tells me this equation can't be found. Can I have some advice please?
Thank you very much!!!
For your first question,
MASS::fractions(solve(A,b))
gives {4/21, 32/21} (note that you won't always be guaranteed the correct answer, as R does floating-point calculation unlike e.g. Mathematica)
For your second question, it looks like the plotEqn() function is in the matlib package: if you have that package installed, then either first loading the package (with library("matlib")) or matlib::plotEqn(A,b) should work.
On closer inspection it looks like you want matlib::Solve() for the first question (note that R is case-sensitive, so solve and Solve are different):
library(matlib)
Solve(A,b, fraction=TRUE)
## x1 = 4/21
## x2 = 32/21
I want to multiply a this number 4.193215e+12 with a dataframe. My code is
df <- cbind(Dataset = df$Dataset, df[,2:4] * 4.193215e^12
However an error appears. What is the proper way to code this number 4.193215e+12 in R?
While this is found in the not-quite-obvious location ?NumericConstants , I am hard-pressed to think of a language in which Xe^Y is syntactically correct. Always use either e or ^ for powers.
I am still quite new to r (used to program in Matlab) and I am trying use the parallel package to speed up some calculations. Below is an example which I am trying to calculate the rolling standard deviation of a matrix (by column) with the use of zoo package, with and without parallelising the codes. However, the shape of the outputs came out to be different.
# load library
library('zoo')
library('parallel')
library('snow')
# Data
z <- matrix(runif(1000000,0,1),100,1000)
#This is what I want to calculate with timing
system.time(zz <- rollapply(z,10,sd,by.column=T, fill=NA))
# Trying to achieve the same output with parallel computing
cl<-makeSOCKcluster(4)
clusterEvalQ(cl, library(zoo))
system.time(yy <-parCapply(cl,z,function(x) rollapplyr(x,10,sd,fill=NA)))
stopCluster(cl)
My first output zz has the same dimensions as input z, whereas output yy is a vector rather than a matrix. I understand that I can do something like matrix(yy,nrow(z),ncol(z)) however I would like to know if I have done something wrong or if there is a better way of coding to improve this. Thank you.
From the documentation:
parRapply and parCapply always return a vector. If FUN always returns
a scalar result this will be of length the number of rows or columns:
otherwise it will be the concatenation of the returned values.
And:
parRapply and parCapply are parallel row and column apply functions
for a matrix x; they may be slightly more efficient than parApply but
do less post-processing of the result.
So, I'd suggest you use parApply.
In R, how can I produce all the permutation of a group, but in this group there are some repetitive elements.
Example :
A = {1,1,2,2,3}
solution :
1,1,2,2,3
1,1,2,3,2
1,1,3,2,2
1,2,1,2,3
1,2,2,1,3
1,2,2,3,1
.
.
using the gtools package,
library(gtools)
x <- c(1,1,2,2,3)
permutations(5, 5, x, set = FALSE)
Just use the combinat package:
A = c(1,1,2,2,3)
library(combinat)
permn(A)
If you want to do it with built-in R:
permute <- function(vec,n=length(vec)) {
permute.index <- sample.int(length(vec),n)
return(vec[permute.index])
}
permute(A)
Using the permute package:
x <- c(1,1,2,2,3)
require(permute)
allPerms(x, observed = TRUE)
I have done extensive research on combination and permutation. This result which I have found is written on a book Known as Junction (an art of counting combination and permutation. To view my site then log on to https://sites.google.com/site/junctionslpresentation/home
I have also have solution for your question. I have also found to order a multiple object permutation. This multiple object permutation I call it (CON of MSNO) which means Combination Order Number of Multiple Same Number of Objects.
To view this method of ordering then go to the site https://sites.google.com/site/junctionslpresentation/proof-for-advance-permutation
at the bottom of this site I have attached some word documents. Your required solution is written on the word document 12 Proof (CON of MSNO) and 13 Proof (Converse of CON of MSNO). Download this word document for the proper view of the written matters.