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 \( y = ab^x \).
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:
\[
Y = A + Bx
\]
Calculate the transformed \( Y \) values for each \( y \) in the table:
\[
\begin{align_}
Y_1 &= \ln(2) \approx 0.6931 \\
Y_2 &= \ln(4) \approx 1.3863 \\
Y_3 &= \ln(11) \approx 2.3979 \\
Y_4 &= \ln(28) \approx 3.3322 \\
Y_5 &= \ln(65) \approx 4.1744 \\
\end{align_}
\]
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).
Calculate the necessary sums:
\[
\begin{align_}
\sum x &= 1 + 2 + 3 + 4 + 5 = 15 \\
\sum Y &= 0.6931 + 1.3863 + 2.3979 + 3.3322 + 4.1744 = 11.9839 \\
\sum xY &= 1 \cdot 0.6931 + 2 \cdot 1.3863 + 3 \cdot 2.3979 + 4 \cdot 3.3322 + 5 \cdot 4.1744 = 39.7321 \\
\sum x^2 &= 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55 \\
\end{align_}
\]
Substitute the sums into the normal equations:
\[
\begin{align_}
11.9839 &= 5A + 15B \\
39.7321 &= 15A + 55B \\
\end{align_}
\]
Solve for \( A \) and \( B \):
- Multiply the first equation by 3:
\[
35.9517 = 15A + 45B
\]
- Subtract from the second equation:
\[
39.7321 - 35.9517 = 10B \\
3.7804 = 10B \\
B = 0.3780
\]
- Substitute \( B \) back into the first equation:
\[
11.9839 = 5A + 15(0.3780) \\
11.9839 = 5A + 5.6700 \\
5A = 6.3139 \\
A = 1.2628
\]
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:
\[
y = 3.5355 \cdot 1.4595^x
\]
The exponential function that best fits the data is:
\[
\boxed{y = 3.5355 \cdot 1.4595^x}
\]