Notes, summaries, assignments, exams, and problems for Mathematics

Functions and Objectives

(a) f(z) = |z|, where z is a complex number.

(b) f(x, y) = x²y / (x² + y²)

Proof Objective

• (a) Prove that f(z) = |z| is continuous everywhere but nowhere differentiable except at the origin.
• (b) Find the iterative limit and simultaneous limit of f(x, y) = x²y / (x² + y²) as (x, y) → (0, 0).

Proof Process

(a) Continuity of f(z) = |z|

[Step 1]: Show that f(z) = |z| is continuous everywhere.

Let z₀ be an arbitrary complex number. We want to show that for any ε > 0, there exists a δ > 0 such that if |z - z₀| < δ, then |f(z) - f(z₀)| < ε.

We have f(z) = |z| and f(z₀) = |z₀|. Then |f(z) - f(z₀)| = ||z| - |z₀||.

By the reverse triangle inequality, we know that ||z| - |z₀|| ≤ |z - z₀|.

So, if... Continue reading "Complex Analysis: Continuity, Differentiability, and Limits" »

Matrices (Question 6a)

Verify that A · (\text{adj } A) = (\text{adj } A) · A = |A| · I_3 for
A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 0 & 1 \\ 2 & 1 & 5 \end{bmatrix}.

Tasks:

• Find the determinant |A|:
• Find the Adjoint (\text{adj } A): This involves finding the cofactor of each element and then transposing the resulting matrix.
• Cofactors: C₁₁ = -1, C₁₂ = -13, C₁₃ = 3, C₂₁ = -11, C₂₂ = 2, C₂₃ = 4, C₃₁ = 3, C₃₂ = 10, C₃₃ = -9
• Multiply A · (\text{adj } A)

4. Financial Arithmetic (Question 2g)

Find the compound interest on Rs. 8,000 for 1 1/2 years at 10% per annum, compounded annually.

• Amount for the first year:
• Interest for the next half year: Use simple interest on the new principal.
• Total Compound Interest:

Answers use standard... Continue reading "Matrix Determinant and Adjoint Verification with AP/GP and CI" »

Calculating the Future Value of an Annuity Due

Step 1: Determine the Variables

The problem provides the following details:

• Annual payment: Rs. 200. Therefore, the half-yearly payment (Pmt) is:
  
  Rs. 200 / 2 = Rs. 100
  Rs. 200 / 2 = Rs. 100
• Annual interest rate (r): 4% or 0.04. Since the interest is compounded half-yearly, the interest rate per period (i) is:
  
  0.04 / 2 = 0.02
  0.04 / 2 = 0.02
• Term: 20 years. Payments are made half-yearly, so the total number of periods (n) is:
  
  20 × 2 = 40
  20 × 2 = 40
• The annuity type is an annuity due, meaning payments are made at the beginning of each period.

Step 2: Apply the Future Value Formula

The formula for the Future Value (FV) of an annuity due is given by:

FV = Pmt × [((1 + i)^n - 1) / i] × (1 + i)

FV = Pmt × [((1

Solving Polynomials Using the Remainder Theorem

We are given the function:

f(x) = mx³ − 3x² + nx + 2

• When divided by (x + 3), the remainder is −3.
• When divided by (x − 2), the remainder is −4.

By the Remainder Theorem, if f(x) is divided by (x − a), the remainder is f(a).

Step 1: Substitute x = −3

f(−3) = m(−3)³ − 3(−3)² + n(−3) + 2

= −27m − 27 − 3n + 2

= −27m − 3n − 25

Given remainder −3:

−27m − 3n − 25 = −3

−27m − 3n = 22

27m + 3n = −22 (Equation 1)

Step 2: Substitute x = 2

f(2) = m(2)³ − 3(2)² + n(2) + 2

= 8m − 12 + 2n + 2

= 8m + 2n − 10

Given remainder −4:

8m + 2n − 10 = −4

8m + 2n = 6 (Equation 2)

Step 3: Solve the System of Equations

Multiply Equation 2 by 3 → 24m + 6n = 18

Multiply Equation... Continue reading "Solving Polynomial Remainder Theorem and Transformations" »

Customer Value: (Quality, Time, Flexibility, Customer Experience, Innovation) / Price

Sustainability Paradigms: Economic (Viable, Equitable), Environment (Viable, Bearable), Social (Bearable, Equitable)

- Efficiency (Optimization), Differentiator (Innovation), Driver (Motivation) | - Audits (Assess Sustainability Performance)

Table 7.3: Factors for Calculating Three-Sigma Limits for the X (Bar) Chart and R-Chart

*A sample is out of control if its value falls below the LCL or above the UCL*

lim x->0 sinx/x = 1 | H.A.: compare degrees | V.A.: denom = 0 | Continuous if: f(a), lim, equal

DERIVATIVES: (x^n)'=nx^(n-1), (e^x)'=e^x, (a^x)'=a^x ln a, (lnx)'=1/x | (uv)'=u'v+uv', (u/v)'=(u'v-uv')/v^2 |chain: (f(g(x)))'=f'(g(x))g'(x)

TRIG DERIVATIVES: (sin)'=cos, (cos)'=-sin, (tan)'=sec^2 | (sec)'=sec·tan, (csc)'=-csc·cot, (cot)'=-csc^2

CRITICAL POINTS:f'=0 or DNE ⇒ crit pt | f'>0 inc | f'<0 dec | f''>0 conc up | f''<0 conc down | inflec = f'' signchange

INTEGRATION: ∫x^n dx = x^(n+1)/(n+1)+C | ∫e^x dx = e^x+C | ∫a^x dx = a^x/ln a+C | ∫1/x dx = ln|x|+C | ∫sin x dx

= -cos x+C | ∫cos x dx = sin x+C

FTC: Part 1: d/dx ∫_a^x f(t) dt = f(x) | Part 2: ∫_a^b f(x) dx = F(b)-F(a)

AREA & VOLUME: A = ∫_a^b (top - bot)... Continue reading "5 hr 20 min 20 sec corresponds to a longitude difference of" »

Section A: R Basics and Data Types (Weeks 1-4)

Model Questions and Answers

Q: Create a vector v (3, NA, Inf, -Inf). Explain adding 5 to v.

A: The operation is element-wise. Missing values (NA) propagate, resulting in NA. Infinite values (Inf, -Inf) remain infinite.

v <- c(3, NA, Inf, -Inf)
print(v + 5)
# Output: [1]  8 NA Inf -Inf

Q: Given vector a, write code to count and replace NAs.

A: Assuming a <- c(10, 15, NA, 20).

• Count NAs: sum(is.na(a)) → 1.
• Replace NAs (e.g., with 0): a[is.na(a)] <- 0.

Q: Explain the difference between a[a > 12] and a[which(a > 12)].

A: Both select elements greater than 12 (15, 20). However:

• a[a > 12] → [1] 15 NA (Uses logical indexing; preserves the position of NA in the original vector as NA).
• a[which(

Two-State Actuarial Model

The Two-State Model (also known as the Dead-Alive or Binary Model) is a fundamental actuarial framework used to represent processes that exist in one of two possible states, such as Alive/Dead or Working/Retired. It is widely utilized in life insurance and pension modeling to estimate the probability of transition between states. The model assumes that at any given time, an individual occupies only one state, allowing actuaries to calculate premiums, reserves, and expected present values by simplifying complex uncertainties into binary outcomes.

Core Assumptions

• Binary States: The system exists in only one of two states at any time.
• Markov Property: Transitions depend solely on the current state.
• Constant Probabilities: