The Simplest Way to Understand the Concept of Taylor Series Expansion

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Hello. I am going to give you the simplest way to understand the concept of Taylor series expansion, or simply Taylor series.

With this video and text, I believe you will understand the concept of Taylor series intuitively.

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In engineering and science, we deal with waveforms. A waveform may represent a characteristic of a phenomenon with respect to a certain variable, or variation of a quantity with the evolution of a certain variable for instance time.

In order to apply a mathematical method to a given waveform, the waveform must be represented by a simple mathematical expression. If a given waveform is represented by a complicated equation or by just numerical data, then applying a certain mathematical method would be quite complicated.

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For representing a given waveform by a simple mathematical expression, note that there are basically three different categories. If you are interested in local behavior of a waveform, you should use Taylor series. Local behavior means if the quantity goes up or goes down from a local point. If this is the case, you use Taylor series.

If a given waveform is periodic and you want to capture the entire behavior of the waveform, then use Fourier series. If a given waveform is single-shot and you want to capture the entire behavior of the waveform, then use Fourier or Laplace transform.

So, when you want to describe local behavior of a given waveform, then you should use Taylor series.

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For example, consider a waveform with respect to time. It is written as $y = f(t)$. Let's assume that you are interested in the local behavior of the waveform at $t = 0$.

Taylor series assumes that $f(t)$ can be represented by the sum of all powers of $t$, or the power series expansion with respect to $t$.

$$f(t) = a_0 + a_1t + a_2t^2 + a_3t^3 + \cdots \nonumber$$

Why? This is not the point to ask why. This is Taylor series. So, Taylor series tries to capture the local behavior of $f(t)$ around $t=0$ by representing or approximating the given waveform $f(t)$ by the sum of power series. Note that it tries to capture the local behavior only. Taylor series, in this case, focuses on the vicinity of $t=0$ only and does not care the behavior far from $t=0$.

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In order to represent $f(t)$ around $t=0$ by the sum of power series, you want to identify the values of the coefficients $a_0, a_1, a_2, a_3, \cdots$ so that the right-hand side of the equation well approximates the behavior of $f(t)$ around $t=0$.

How can we identify those numbers? Taylor series assumes that you know the value of $f(t)$ at $t=0$, the value of $f'(t)$ the first derivative of $f(t)$ at $t=0$, the value of $f''(t)$ the second derivative of $f(t)$ at $t=0$, the value of $f'''(t)$ the third derivative of $f(t)$ at $t=0$, and so on. So, let's assume somehow we know the value of the waveform and the values of its derivatives at $t=0$. With these values, let's identify the values of the coefficients of the power series. Don't worry. The identification is quite easy.

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Ok, now we are trying to approximate a given waveform by this function.

$$f(t) = a_0 + a_1t + a_2t^2 + a_3t^3 + \cdots \nonumber$$

Substituting $t=0$ into this function gives

$$f(0) = a_0 \quad \Rightarrow \quad a_0 = f(0) \nonumber$$

We get the value of $a_0$, which is the value of the given waveform $f(t)$ at $t=0$.

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Then, let's take the derivative of $f(t)$.

$$f'(t) = a_1 + 2a_2t + 3a_3t^2 + 4a_4t^3 + \cdots \nonumber$$

and substituting $t=0$ into this equation gives

$$f'(0) = a_1 \quad \Rightarrow \quad a_1 = f'(0) \nonumber$$

We get the value of $a_1$, which is the value of the first derivative $f'(t)$ at $t=0$.

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Let's take the derivative of $f'(t)$ which is the second derivative of $f(t)$

$$f''(t) = 2a_2 + 6a_3t + 12a_4t^2 + 20a_5t^3 + \cdots \nonumber$$

If you substitute $t=0$ into this equation, then you get

$$f''(0) = 2a_2 \quad \Rightarrow \quad a_2 = \frac{1}{2}f''(0) \nonumber$$

Yes, we can obtain the value of $a_2$ from the value of the second derivative of $f(t)$ at $t=0$.

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If we further take the derivative of $f''(t)$ which is the third derivative of $f(t)$, then we get

$$f'''(t) = 6a_3 + 24a_4t + 60a_5t^2 + 120a_6t^3 + \cdots \nonumber$$

Substituting $t=0$ into this equation gives

$$f'''(0) = 6a_3 \quad \Rightarrow \quad a_3 = \frac{1}{6}f'''(0) \nonumber$$

We obtain the value of $a_3$ from the value of the third derivative of $f(t)$ at $t=0$.

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So, as you may have figured out, the general term of the power series coefficients is obtained in this form.

$$a_n = \frac{1}{n!}f^{(n)}(0) \nonumber$$

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Now, we have identified the values of all coefficients in the power series expansion only from the information at $t=0$. So, the power series expansion

$$f(t) = a_0 + a_1t + a_2t^2 + a_3t^3 + \cdots \nonumber$$

should represent the local behavior of the given waveform $f(t)$ well in the vicinity of $t=0$.

In actual applications, we cannot take infinite terms, so we use only the first few terms. Such Taylor series approximation is quite accurate in the vicinity of $t=0$, but its accuracy should become worse as you move away from $t=0$.

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Although we have been working on a waveform with respect to time, the variable does not have to be time. It can be any variable, distance $x$ for instance. So, let's write the Taylor series formula again for $x$.

$$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \nonumber$$

where

$$a_n = \frac{1}{n!}f^{(n)}(0) \nonumber$$

We have been focusing on $x=0$. This is actually a special case of Taylor series. This special case is called Maclaurin series.

In order to focus on an arbitrary point $x=x_0$ as a general case, we replace $x$ in the right-hand side of the equation with $x-x_0$, then we get the general Taylor series formula.

$$f(x) = a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + a_3(x-x_0)^3 + \cdots \nonumber$$

where

$$a_n = \frac{1}{n!} f^{(n)}(x_0) \nonumber$$

This looks a bit more complicated when compared with the Maclaurin series formula. But, if you notice the fact that your are now focusing on $x=x_0$ instead of $x=0$, this equation is essentially the same as that of Maclaurin series.

We are simply shifting the origin to $x_0$. In the Maclaurin case, $x$ means a small excursion from $x=0$. In the Taylor case, on the other hand, $x-x_0$ means a small excursion from $x=x_0$. That's it! Don't be confused by the complexity of the equation.

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Finally, let's look at some examples. I take sine and cosine functions as examples. This is a simple sine function.

$$f(x) = \sin x \nonumber$$

Let's calculate its Taylor series coefficients up to the third order. The constant term is

$$a_0 = \sin 0 = 0 \nonumber$$

The first-order term is

$$a_1 = \left.f'(x)\right\vert_{x=0} = \left.\cos x \right\vert_{x=0} = 1 \nonumber$$

The second-order term is

$$a_2 = \frac{1}{2} \left. f''(x) \right\vert_{x=0} = -\frac{1}{2} \left. \sin x \right\vert_{x=0} = 0 \nonumber$$

The third-order term is

$$a_3 = \frac{1}{6} \left. f'''(x) \right\vert_{x=0} = -\frac{1}{6} \left. \cos x \right\vert_{x=0} = -\frac{1}{6} \nonumber$$

So, the Taylor series expansion of the sine function up to the third-order term is

$$f(x) \simeq x - \frac{1}{6}x^3 \nonumber$$

Look at the figure. The Taylor series expansion up to the first-order term is a straight line and simply represents the slope of $\sin x$ at the origin. If you include up to the third-order term, then the Taylor series expansion better approximates the sine function in a wider region. The more terms you include, the better the Taylor series expansion approximates the given sine function in the wider range. If you include infinite terms, the Taylor series expansion should match the given sine function.

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Next one is a simple cosine function.

$$f(x) = \cos x \nonumber$$

Let's calculate its Taylor series coefficients up to the third order in the same way as the previous example. The constant term is

$$a_0 = \cos 0 = 1 \nonumber$$

The first-order term is

$$a_1 = \left. f'(x) \right\vert_{x=0} = - \left. \sin x \right\vert_{x=0} = 0 \nonumber$$

The second-order term is

$$a_2 = \frac{1}{2} \left. f''(x) \right\vert_{x=0} = -\frac{1}{2} \left. \cos x \right\vert_{x=0} = -\frac{1}{2} \nonumber$$

The third-order term is

$$a_3 = \frac{1}{6} \left. f'''(x) \right\vert_{x=0} = \frac{1}{6} \left. \sin x \right\vert_{x=0} = 0 \nonumber$$

So, the Taylor series expansion of the cosine function up to the third-order term, actually up to the second-order term since the third-order term is zero, is

$$f(x) \simeq 1 - \frac{1}{2}x^2 \nonumber$$

Look at the figure. The Taylor series expansion up to the zeroth-order term is constant and simply represents the value at $x=0$. If you include up to the second-order term, then the Taylor series expansion better approximates the cosine function in a wider region. Of course, the more terms you include, the better the Taylor series expansion approximates the given cosine function in the wider range. If you include infinite terms, the Taylor series expansion should match the given cosine function.

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Finally, you should note the following points.

Since the sine function is an odd function, which has point symmetry about the origin, its Taylor series consists only of odd-order terms. On the other hand, since the cosine function is an even function, which has line symmetry about the vertical axis, its Taylor series consists only of even-order terms. Actually, this is the reason why the former is called an odd function and the latter is called an even function.

When you actually apply Taylor series expansion in your scientific or engineering work, the first two terms should sufficient in most cases. Since you focus on local behavior of a function at a certain point, you are interested in its behavior only in the vicinity of that point. So, two terms should well represent the local behavior. As we saw in the two examples, the two terms well represent the local behaviors.

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I hope you now have reached real understanding of Taylor series. Thank you.