Derivatives

Exercise 1

\[ lambda * exp(-(lambda * x)) \]

Exercise 2

\[ 1/(b - a) \]

Exercise 3

\[ 2 * ((x - a)/((b - a) * (c - a))) \]

Exercise 4

\[ 2 * ((b - x)/((b - a) * (c - a))) \]

Integrals

Exercise 5

\[ 7500 \]

Exercise 6

integration by parts using derivative from above
u = x, dv =

Deriv(function (x) 1 - exp(-1 * lambda * x))
## function (x) 
## lambda * exp(-(lambda * x))

du = dx, v = 1 - exp(-1 * lambda * x)

integral of vdx:

antiD(1-exp(-1 * lambda * x) ~ x)
## function (x, C = 0, lambda) 
## 1 * x - 1/(-1 * lambda) * exp(-1 * lambda * x) + C

subtract that from vu to get final answer:

\[ e^{-\lambda x} * (\frac{1}{-\lambda} - x) \]

Exercise 7

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

\[ (0.5) / (b-a) \]

Exercise 8

integration by parts by multiplying the given equation by \(\frac{1}{(B^\alpha)^2}\)
u = \(x * \frac{1}{(B^\alpha)^2}\), dv = gamma_pdf

Deriv(function (x) x * 1/(B^a)^2)
## function (x) 
## 1/B^(2 * a)

du = \(\frac{1}{B^{2\alpha}}\) dx, v = gamma_cdf [because cdf is the integral of pdf]

vu - \(\int\)vdu =

gamma_cdf \(* x * \frac{1}{(B^\alpha)^2}\) - \(\int \frac{1}{B^{2\alpha}} *\) gamma_cdf dx

Linear Algebra

Exercise 9

\[ X = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 3 & 1 \\ 4 & 6 & 8 \\ \end{bmatrix} \] Augment with \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \] and perform row reduction until left with: \[ \begin{bmatrix} -4.5 & -0.5 & 1.75 \\ 5 & 1 & -2 \\ -1.5 & -0.5 & 0.75 \\ \end{bmatrix} \]

Exercise 10

det(X)
## [1] -4

Exercise 11

We can find U easily through elimination. This results in U= \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -8 \\ 0 & 0 & 1.33 \\ \end{bmatrix} \] And since LU = X that means
\[ (L)(U)(U^{-1})(L^{-1}) = (X)(X^{-1})\] and therefore
\[ (U^{-1})(L^{-1}) = (X^{-1})\]

So we can find L:

U <- rbind(c(1, 2, 3),
           c(0, -3, -8),
           c(0, 0, 1+1/3))
L <- solve(solve(solve(U), solve(X)))

L = \[ \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 4 & 0.67 & 1 \\ \end{bmatrix} \]

Exercise 12

solution12 <- round(X %*% solve(X))

\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \]