I was trying to knit following code to Pdf (knit without code is working well)
But i got following error
ERROR
processing file: pdf-test.Rmd
output file: pdf-test.knit.md
"C:/RStudio/resources/app/bin/quarto/bin/tools/pandoc" +RTS -K512m -RTS pdf-test.knit.md --to latex --from markdown+autolink_bare_uris+tex_math_single_backslash --output pdf-test.tex --lua-filter "C:\Users\Sava\AppData\Local\R\win-library\4.3\rmarkdown\rmarkdown\lua\pagebreak.lua" --lua-filter "C:\Users\Sava\AppData\Local\R\win-library\4.3\rmarkdown\rmarkdown\lua\latex-div.lua" --embed-resources --standalone --highlight-style tango --pdf-engine pdflatex --variable graphics --variable "geometry:margin=1in"
! Illegal parameter number in definition of \GetTitleStringResult.
Error: LaTeX failed to compile pdf-test.tex. See The R package tinytex - Helper Functions to Manage TinyTeX, and Compile LaTeX Documents - Yihui Xie | 谢益辉 for debugging tips. See pdf-test.log for more info.
Execution halted
CODE
title: "NEWTEST2"
output: pdf_document
date: "2023-11-21"
knitr::opts_chunk$set(echo = TRUE)
library(microbenchmark)
#AUFGABE 25 A
create_hilbert_matrix <- function(n) {
hilbert_matrix <- matrix(NA, nrow = n, ncol = n)
for (i in 1:n) {
for (j in 1:n) {
hilbert_matrix[i, j] <- 1 / (i + j - 1)
}
}
return(hilbert_matrix)
}
# function with n = 6
n <- 6
result <- create_hilbert_matrix(n)
print(result)
#AUFGABE 25 B
library(microbenchmark)
# Function to create a Hilbert matrix
create_hilbert_matrix <- function(n) {
outer(1:n, 1:n, function(x, y) 1 / (x + y - 1))
}
# Function to compare Cholesky decomposition and solve() for matrix inversion
compare_inversion_methods <- function(n) {
# Generate Hilbert
hilbert <- create_hilbert_matrix(n)
# Compute inverse using Cholesky decomposition
chol_inv <- function() solve(chol(hilbert))
# Compute inverse using solve()
solve_inv <- function() solve(hilbert)
# Perform the microbenchmark comparison
mb <- microbenchmark(
chol_inv(),
solve_inv(),
times = 100
)
return(mb)
}
# Compare for n = 3
result_n3 <- compare_inversion_methods(3)
print(result_n3)
# Compare for n = 10
result_n10 <- compare_inversion_methods(10)
print(result_n10)
library(microbenchmark)
# Function to create a Hilbert matrix
hilbert_matrix <- function(n) {
outer(1:n, 1:n, function(x, y) 1 / (x + y - 1))
}
#Matrices
hilbert1 <- hilbert_matrix(3)
hilbert2 <- hilbert_matrix(10)
result1 <- microbenchmark(solve(hilbert1),
chol2inv(chol(hilbert1)),
setup = set.seed(120))
result1
###For the 3×3
#solve(hilbert1) had a higher average computation time (mean = 40.786 microseconds) compared to chol2inv(chol(hilbert1)) (mean = 21.436 microseconds). This suggests that for a 3×3 Hilbert matrix, using Cholesky decomposition followed by inversion (chol2inv()) is faster than direct inversion using solve()
res2 <- microbenchmark(solve(hilbert2),
chol2inv(chol(hilbert2)),
setup = set.seed(120))
res2
###For the 10×10
#chol2inv(chol(hilbert2)) showed lower average computation time (mean = 34.093 microseconds) compared to solve(hilbert2) (mean = 48.82 microseconds). Again, indicating that for a larger 10×10 Hilbert matrix, using Cholesky decomposition followed by inversion is faster than direct inversion using solve().
###AUFGABE 26 A
# Given matrix X
X <- matrix(c(1, 2, 3, 1, 4, 9), ncol = 2)
# Compute Y = 1.5 * X
Y <- 1.5 * X
# Compute X as Y divided by 1.5
X <- Y / drop(1.5)
# Compute transposed Y
t_Y <- t(Y)
# Display transposed Y
t_Y
# Calculate cross-product of Y
first_transposed <- crossprod(Y)
# Calculate cross-product using tcrossprod()
second_transposed <- tcrossprod(Y)
# Benchmarking with microbenchmark
microbenchmark(
crossprod(Y),
tcrossprod(Y),
times = 100L,
setup = set.seed(120)
)
Explanation of crossprod and tcrossprod functions in R:
The crossprod function computes the cross-product of matrices by multiplying the transposed matrix with itself.
The tcrossprod function calculates the cross-product of matrices, but in a different way.
When computing Y^T * Y and Y * Y^T, where Y is a matrix:
- tcrossprod(Y) is faster for computing Y * Y^T as it performs the multiplication without explicitly transposing Y.
- crossprod(t(Y)) is used for computing Y^T * Y but requires explicitly transposing Y before the cross-product.
#tcrossprod(Y) is more efficient for matrix multiplication involving the matrix and its transpose due to its direct calculation.
#AUFGABE 26 B
A <- matrix(rep(1, 1000000), nrow = 1000)
v <- rep(1, 1000)
# Benchmarking the matrix multiplications for 10 times
microbenchmark(
A2v = A %*% A %*% v,
AA_v = (A %*% A) %*% v,
A_Av = A %*% (A %*% v),
times = 10L,
setup = set.seed(120)
)
For operation A^2v:
The attempt to directly compute A^2v using the %^% operator resulted in an error,
as this operator is not a standard operation in R for matrix exponentiation.
To emulate A^2v, the code was adjusted to perform A * A * v using %*% for matrix multiplication.
For operation (AA)v:
This operation calculated (AA)v by performing A * A (matrix multiplication)
and then multiplying the resulting matrix with vector v.
For operation A(Av):
A(Av) was calculated by first computing A * v (matrix-vector multiplication)
and then multiplying the resulting vector with matrix A.
The results indicated differences in computational times among these operations:
- A(Av) operation demonstrated significantly lower execution times compared to the other operations.
- A^2v and (AA)v operations displayed similar execution times, which were notably higher than the A(Av) operation.
##AUFGABE 26 C
B <- diag(1, 1000)
A <- matrix(rep(1, 1000000), nrow = 1000)
v <- rep(1, 1000)
microbenchmark(
Av = A %*% v,
ABv = A %*% B %*% v,
AB_v = (A %*% B) %*% v,
A_Bv = A %*% (B %*% v),
times = 10L,
setup = set.seed(120)
)
Discussion of Benchmarking Results for Matrix Operations Av, ABv, (AB)v, and A(Bv):
Av Operation:
The operation A %*% v represents a straightforward matrix-vector multiplication.
It exhibited the fastest execution time among the four operations, taking approximately 5.02 milliseconds.
This operation involves multiplying matrix A with vector v, which is less computationally complex and faster.
ABv and (AB)v Operations:
Both ABv and (AB)v involve matrix multiplication with A and B followed by a matrix-vector multiplication with v.
ABv took approximately 2182.66 milliseconds, while (AB)v took around 2231.87 milliseconds.
These operations are more computationally intensive due to the additional step of matrix multiplication involving both matrices (A and B).
A(Bv) Operation:
The A(Bv) operation involves a matrix-vector multiplication of matrix B with vector v,
followed by a matrix-vector multiplication with matrix A.
It took about 9.18 milliseconds, which is longer than Av but shorter than ABv and (AB)v.
This operation includes the multiplication of matrix B with vector v before further matrix-vector multiplication with matrix A.
Conclusion:
- Av, being the simplest operation, executed the fastest.
- ABv and (AB)v, involving matrix multiplications with both A and B, took longer due to their computational complexity.
- A(Bv) took longer than Av due to the additional multiplication of B with v before the final matrix-vector multiplication.
###AUFGABE 27 A
X <- hilbert_matrix(6)
H <- X%*% solve(t(X) %*% X) %*% t(X)
###AUFGABE 27 B
eig_H <- eigen(H, symmetric=TRUE)
str(eig_H)
eigenvalues <- eig_H$values
eigenvalues
The command "str(my_eigen)" returns a list consisting of 2 elements.
The first element of the list is a vector containing eigenvalues,
and the second element is a vector containing eigenvectors.
Accessing eigenvalues can be done using the "$" operator.
"my_eigen$values" provides us with all the eigenvalues.
###AUFGABE 27 C
trace <- function(x) sum(diag(x))
trace(H)
sum(eigenvalues)
###AUFGABE 27 D
det(H)
prod(eigenvalues)
###AUFGABE 27 E
#eigenvalues of matrix X
x_eigen <- eigen(X)
x_val <- x_eigen$values
#inverse of matrix X
inv_X <- solve(X)
#eigenvalues of the inverse matrix
xinv_eigen <- eigen(inv_X)
xinv_val <- xinv_eigen$values
#reciprocal of eigenvalues of matrix X
to_check <- 1 / x_val
#comparison
sorted_xinv_val <- sort(xinv_val)
sorted_to_check <- sort(to_check)
print(sorted_xinv_val)
print(sorted_to_check)
###AUFGABE 28 A
#Hilbert matrix for n = 20
hilb20 <- hilbert_matrix(20)
# Explanation of the situation:
# The Cholesky decomposition function (chol()) requires a matrix to be positive definite.
# A symmetric matrix, like the Hilbert matrix, is considered positive definite if all its eigenvalues are positive.
# However, for the Hilbert matrix hilb20 with n = 20, its entries become exceedingly small.
# Upon analyzing the eigenvalues of hilb20, some are significantly smaller than the machine epsilon (.Machine$double.eps).
# These extremely small eigenvalues, below the machine epsilon, cause issues:
# R interprets these values as close to zero or potentially negative, making the matrix non-positive definite.
hilb20_eig <- eigen(hilb20)
hilb20_eig_values <- hilb20_eig$values
# Check eigenvalues smaller or equal to the machine epsilon
hilb20_small_eig <- hilb20_eig_values <= .Machine$double.eps
hilb20_small_eig
#This "issue" occurs with all Hilbert matrices where n>12, and Cholesky decomposition cannot be performed
###AUFGABE 28 B
# Calculate condition numbers for Hilbert matrices of different sizes
cond_num_3x3 <- kappa(hilbert_matrix(3))
cond_num_5x5 <- kappa(hilbert_matrix(5))
cond_num_7x7 <- kappa(hilbert_matrix(7))
cond_num_20x20 <- kappa(hilbert_matrix(20))
# Display the computed condition numbers
print(cond_num_3x3)
print(cond_num_5x5)
print(cond_num_7x7)
print(cond_num_20x20)
#condition numbers for Hilbert matrices of sizes 1x1 to 20x20
n <- 20
conditionHilbert <- rep(NA, n)
for (i in 1:n) {
conditionHilbert[i] <- kappa(hilbert_matrix(i))
}
#condition numbers against the size of Hilbert matrices
plot(1:n, conditionHilbert, pch = 20, col = "green",
xlab = "Size of Hilbert Matrix",
ylab = "Condition number",
log = "y",
main = "Condition Number of Hilbert Matrices")
# Explanation of the plot
# The y-axis is logarithmically scaled to display logarithms of condition numbers in the plot.
# Logarithmic scaling helps visualize linear growth in log-transformed condition numbers,
# indicating exponential growth in the original condition numbers.
# Therefore, as the size of the Hilbert matrix increases, its condition number also increases exponentially.
# A peculiarity is observed after n = 13 where condition numbers almost stabilize and remain close to each other.
# Selecting condition numbers for specific Hilbert matrix sizes
interesting_values <- conditionHilbert[c(3, 5, 7, 20)]
interesting_values
###AUFGABE 29 A
LmNaive <- function(y, X) {
XtX <- t(X) %*% X
# Check if XtX is full rank before inverting
if (qr(XtX)$rank < ncol(XtX)) {
stop("The matrix (X^TX) is not full rank. OLS estimation failed.")
}
beta_hat <- solve(XtX) %*% t(X) %*% y
return(beta_hat)
}
LmNaive <- function(y, X) {
XtX <- t(X) %*% X
XtX_inv <- solve(XtX)
Xty <- t(X) %*% y
beta_hat <- XtX_inv %*% Xty
return(beta_hat)
}
###AUFGABE 29 B
LmCP <- function(y, X) {
# Compute X'X and its inverse using crossprod and solve
XtX_inv <- solve(crossprod(X)) # Compute (X'X)^(-1)
# Compute βˆ using tcrossprod and solve
beta_hat <- solve(tcrossprod(X, y)) %*% XtX_inv %*% tcrossprod(X, y)
return(beta_hat) # Return the estimated coefficients βˆ
}
###AUFGABE 29 c
# LmChol function computes OLS coefficients using Cholesky decomposition
LmChol <- function(y, X) {
# X'X and its Cholesky decomposition
XtX <- t(X) %*% X # Compute X'X
XtX_chol <- chol(XtX) # Cholesky decomposition of X'X
# X'y
Xty <- t(X) %*% y # Calculate X'y
# Solving Lz = X'y using forward substitution
z <- forwardsolve(t(XtX_chol), Xty) # Solve Lz = X'y
# Solving L'β = z using back substitution
beta_hat <- backsolve(XtX_chol, z) # Solve L'β = z
return(beta_hat)
}
###AUFGABE 29 D
LmSvd <- function(y, X){
svd_output <- svd(X)
U <- svd_output[["u"]]
V <- svd_output[["v"]]
sigma <- svd_output[["d"]]
beta <- V %*% diag(1 / sigma) %*% t(U) %*% y
return(beta)
}
###AUFGABE 29 E
#βˆ = R−1QT y
LmQR <- function(y, X){
QR <- qr(X)
Q <- qr.Q(QR)
R <- qr.R(QR)
beta_hat <- backsolve(R, crossprod(Q, y))
return(beta_hat)
}
###AUFGABE 29 F
?lm.fit()