Find the derivatives with the respect to x of the following.
\(F(x|x \ge 0)=1-e^{\lambda*x}\)
# in order to find the derivative in R, we will install the "Deriv" library.
# we need to include the "repos" so it doesnt give error for R markdowns.
install.packages('Deriv', repos="http://cran.us.r-project.org")
## Installing package into 'C:/Users/Anil Akyildirim/Documents/R/win-library/3.6'
## (as 'lib' is unspecified)
## package 'Deriv' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\Anil Akyildirim\AppData\Local\Temp\Rtmp4Y5XdH\downloaded_packages
library('Deriv')
# creating the function in R
f1=function(x)1-(exp(-lambda*x))
# derivative of the function
der_f1=Deriv(f1)
der_f1
## function (x)
## lambda * exp(-(lambda * x))
\(F(x|b > a)=(x-a)/(b-a)\)
# creating the function in R
f2=function(x)(x-a)/(b-a)
# finding the derivative of the function - f2-
der_f2=Deriv(f2)
der_f2
## function (x)
## 1/(b - a)
\(F(x|a < x \le c \le b)=(x-a)^2/((b-a)*(c-a))\)
# creating the function in R
f3=function(x)(x-a)^2/((b-a)*(c-a))
# finding the derivative of the function -f3-
der_f3=Deriv(f3)
der_f3
## function (x)
## 2 * ((x - a)/((b - a) * (c - a)))
\(F(x|a \le c < x < b)=1-(((b-x)^2)/((b-a)*(c-a)))\)
# creating the function in R
f4=function(x)1-(((b-x)^2)/((b-a)*(c-a)))
# finding the derivative of the function
der_f4=Deriv(f4)
der_f4
## function (x)
## 2 * ((b - x)/((b - a) * (c - a)))
Solve the following definite and indefinite integrals
\(\int_{0}^{10}\)\(3x^3dx\)
# creating the function in R
f5=function(x)3*x^3
integrate(Vectorize(f5), 0, 10) # it is a definite integral
## 7500 with absolute error < 8.3e-11
\(\int_{a}^{b}\)\(x*\lambda*e^(-\lambda*x)dx\)
This is an indefinite integral. We installed mosaic package(please be aware if you try to install the package via script it does not include mosaicCalc which is neccessary to install in order to use D() or antiD() functions. We install this package via RStudio Packages section.
library('mosaicCalc')
## Loading required package: mosaicCore
## Registered S3 method overwritten by 'mosaic':
## method from
## fortify.SpatialPolygonsDataFrame ggplot2
##
## Attaching package: 'mosaicCalc'
## The following object is masked from 'package:stats':
##
## D
# creating the function in R
f6=function(x)x*lambda*exp(-lambda*x)
#using antiD function for indefinite integral
antiD(x*lambda*exp(-lambda*x)~x)
## function (x, lambda, C = 0)
## {
## numerical_integration(.newf, .wrt, as.list(match.call())[-1],
## formals(), from, ciName = intC, .tol)
## }
## <environment: 0x000000002a74f630>
\(\int_{0}^{5}\)\(1/(b-a)dx\)
# in this case a and b are constant and i need to provide them a number to get a result.
# creating a function
a=1
b=2
fx2=function(x)(1/(b-a))
integrate(Vectorize(fx2), 0, 5)
## 5 with absolute error < 5.6e-14
# indefinite integral
antiD(x*1/gamma(x)*x^(sigma-1)*exp(-beta*x)~x)
## function (x, sigma, beta, C = 0)
## {
## numerical_integration(.newf, .wrt, as.list(match.call())[-1],
## formals(), from, ciName = intC, .tol)
## }
## <environment: 0x000000002afeb358>
With the following matrix
Invert it using Gaussian row reduction.
# creating the matrix first
A <- matrix(c(1,2,3,3,3,1,4,6,8), nrow =3, byrow=TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 3 3 1
## [3,] 4 6 8
# we can use solve() function to find the inverse of a matrix
B <- solve(A)
B
## [,1] [,2] [,3]
## [1,] -4.5 -0.5 1.75
## [2,] 5.0 1.0 -2.00
## [3,] -1.5 -0.5 0.75
Find the determinant.
# to find the determinant i can use the det() function in R
det(A)
## [1] -4
Conduct LU decomposition
# we can use lu() or ln.decomposition() functions for LU decomposition of the matrix
# in order to do that, we need to install pracma v1.9.9
# I have done that from the "Packages" section instead of a running the install.packages('') script.
# i have to include library('pracma') otherwise this code will not knit to HTML
library('pracma')
##
## Attaching package: 'pracma'
## The following object is masked from 'package:mosaicCore':
##
## logit
lu(A)
## $L
## [,1] [,2] [,3]
## [1,] 1 0.0000000 0
## [2,] 3 1.0000000 0
## [3,] 4 0.6666667 1
##
## $U
## [,1] [,2] [,3]
## [1,] 1 2 3.000000
## [2,] 0 -3 -8.000000
## [3,] 0 0 1.333333
Multiply the matrix by itβs inverse.
# in order to multiply our matrix (which is A) and inverse of our matrix (which is B) we can use %%
A%%B
## [,1] [,2] [,3]
## [1,] -3.5 -8.881784e-16 1.25
## [2,] 3.0 1.000000e+00 -1.00
## [3,] -0.5 -1.776357e-15 0.50