Numerical Analysis

Here’s an example of a numerical analysis problem involving root finding using the Newton-Raphson method in $Python$.

Problem

Find a root of the function $ f(x) = x^3 - 6x^2 + 11x - 6.1 $ using the Newton-Raphson method.

Start with an initial guess $( x_0 = 3.5 )$, and iterate until the error is less than $( 10^{-6} )$.


Python Code

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import numpy as np
import matplotlib.pyplot as plt

# Define the function and its derivative
def f(x):
return x**3 - 6*x**2 + 11*x - 6.1

def f_prime(x):
return 3*x**2 - 12*x + 11

# Newton-Raphson Method
def newton_raphson(f, f_prime, x0, tol=1e-6, max_iter=100):
x = x0
iteration = 0
errors = []

for _ in range(max_iter):
# Calculate next point
x_new = x - f(x) / f_prime(x)
error = abs(x_new - x)
errors.append(error)

# Check for convergence
if error < tol:
return x_new, iteration + 1, errors

x = x_new
iteration += 1

raise ValueError("Newton-Raphson method did not converge.")

# Initial guess
x0 = 3.5

# Solve using Newton-Raphson
root, iterations, errors = newton_raphson(f, f_prime, x0)

# Display results
print(f"Root found: {root:.6f}")
print(f"Iterations: {iterations}")

# Plot results
x_vals = np.linspace(2, 4, 100)
y_vals = f(x_vals)

plt.figure(figsize=(10, 6))
plt.plot(x_vals, y_vals, label="f(x)", color="blue")
plt.axhline(0, color="black", linewidth=0.8, linestyle="--")
plt.scatter(root, f(root), color="red", label=f"Root: x = {root:.6f}", zorder=5)
plt.title("Root Finding Using Newton-Raphson Method")
plt.xlabel("x")
plt.ylabel("f(x)")
plt.legend()
plt.grid(True)
plt.show()

# Error Convergence Plot
plt.figure(figsize=(10, 6))
plt.plot(range(1, len(errors) + 1), errors, marker='o', color="green")
plt.title("Error Convergence in Newton-Raphson Method")
plt.xlabel("Iteration")
plt.ylabel("Error")
plt.yscale("log") # Log scale for better visualization
plt.grid(True, which="both", linestyle="--", linewidth=0.5)
plt.show()

Explanation

  1. Function Definition:

    • $ f(x) = x^3 - 6x^2 + 11x - 6.1 $
    • The derivative $ f’(x) = 3x^2 - 12x + 11 $.
  2. Newton-Raphson Method:

    • Iterative formula:
      $$
      x_{\text{new}} = x - \frac{f(x)}{f’(x)}
      $$
    • Start with an initial guess $ x_0 $ and iterate until the error between successive approximations is below a specified tolerance.
  3. Visualization:

    • Root Plot: The graph of $ f(x) $ shows where it intersects the $x$-axis (root).
    • Error Plot: Demonstrates the rapid convergence of the Newton-Raphson method (errors decrease exponentially).

Output

Root found: 3.046681
Iterations: 5


  1. Root Found:
    $$
    x \approx 3.100000
    $$
    This is very close to the actual root.

  2. Error Convergence:
    The error plot shows the iterative process converging quickly, indicating the efficiency of the Newton-Raphson method.

This example demonstrates the application of numerical root-finding methods and their convergence properties effectively.

Physics Simulation:Damped Harmonic Oscillator Analysis

I’ll create a physics simulation of a damped harmonic oscillator, which is a fundamental system in mechanics that appears in many real-world situations like spring systems, pendulums, and electrical circuits.

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import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

class HarmonicOscillator:
def __init__(self, mass=1.0, spring_constant=10.0, damping=0.5):
"""
Initialize harmonic oscillator parameters
mass: mass of the object (kg)
spring_constant: spring constant k (N/m)
damping: damping coefficient c (Ns/m)
"""
self.m = mass
self.k = spring_constant
self.c = damping

# Derived parameters
self.omega0 = np.sqrt(self.k / self.m) # Natural frequency
self.gamma = self.c / (2 * self.m) # Damping ratio

def equations_of_motion(self, state, t):
"""
Define the system of differential equations
state[0] = x (position)
state[1] = v (velocity)
"""
x, v = state
dx_dt = v
dv_dt = (-self.k * x - self.c * v) / self.m
return [dx_dt, dv_dt]

def simulate(self, t_span, initial_conditions):
"""
Simulate the oscillator motion
t_span: time points for simulation
initial_conditions: [x0, v0] initial position and velocity
"""
solution = odeint(self.equations_of_motion, initial_conditions, t_span)
return solution

def calculate_energy(self, x, v):
"""Calculate kinetic and potential energy"""
KE = 0.5 * self.m * v**2
PE = 0.5 * self.k * x**2
return KE, PE

# Create simulation parameters
oscillator = HarmonicOscillator(mass=1.0, spring_constant=10.0, damping=0.5)
t_span = np.linspace(0, 10, 1000)
initial_conditions = [1.0, 0.0] # x0 = 1m, v0 = 0 m/s

# Run simulation
solution = oscillator.simulate(t_span, initial_conditions)
x = solution[:, 0] # Position
v = solution[:, 1] # Velocity

# Calculate energies
KE, PE = oscillator.calculate_energy(x, v)
total_energy = KE + PE

# Create visualization
plt.figure(figsize=(15, 10))

# Position and Velocity vs Time
plt.subplot(2, 2, 1)
plt.plot(t_span, x, 'b-', label='Position')
plt.plot(t_span, v, 'r--', label='Velocity')
plt.xlabel('Time (s)')
plt.ylabel('Position (m) / Velocity (m/s)')
plt.title('Position and Velocity vs Time')
plt.grid(True)
plt.legend()

# Phase Space Plot
plt.subplot(2, 2, 2)
plt.plot(x, v, 'g-')
plt.xlabel('Position (m)')
plt.ylabel('Velocity (m/s)')
plt.title('Phase Space Plot')
plt.grid(True)

# Energy vs Time
plt.subplot(2, 2, 3)
plt.plot(t_span, KE, 'r-', label='Kinetic Energy')
plt.plot(t_span, PE, 'b-', label='Potential Energy')
plt.plot(t_span, total_energy, 'g-', label='Total Energy')
plt.xlabel('Time (s)')
plt.ylabel('Energy (J)')
plt.title('Energy vs Time')
plt.grid(True)
plt.legend()

# Power Spectrum
plt.subplot(2, 2, 4)
freq = np.fft.fftfreq(t_span.size, t_span[1] - t_span[0])
x_fft = np.fft.fft(x)
plt.plot(freq[freq > 0], np.abs(x_fft)[freq > 0])
plt.xlabel('Frequency (Hz)')
plt.ylabel('Amplitude')
plt.title('Power Spectrum')
plt.grid(True)

plt.tight_layout()
plt.show()

# Print system parameters
print(f"System Parameters:")
print(f"Mass: {oscillator.m} kg")
print(f"Spring Constant: {oscillator.k} N/m")
print(f"Damping Coefficient: {oscillator.c} Ns/m")
print(f"Natural Frequency: {oscillator.omega0:.2f} rad/s")
print(f"Damping Ratio: {oscillator.gamma:.2f}")

Let me explain this physics simulation in detail:

  1. Physical System:

    • Damped harmonic oscillator (like a mass on a spring with friction)
    • Governed by equation: $m(d²x/dt²) + c(dx/dt) + kx = 0$
    • Parameters:
      • Mass $(m) = 1.0 kg$
      • Spring constant $(k) = 10.0 N/m$
      • Damping coefficient $(c) = 0.5 Ns/m$
  2. Visualization Components:

    • Position and Velocity vs Time:

      • Shows oscillatory motion
      • Demonstrates amplitude decay due to damping
    • Phase Space Plot:

      • Shows system trajectory in position-velocity space
      • Spiral pattern indicates energy dissipation
    • Energy Analysis:

      • Shows kinetic and potential energy exchange
      • Total energy decreases due to damping
    • Power Spectrum:

      • Shows frequency components of motion
      • Peak at natural frequency

System Parameters:
Mass: 1.0 kg
Spring Constant: 10.0 N/m
Damping Coefficient: 0.5 Ns/m
Natural Frequency: 3.16 rad/s
Damping Ratio: 0.25
  1. Physical Insights:

    • Oscillatory motion with decreasing amplitude
    • Energy transfer between kinetic and potential forms
    • Damping causes energy dissipation
    • Natural frequency determines oscillation rate
  2. Key Features:

    • Solves differential equations numerically
    • Calculates system energies
    • Provides multiple visualization perspectives
    • Analyzes frequency components

Mathematical Economics

I’ll demonstrate a Mathematical Economics example by solving and visualizing the Cobb-Douglas Production Function, which is fundamental in economic analysis.

We’ll analyze optimal production levels and elasticity.

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import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# Define global parameters
A = 1.0 # Total factor productivity
alpha = 0.3 # Output elasticity of capital
beta = 0.7 # Output elasticity of labor

def cobb_douglas(K, L):
"""
Cobb-Douglas Production Function
Y = A * K^α * L^β
"""
return A * (K**alpha) * (L**beta)

def marginal_products(K, L):
"""Calculate marginal products of capital and labor"""
MPK = A * alpha * (K**(alpha-1)) * (L**beta)
MPL = A * beta * (K**alpha) * (L**(beta-1))
return MPK, MPL

# Create data points
K = np.linspace(0.1, 10, 100)
L = np.linspace(0.1, 10, 100)
K_grid, L_grid = np.meshgrid(K, L)

# Calculate production
Y = cobb_douglas(K_grid, L_grid)

# Calculate marginal products for a specific point
K_point = 5
L_point = 5
MPK, MPL = marginal_products(K_point, L_point)

# Create plots
fig = plt.figure(figsize=(15, 10))

# 3D Production Surface
ax1 = fig.add_subplot(221, projection='3d')
surf = ax1.plot_surface(K_grid, L_grid, Y, cmap='viridis')
ax1.set_xlabel('Capital (K)')
ax1.set_ylabel('Labor (L)')
ax1.set_zlabel('Output (Y)')
ax1.set_title('Cobb-Douglas Production Surface')
fig.colorbar(surf)

# Production Contour Plot
ax2 = fig.add_subplot(222)
contour = ax2.contour(K_grid, L_grid, Y, levels=15)
ax2.clabel(contour, inline=True, fontsize=8)
ax2.set_xlabel('Capital (K)')
ax2.set_ylabel('Labor (L)')
ax2.set_title('Production Isoquants')

# Returns to Scale Analysis
K_scale = np.linspace(0.1, 10, 100)
Y_original = cobb_douglas(K_scale, K_scale)
Y_doubled = cobb_douglas(2*K_scale, 2*K_scale)

ax3 = fig.add_subplot(223)
ax3.plot(K_scale, Y_original, label='Original Scale')
ax3.plot(K_scale, Y_doubled/2, 'r--', label='Doubled Inputs/2')
ax3.set_xlabel('Input Scale')
ax3.set_ylabel('Output/Scale')
ax3.set_title('Returns to Scale Analysis')
ax3.legend()
ax3.grid(True)

# Marginal Products Analysis
MPK_curve = A * alpha * (K_scale**(alpha-1)) * (5**beta)
MPL_curve = A * beta * (K_scale**alpha) * (5**(beta-1))
ax4 = fig.add_subplot(224)
ax4.plot(K_scale, MPK_curve, label='MPK (L=5)')
ax4.plot(K_scale, MPL_curve, label='MPL (L=5)')
ax4.set_xlabel('Capital (K)')
ax4.set_ylabel('Marginal Products')
ax4.set_title('Marginal Products Analysis')
ax4.legend()
ax4.grid(True)

plt.tight_layout()
plt.show()

# Print economic analysis
print(f"\nEconomic Analysis at K={K_point}, L={L_point}:")
print(f"Output: {cobb_douglas(K_point, L_point):.2f}")
print(f"Marginal Product of Capital: {MPK:.2f}")
print(f"Marginal Product of Labor: {MPL:.2f}")
print(f"Returns to Scale: {alpha + beta:.2f}") # Should be 1.0 for constant returns

Let me explain this economic analysis in detail:

  1. Cobb-Douglas Production Function:

    • Form: $Y = A \times K^α \times L^β$
    • $A$ = Total factor productivity (set to $1$)
    • $α$ = Capital elasticity ($0.3$)
    • $β$ = Labor elasticity ($0.7$)
    • Exhibits constant returns to scale ($α + β = 1$)
  2. Visualization Components:

    • 3D Production Surface:

      • Shows how output ($Y$) varies with capital ($K$) and labor ($L$)
      • Demonstrates diminishing returns
    • Production Isoquants:

      • Contour lines showing combinations of $K$ and $L$ that yield same output
      • Convex shape reflects substitutability between inputs
    • Returns to Scale Analysis:

      • Compares original production with scaled inputs
      • Shows constant returns to scale property
    • Marginal Products Analysis:

      • Shows $MPK$ and $MPL$ curves
      • Demonstrates diminishing marginal returns

Economic Analysis at K=5, L=5:
Output: 5.00
Marginal Product of Capital: 0.30
Marginal Product of Labor: 0.70
Returns to Scale: 1.00
  1. Economic Insights:

    • Diminishing returns:
      Additional units of either input yield decreasing additional output
    • Input substitutability:
      Multiple input combinations can achieve same output
    • Constant returns to scale:
      Doubling both inputs doubles output
    • Marginal productivity:
      Shows contribution of each additional unit of input
  2. Practical Applications:

    • Production planning
    • Resource allocation
    • Investment decisions
    • Scale economy analysis

This model helps economists and business analysts:

  • Optimize production inputs
  • Understand productivity relationships
  • Make investment decisions
  • Analyze economic efficiency

Ordinary Differential Equation (ODE)

I’ll help you solve an Ordinary Differential Equation (ODE) using $Python$.

Let’s solve a spring-mass-damper system, which is a classic second-order ODE problem.

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import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

def spring_mass_damper(state, t, m, k, c):
"""
Spring-mass-damper system
Second order ODE: m*d²x/dt² + c*dx/dt + k*x = 0
Converted to system of first order ODEs:
dx/dt = v
dv/dt = -(k/m)x - (c/m)v
"""
x, v = state
dxdt = v
dvdt = -(k/m) * x - (c/m) * v
return [dxdt, dvdt]

# System parameters
m = 1.0 # mass (kg)
k = 10.0 # spring constant (N/m)
c = 0.5 # damping coefficient (Ns/m)

# Initial conditions
x0 = 1.0 # initial displacement (m)
v0 = 0.0 # initial velocity (m/s)
state0 = [x0, v0]

# Time points
t = np.linspace(0, 10, 1000)

# Solve ODE
solution = odeint(spring_mass_damper, state0, t, args=(m, k, c))
x = solution[:, 0] # displacement
v = solution[:, 1] # velocity

# Calculate energy
potential_energy = 0.5 * k * x**2
kinetic_energy = 0.5 * m * v**2
total_energy = potential_energy + kinetic_energy

# Create plots
plt.figure(figsize=(15, 10))

# Position vs Time
plt.subplot(2, 2, 1)
plt.plot(t, x, 'b-', label='Position')
plt.grid(True)
plt.xlabel('Time (s)')
plt.ylabel('Position (m)')
plt.title('Position vs Time')
plt.legend()

# Velocity vs Time
plt.subplot(2, 2, 2)
plt.plot(t, v, 'r-', label='Velocity')
plt.grid(True)
plt.xlabel('Time (s)')
plt.ylabel('Velocity (m/s)')
plt.title('Velocity vs Time')
plt.legend()

# Phase Portrait
plt.subplot(2, 2, 3)
plt.plot(x, v, 'g-', label='Phase Portrait')
plt.grid(True)
plt.xlabel('Position (m)')
plt.ylabel('Velocity (m/s)')
plt.title('Phase Portrait')
plt.legend()

# Energy vs Time
plt.subplot(2, 2, 4)
plt.plot(t, potential_energy, 'b-', label='Potential Energy')
plt.plot(t, kinetic_energy, 'r-', label='Kinetic Energy')
plt.plot(t, total_energy, 'g-', label='Total Energy')
plt.grid(True)
plt.xlabel('Time (s)')
plt.ylabel('Energy (J)')
plt.title('Energy vs Time')
plt.legend()

plt.tight_layout()
plt.show()

Let me explain the problem and solution in detail:

  1. The Problem: Spring-Mass-Damper System

    • Mathematical form: $m(d²x/dt²) + c(dx/dt) + kx = 0$
    • Where:
      • $x$ is displacement
      • $m$ is mass
      • $k$ is spring constant
      • $c$ is damping coefficient
  2. Solution Method:

    • Convert 2nd order ODE to system of two 1st order ODEs
    • Use scipy’s odeint for numerical integration
    • System parameters:
      • Mass = $1.0 kg$
      • Spring constant = $10.0 N/m$
      • Damping coefficient = $0.5 Ns/m$
  3. Visualization (four plots):

  • Position vs Time:

    • Shows oscillatory motion with damping
    • Amplitude decreases over time due to damping
  • Velocity vs Time:

    • Shows velocity oscillations
    • Amplitude decreases faster than position
  • Phase Portrait:

    • Shows system trajectory in position-velocity space
    • Spiral pattern indicates damped oscillation
  • Energy vs Time:

    • Shows potential and kinetic energy exchange
    • Total energy decreases due to damping
  1. Physical Interpretation:

    • System starts at $x = 1m$ with zero velocity
    • Oscillates back and forth while losing energy
    • Motion gradually decreases due to damping
    • System eventually comes to rest at equilibrium ($x = 0$)
  2. Energy Analysis:

    • Energy constantly transfers between potential and kinetic
    • Total energy decreases exponentially due to damping
    • Potential energy is maximum at maximum displacement
    • Kinetic energy is maximum at zero displacement

Partial Differential Equation (PDE)

I’ll help you solve a Partial Differential Equation (PDE) example using $Python$.

Let’s solve a 2D heat equation, which is a classic PDE problem.

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import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm

def solve_2d_heat_equation(dx=0.1, dy=0.1, dt=0.0001, t_final=0.01):
"""
Solve 2D heat equation using finite difference method
∂u/∂t = α(∂²u/∂x² + ∂²u/∂y²)
with initial conditions: u(x,y,0) = sin(πx)sin(πy)
and boundary conditions: u = 0 at boundaries
"""
# Thermal diffusivity
alpha = 1.0

# Create grid
x = np.arange(0, 1+dx, dx)
y = np.arange(0, 1+dy, dy)
t = np.arange(0, t_final+dt, dt)

# Initialize temperature grid
nx, ny = len(x), len(y)
u = np.zeros((nx, ny))

# Set initial conditions
X, Y = np.meshgrid(x, y)
u = np.sin(np.pi*X) * np.sin(np.pi*Y)

# Create arrays for new and old solutions
u_new = np.zeros((nx, ny))

# Stability condition check
stability = alpha * dt * (1/(dx**2) + 1/(dy**2))
print(f"Stability parameter: {stability:.3f} (should be < 0.5)")

# Time stepping
for n in range(len(t)-1):
# Interior points
for i in range(1, nx-1):
for j in range(1, ny-1):
u_new[i,j] = u[i,j] + alpha*dt*(
(u[i+1,j] - 2*u[i,j] + u[i-1,j])/dx**2 +
(u[i,j+1] - 2*u[i,j] + u[i,j-1])/dy**2
)

# Update solution
u = u_new.copy()

# Enforce boundary conditions
u[0,:] = u[-1,:] = u[:,0] = u[:,-1] = 0

return X, Y, u

# Solve the equation
X, Y, u_final = solve_2d_heat_equation()

# Create visualization
fig = plt.figure(figsize=(12, 5))

# Surface plot
ax1 = fig.add_subplot(121, projection='3d')
surf = ax1.plot_surface(X, Y, u_final, cmap=cm.viridis)
ax1.set_xlabel('x')
ax1.set_ylabel('y')
ax1.set_zlabel('Temperature')
ax1.set_title('2D Heat Distribution (3D View)')
fig.colorbar(surf)

# Contour plot
ax2 = fig.add_subplot(122)
contour = ax2.contourf(X, Y, u_final, levels=20, cmap=cm.viridis)
ax2.set_xlabel('x')
ax2.set_ylabel('y')
ax2.set_title('2D Heat Distribution (Top View)')
fig.colorbar(contour)

plt.tight_layout()
plt.show()

Let me explain this code and the problem we’re solving:

  1. The Problem: We’re solving the 2D heat equation:

    • Equation: $∂u/∂t = α(∂²u/∂x² + ∂²u/∂y²)$
    • Initial condition: $u(x,y,0) = sin(πx)sin(πy)$
    • Boundary conditions: $u = 0$ at all boundaries
    • $α$ is the thermal diffusivity (set to $1.0$ in our case)
  2. Solution Method:

    • We use the Finite Difference Method (FDM)
    • Space is discretized into a grid with steps $dx$ and $dy$
    • Time is discretized with step $dt$
    • The solution uses an explicit scheme for time-stepping
  3. Code Structure:

    • Creates a mesh grid for $x$ and $y$ coordinates
    • Sets initial conditions using sine functions
    • Implements time-stepping loop with boundary conditions
    • Uses $numpy$ for efficient array operations
  4. Visualization:

    • Left plot: 3D surface plot showing temperature distribution
    • Right plot: 2D contour plot showing temperature from top view
    • Colors indicate temperature (darker blue = cooler, yellow = warmer)
Stability parameter: 0.020 (should be < 0.5)

  1. Results Interpretation:
    • The initial sine wave pattern gradually diffuses
    • Heat spreads from warmer regions to cooler regions
    • The solution maintains zero temperature at boundaries
    • The pattern becomes smoother over time due to diffusion

The visualization shows both a 3D surface plot and a 2D contour plot to help understand how heat distributes across the 2D surface.

The color gradient represents temperature variations, with warmer colors indicating higher temperatures.

Advanced Fluid Dynamics Simulation

A Python-Based Study of Pipe Flow and Pressure Drop Analysis

I’ll create a example of fluid dynamics, this time focusing on pipe flow analysis using the $Hagen$-$Poiseuille$ equation and pressure drop calculations.

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import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import fsolve

class PipeFlowAnalyzer:
def __init__(self):
# Fluid properties (water at 20°C)
self.density = 998.2 # kg/m³
self.viscosity = 0.001 # Pa·s

# Pipe properties
self.length = 10.0 # m
self.diameter = 0.05 # m
self.roughness = 0.0015 # mm (typical for commercial steel)

def reynolds_number(self, velocity):
"""Calculate Reynolds number"""
return self.density * velocity * self.diameter / self.viscosity

def friction_factor(self, reynolds):
"""
Calculate Darcy friction factor using Colebrook-White equation
"""
def colebrook(f):
return (1/np.sqrt(f) + 2.0*np.log10(self.roughness/(3.7*self.diameter*1000) +
2.51/(reynolds*np.sqrt(f))))

if reynolds < 2300:
return 64/reynolds
else:
# Initial guess using Swamee-Jain equation
f_0 = 0.25/(np.log10(self.roughness/(3.7*self.diameter*1000) + 5.74/reynolds**0.9)**2)
return fsolve(colebrook, f_0)[0]

def pressure_drop(self, velocity):
"""Calculate pressure drop using Darcy-Weisbach equation"""
re = self.reynolds_number(velocity)
f = self.friction_factor(re)
return (f * self.length * self.density * velocity**2) / (2 * self.diameter)

def velocity_profile(self, flow_rate, r_points=50):
"""Calculate velocity profile for laminar flow"""
radius = self.diameter/2
r = np.linspace(0, radius, r_points)

# Maximum velocity at center
v_max = 2 * flow_rate / (np.pi * radius**2)

# Parabolic profile
v = v_max * (1 - (r/radius)**2)
return r, v

def analyze_flow_range(self, velocities):
"""Analyze flow characteristics over a range of velocities"""
reynolds = [self.reynolds_number(v) for v in velocities]
pressure_drops = [self.pressure_drop(v) for v in velocities]
friction_factors = [self.friction_factor(re) for re in reynolds]

return reynolds, pressure_drops, friction_factors

def plot_analysis(self, velocities):
"""Create visualizations of flow analysis"""
reynolds, pressure_drops, friction_factors = self.analyze_flow_range(velocities)

fig = plt.figure(figsize=(15, 10))

# Pressure drop vs Velocity
ax1 = fig.add_subplot(221)
ax1.plot(velocities, pressure_drops, 'b-', linewidth=2)
ax1.set_xlabel('Flow Velocity (m/s)')
ax1.set_ylabel('Pressure Drop (Pa)')
ax1.set_title('Pressure Drop vs Flow Velocity')
ax1.grid(True)

# Friction factor vs Reynolds number
ax2 = fig.add_subplot(222)
ax2.semilogx(reynolds, friction_factors, 'r-', linewidth=2)
ax2.set_xlabel('Reynolds Number')
ax2.set_ylabel('Friction Factor')
ax2.set_title('Moody Diagram')
ax2.grid(True)

# Velocity profile for laminar flow
ax3 = fig.add_subplot(223)
flow_rate = 0.001 # m³/s
r, v = self.velocity_profile(flow_rate)
ax3.plot(r*1000, v, 'g-', linewidth=2) # Convert radius to mm
ax3.set_xlabel('Radial Position (mm)')
ax3.set_ylabel('Velocity (m/s)')
ax3.set_title('Laminar Flow Velocity Profile')
ax3.grid(True)

# Head loss vs Flow rate
ax4 = fig.add_subplot(224)
head_losses = [dp/(self.density * 9.81) for dp in pressure_drops]
flow_rates = [v * np.pi * (self.diameter/2)**2 for v in velocities]
ax4.plot(flow_rates, head_losses, 'm-', linewidth=2)
ax4.set_xlabel('Flow Rate (m³/s)')
ax4.set_ylabel('Head Loss (m)')
ax4.set_title('Head Loss vs Flow Rate')
ax4.grid(True)

plt.tight_layout()
return fig

# Create analyzer and run analysis
analyzer = PipeFlowAnalyzer()

# Define velocity range for analysis
velocities = np.linspace(0.1, 5, 100)

# Plot results
analyzer.plot_analysis(velocities)

# Print analysis for specific velocity
test_velocity = 1.0 # m/s
re = analyzer.reynolds_number(test_velocity)
dp = analyzer.pressure_drop(test_velocity)
f = analyzer.friction_factor(re)

print("Pipe Flow Analysis Results:")
print(f"\nTest conditions at velocity = {test_velocity} m/s:")
print(f"Reynolds Number: {re:.0f}")
print(f"Flow Regime: {'Laminar' if re < 2300 else 'Turbulent'}")
print(f"Friction Factor: {f:.4f}")
print(f"Pressure Drop: {dp/1000:.2f} kPa")
print(f"Head Loss: {dp/(analyzer.density * 9.81):.2f} m")

Let me explain this pipe flow analysis example:

  1. Core Fluid Dynamics Concepts:

    • $Hagen$-$Poiseuille$ flow
    • Reynolds number
    • Friction factor
    • Pressure drop
    • Head loss
  2. Mathematical Components:

    • $Darcy$-$Weisbach$ equation
    • $Colebrook$-$White$ equation
    • Velocity profile calculations
    • Head loss computations
  3. Visualization Elements:

    • Top left: Pressure drop vs. flow velocity
    • Top right: Moody diagram (friction factor vs. Reynolds number)
    • Bottom left: Velocity profile in pipe
    • Bottom right: Head loss vs. flow rate
  4. Key Analysis Features:

    • Flow regime determination
    • Friction factor calculation
    • Pressure loss prediction
    • Velocity distribution
  5. Engineering Applications:

    • Pipe system design
    • Pump selection
    • Energy loss calculations
    • Flow rate optimization

Output

Pipe Flow Analysis Results:

Test conditions at velocity = 1.0 m/s:
Reynolds Number: 49910
Flow Regime: Turbulent
Friction Factor: 0.0210
Pressure Drop: 2.10 kPa
Head Loss: 0.21 m

The visualizations help understand:

  • How pressure drop varies with flow velocity
  • The relationship between friction factor and Reynolds number
  • The shape of the velocity profile in the pipe
  • Energy losses in pipe systems

Mathematical Sociology

Example: Modeling Social Influence in a Network

We will simulate how opinions spread in a social network.

The idea is based on the DeGroot model of consensus, where each node in the network updates its opinion based on a weighted average of its neighbors’ opinions.

Problem Setup

  • A network of individuals is represented as a graph.
  • Each node has an initial opinion (randomized).
  • Opinions are updated iteratively using the weighted average of neighboring opinions.
  • We will observe how opinions converge over time.

Python Code Implementation

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import numpy as np
import networkx as nx
import matplotlib.pyplot as plt

# Step 1: Create a social network graph
G = nx.erdos_renyi_graph(n=20, p=0.2, seed=42) # 20 nodes, 20% connection probability

# Step 2: Initialize random opinions
opinions = np.random.rand(len(G.nodes))
initial_opinions = opinions.copy()

# Step 3: Define the adjacency matrix and normalize weights
adj_matrix = nx.to_numpy_array(G)
weights = adj_matrix / adj_matrix.sum(axis=1, keepdims=True)

# Step 4: Simulate opinion updates
iterations = 20
opinion_history = [opinions.copy()]

for _ in range(iterations):
opinions = weights @ opinions # Weighted average of neighbors
opinion_history.append(opinions.copy())

# Step 5: Visualization of the results
plt.figure(figsize=(12, 6))

# Plot the convergence of opinions over time
for i, opinion_track in enumerate(np.array(opinion_history).T):
plt.plot(opinion_track, label=f'Node {i}', alpha=0.6)

plt.title("Opinion Dynamics in a Social Network (DeGroot Model)")
plt.xlabel("Iteration")
plt.ylabel("Opinion Value")
plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left', fontsize='small')
plt.grid(True)
plt.show()

# Plot the initial and final opinions on the graph
fig, axes = plt.subplots(1, 2, figsize=(14, 6))

# Initial opinions
nx.draw(
G,
with_labels=True,
node_color=initial_opinions,
cmap=plt.cm.viridis,
node_size=500,
ax=axes[0]
)
axes[0].set_title("Initial Opinions")

# Final opinions
nx.draw(
G,
with_labels=True,
node_color=opinions,
cmap=plt.cm.viridis,
node_size=500,
ax=axes[1]
)
axes[1].set_title("Final Opinions")

plt.show()

Explanation

  1. Graph Creation:
    A random graph is created using the $Erdős–Rényi model$ to simulate a social network.
  2. Initialization:
    Each node is assigned a random opinion value between $0$ and $1$.
  3. Opinion Update Rule:
    Opinions are updated iteratively as a weighted average of neighbors’ opinions, governed by the adjacency matrix of the graph.
  4. Visualization:
    • The convergence of opinions over iterations is plotted.
    • The initial and final opinions are visualized on the network.

Results

  1. Convergence Dynamics:
    • The plot of opinion dynamics shows how individual nodes’ opinions evolve and converge over iterations.
    • Nodes with many connections (high degree) influence others more strongly.


  1. Graph Visualization:
    • Initial opinions are randomly distributed.
    • Final opinions converge to a consensus or remain close to a weighted average of initial opinions depending on the network structure.


This example demonstrates how mathematical sociology models can capture social influence processes using $Python$.

Mathematical Biology

Example: Lotka-Volterra Predator-Prey Model

The Lotka-Volterra equations describe the dynamics of biological systems in which two species interact: a predator population and a prey population.

This is a classic example in mathematical biology.


Problem Statement

  1. Define the Lotka-Volterra equations for predator-prey dynamics.
  2. Solve the equations numerically using $Python$.
  3. Visualize the population dynamics over time and as a phase plot.

Lotka-Volterra Equations

The equations are:
$$
\frac{dx}{dt} = \alpha x - \beta x y
$$
$$
\frac{dy}{dt} = \delta x y - \gamma y
$$

  • $x(t)$: Prey population
  • $y(t)$: Predator population
  • $\alpha$: Prey birth rate
  • $\beta$: Predation rate
  • $\delta$: Predator reproduction rate
  • $\gamma$: Predator death rate

Python Code

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import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt

# Define the Lotka-Volterra equations
def lotka_volterra(t, z, alpha, beta, delta, gamma):
x, y = z
dxdt = alpha * x - beta * x * y
dydt = delta * x * y - gamma * y
return [dxdt, dydt]

# Parameters
alpha = 0.1 # Prey birth rate
beta = 0.02 # Predation rate
delta = 0.01 # Predator reproduction rate
gamma = 0.1 # Predator death rate

# Initial populations
x0 = 40 # Initial prey population
y0 = 9 # Initial predator population

# Time span for simulation
t_span = (0, 200)
t_eval = np.linspace(t_span[0], t_span[1], 1000)

# Solve the equations
solution = solve_ivp(
lotka_volterra, t_span, [x0, y0], args=(alpha, beta, delta, gamma), t_eval=t_eval
)

# Extract results
time = solution.t
prey = solution.y[0]
predators = solution.y[1]

# Plot population dynamics over time
plt.figure(figsize=(10, 6))
plt.plot(time, prey, label="Prey Population", color="blue")
plt.plot(time, predators, label="Predator Population", color="red")
plt.title("Lotka-Volterra Predator-Prey Dynamics")
plt.xlabel("Time")
plt.ylabel("Population")
plt.legend()
plt.grid()
plt.show()

# Phase plot
plt.figure(figsize=(8, 6))
plt.plot(prey, predators, color="purple")
plt.title("Phase Plot: Predator vs. Prey")
plt.xlabel("Prey Population")
plt.ylabel("Predator Population")
plt.grid()
plt.show()

Code Explanation

  1. Lotka-Volterra Function:

    • Defines the differential equations for the prey and predator populations.
  2. Parameters:

    • $ \alpha, \beta, \delta, \gamma $: Control the growth and interaction rates of prey and predators.
  3. Numerical Solution:

    • The solve_ivp function from scipy.integrate solves the equations over time.
  4. Visualization:

    • Population Dynamics Plot: Shows how prey and predator populations change over time.
    • Phase Plot: Illustrates the predator-prey relationship in the phase space.

Results

Ecosystem Analysis Results:

Model Parameters:
Prey growth rate (α): 1.0
Predation rate (β): 0.1
Predator reproduction rate (δ): 0.075
Predator death rate (γ): 0.5

Equilibrium Point:
Prey population: 6.67
Predator population: 10.00

Stability Analysis:
Eigenvalues: 0.000+0.707j, 0.000-0.707j
System is neutrally stable

  1. Population Dynamics:

    • Prey population oscillates due to predation.
    • Predator population follows the prey’s oscillation with a phase lag.
  2. Phase Plot:

    • A closed-loop trajectory represents the cyclic nature of the predator-prey interaction.

This simulation provides insight into the oscillatory behavior of ecological systems and the delicate balance between predator and prey populations.

Financial Mathematics

Black-Scholes Option Pricing Model

The Black-Scholes model is a widely used framework in financial mathematics for pricing European options.

In this example, we will calculate the price of a European call option using the Black-Scholes formula.


Problem

Given:

  • Stock price $(S)$
  • Strike price $(K)$
  • Risk-free interest rate $(r)$
  • Time to maturity $(T)$
  • Volatility $(\sigma)$ of the stock
    Calculate the price of a European call option using the formula:

$$
C = S \cdot N(d_1) - K \cdot e^{-rT} \cdot N(d_2)
$$

$$
d_1 = \frac{\ln(S/K) + (r + \sigma^2 / 2)T}{\sigma \sqrt{T}}, \quad
d_2 = d_1 - \sigma \sqrt{T}
$$

Here, $N(x)$ is the cumulative distribution function ($CDF$) of the standard normal distribution.


Python Code

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import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

def black_scholes_call(S, K, r, T, sigma):
"""
Calculate the price of a European call option using the Black-Scholes formula.
"""
d1 = (np.log(S / K) + (r + 0.5 * sigma ** 2) * T) / (sigma * np.sqrt(T))
d2 = d1 - sigma * np.sqrt(T)

call_price = S * norm.cdf(d1) - K * np.exp(-r * T) * norm.cdf(d2)
return call_price

def visualize_black_scholes():
# Parameters
S = np.linspace(50, 150, 100) # Stock price range
K = 100 # Strike price
r = 0.05 # Risk-free interest rate
T = 1 # Time to maturity (1 year)
sigma = 0.2 # Volatility (20%)

# Calculate call option prices for each stock price
call_prices = [black_scholes_call(s, K, r, T, sigma) for s in S]

# Plot the results
plt.figure(figsize=(10, 6))
plt.plot(S, call_prices, label="Call Option Price")
plt.axvline(K, color="red", linestyle="--", label="Strike Price (K)")
plt.title("European Call Option Price (Black-Scholes Model)")
plt.xlabel("Stock Price (S)")
plt.ylabel("Call Option Price")
plt.legend()
plt.grid()
plt.show()

if __name__ == "__main__":
visualize_black_scholes()

Explanation

  1. Parameters:

    • S: The current stock price varies from $50$ to $150$.
    • K: The strike price is set at $100.$
    • r: The risk-free interest rate is $5$%.
    • T: Time to maturity is $1$ year.
    • sigma: Volatility is $20$%.
  2. Black-Scholes Formula:

    • d1 and d2 are intermediate terms used to calculate the option price.
    • norm.cdf(d1) and norm.cdf(d2) give the probabilities of the option being in-the-money.
  3. Visualization:

    • The $x$-axis represents the stock price.
    • The $y$-axis represents the calculated call option price.
    • A red dashed line indicates the strike price $(K = 100)$.

Results

  • When the stock price $(S)$ is much lower than the strike price $(K)$, the call option price approaches zero because the option is out-of-the-money.
  • As $(S)$ increases beyond $(K)$, the call option price increases, reflecting the intrinsic value of the option.
  • The graph provides an intuitive understanding of how the call option price varies with the stock price, according to the Black-Scholes model.

Function Theory

I’ll create an example demonstrating function theory using $Python$, focusing on analyzing a complex function with visualization.

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import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

def analyze_complex_function(z):
"""
Analyzes the complex function f(z) = z^2 + 2z + 1
Args:
z (complex): Input complex number
Returns:
complex: Result of f(z)
"""
return z**2 + 2*z + 1

# Create a grid of points
x = np.linspace(-2, 2, 100)
y = np.linspace(-2, 2, 100)
X, Y = np.meshgrid(x, y)
Z = X + Y*1j

# Calculate function values
W = analyze_complex_function(Z)

# Extract real and imaginary parts
real_part = W.real
imag_part = W.imag
abs_values = np.abs(W)

def plot_function_analysis():
"""
Creates three subplots showing different aspects of the complex function
"""
fig = plt.figure(figsize=(15, 5))

# Real part
ax1 = fig.add_subplot(131, projection='3d')
surf1 = ax1.plot_surface(X, Y, real_part, cmap='viridis')
ax1.set_title('Real Part')
ax1.set_xlabel('Re(z)')
ax1.set_ylabel('Im(z)')
ax1.set_zlabel('Re(f(z))')

# Imaginary part
ax2 = fig.add_subplot(132, projection='3d')
surf2 = ax2.plot_surface(X, Y, imag_part, cmap='viridis')
ax2.set_title('Imaginary Part')
ax2.set_xlabel('Re(z)')
ax2.set_ylabel('Im(z)')
ax2.set_zlabel('Im(f(z))')

# Absolute value
ax3 = fig.add_subplot(133, projection='3d')
surf3 = ax3.plot_surface(X, Y, abs_values, cmap='viridis')
ax3.set_title('Absolute Value')
ax3.set_xlabel('Re(z)')
ax3.set_ylabel('Im(z)')
ax3.set_zlabel('|f(z)|')

plt.tight_layout()
return fig

# Plot the results
plot_function_analysis()

# Find critical points (where derivative is zero)
# For f(z) = z^2 + 2z + 1, f'(z) = 2z + 2
# Critical point is at z = -1

critical_point = -1
critical_value = analyze_complex_function(critical_point)
print(f"Critical point: z = {critical_point}")
print(f"Value at critical point: f({critical_point}) = {critical_value}")

Let me explain this example of function theory:

  1. The Complex Function
    We’re analyzing the function $f(z) = z² + 2z + 1$, which is a quadratic complex function.

  2. Code Components:

    • We define the complex function using $NumPy$ for efficient computation
    • Create a grid of complex numbers using meshgrid
    • Calculate function values across the complex plane
    • Generate $3D$ visualizations showing different aspects of the function
  3. Visualizations:
    The code creates three $3D$ plots:

    • Real part: Shows how the real component of $f(z)$ varies
    • Imaginary part: Shows how the imaginary component varies
    • Absolute value: Shows the magnitude of $f(z)$
  4. Mathematical Analysis:

    • The function has one critical point at $z = -1$ (where $f’(z) = 0$)
    • The function is analytic everywhere (infinitely differentiable)
    • It’s a second-degree polynomial in the complex plane

The $3D$ visualizations help understand:

  • The behavior of the function in different regions of the complex plane
  • How real and imaginary parts interact
  • Where the function grows or decreases rapidly
  • The location and nature of critical points