Question

Often in practice, an approximation of the form $$ u(t)=\gamma_{1} e^{\gamma_{2} t} $$ is sought for a data fitting problem, where \(\gamma_{1}\) and \(\gamma_{2}\) are constants. Assume given data \(\left(t_{1}, z_{1}\right),\left(t_{2}, z_{2}\right), \ldots,\left(t_{m}, z_{m}\right)\), where \(z_{i}>0, i=1,2, \ldots, m\), and \(m>0\) (a) Explain in one brief sentence why the techniques introduced in the present chapter cannot be directly applied to find this \(u(t)\). (b) Considering instead $$ v(t)=\ln u(t)=\left(\ln \gamma_{1}\right)+\gamma_{2} t $$ it makes sense to define \(b_{i}=\ln z_{i}, i=1,2, \ldots, m\), and then find coefficients \(x_{1}\) and \(x_{2}\) such that \(v(t)=x_{1}+x_{1} t\) is the best least squares fit for the data $$ \left(t_{1}, b_{1}\right),\left(t_{2}, b_{2}\right), \ldots,\left(t_{m}, b_{m}\right) $$ Using this method, find \(u(t)\) for the data $$ \begin{array}{|c|c|c|c|} \hline i & 1 & 2 & 3 \\ \hline t_{i} & 0.0 & 1.0 & 2.0 \\ \hline z_{i} & e^{0.1} & e^{0.9} & e^{2} \\ \hline \end{array} $$

Short answer

Question: Find the exponential function that best fits the given data using the least squares method: (0.0, e^0.1), (1.0, e^0.9), (2.0, e^2). Answer: The exponential function that best fits the given data is \(u(t) = e^t\).

Step-by-step solution

1. Transformation of data into linear form

Take natural logarithm of \(z_i\), i.e., \(b_i = \ln z_i\) for \(i = 1,2,...,m\). \(b_1 = \ln e^{0.1} = 0.1\) \(b_2 = \ln e^{0.9} = 0.9\) \(b_3 = \ln e^{2} = 2\) Thus, the transformed data set is \((t_1, b_1) = (0.0, 0.1), (t_2, b_2) = (1.0, 0.9), (t_3, b_3) = (2.0, 2)\).

2. Use least squares for linear fitting

To find \(x_1\) and \(x_2\) for the best least squares fit, define \(\bar{t}\) and \(\bar{b}\) as the average of \(t_i\)'s and \(b_i\)'s, respectively. Also, compute \((t_i - \bar{t})(b_i - \bar{b})\) and \((t_i - \bar{t})^2\): \(\bar{t} = \frac{0.0 + 1.0 + 2.0}{3} = 1.0\) \(\bar{b} = \frac{0.1 + 0.9 + 2}{3} = 1.0\) \((t_1 - \bar{t})(b_1 - \bar{b}) = (-1.0)(-0.9) = 0.9\) \((t_2 - \bar{t})(b_2 - \bar{b}) = (0)(-0.1) = 0\) \((t_3 - \bar{t})(b_3 - \bar{b}) = (1.0)(1) = 1\) \((t_1 - \bar{t})^2 = (-1.0)^2 = 1\) \((t_2 - \bar{t})^2 = (0)^2 = 0\) \((t_3 - \bar{t})^2 = (1.0)^2 = 1\) Finally, compute \(x_1\) and \(x_2\) using least squares formulas: \(x_2 = \frac{\sum(t_i - \bar{t})(b_i - \bar{b})}{\sum(t_i - \bar{t})^2} = \frac{0.9 + 0 + 1}{1 + 0 + 1} = 1\) \(x_1 = \bar{b} - x_2\bar{t} = 1 - 1(1) = 0\) Thus, the best least squares fit for the transformed data is \(v(t) = x_1 + x_2t = 1t\).

3. Transform back to exponential form

To obtain the approximating function \(u(t)\), raise the base e to the power of our linear fit function \(v(t)\): \(u(t) = e^{v(t)} = e^{1t}\) Therefore, the exponential function that best fits the given data is \(u(t) = e^t\).

Key concepts

Exponential Data Fitting

Exponential data fitting is a statistical technique used to model data that follows an exponential trend. This type of fitting is especially useful in fields such as biology, economics, and engineering, where growth or decay processes are exponential in nature. The general form of an exponential function can be expressed as:
\[ u(t) = \text{'} + e^{\text{'} t} \]
where \text{'} and \text{'} are constants that define the specific characteristics of the exponential trend—\text{'} as the initial amount or magnitude and \text{'} as the growth or decay rate.
Fitting exponential models to a set of data points involves estimating these constants such that the curve formed by the exponential function aligns closely with the data. However, due to the nonlinear nature of exponential functions, traditional linear least squares fitting methods cannot be directly applied. This necessitates a transformation of the data, allowing for the use of linear least squares techniques—improving the accessibility of the solution and simplifying the computational process.
For an enhanced understanding, students should recognize that exponential data fitting is sensitive to large values due to the exponential function's growth behavior. This understanding is crucial when interpreting the results of an exponential fit, especially in predicting future behavior based on past data.

Transformation of Data

The transformation of data is an essential step in statistical analysis when dealing with nonlinear models. Transformation involves converting the original data into a different format that allows for simpler analysis methods to be used. In the context of exponential data fitting, data is often transformed from a nonlinear to a linear form to enable the application of linear least squares fitting.
The transformation commonly applied to exponential data utilizes natural logarithms, converting the original exponential equation \( u(t) = \text{'} e^{\text{'} t} \) to a linear form:
\[ v(t) = \text{'} t + \text{'} \]
Here, \text{'} is the natural logarithm of the initial amount \text{'}(i.e., \text{'} = \text{'}), and \text{'} is the growth rate constant from the original equation.
By taking the natural logarithm of the data points, the relationship between \text{'} and \text{'} becomes linear, with \text{'} serving as the slope and \text{'} as the y-intercept. After performing a least squares fit on the transformed data, the resulting linear model parameters can be used to reconstruct the original exponential model, leading to a more intuitive understanding of the data's underlying patterns. Practical exercises that involve actual data transformation solidify the student's comprehension of the concept and clarify how transformation enables the use of linear fitting algorithms on fundamentally nonlinear problems.

Least Squares Method

The least squares method is a foundational statistical technique used to find the best-fitting curve to a set of data points by minimizing the sum of the squares of the vertical distances (residuals) between the points and the curve. In simple terms, it's a way to draw the 'best' line through a scatter plot of points, such that the line represents the underlying trend of the data as accurately as possible.
For linear data fitting, this involves determining the optimal slope and y-intercept for a line, denoted as \text{'} and \text{'} respectively in a linear equation: \[ v(t) = \text{'} + \text{'}t \]
The least squares procedure calculates these parameters such that the sum of squared differences between the observed data points and the values predicted by the line is minimized. To perform this calculation, the following formulas are employed, based on the means of the original data points \text{'} and \text{'}(\text{'} is the average of the data 'time' values and \text{'} is the average of the transformed data values): \[ \text{'} = \frac{\text{'}(t_i - \text{'})(b_i - \text{'}))}{\text{'}(t_i - \text{'}\text{'}))} \] \[ \text{'} = \text{'} - \text{'}\text{'} \]
By executing these steps, the parameters for the linear model are obtained, which can later be used to reconstitute the exponential model through reverse transformation. The least squares method is essential for students to grasp, not only for its frequent use in exponential data fitting but because it serves as a cornerstone of many other statistical procedures and predictive modeling techniques.