Questions: Find the exponential function that is the best fit for (f(x)) defined by the table below. x 1 2 3 4 5 y 2 4 11 28 65

Find the exponential function that is the best fit for (f(x)) defined by the table below.

x  1  2  3  4  5
y  2  4  11  28  65
Transcript text: Find the exponential function that is the best fit for $f(x)$ defined by the table below. \begin{tabular}{|c|c|c|c|c|c|} \hline $\mathbf{x}$ & 1 & 2 & 3 & 4 & 5 \\ \hline $\mathbf{y}$ & 2 & 4 & 11 & 28 & 65 \\ \hline \end{tabular}
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Solution

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Solution Steps

To find the exponential function that best fits the given data, we can use a method called exponential regression. This involves finding the parameters of the exponential function \( y = a \cdot b^x \) that minimize the difference between the observed values and the values predicted by the model. We can use Python's libraries such as NumPy and SciPy to perform this regression and find the best-fit parameters.

To find the exponential function that best fits the given data, we will use the method of least squares to fit an exponential model of the form \( y = ab^x \).

Step 1: Transform the Data

First, we transform the exponential model into a linear form. Taking the natural logarithm of both sides, we have:

\[ \ln(y) = \ln(a) + x \ln(b) \]

Let \( Y = \ln(y) \), \( A = \ln(a) \), and \( B = \ln(b) \). The equation becomes:

\[ Y = A + Bx \]

Step 2: Calculate Transformed Values

Calculate the transformed \( Y \) values for each \( y \) in the table:

\[ \begin{align_} Y_1 &= \ln(2) \approx 0.6931 \\ Y_2 &= \ln(4) \approx 1.3863 \\ Y_3 &= \ln(11) \approx 2.3979 \\ Y_4 &= \ln(28) \approx 3.3322 \\ Y_5 &= \ln(65) \approx 4.1744 \\ \end{align_} \]

Step 3: Set Up the System of Equations

Using the least squares method, we need to solve for \( A \) and \( B \) in the linear equation \( Y = A + Bx \). The normal equations are:

\[ \begin{align_} \sum Y &= nA + B\sum x \\ \sum xY &= A\sum x + B\sum x^2 \\ \end{align_} \]

Where \( n = 5 \) (the number of data points).

Step 4: Calculate Sums

Calculate the necessary sums:

\[ \begin{align_} \sum x &= 1 + 2 + 3 + 4 + 5 = 15 \\ \sum Y &= 0.6931 + 1.3863 + 2.3979 + 3.3322 + 4.1744 = 11.9839 \\ \sum xY &= 1 \cdot 0.6931 + 2 \cdot 1.3863 + 3 \cdot 2.3979 + 4 \cdot 3.3322 + 5 \cdot 4.1744 = 39.7321 \\ \sum x^2 &= 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55 \\ \end{align_} \]

Step 5: Solve the System of Equations

Substitute the sums into the normal equations:

\[ \begin{align_} 11.9839 &= 5A + 15B \\ 39.7321 &= 15A + 55B \\ \end{align_} \]

Solve for \( A \) and \( B \):

  1. Multiply the first equation by 3:

\[ 35.9517 = 15A + 45B \]

  1. Subtract from the second equation:

\[ 39.7321 - 35.9517 = 10B \\ 3.7804 = 10B \\ B = 0.3780 \]

  1. Substitute \( B \) back into the first equation:

\[ 11.9839 = 5A + 15(0.3780) \\ 11.9839 = 5A + 5.6700 \\ 5A = 6.3139 \\ A = 1.2628 \]

Step 6: Convert Back to Exponential Form

Convert \( A \) and \( B \) back to \( a \) and \( b \):

\[ a = e^A = e^{1.2628} \approx 3.5355 \\ b = e^B = e^{0.3780} \approx 1.4595 \]

Thus, the exponential function is:

\[ y = 3.5355 \cdot 1.4595^x \]

Final Answer

The exponential function that best fits the data is:

\[ \boxed{y = 3.5355 \cdot 1.4595^x} \]

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