Least squares and integration problem

In summary, the conversation discusses the vector space C[-1,1] and its inner product, and explores finding an orthonormal set of vectors and the best least squares approximation to h(x)= x^(1/3) + x^(2/3) by a linear function. The conversation also touches on approaching the problem with integrals and potential difficulties with complex numbers.
  • #1
Ghost of Progress
5
0
Question states
Consider the vector space C[-1,1] with an inner product defined by
<f,g> = the integral from 1 to -1 of f(x)g(x) dx
a)
Show that
u1(x)= 1/(2^.5) u2(x)= ((6^.5)/2)x
form an orthonormal set of vectors
b)
Use the result from a) to find the best least squates approximation to
h(x)= x^(1/3) + x^(2/3)
by a linear function.
For part a) I have shown that u1 and u2 each have an inner product of zero and a length of one.
I've been trying to find a solution to b) in the form
p(x) = (c1)(u1(x)) + (c2)(u2(x))
where ci = the integral from -1 to 1 of (ui(x))h(x)
evaluating this integral for c1 produces
(1/(2^.5))[(3/4)x^(4/3) + (3/5)x^(5/3)] evaluated from -1 to 1
Now I've finally got to the problem
Putting -1 in would produce complex numbers and I don't know how to proceed.
I'm not sure weather the problem here is that I'm not aproaching the least squares problem correctly or weather I'm not approaching the integral correctly. Any help would be appreciated, thanks.
 
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  • #2
Putting -1 in for x? That doesn't produce complex numbers. The cube root of -1 is -1. Its value at x= 1 is [itex]\frac{1}{\sqrt{2}}\left(\frac{3}{4}+\frac{3}{5}\right)[/itex] and its value at x= -1 is [itex]\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{3}{5}\right)[/itex] Subtracting gives just [itex]\frac{1}{\sqrt{2}}\frac{6}{5}[/itex].
 

1. What is the least squares method?

The least squares method is a mathematical technique used to find the line of best fit for a set of data points. It minimizes the sum of the squared differences between the actual data points and the predicted values from the line of best fit. This method is commonly used in regression analysis and curve fitting.

2. How is the least squares method used in data analysis?

The least squares method is used to find the relationship between two variables by fitting a line to a set of data points. This line can then be used to make predictions or analyze the relationship between the two variables. It is also used to determine the coefficients in a linear regression model.

3. What is the difference between simple linear regression and multiple linear regression?

Simple linear regression involves finding the relationship between two variables, while multiple linear regression involves finding the relationship between more than two variables. In multiple linear regression, the least squares method is used to find the line of best fit that minimizes the sum of the squared differences between the actual data points and the predicted values from the line.

4. What is the integration problem in least squares?

The integration problem in least squares refers to the issue of integrating the residuals (the difference between the actual data points and the predicted values from the line of best fit) to determine the overall fit of the model. This problem can be solved using numerical integration methods, such as the trapezoidal rule or Simpson's rule.

5. How do you determine the quality of the fit using least squares?

The quality of the fit using least squares can be determined by calculating the coefficient of determination (R-squared). This value represents the proportion of the variance in the dependent variable that can be explained by the independent variable. A higher R-squared value indicates a better fit, while a lower R-squared value indicates a poorer fit.

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