How to solve this integral in R? - r

I need to find evaluate the following function, including an integral, in R:
Probability density functions involving multiple variables, where u = t - y.
The problem I'm running into is that while the input variables of the function as a whole are x and t, the integral needs to be evaluated over the variable u = t - y. The functions f and m' both return values, but I don't know how to make it so that R evaluates the integral over this u rather than x or T.
I currently have the following, but this doesn't return the values I'm supposed to be getting, so I'm wondering if I did it properly?
Thank you in advance!
a = 3
b = 10
T = 2.6
mprime = function(x){
return (1/x)
}
f = function(x){
if (a <= x & x <= b){
return (1/(b-a))
}
else{
return (0)
}
}
toIntegrate = function(u){
return (f(u + x)*mprime(T-u))
}
solution = function (x, T){
return (f(T + x)) + (integrate(toIntegrate(T-y), 0, T))
}
solution(5,T)

There are a few errors in your code:
f and other functions you'll be using in your integration need to be vectorised, i.e. it should take a vector as an input and return a vector as an output.
toIntegrate uses x which is neither a global variable nor an argument.
return is a function, so only the expression between parentheses are returned. As a result, your integral would not be evaluated because your function would return f(T+x)
The first argument to integrate should be toIntegrate, not toIntegrate(T-y)
mprime will return infinity for u=t, so the limits may need to be adjusted.
a = 3
b = 10
t = 2.6
mprime = function(x){
return (1/x)
}
f = function(x){
ifelse(a <= x & x <= b,1/(b-a),0)
}
toIntegrate = function(u,x,t){
return (f(u + x)*mprime(t-u))
}
solution = function (x, t){
return(f(t + x) + integrate(toIntegrate, 0, t,x=x,t=t,stop.on.error = F)$value)
}
solution(5,T)

Related

Double integration with a differentiation inside in R

I need to integrate the following function where there is a differentiation term inside. Unfortunately, that term is not easily differentiable.
Is this possible to do something like numerical integration to evaluate this in R?
You can assume 30,50,0.5,1,50,30 for l, tau, a, b, F and P respectively.
UPDATE: What I tried
InnerFunc4 <- function(t,x){digamma(gamma(a*t*(LF-LP)*b)/gamma(a*t))*(x-t)}
InnerIntegral4 <- Vectorize(function(x) { integrate(InnerFunc4, 1, x, x = x)$value})
integrate(InnerIntegral4, 30, 80)$value
It shows the following error:
Error in integrate(InnerFunc4, 1, x, x = x) : non-finite function value
UPDATE2:
InnerFunc4 <- function(t,L){digamma(gamma(a*t*(LF-LP)*b)/gamma(a*t))*(L-t)}
t_lower_bound = 0
t_upper_bound = 30
L_lower_bound = 30
L_upper_bound = 80
step_size = 0.5
integral = 0
t <- t_lower_bound + 0.5*step_size
while (t < t_upper_bound){
L = L_lower_bound + 0.5*step_size
while (L < L_upper_bound){
volume = InnerFunc4(t,L)*step_size**2
integral = integral + volume
L = L + step_size
}
t = t + step_size
}
Since It seems that your problem is only the derivative, you can get rid of it by means of partial integration:
Edit
Not applicable solution for lower integration bound 0.

Complex numbers and missing arguments in R function

I am solving a task for my R online course. The task is to write a function, that solves the quadratic equation with the Lagrange resolvents, or:
x1<--p/2+sqrt((p/2)^2-q)
x2<--p/2-sqrt((p/2)^2-q)
1) If the arguments are non-numeric, the function should return an explained error (or why the error has happend). 2) If there are missing arguments, the function should return an explained error (different from the default). 3) If x1 and x2 are complex numbers (for example if p=-4 and q=7, then x1=2+i*1.73 and x2=2-i*1.73), the function should should also solve the equation instead of generating NaNs and return a warning message, that the numbers are complex. Maybe if I somehow cast it to as.complex, but I want this to be a special case and don't want to cast the basic formula.
My function looks like this:
quadraticEquation<-function(p,q){
if(!is.numeric(c(p,q)))stop("p and q are not numeric") #partly works
if(is.na(c(p,q)))stop("there are argument/s missing") #does not work
x1<--p/2+sqrt((p/2)^2-q)
x2<--p/2-sqrt((p/2)^2-q)
#x1<--p/2+sqrt(as.complex((p/2)^2-q)) works, but I want to perform this only in case the numbers are complex
#x2<--p/2-sqrt(as.complex((p/2)^2-q))
return (c(x1,x2))
}
When testing the function:
quadraticEquation(4,3) #basic case is working
quadraticEquation(TRUE,5) #non-numeric, however the if-statement is not executed, because it assumes that TRUE==1
quadraticEquation(-4,7) #complex number
1) how to write the function, so it assumes TRUE (without "") and anything that is non-numeric as non-numeric?
2) basic case, works.
3) how can I write the function, so it solves the equation and prints the complex numbers and also warns that the numbers are complex (warning())?
Something like this?
quadraticEquation <- function(p, q){
## ------------------------% chek the arguments %---------------------------##
if(
missing(p) | missing(q) # if any of arguments is
){ # missing - stop.
stop("[!] There are argument/s missing")
}
else if(
!is.numeric(p) | !is.numeric(q) | any(is.na(c(p, q))) # !is.numeric(c(1, T))
){ # returns TRUE - conver-
stop("[!] Argument/s p or/and q are not numeric") # tion to the same type
}
## --------------------% main part of the function %--------------------------##
r2 <- p^2 - 4*q # calculate r^2,
if(r2 < 0){ # if r2 < 0 (convert) it
warning("equation has complex roots") # to complex and warn
r2 <- as.complex(r2)
}
# return named roots
setNames(c(-1, 1) * sqrt(r2)/2 - p/2, c("x1", "x2"))
}
quadraticEquation() # No arguments provided
#Error in quadraticEquation() : [!] There are argument/s missing
quadraticEquation(p = 4) # Argument q is missing
#Error in quadraticEquation(p = 4) : [!] There are argument/s missing
quadraticEquation(p = TRUE, q = 7) # p is logical
#Error in quadraticEquation(p = TRUE, q = 7) :
#[!] Argument/s p or/and q are not numeric
quadraticEquation(p = NA, q = 7) # p is NA
#Error in quadraticEquation(p = NA, q = 7) :
#[!] Argument/s p or/and q are not numeric
quadraticEquation(p = 7, q = -4) # real roots
# x1 x2
#-7.5311289 0.5311289
quadraticEquation(p = -4, q = 7) # complex roots
# x1 x2
#2-1.732051i 2+1.732051i
#Warning message:
#In quadraticEquation(p = -4, q = 7) : equation has complex roots
When you write is.numeric(c(p, q)), R first evaluates c(p, q) before determining whether it is numeric or not. In particular if p = TRUE and q = 3, then c(p, q) is promoted to the higher type: c(1, 3).
Here is a vectorized solution, so if p and q are vectors instead of scalars the result is also a vector.
quadraticEquation <- function(p, q) {
if (missing(p)) {
stop("`p` is missing.")
}
if (missing(q)) {
stop("`q` is missing.")
}
if (!is.numeric(p)) {
stop("`p` is not numeric.")
}
if (!is.numeric(q)) {
stop("`q` is not numeric.")
}
if (anyNA(p)) {
stop("`p` contains NAs.")
}
if (anyNA(q)) {
stop("`q` contains NAs.")
}
R <- p^2 / 4 - q
if (min(R) < 0) {
R <- as.complex(R)
warning("Returning complex values.")
}
list(x1 = -p / 2 + sqrt(R),
x2 = -p / 2 - sqrt(R))
}
Also, you should never write x1<--p/2. Keep spaces around infix operators: x1 <- -p/2.

Integrate a sum in R

I am trying to compute the MISE of an estimator and for that i need to do the integral of :
(fp(x) - f(x))^2 where f(x) is exp(-x) and fp(x) is : sum_{i}^n { (1/n)*((K((x - X[i])/h))/h) }
The problem here is that X is a matrix, and i don't know how integrate this sum.
I've tried this :
Kgauss <- function(u) dnorm(u) #Gaussian kernel
func = function(x, n, h, X){ ((1/n) * sum(Kgauss((x-X[0:n])/h)/h) - exp(-x))^2 } # h, n are constants
vfunc = Vectorize(func)
integrate(vfunc, n = 3, K = Kgauss, h = 0.25, X = rexp(3), lower = 0, Inf)
But sadly it didn't work out. The big problem here is fp(x), it consists of the sum of multiple functions .
I hope you can help me with this one, I've been struggling for a while now.
Basically i want to make : integral((K(X1) + ... + K(Xn) - exp(-x))²)
You can define the n, h, and K outside the func and then have x as the only parameter:
n = 3; h = 0.25; X = rexp(3)
func = function(x){ ((1/n) * sum(dnorm((x-X[0:n])/h)/h) - exp(-x))^2 }
vfunc = function(x) { sapply(x, func)}
integrate(vfunc, lower = 0, Inf)
# 0.2070893 with absolute error < 1.7e-05
(I'm not sure that you even need to vectorize func. It's built with vectorized functions already.)

R - Change return values of existing function

I am using the approxfun() function to get linear interpolation. I want to write a function which takes the results of approxfun() then shifts and scales it by an amount that I specify. I need to be able to call this new function just like I would call any other function.
Simplified version of my attempt:
set.seed(42)
x = rnorm(50)
y = rnorm(50, 5, 2)
fhat = approxfun(x, y, rule = 2)
new_function = function(fhat, a, b){
new_fhat <- (function(fhat, a, b) a * fhat() + b)()
return(new_fhat)
}
I expect the results to be the same as
2 * fhat(1) + 3
but instead when I run my function
new_function(fhat, a = 2, b = 3)
I get an error message:
*Error in (function(fhat, a, b) a * fhat() + b)() :
argument "a" is missing, with no default*
You have four problems:
new_fhat isn't passed the values of a and b from the new_function call, and can't see them because you are creating new ones in your function definition. This is actually a bit of a red herring because...
The function you return should only have a single argument - the point at which you want to evaluate it.
You are attempting to evaluate new_fhat immediately.
You are trying to call fhat without an argument.
The solution is:
new_function = function(fhat, a, b){
new_fhat <- function(v) a * fhat(v) + b
return(new_fhat)
}
Results:
fhat(1)
[1] 5.31933
new_function(fhat,a=2,b=3)(1)
[1] 13.63866
2 * fhat(1) + 3
[1] 13.63866
This is equivalent to the code in the question, but simplified:
new_function = function(fhat, a, b) {
a * fhat() + b
}
But this is not correct, because in the posted code fhat requires an argument. For example, to make new_function(fhat, a = 2, b = 3) return the same results as 2 * fhat(1) + 3, you would have to add the parameter 1 inside the function:
new_function = function(fhat, a, b) {
a * fhat(1) + b
}

Can Arg function in R give values greater than pi?

In R, the function Arg applied to a complex returns an angle between -pi and +pi. For example :
Arg((1+i)^5) = -2.356... (-3*pi/4 radians)
Could I have a function returning the positive angle, even greater than +pi? For example, I would like to have :
"function"((1+i)^5) = 3.926... (5*pi/4 radians)
It seems Arg is not adaptable ; maybe some other function exits in some package ?
Thanks for any help.
5*pi/4 is the same as -3*pi/4. So you could do:
Arg_positive = function(complex) {
initial = Arg(complex)
ifelse(initial < 0,
initial + 2*pi,
initial)
}
Just take the remainder of the division to 2pi:
Arg((1+1i)^5) %% (2*pi)
#[1] 3.926991
You could create your own function to address this issue.
pos.arg <- function(num)
{
arg = Arg(num)
if (arg < 0)
arg <- arg + 2*pi
arg
}
x <- complex(length.out = 1, 1, 1)
y <- x^5
pos.arg(y)
[1] 3.926991

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