Welcome to the World of Modelling and Simulation

What is Modelling?

This blog is all about system dynamics modelling and simulation applied in the engineering field, especially mechanical, electrical, and ...

Polynomial Curve Fitting to Approximate a Function

In this tutorial, we will see the application of the polynomial curve fitting method to approximate a function. Let us consider the following differential equation.

t = m ∂f/∂x

When, x is at zero and m is 0.00024. The experimental data is given for the function f at four different points of x.

x     |    0     |           1      |            2  |        3
----------------------------------------------
     |   0      |   55.56     |   88.89   |    100

Plotting the values of f with respect to x gives following graph:
Function f is plotted against four given points














Our goal is to find the value of t by fitting the given experimental data through a polynomial. The most general nth degree polynomial is,

Pₙ (x) = aₒ + a₁ x + a₂ x² + a₃ x³ + ... .... ... + aₙ xⁿ

Where, aₒa₁, a₂, a₃, ... ... aₙ are constant coefficients of the polynomial equation. As it is mentioned in the experimental data where the function values are known for four respective points, so a third degree polynomial would fit the data. The third degree polynomial is,

f (x) = aₒ + a₁ x + a₂ x² + a₃ x³

Where, the coefficients are to be determined. According to the experimental data, we have,

f (0) = aₒ + a₁ (0) + a₂ (0)² + a₃ (0)³

⇒ aₒ = 0

Then,

f (1) = aₒ + a₁ (1) + a₂ (1)² + a₃ (1)³

⇒ a₁ + a₂ + a₃ = 55.56 .....................................................(1)

Then,

f (2) = aₒ + a₁ (2) + a₂ (2)² + a₃ (2)³

⇒ 2a₁ + 4a₂ + 8a₃ = 88.89 .............................................(2)

Then,

f (3) = aₒ + a₁ (3) + a₂ (3)² + a₃ (3)³

⇒ 3a₁ + 9a₂ + 27a₃ = 100 ...............................................(3)

Now, from equation (1):

a₁ = 55.56 - a₂ - a₃

Substituting a₁ into equation (2):

a₂ + 3a₃ = -11.115 ..............................................................(4)

Again, substituting a₁ into equation (3):

a₂ + 4a₃ = -11.113 ..............................................................(5)

Subtracting equation (5) from (4), we have,

a₃ = 0.002

Now, substituting a₃ in equation (4), we get,

a₂ = - 11.121

And finally we get, a₁ = 66.679

So, our polynomial function is,

f (x) = 66.679 - 11.121 x² + 0.002 x³

A plot is obtained to visualize the approximated function’s shape for the corresponding values of x where function values are determined for x, which is started from 0 with an interval 0.005 between points until 3. Following figure shows a smooth graph compared to the aforementioned one.

Plot of the approximated polynomial for different values of x















Now, we can determine the value of  ∂f/∂x:

∂f/∂x = 66.679  - 22.242 x + 0.006 

At x = 0, we get,

∂f/∂x = 66.679

Now, to evaluate t, according to the given equation:

t = m ∂f/∂x

t = 0.00024 × 66.679 = 0.016



#CurveFitting #Matlab #PolynomialFunction #Blog #Blogger

No comments:

Post a Comment