import matplotlib
if not hasattr(matplotlib.RcParams, "_get"):
matplotlib.RcParams._get = dict.get
II.ii. Taylor Series Expansion#
Note
Important things to retain from this block:
Understand how to compute Taylor series expansion (TSE) of a given function around a given point and its limitations
Things you do not need to know:
Any kind of Taylor expansion of a function by heart
Definition#
The basic idea behind Taylor series is to approximate any kind of function in the neighborhood of a certain point with a polynomial. This approximation is useful because polynomials are “friendly” functions: it is easy to evaluate them, compute their derivatives or integrals.
The Taylor series expansion (TSE) of an arbitrary function \(f(x)\) around \(x=x_i\) is a polynomial given by
Let’s have a look at the structure and meaning of the individual terms.
The first terms just equals the value of the function that we want to approximate. The following terms include derivatives of the function with increasing order.
The second term makes the slope of the polynomial equal to the slope of our function \(f\). (You can verify this by taking the first derivative of the Taylor series expansion with respect to \(x\) and evaluating it at \(x_i\).)
Each additional term sets a higher derivative of the Taylor series equal to the derivative of the function we want to approximate.
The interactive plot below visualizes this concept. It shows the Taylor series approximation of a sine function for different \(x_i\), with a varying number of terms. Change the order of the Taylor series approximation and the value of \(x_i\) and answer the following questions:
How well does the polynomial approximate \(\sin(x)\)?
How does the result depend on the number of terms used?
How does it vary with distance from \(x_i\)?
import numpy as np
import math
import matplotlib.pyplot as plt
from ipywidgets import widgets, interact
def sine_derivative(x, n):
"Compute the nth derivative of sin(x)"
match np.mod(n, 4):
case 0:
return np.sin(x)
case 1:
return np.cos(x)
case 2:
return -np.sin(x)
case 3:
return -np.cos(x)
def taylor_poly_sine(x, xi, order):
"Compute the Taylor series expansion of sin(x) around xi"
return sum(
(x - xi) ** n / math.factorial(n) * sine_derivative(xi, n)
for n in range(order + 1)
)
def taylor_plot(order, xi):
x = np.linspace(-2 * np.pi, 2 * np.pi, 100)
tse = taylor_poly_sine(x, xi=xi, order=order)
plt.plot(x, np.sin(x), label="sin(x)")
plt.plot(x, tse, label=f"Taylor series expansion (order {order})")
plt.scatter(xi, np.sin(xi), marker="o", color="k")
plt.annotate("$x_i$", xy=(xi + 0.2, np.sin(xi) - 0.1))
plt.ylim(-5, 5)
# plt.axis("off")
plt.legend(loc="upper right")
plt.show();
interact(
taylor_plot,
order=widgets.IntSlider(
value=0,
min=0,
max=10,
step=1,
description="order",
),
xi=widgets.FloatSlider(
value=0,
min=-2,
max=2,
step=0.1,
description="xi",
),
);
Relevant conclusions
The 1st order, which depends only on the first derivative evaluation, is a straight line.
The more terms used (larger order) the better we can approximate the function.
The further from the starting point (e.g., in the plots \(x_i=0\)), the larger the error.
The Taylor series expansion is exact as long as we include infinite terms. We, however, are limited to a truncated expression: an approximation of the real function. For example, we can write the Taylor series approximation with only 3 terms as:
Here, we defined \(\Delta x=x-x_i\), where \(\Delta x\) is the distance between the point we “know” and the desired point. \(\mathcal{O}(\Delta x^4)\) means that we do not take into account the terms associated to \(\Delta x^4\) and therefore that is the truncation error order. From here we can also conclude that the larger the step \(\Delta x\), the larger the error!
Tip
We will use \(\Delta x\) more frequently from this point on, so it is good to recognize now, using the equations above, that it is a different way of representing the differential increment, for example, \(f(x_{i+1})=f(x_i+\Delta x)\) or \(f(x_{i+2})=f(x_i+2\Delta x)\).
Now let’s see some examples how we can use the Taylor series expansion to approximate specific functions.
Compute \(e^{0.2}\) using 3 terms of TSE around \(x_i=0\).
Solution
We want to evaluate \(e^x=e^{x_i+\Delta x}=e^{0.2}\). Therefore, \(\Delta x=0.2\). The value of \(f(x_i=0)=e^0=1\). For the case of this exponential, the derivatives have the same value \(f'(x_i=0)=f''(x_i=0)=f'''(x_i=0)=e^0=1\). The TSE looks like this:
Compute the TSE polynomial truncated until \(\mathcal{O}(\Delta x^6)\) for \(f(x)=\sin(x)\) around \(x_i=0\).
Solution
Applying the definition of TSE:
Note that \(x\) and \(\Delta x\) can be written interchangeably when \(x_i=0\) as \(\Delta x=x-x_i\). If this would not be the case, for example if \(x_i=\pi\) the result is completely different.
Compute the TSE polynomial truncated until \(\mathcal{O}(\Delta x^6)\) for \(f(x)=\sin(x)\) around \(x_i=\frac{\pi}{2}\).
Solution
Applying the definition of TSE:
Attribution
This chapter is written by Jaime Arriaga Garcia, Anna Störiko, Justin Pittman and Robert Lanzafame. Find out more here.