Why Curve fitting (polyfit, fit) in MATLAB? - Purpose & Use Cases

The Big Idea

What if your computer could instantly draw the perfect curve through your messy data points?

The Scenario

Imagine you have a set of scattered points from an experiment, and you want to draw a smooth curve that best represents the trend. Doing this by hand means guessing where the curve should go, drawing it, and hoping it fits well.

The Problem

Manually drawing or calculating the curve is slow and often inaccurate. It's easy to make mistakes, and you can't quickly adjust the curve if new data comes in. This makes understanding the data's true pattern very hard.

The Solution

Curve fitting functions like polyfit and fit in MATLAB automatically find the best curve that matches your data points. They do the heavy math for you, giving precise results quickly and allowing easy adjustments.

Before vs After

✗ Before

x = [1 2 3 4]; y = [2 4 5 7]; % Guess coefficients manually

✓ After

p = polyfit(x, y, 2); % Automatically find best quadratic fit

What It Enables

It lets you quickly understand and predict trends in data by finding the best mathematical curve that fits your points.

Real Life Example

Scientists measuring temperature changes over time can use curve fitting to model the trend and predict future temperatures without guessing.

Key Takeaways

Manual curve drawing is slow and error-prone.

Curve fitting automates finding the best match to data.

This helps in analysis, prediction, and understanding data trends.

Practice

1. What does the MATLAB function polyfit(x, y, n) do?

easy

A. It returns a fit object that can be used to predict new values.

B. It plots the data points x and y on a graph.

C. It finds the coefficients of a polynomial of degree n that fits the data points x and y.

D. It calculates the correlation coefficient between x and y.

Solution

1. Step 1: Understand the purpose of polyfit

   polyfit calculates polynomial coefficients that best fit the data points given by x and y.

2. Step 2: Differentiate from other functions

   Unlike fit, polyfit returns just coefficients, not a fit object or plot.

3. Final Answer:

   It finds the coefficients of a polynomial of degree n that fits the data points x and y. -> Option C

4. Quick Check:

   polyfit = polynomial coefficients [OK]

Hint: Remember: polyfit returns coefficients, fit returns a fit object [OK]

Common Mistakes:

• Confusing polyfit with fit function
• Thinking polyfit plots data
• Assuming polyfit returns a fit object

2. Which of the following is the correct syntax to fit a quadratic polynomial to data vectors x and y using polyfit?

easy

A. p = polyfit(x, y, 2);

B. p = polyfit(y, x, 2);

C. p = polyfit(x, y, 3);

D. p = polyfit(x, 2, y);

Solution

1. Step 1: Identify correct argument order

   The syntax is polyfit(x, y, n) where n is the polynomial degree.

2. Step 2: Confirm degree for quadratic

   Quadratic means degree 2, so the third argument must be 2.

3. Final Answer:

   p = polyfit(x, y, 2); -> Option A

4. Quick Check:

   polyfit(x, y, 2) fits quadratic [OK]

Hint: Remember order: x, y, degree in polyfit [OK]

Common Mistakes:

• Swapping x and y arguments
• Using wrong polynomial degree
• Incorrect argument order

3. What is the output of the following MATLAB code?

x = [1 2 3 4 5];
y = [2.2 4.1 6.0 8.1 10.2];
p = polyfit(x, y, 1);
y_fit = polyval(p, 6);

medium

A. y_fit = 12.1

B. y_fit = 12.3

C. y_fit = 13.0

D. y_fit = 10.2

Solution

1. Step 1: Fit a line to data

   Using polyfit with degree 1 fits a line y = mx + b to points.

2. Step 2: Calculate predicted value at x=6

   Using polyval, we evaluate the fitted line at x=6. The slope and intercept from polyfit give approximately y_fit = 12.1.

3. Final Answer:

   y_fit = 12.1 -> Option A

4. Quick Check:

   polyval(polyfit(x,y,1),6) ≈ 12.1 [OK]

Hint: polyval evaluates polynomial at given x [OK]

Common Mistakes:

• Confusing polyfit output with predicted value
• Using wrong degree in polyfit
• Calculating y_fit without polyval

4. Identify the error in this MATLAB code snippet for fitting data with fit:

x = (1:5)';
y = [2 4 6 8 10]';
f = fit(x, y, 'poly2');

medium

A. The fit type string should be 'poly1' for linear fit, not 'poly2'.

B. The fit function requires the Curve Fitting Toolbox to be installed.

C. The input vectors x and y must be row vectors, not column vectors.

D. The syntax is correct; there is no error.

Solution

1. Step 1: Check syntax and input shapes

   Vectors x and y as column vectors are valid inputs for fit.

2. Step 2: Verify toolbox requirement

   The fit function requires the Curve Fitting Toolbox; without it, the code will error.

3. Final Answer:

   The fit function requires the Curve Fitting Toolbox to be installed. -> Option B

4. Quick Check:

   fit needs toolbox installed [OK]

Hint: fit needs Curve Fitting Toolbox installed [OK]

Common Mistakes:

• Assuming fit works without toolbox
• Confusing poly1 and poly2 as errors
• Thinking vector orientation causes error

5. You have noisy data vectors x and y. You want to fit a cubic polynomial and then predict the value at x = 7. Which sequence of MATLAB commands correctly does this?

hard

A. p = polyfit(x, y, 3); y_pred = p(7);

B. f = fit(x, y, 'poly2'); y_pred = feval(f, 7);

C. p = polyfit(y, x, 3); y_pred = polyval(p, 7);

D. f = fit(x, y, 'poly3'); y_pred = feval(f, 7);

Solution

1. Step 1: Choose correct fit function and degree

   Using fit with 'poly3' fits a cubic polynomial to x and y.

2. Step 2: Predict value at x=7 using fit object

   Use feval(f, 7) or f(7) to get predicted y value from the fit object.

3. Final Answer:

   f = fit(x, y, 'poly3'); y_pred = feval(f, 7); -> Option D

4. Quick Check:

   fit with 'poly3' and feval predicts y at 7 [OK]

Hint: Use fit with 'poly3' and feval to predict [OK]

Common Mistakes:

• Swapping x and y in polyfit
• Using wrong polynomial degree
• Using poly2 instead of poly3 for cubic fit