Least squares | MIT 18.02SC Multivariable Calculus, Fall 2010

In this introductory section, Christine Breiner welcomes the audience back to recitation and outlines the task at hand: using the least squares method to fit a line to a given set of three data points (0, 1), (2, 1), and (3, 4). She encourages the audience to attempt the problem first, based on the techniques learned in class, and then return to the video for a detailed explanation and comparison of solutions. Breiner then explains the concept of least squares, emphasizing the goal of minimizing the sum of the squares of the differences between the actual and expected values. She visually describes this process on a graph, highlighting the trade-off between the sizes of the squares formed by the differences in values. The explanation progresses to discuss how adjusting the line's slope and intercept affects the sum of the areas of these squares, which is the quantity being minimized. Breiner also touches on the optimization process involving both the slope and y-intercept, which is why it's studied in multivariable calculus rather than single-variable calculus. Finally, she presents the equations derived from optimizing the least squares line and explains the values needed for the calculation, which are the sums of the x-values, y-values, x-squared values, and x*y product from the given data points.

In this segment, Breiner continues the discussion on least squares by diving into the specifics of solving the system of equations derived from the method. She presents the two equations that result from setting the partial derivatives of the function with respect to the variables (a and b) equal to zero. The equations, which involve the sums of the x_i's, y_i's, x_i squared, and x_i*y_i product from the data points, are used to solve for the slope (a) and the y-intercept (b). Breiner calculates the values of a and b, finding them to be 6/7 and 4/7, respectively. This leads to the equation of the least squares line, y = (6/7)x + 4/7, which is the best fit for the given data points. She concludes this part by hinting at a more advanced application of least squares for fitting higher-degree polynomials, such as a parabola, to the data. Breiner outlines the function form for the challenge problem, which involves minimizing a different function that accounts for the squared differences between the actual and expected values based on a quadratic equation. She explains the process of taking partial derivatives with respect to each variable (a, b, and c) and setting them to zero to solve for the coefficients of the quadratic equation, which will give the parabola of best fit for the three data points.