0.1 Install Dependent Library Packages

#install.packages("mosaic",repos = "http://cran.us.r-project.org")
#library(mosaic)
install.packages("mosaicCalc",repos = "http://cran.us.r-project.org")

## package 'mosaicCalc' successfully unpacked and MD5 sums checked
## 
## The downloaded binary packages are in
##  C:\Users\Debabrata\AppData\Local\Temp\RtmpQbn5X6\downloaded_packages

library(mosaicCalc)

## Loading required package: mosaicCore

## 
## Attaching package: 'mosaicCalc'

## The following object is masked from 'package:stats':
## 
##     D

1 Derivatives

Find the derivatives with the respect to x of the following.

1.1 (1) F(x|x≥0) = 1−(e^−λx)

f(x)=1−(e^(−λx))
F(x)=f’(x)=df(x)/dx=d(1−(e^(−λx)))/dx=d1/dx-d(e^(−λx))/dx= 0 - ((-λdx/dx)e^(−λx) = λe^(−λx)

fx <- function(x){1-e^-(lambda*x)}
#Deriv(fx)
Fx <- D((1-e^-(l*x))~x)
Fx

## function (x, e, l) 
## e^-(l * x) * (log(e) * l)

1.2 (2) F(x|b>a) = (x−a)/(b−a)

fx=(x−a)/(b−a)
F(x)=f’(x)=df(x)/dx=d((x−a)/(b−a))/dx=(1/(b-a))d(x-a)/dx=1/(b-a)

Fx <- D(((x−a)/(b−a))~x)
Fx

## function (x, a, b) 
## 1/(b - a)

1.3 (3) F(x|a < x ≤ c ≤ b) = (x−a)²/((b−a)(c−a))

fx=(x−a)²/((b−a)(c-a))
F(x)=f’(x)=df(x)/dx = d((x−a)²/((b−a)(c-a)))/dx = (1/((b−a)(c-a)))d((x-a)²)/dx = (1/((b−a)(c-a)))*2(x-a)

Fx <- D(((x−a)^2)/((b−a)*(c−a))~x)
Fx

## function (x, a, b, c) 
## 2 * (x - a)/((b - a) * (c - a))

1.4 (4) F(x|a ≤ c < x < b) = 1−((b−x)²/((b−a)(c−a)))

fx=1−((b−x)²/((b−a)(c−a)))
F(x)=f’(x)=df(x)/dx=d(1−((b−x)²/((b−a)(c−a))))/dx=d1/dx - (1/((b−a)(c-a)))d((b-x)²) = 0 - (2(b-x)/((b−a)(c−a)))d(b-x)/dx = (2(b-x)/((b−a)(c−a)))

Fx <- D(1−((b−x)^2/((b−a)*(c−a)))~x)
Fx

## function (x, b, a, c) 
## 2 * (b - x)/((b - a) * (c - a))

2 Integrals

2.1 (5) ₀∫¹⁰3x³dx

₀∫¹⁰3x³dx = 3₀∫¹⁰x³dx = 3x⁽³⁺¹⁾/(3+1) = 3x⁴/4 + C = 3*10⁴/4 = 7500

integration <- antiD(3*x^3~x, x=10)
integration

## function (x = 10, C = 0) 
## 3/4 * x^4 + C

integration()

## [1] 7500

2.2 (6) ₀∫^xλxe^−λxdx

integration <- antiD(lambda*x*e^-(lambda*x)~x)
integration

## function (x, lambda, e, C = 0) 
## {
##     numerical_integration(.newf, .wrt, as.list(match.call())[-1], 
##         formals(), from, ciName = intC, .tol)
## }
## <environment: 0x000000002a471450>

2.3 (7) ₀∫^0.51/(b−a)dx

0∫^0.51/(b−a)dx = x/(b-a) = 0.5/(b-a)

integration <- antiD((1/(b-a))~x,x=0.5)
integration

## function (x = 0.5, C = 0, b, a) 
## (1/(b - a)) * x + C

2.4 (8) ₀∫^xx(1/(Γ(α)β^α))x^(α−1)e^(−βx)dx

integration <- antiD(x*(1/(Γ(α)*B^α))*x^(α−1)*e^(−Bx)~x)
integration

## function (x, a, B, e, Bx, C = 0) 
## {
##     numerical_integration(.newf, .wrt, as.list(match.call())[-1], 
##         formals(), from, ciName = intC, .tol)
## }
## <environment: 0x000000002aa0fd90>

3 Linear Algebra

X = matrix(c(1,3,4,2,3,6,3,1,8),3,3)
print(X)

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    3    3    1
## [3,]    4    6    8

3.1 (9) Invert it using Gaussian row reduction.

I = matrix(c(1,0,0,0,1,0,0,0,1),3,3)
print(I)

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

XI <- cbind(X,I)
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    1    0    0
## [2,]    3    3    1    0    1    0
## [3,]    4    6    8    0    0    1

XI[2,] <- XI[2,] - 3*XI[1,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    1    0    0
## [2,]    0   -3   -8   -3    1    0
## [3,]    4    6    8    0    0    1

XI[3,] <- XI[3,] - 4*XI[1,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    1    0    0
## [2,]    0   -3   -8   -3    1    0
## [3,]    0   -2   -4   -4    0    1

XI[2,] <- XI[2,] - XI[3,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    1    0    0
## [2,]    0   -1   -4    1    1   -1
## [3,]    0   -2   -4   -4    0    1

XI[3,] <- XI[3,] - 2*XI[2,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    1    0    0
## [2,]    0   -1   -4    1    1   -1
## [3,]    0    0    4   -6   -2    3

XI[2,] <- XI[2,] + XI[3,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    1    0    0
## [2,]    0   -1    0   -5   -1    2
## [3,]    0    0    4   -6   -2    3

XI[3,] <- XI[3,] / 4
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3  1.0  0.0 0.00
## [2,]    0   -1    0 -5.0 -1.0 2.00
## [3,]    0    0    1 -1.5 -0.5 0.75

XI[1,] <- XI[1,] - 3*XI[3,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5]  [,6]
## [1,]    1    2    0  5.5  1.5 -2.25
## [2,]    0   -1    0 -5.0 -1.0  2.00
## [3,]    0    0    1 -1.5 -0.5  0.75

XI[1,] <- XI[1,] + 2*XI[2,]
print(XI)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    0    0 -4.5 -0.5 1.75
## [2,]    0   -1    0 -5.0 -1.0 2.00
## [3,]    0    0    1 -1.5 -0.5 0.75

XI[2,] <- XI[2,] * (-1)
print(XI)

##      [,1] [,2] [,3] [,4] [,5]  [,6]
## [1,]    1    0    0 -4.5 -0.5  1.75
## [2,]    0    1    0  5.0  1.0 -2.00
## [3,]    0    0    1 -1.5 -0.5  0.75

Xi=XI[1:3,4:6]
print(Xi)

##      [,1] [,2]  [,3]
## [1,] -4.5 -0.5  1.75
## [2,]  5.0  1.0 -2.00
## [3,] -1.5 -0.5  0.75

solve(X)

##      [,1] [,2]  [,3]
## [1,] -4.5 -0.5  1.75
## [2,]  5.0  1.0 -2.00
## [3,] -1.5 -0.5  0.75

3.2 (10) Find the determinant.

Xd=(X[1,1]*(X[2,2]*X[3,3]-X[2,3]*X[3,2]))+(X[1,2]*(X[2,3]*X[3,1]-X[2,1]*X[3,3]))+(X[1,3]*(X[2,1]*X[3,2]-X[2,2]*X[3,1]))
print(Xd)

## [1] -4

det(X)

## [1] -4

3.3 (11) Conduct LU decomposition

U=X
print(U)

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    3    3    1
## [3,]    4    6    8

#U[2,] <- -(U[2,1]/U[1,1])*U[1,] + U[2,]
U[2,] <- -3*U[1,] + U[2,]
print(U)

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0   -3   -8
## [3,]    4    6    8

#U[3,] <- -(U[3,1]/U[1,1])*U[1,] + U[3,]
U[3,] <- -4*U[1,] + U[3,]
print(U)

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0   -3   -8
## [3,]    0   -2   -4

#U[3,] <- -(U[3,2]/U[2,2])*U[2,] + U[3,]
U[3,] <- -(2/3)*U[2,] + U[3,]
print(U)

##      [,1] [,2]      [,3]
## [1,]    1    2  3.000000
## [2,]    0   -3 -8.000000
## [3,]    0    0  1.333333

L=round(X%*%solve(U),2) # LU=X, hence L=X(Ui)
print(L)

##      [,1] [,2] [,3]
## [1,]    1 0.00    0
## [2,]    3 1.00    0
## [3,]    4 0.67    1

3.4 (12) Multiply the matrix by it’s inverse.

I=X%*%Xi
print(I)

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

CUNY Assignment : Week 1 [Dec 18-Dec 24] Data Science Math Bridge Week 1 Assignment

Debabrata Kabiraj

December 29, 2018

0.1 Install Dependent Library Packages

1 Derivatives

1.1 (1) F(x|x≥0) = 1−(e^−λx)

1.2 (2) F(x|b>a) = (x−a)/(b−a)

1.3 (3) F(x|a < x ≤ c ≤ b) = (x−a)²/((b−a)(c−a))

1.4 (4) F(x|a ≤ c < x < b) = 1−((b−x)²/((b−a)(c−a)))

2 Integrals

2.1 (5) ₀∫¹⁰3x³dx

2.2 (6) ₀∫^xλxe^−λxdx

2.3 (7) ₀∫^0.51/(b−a)dx

2.4 (8) ₀∫^xx(1/(Γ(α)β^α))x^(α−1)e^(−βx)dx

3 Linear Algebra

3.1 (9) Invert it using Gaussian row reduction.

3.2 (10) Find the determinant.

3.3 (11) Conduct LU decomposition

3.4 (12) Multiply the matrix by it’s inverse.

CUNY Assignment : Week 1 [Dec 18-Dec 24] Data Science Math Bridge Week 1 Assignment

Debabrata Kabiraj

December 29, 2018

0.1 Install Dependent Library Packages

1 Derivatives

1.1 (1) F(x|x≥0) = 1−(e−λx)

1.2 (2) F(x|b>a) = (x−a)/(b−a)

1.3 (3) F(x|a < x ≤ c ≤ b) = (x−a)2/((b−a)(c−a))

1.4 (4) F(x|a ≤ c < x < b) = 1−((b−x)2/((b−a)(c−a)))

2 Integrals

2.1 (5) 0∫103x3dx

2.2 (6) 0∫xλxe−λxdx

2.3 (7) 0∫0.51/(b−a)dx

2.4 (8) 0∫xx(1/(Γ(α)βα))x(α−1)e(−βx)dx

3 Linear Algebra

3.1 (9) Invert it using Gaussian row reduction.

3.2 (10) Find the determinant.

3.3 (11) Conduct LU decomposition

3.4 (12) Multiply the matrix by it’s inverse.

1.1 (1) F(x|x≥0) = 1−(e^−λx)

1.3 (3) F(x|a < x ≤ c ≤ b) = (x−a)²/((b−a)(c−a))

1.4 (4) F(x|a ≤ c < x < b) = 1−((b−x)²/((b−a)(c−a)))

2.1 (5) ₀∫¹⁰3x³dx

2.2 (6) ₀∫^xλxe^−λxdx

2.3 (7) ₀∫^0.51/(b−a)dx

2.4 (8) ₀∫^xx(1/(Γ(α)β^α))x^(α−1)e^(−βx)dx