# Linear Regression


###

## Linear Regression
### What is Linear Regression?
Linear regression is one of the most fundamental techniques in data analysis.  
It models the relationship between two (or more) variables by fitting a **straight line** that best describes the trend in the data.

Mathematically, the model assumes a linear equation:
$$
y = m x + b
$$
where  
- $y$ = dependent variable 
- $x$ = independent variable
- $m$ = slope (rate of change)  
- $b$ = intercept (value of $y$ when $x = 0$) 

Linear regression helps identify proportional relationships, estimate calibration constants, or model linear system responses.

### Problem 1: Stress–Strain Relationship
Let’s assume we’ve measured the stress (σ) and strain (ε) for a material test and want to estimate Young’s modulus (E) from the slope.

```python
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# Example data (strain, stress)
strain = np.array([0.000, 0.0005, 0.0010, 0.0015, 0.0020, 0.0025])
stress = np.array([0.0, 52.0, 104.5, 157.2, 208.1, 261.4])  # MPa

# Fit a linear regression line using NumPy
coeffs = np.polyfit(strain, stress, deg=1)
m, b = coeffs
print(f"Slope (E) = {m:.2f} MPa, Intercept = {b:.2f}")

# Predicted stress
stress_pred = m * strain + b

# Plot
plt.figure()
plt.scatter(strain, stress, label="Experimental Data", color="navy")
plt.plot(strain, stress_pred, color="red", label="Linear Fit")
plt.xlabel("Strain (mm/mm)")
plt.ylabel("Stress (MPa)")
plt.title("Linear Regression – Stress–Strain Curve")
plt.legend()
plt.grid(True)
plt.show()

```

The slope `m` represents the Young’s Modulus (E), showing the stiffness of the material in the linear elastic region. 

```python
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.metrics import r2_score, mean_squared_error

# ------------------------------------------------
# 1. Example Data: Stress vs. Strain
# (Simulated material test data)
strain = np.array([0.000, 0.0005, 0.0010, 0.0015, 0.0020, 0.0025])
stress = np.array([0.0, 52.0, 104.5, 157.2, 208.1, 261.4])  # MPa

# Reshape strain for scikit-learn (expects 2D input)
X = strain.reshape(-1, 1)
y = stress

# ------------------------------------------------
# 2. Fit Linear Regression Model
model = LinearRegression()
model.fit(X, y)

# Extract slope and intercept
m = model.coef_[0]
b = model.intercept_
print(f"Linear model: Stress = {m:.2f} * Strain + {b:.2f}")

# ------------------------------------------------
# 3. Predict Stress Values and Evaluate the Fit
y_pred = model.predict(X)

# Coefficient of determination (R²)
r2 = r2_score(y, y_pred)

# Root mean square error (RMSE)
rmse = np.sqrt(mean_squared_error(y, y_pred))

print(f"R² = {r2:.4f}")
print(f"RMSE = {rmse:.3f} MPa")

# ------------------------------------------------
# 4. Visualize Data and Regression Line
plt.figure(figsize=(6, 4))
plt.scatter(X, y, color="navy", label="Experimental Data")
plt.plot(X, y_pred, color="red", label="Linear Fit")
plt.xlabel("Strain (mm/mm)")
plt.ylabel("Stress (MPa)")
plt.title("Linear Regression – Stress–Strain Relationship")
plt.legend()
plt.grid(True)
plt.tight_layout()
plt.show()

```