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⇦ Back If you have a set of data points that look like they’re increasing rapidly, it might be useful to fit them with a smooth, exponentially increasing line in order to describe the general shape of the data: The line that you need to fit in order to achieve this shape will be one that is described by an exponential function, that is any function of the form: \(y = AB^x + C\) or \(y = ae^{bx} + c\) (these two are mathematically equivalent because \(AB^x = Ae^{x\ln(B)}\)). The important thing to realise is that an exponential function can be fully defined with three constants. We will use the second of these formulations, which can be written in Python as Example DataFor this tutorial, let’s create some fake data to use as an example. This should be a set of points that increase exponentially (or else our attempts to fit an exponential curve to them won’t work well!) with some random noise thrown in to mimic real-world data:
The random noise is being added with the
Method 1: polyfit
The \(y = ae^{bx} \implies \ln(y) = \ln(a) + bx\) because we can take the natural logarithm of both sides. This creates a linear equation \(f(x) = mx + c\) where:
So
This polynomial can now be converted back into an exponential:
Let’s take a look at the fit:
This method has the disadvantage of over-emphasising small values: points that have large values and which are
relatively close to the linear line of best fit created by
Using a weight has improved the fit. Method 2: curve_fitFrom the Scipy pacakge we can get the Let’s use our original example data (with \(c \neq 0\)):
Now let’s fit the function \(y = ae^{bx} + c\). This is done by defining it as a lambda function (ie as an object rather than as a command) of a dummy variable \(t\) and using the
The first output,
Let’s see what this looks like:
This looks really
good, and we didn’t need to provide an initial guess! This is because the example data we are using is close enough to exponential in nature that the optimisation algorithm behind
Comparison of MethodsLet’s plot all three methods against one another using the same example data (\(c = 0\)) for each:
As you can see, the Interpolation and Extrapolation (Forecasting/Predicting/Estimating)We can use the fitted curve to estimate what our data would be for other values of \(x\) that are not in our raw dataset: what would the value be at \(x=11\) (which is outside our domain and thus requires us to forecast into the future) or \(x = 8.5\) (which is inside our domain and thus requires us to ‘fill in a gap’ in our data)? To answer these questions, we simply plug these x-values as numbers into the equation of the fitted curve:
More explicitly:
Using a Bar PlotIf you want to use a bar plot instead of a scatter plot:
⇦ Back Which of the following is one of the normal equations of ya bX?The normal equations for a second degree parabola y = ax2 + bx + c are Σy = aΣx2 + bΣx + nc, Σxy = aΣx3 + bΣx2 + cΣx and Σx2y = aΣx4 + bΣx3 + cΣx2..
How do you write the equation of least square method?Least Square Method Formula. Suppose when we have to determine the equation of line of best fit for the given data, then we first use the following formula.. The equation of least square line is given by Y = a + bX.. Normal equation for 'a':. ∑Y = na + b∑X.. Normal equation for 'b':. ∑XY = a∑X + b∑X2. What is least square method example?Least Square Method Examples
Example 1: Consider the set of points: (1, 1), (-2,-1), and (3, 2). Plot these points and the least-squares regression line in the same graph. Now, find the value of m, using the formula. Therefore, the equation of regression line is y = 23/38x + 5/19.
Which method gives a unique set of values to the constants in the equation of the fitting curve Mcq?The graphical method has the drawback in that the straight line drawn may not be unique but principle of least squares provides a unique set of values to the constants and hence suggests a curve of best fit to the given data.
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