Exponential Fit: Find Value At X=2.5

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Alright, guys, let's dive into how we can fit an exponential function to a given set of data and then figure out the value of that function at a specific point, like x = 2.5. This is a common task in various fields, from data analysis to predictive modeling. So, buckle up, and let's get started!

Understanding the Data

First, let's take a look at the data we've got. We have a table of x and y values:

x 0 1 2 3 4 5
y 8.03 3.01 1.10 0.40 0.15 0.05

Our goal is to find an exponential function that best represents this data. An exponential function generally looks like this:

y = A * e^(Bx)

Where:

  • y is the dependent variable.
  • x is the independent variable.
  • A is the initial value (the value of y when x is 0).
  • e is the base of the natural logarithm (approximately 2.71828).
  • B is the rate of growth or decay.

Fitting an Exponential Function

To fit an exponential function, we need to determine the values of A and B. There are a few ways to do this, including using logarithmic transformation and linear regression or using non-linear regression techniques.

Method 1: Logarithmic Transformation and Linear Regression

Since the exponential function is y = A * e^(Bx), we can take the natural logarithm of both sides to linearize the equation:

ln(y) = ln(A * e^(Bx)) ln(y) = ln(A) + ln(e^(Bx)) ln(y) = ln(A) + Bx

Now, we have a linear equation of the form Y = a + Bx, where Y = ln(y) and a = ln(A). We can use linear regression to find the values of a and B.

  1. Transform the y-values:

    Take the natural logarithm of each y-value:

    x 0 1 2 3 4 5
    ln(y) 2.0833 1.1019 0.0953 -0.9163 -1.8971 -2.9957

  2. Perform Linear Regression:

    Use the x-values and the transformed ln(y) values to perform a linear regression. You can use a calculator, spreadsheet software (like Excel), or a programming language (like Python with libraries such as NumPy and SciPy) to do this. The goal is to find the best-fit line Y = a + Bx.

    The formulas for a and B are:

    B = (n * Σ(xᵢ * Yᵢ) - Σxᵢ * ΣYᵢ) / (n * Σ(xᵢ²) - (Σxᵢ)²) a = (ΣYᵢ - B * Σxᵢ) / n

    Where:

    • n is the number of data points.
    • xáµ¢ and Yáµ¢ are the individual x and ln(y) values.

    Let's calculate these values:

    • n = 6
    • Σxáµ¢ = 0 + 1 + 2 + 3 + 4 + 5 = 15
    • ΣYáµ¢ = 2.0833 + 1.1019 + 0.0953 - 0.9163 - 1.8971 - 2.9957 = -2.5286
    • Σ(xáµ¢ * Yáµ¢) = (0*2.0833) + (1*1.1019) + (2*0.0953) + (3*-0.9163) + (4*-1.8971) + (5*-2.9957) = -25.3274
    • Σ(xᵢ²) = 0² + 1² + 2² + 3² + 4² + 5² = 55

    Now, plug these values into the formulas:

    B = (6 * -25.3274 - 15 * -2.5286) / (6 * 55 - 15²) = (-151.9644 + 37.929) / (330 - 225) = -114.0354 / 105 = -1.0861 a = (-2.5286 - (-1.0861 * 15)) / 6 = (-2.5286 + 16.2915) / 6 = 13.7629 / 6 = 2.2938

    So, we have a = 2.2938 and B = -1.0861.

  3. Convert back to the Exponential Function:

    Since a = ln(A), we can find A by taking the exponential of a:

    A = e^a = e^2.2938 ≈ 9.908

    Thus, our exponential function is:

    y = 9.908 * e^(-1.0861x)

Method 2: Using Non-Linear Regression

Non-linear regression is a more direct approach, especially when you have software or libraries that can handle it. Software like R, Python (with libraries like SciPy), or even Excel can perform non-linear regression. The advantage of this method is that it directly fits the exponential function to the data without the need for logarithmic transformation.

  1. Choose a Software/Library:

    For example, in Python with SciPy, you can use the curve_fit function.

  2. Define the Exponential Function:

    def exponential_func(x, A, B): return A * np.exp(B * x)

  3. Use curve_fit:

    from scipy.optimize import curve_fit x_data = np.array([0, 1, 2, 3, 4, 5]) y_data = np.array([8.03, 3.01, 1.10, 0.40, 0.15, 0.05]) popt, pcov = curve_fit(exponential_func, x_data, y_data, p0=[1, -1]) A, B = popt

    Here, p0=[1, -1] provides initial guesses for A and B. The curve_fit function returns the optimized values for A and B.

    After running this, you might get values like A ≈ 8.02 and B ≈ -1.09.

    So, our exponential function is:

    y = 8.02 * e^(-1.09x)

Determining the Value at x = 2.5

Now that we have our exponential function, we can find the value of y when x = 2.5.

Using the function y = 9.908 * e^(-1.0861x):

y = 9.908 * e^(-1.0861 * 2.5) y = 9.908 * e^(-2.71525) y = 9.908 * 0.0662 y ≈ 0.656

Using the function y = 8.02 * e^(-1.09x):

y = 8.02 * e^(-1.09 * 2.5) y = 8.02 * e^(-2.725) y = 8.02 * 0.0654 y ≈ 0.524

Conclusion

So, based on our exponential function fitting, the value of y when x = 2.5 is approximately 0.656 using the logarithmic transformation method and 0.524 using the non-linear regression method. Keep in mind that the slight differences arise from the approximations and methods used. For higher precision, non-linear regression directly applied to the exponential function is often preferred, especially with the aid of computational tools.

I hope this detailed explanation helps you understand how to fit an exponential function to data and determine values at specific points. Let me know if you have any more questions, and happy fitting!