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Maclaurin Series

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For many years one of the most famous Formula One teams was McLaren, winning several championships during the '70s and '80s. The name McLaren was for a long time synonym for power and technology. But do not fool yourself! This article will talk about the Maclaurin series, which is also as unique as the McLaren team, but the Maclaurin series will help you write functions in a more beautiful way; as in Taylor series, you will also be writing a function as a power series using its own derivatives.

In the Taylor series article, you can see how to write a function as a power series using its own derivatives, but then what is the point of a Maclaurin series if we can already do this using the Taylor series?

Long story short, Colin Maclaurin studied the particular case of the Taylor series so much that this special case was named after him. But first, let's remember the Taylor series:

Let \( f \) be a function that has derivatives of all orders at \( x=a \).

The **Taylor Series** for \( f \) at \( x=a \) is

\[ T_f(x) = f(a) + f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\cdots +\dfrac{f^{(n)}(a)}{n!}(x-a)^n+\cdots, \]

where \(T_f\) means the Taylor series of \(f\), and \( f^{(n)} \) indicates the \( n\)-th derivative of \( f \).

So as you can see, the Taylor series is always centered in a given value \( x=a\), so whenever we center it at \( x=0\), we call this series a Maclaurin series, let's see:

Let \( f \) be a function that has derivatives of all orders at \( x=0 \).

The **Maclaurin Series** (expanded form) for \( f \) is

\[ M_f(x) = f(0) + f'(0)x+\dfrac{f''(0)}{2!}x^2+\cdots +\dfrac{f^{(n)}(0)}{n!}x^n+\cdots, \]

where \(M_f\) means the Maclaurin series of \(f\), and \( f^{(n)} \) indicates the \( n\)-th derivative of \( f \).

The Maclaurin series can be presented in many forms: by writing the terms of the series or by showing the sigma notation of it. Depending on each case, one or the other will be the best way to present the Maclaurin series formula. Before we saw the **expanded form** of the series, let's see now the **sigma notation**:

Let \( f \) be a function that has derivatives of all orders at \( x=0 \).

The **Maclaurin Series** (sigma notation) for \( f \) is

\[ M_f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n , \]

where \( f^{(n)} \) indicates the \( n\)-th derivative of \( f \), and \( f^{(0)}\) is the original function \( f\).

In the end, the process is the same as the Taylor series:

** Step 1:** find the derivatives;

** Step 2:** evaluate them at \( x=0 \);

** Step 3:** and then set up the power series.

Let's see an example:

Write the Maclaurin series for the function \( f(x)=\ln(1+x)\).

**Solution**

**Step 1:** Start this by taking the derivatives of \(f(x)\):

\[ \begin{align} f(x)&=\ln(1+x) \\ \\ f'(x)&=\dfrac{1}{1+x} \\ \\ f''(x)&=-\dfrac{1}{(1+x)^2} \\ \\ f'''(x)&=\dfrac{2}{(1+x)^3} \\ \\ f^{(4)}(x)&=-\dfrac{6}{(1+x)^4} \end{align}\]

Analyzing the derivatives, we can identify the following pattern for \(n>0\):

\[f^{(n)}(x)=(-1)^{n-1}\dfrac{(n-1)!}{(1+x)^n}\]

Notice that:

- each consecutive derivative changes sign in relation to the previous derivative, hence the factor \( (-1)^{n-1} \);
- the numerators form a sequence of rule \( (n-1)! \);
- the denominators are just powers of \( (1+x) \).

You can always check this formula by replacing n with positive integer values (1, 2, 3, ...)

** Step 2:** Evaluate each derivative at \(x=0\)

\[ \begin{align} f(0)&=0 \\ \\ f'(0)&=1 \\ \\ f''(0)&=-1 \\ \\ f'''(0)&=2 \\ \\ f^{(4)}(0)&=-6 \\ \\ f^{(n)}(0)&=(-1)^{n-1}(n-1)! \end{align}\]

** Step 3:** Apply these results to the Maclaurin series formula:

\[ M_f(x) = 0+ 1\cdot x+\dfrac{-1}{2!}x^2+\dfrac{2!}{3!}x^3+\dfrac{-3!}{4!}x^4+\cdots \]

- Simplifying it:

\[ M_f(x) = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots \]

- In sigma notation, we have

\[ M_f(x) = \sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}, \]

Notice that this series starts at \( n=1\) because \(f(0)=0\).

The proof of the Maclaurin series is the same as the proof of the Taylor series. This is an interesting and challenging proof to write!

In short, the proof shows that

inside the interval of convergence, the Taylor series (or Maclaurin series) converges to the function itself;

it is based on showing that the difference between the original function and the series gets smaller and smaller for each term added to the series.

Although this is an important result for the math world, let's focus on its application. First, let's compare the Maclaurin series with the original function.

Consider a function \( f(x) \) that has derivatives of all orders at \( x=0 \) and consider \(M_f(x)\) as the Maclaurin series of \( f\), let's evaluate the derivatives of \(M_f(x)\) at \(x=0\):

\[ \begin{align} M_f(x) &= f(0) + f'(0)x+\dfrac{f''(0)}{2!}x^2+\dfrac{f'''(0)}{3!}x^3+\cdots +\dfrac{f^{(n)}(0)}{n!}x^n+\cdots \\ \\ M'_f(x) &= f'(0)+\dfrac{f''(0)}{2!}2x+\dfrac{f'''(0)}{3!}3x^2+\cdots +\dfrac{f^{(n)}(0)}{n!}nx^{n-1}+\cdots \\ \\ M''_f(x) &= f''(0)+\dfrac{f'''(0)}{3!}6x+\cdots +\dfrac{f^{(n)}(0)}{n!}n(n-1)x^{n-2}+\cdots \end{align} \]

If we evaluate each derivative at \( x= 0 \) we will have the following:

\[ \begin{align} M_f(0) &= f(0) \\ \\ M'_f(0) &= f'(0) \\ \\ M''_f(0) &= f''(0) \\ &\vdots \\ M^{(n)}_f(0) &= f^{(n)}(0) \\ &\vdots \end{align} \]

Looking at this you can see that you have two functions \( f(x) \) and \( M_f(x) \) that have the exact same derivatives of all orders at \(x=0\), this can only mean that those two functions are the same. Therefore, inside the interval of convergence, you have that

\[ f(x) = M_f(x).\]

Hence, we have that

\[ f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n . \]

Writing the Maclaurin series given a function is quite easy, you can do it for any function that has derivatives of all orders. As stated before \( f(x) \) is equal to \(M_f(x)\) inside the convergence interval, and that is the expansion of \( f(x)\).

Let \( f \) be a function that has derivatives of all orders at \( x=0 \), and let \(M_f\) be the Maclaurin Series for \( f \).

Then for every value of \(x\) inside the interval of convergence,

\[ f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n . \]

In other words, inside the interval of convergence, the Maclaurin series \(M_f\) and the function \(f\) are precisely the same, and \( M_f \) is a **power series**** ****expansion** of \(f\).

Write the Maclaurin series for \( f(x) = \cos(x) \).

**Solution**

** Step 1:** Start this by taking the derivatives of \(f(x)\):

\[ \begin{align} f(x)&=\cos(x) \\ \\ f'(x)&=-\sin(x) \\ \\ f''(x)&=-\cos(x) \\ \\ f'''(x)&=\sin(x) \\ \\ f^{(4)}(x)&=\cos(x) \end{align}\]

** Step 2:** Before finding a pattern for the derivatives let's evaluate each one at \(x=0\):

\[ \begin{align} f(0)&=\cos(0)=1 \\ \\ f'(0)&=-\sin(0)=0 \\ \\ f''(0)&=-\cos(0)=-1 \\ \\ f'''(0)&=\sin(0)=0 \\ \\ f^{(4)}(0)&=\cos(0)=1 \end{align}\]

Analysing the results we can see that:

- If \(n\) is odd then

\[f^{(n)}(0)=0\]

- If \(n\) is even then

\[f^{(n)}(0)=(-1)^{\tfrac{n}{2}}\]

** Step 3:** Apply these results to the Maclaurin series formula:

\[ M_f(x) = 1 + 0\cdot x+\dfrac{-1}{2!}x^2+\dfrac{0}{3!}x^3+\dfrac{1}{4!}x^4+\dfrac{0}{5!}x^5+\dfrac{-1}{6!}x^6+\cdots \]

- Simplifying it:

\[ M_f(x) = 1 -\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots. \]

- In sigma notation, and considering the convergence interval, we have

\[ f(x) = \sum_{n=0}^{\infty}(-1)^{\tfrac{n}{2}}\dfrac{x^{2n}}{(2n)!}. \]

Maclaurin series can be useful for many other situations, one you know the series expansion for a given function, you can use it to find the series expansion for other related functions, let's see some examples:

Find a power series expansion for the function \( f(x)=x^2e^x\) centered at \(x=0\).

**Solution**

In order to solve this, let's start by writing the Maclaurin series expansion of \( g(x)=e^x\), since this is centered at \(x=0\):

** Step 1:** First, let's consider the derivatives of \( g(x)\), as this is the function \( e^x\) this is easy:

\[ g^{(n)}(x)=e^x, \forall n\ge 0\]

** Step 2:** Evaluate the derivatives at \(x=0\)

\[ g^{(n)}(0)=1\]

** Step 3:** Apply the result in the Maclaurin series formula

\[ M_g(x) = \sum_{n=0}^{\infty}\dfrac{1}{n!}x^n \]

Therefore we have:

\[ g(x) = \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \]

We can easily calculate the interval of convergence, which is \( (-\infty,+\infty)\).

- Now consider that \( f(x)=x^2\cdot g(x) \):

\[ f(x) =x^2 \cdot \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \]

- Simplifying it we have

\[\begin{align} f(x) &=\sum_{n=0}^{\infty}\dfrac{x^2\cdot x^n}{n!} \\ f(x) &=\sum_{n=0}^{\infty}\dfrac{x^{n+2}}{n!} \end{align}\]

Hence the power series expansion for the function \( f(x)=x^2e^x\) centered at \( x=0\) is

\[ f(x) =\sum_{n=0}^{\infty}\dfrac{x^{n+2}}{n!}\]

Here's another example.

Write a power series expansion for \( f(x)=\cosh(x)\) centered at \(x=0\).

**Solution**

To solve this you can either use the definition of Maclaurin series by calculating each derivative of \( f(x)\), or you can apply the definition of \( \cosh(x)=\dfrac{e^x+e^{-x}}{2}\).

Let's check both of them, starting with the **Maclaurin series definition**.

** Step 1:** Calculate the derivatives of \( f(x)\):

\[\begin{align} f(x) &=\cosh(x) \\ \\ f'(x) &=\sinh(x) \\ \\ f''(x) &=\cosh(x) \\ \\ f'''(x) &=\sinh(x) \end{align}\]

** Step 2:** Evaluate each derivative at \( x=0 \):

\[\begin{align} f(0) &=\cosh(0)=1 \\ \\ f'(0) &=\sinh(0)=0 \\ \\ f''(0) &=\cosh(0)=1 \\ \\ f'''(0) &=\sinh(0)=0 \end{align}\]

** Step 3:** Apply these results to the Maclaurin series formula:

\[ M_f(x) = 1 + 0\cdot x+\dfrac{1}{2!}x^2+\dfrac{0}{3!}x^3+\dfrac{1}{4!}x^4+\dfrac{0}{5!}x^5+\dfrac{1}{6!}x^6+\cdots \]

- Simplifying it:

\[ f(x) = 1 +\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\cdots \]

- In sigma notation, and considering the convergence interval, we have

\[ f(x) = \sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!}. \]

Now let's see how can we solve this using the **hyperbolic cosine definition**:

- Looking at the \( \cosh(x) \) definition we have:

\[ \cosh(x)=\dfrac{e^x+e^{-x}}{2} \]

- From the previous example we have:

\[ e^x = \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \]

- Let's evaluate the series expansion with \( -x \):

\[ \begin{align} e^{-x} &= \sum_{n=0}^{\infty}\dfrac{(-x)^n}{n!} \\ e^{-x} &= \sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{n!} \end{align}\]

- Let's expand the terms of the series for \( e^x\) and \( e^{-x}\) and sum it:

\[ \begin{align} e^{x} &= 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\cdots \\ \\ e^{-x} &= 1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\dfrac{x^5}{5!}+\cdots \\ \\ e^x+e^{-x} &= 2+0+2\dfrac{x^2}{2!}+0+2\dfrac{x^4}{4!}+0+\cdots \\ \\ e^x+e^{-x} &= 2+2\dfrac{x^2}{2!}+2\dfrac{x^4}{4!}+\cdots \end{align}\]

- To have the hyperbolic cosine we still need to divide it by two:

\[ \begin{align} \dfrac{e^x+e^{-x}}{2} &= \dfrac{1}{2}\left(2+2\dfrac{x^2}{2!}+2\dfrac{x^4}{4!}+\cdots\right) \\ \\ \dfrac{e^x+e^{-x}}{2} &= 1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots \end{align}\]

- Writing it with sigma notation:

\[ f(x) = \sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!}, \]

Which is the same as the first part.

**Maclaurin Series**of \(f\)\[ M_f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n \]

Inside the convergence interval, the Maclaurin Series is equal to \(f\)

\[ f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n \]

Some Maclaurin series expansions:

\[ \begin{align} e^x &= \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \\ \sin(x) &= \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!} \\ \cos(x) &= \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!} \\ \ln(1+x) &= \sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n} \\ \sinh(x) &= \sum_{n=0}^{\infty}\dfrac{x^{2n+1}}{(2n+1)!} \\ \cosh(x) &= \sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!}\end{align}\]

- To find the
**convergence interval**you need to apply the Ratio Test

\[ \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| <1\]

A Maclaurin series is just a Taylor series centered at \(x=0\).

No, a Maclaurin series is a special case of a Taylor series centered at \( x=0 \).

More about Maclaurin Series

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