Derivatives

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

\[F(x|x >= 0) = 1 - e^-\lambda x\]

library(mosaicCalc)
library(mosaic)

f001<-makeFun(1-exp(-x)~x)
Fn001<-antiD(f001(x)~x)
plotFun(f001(x)~x, x.lim=range(0, 20))

\[F(x|b > a) = x-a/b-a\]

f002 <- makeFun(((x-a)/(b-a)~x))
Fn002=antiD(f002(x)~x)
plotFun(f002(x, a=2, b=5)~x, x.lim=range(0, 20))

\[F(x|a < x <= c <= b) = (x - a)^2/(b-a)(c-a)\]

f003 <- makeFun(((x-a)^2)/((b-a)(c-a))~x)
Fn003=antiD(f003(x)~x)

\[F(x|a <= c < x < b)=1-(b-x)2(b-a)(c-a)\]

Integrals

Solve the following definite and indefinite integrals

\[\int_0^{10} 3x^3 dx\]

fx0001=function(x){3 * x^3}
integrate(Vectorize(fx0001),0,10)
## 7500 with absolute error < 8.3e-11

\[\int_0^{10} xe^xx dx\]

fx0002=function(x){x * exp(x)^x * x}
integrate(Vectorize(fx0002),0,10)
## 1.337304e+44 with absolute error < 2.1e+39

\[\int_0^{5} \frac{1}{b-a} dx\]

#fx0003=function(x){1 / (x-a)}
#integrate(Vectorize(fx0003),0,5)

\[\int_0^{10} x \frac{1}{r( \alpha )\beta \alpha} x^ \alpha -1 e^ -\beta x dx\]

#fx0004=function(x){(1 (r(a)b) * x^(a-1) * exp(x))}
#integrate(Vectorize(fx0004),0,10)

Linear Algebra

With the following matrix,

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 3 & 3 & 1 \\ 4 & 6 & 8 \end{array}\right] \]

Invert it using Gaussian row reduction.

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 3 & 3 & 1 \\ 4 & 6 & 8 \end{array}\right] \] \[\mathbf{I} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

-3R1+R2->R2
-4R1+R3->R3

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & -3 & -8 \\ 0 & -2 & -4 \end{array}\right] \]

\[\mathbf{I} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -4 & 0 & 1 \end{array}\right] \]

1/-3R2->R2 \[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 1 & 2.7 \\ 0 & -2 & -4 \end{array}\right] \]

\[\mathbf{I} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ 1 & 0 & 0 \\ -4 & 0 & 1 \end{array}\right] \]

-2R2+R1->R1 2R2+R3->R3 \[\mathbf{X} = \left[\begin{array} {rrr} 1 & 0 & -2.3 \\ 0 & 1 & 2.7 \\ 0 & 0 & 1.3 \end{array}\right] \]

\[\mathbf{I} = \left[\begin{array} {rrr} -1 & 0.7 & 0 \\ 1 & 0 & 0 \\ -2 & -1 & 1 \end{array}\right] \]

0.769026 x R3->R3 \[\mathbf{X} = \left[\begin{array} {rrr} 1 & 0 & -2.3 \\ 0 & 1 & 2.7 \\ 0 & 0 & 1 \end{array}\right] \]

\[\mathbf{I} = \left[\begin{array} {rrr} -1 & 0.7 & 0 \\ 1 & 0 & 0 \\ -2 & -1 & 0.8 \end{array}\right] \]

2.3 x R3 + R1->R1 -2.6 x R3 + R2->R2 \[\mathbf{X} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

\[\mathbf{I} = \left[\begin{array} {rrr} -5 & -1 & 1.8 \\ 5 & 1 & -2 \\ -2 & -1 & 0.8 \end{array}\right] \]

Find the determinant.

mat001<-matrix(c(1,3,4,2,3,6,3,1,8),3,3)
print(mat001)
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    3    3    1
## [3,]    4    6    8
det001 <- det(mat001)
print(det001)
## [1] -4

Conduct LU decomposition

U

-3R1 + R2

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & -3 & -8 \\ 4 & 6 & 8 \end{array}\right] \]

-4R1 + R3

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & -3 & -8 \\ 0 & -2 & -4 \end{array}\right] \]

-0.666666666667R2 + R3

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & -3 & -8 \\ 0 & 0 & 1.3 \end{array}\right] \]

L

\[\mathbf{X} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 4 & 0.666666666667 & 1 \end{array}\right] \]

Multiply the matrix by it’s inverse.

mat001<-matrix(c(1,3,4,2,3,6,3,1,8),3,3)
print(mat001)
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    3    3    1
## [3,]    4    6    8
#Inverse
invMatrix001 <- solve(mat001)
print(invMatrix001)
##      [,1] [,2]  [,3]
## [1,] -4.5 -0.5  1.75
## [2,]  5.0  1.0 -2.00
## [3,] -1.5 -0.5  0.75
#Multiply Matrix by its Incerse (Correct Answer)
print(mat001 %*% invMatrix001)
##               [,1]          [,2]         [,3]
## [1,]  1.000000e+00 -2.220446e-16 4.440892e-16
## [2,] -4.440892e-16  1.000000e+00 2.220446e-16
## [3,]  0.000000e+00  0.000000e+00 1.000000e+00
#Casewise Multiplication/ Vectorization
print(mat001 * invMatrix001)
##      [,1] [,2]  [,3]
## [1,] -4.5   -1  5.25
## [2,] 15.0    3 -2.00
## [3,] -6.0   -3  6.00