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.
Explore our app and discover over 50 million learning materials for free.
Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken
Jetzt kostenlos anmeldenNie wieder prokastinieren mit unseren Lernerinnerungen.
Jetzt kostenlos anmeldenFor 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:
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:
\[ M_f(x) = 0+ 1\cdot x+\dfrac{-1}{2!}x^2+\dfrac{2!}{3!}x^3+\dfrac{-3!}{4!}x^4+\cdots \]
\[ M_f(x) = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots \]
\[ M_f(x) = \sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}, \]
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:
\[f^{(n)}(0)=0\]
\[f^{(n)}(0)=(-1)^{\tfrac{n}{2}}\]
Step 3: Apply these results to the Maclaurin series formula:
\[ M_f(x) = 1 + 0\cdot x+\dfrac{-1}{2!}x^2+\dfrac{0}{3!}x^3+\dfrac{1}{4!}x^4+\dfrac{0}{5!}x^5+\dfrac{-1}{6!}x^6+\cdots \]
\[ M_f(x) = 1 -\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots. \]
\[ f(x) = \sum_{n=0}^{\infty}(-1)^{\tfrac{n}{2}}\dfrac{x^{2n}}{(2n)!}. \]
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 \( (-\infty,+\infty)\).
\[ f(x) =x^2 \cdot \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \]
\[\begin{align} f(x) &=\sum_{n=0}^{\infty}\dfrac{x^2\cdot x^n}{n!} \\ f(x) &=\sum_{n=0}^{\infty}\dfrac{x^{n+2}}{n!} \end{align}\]
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:
\[ M_f(x) = 1 + 0\cdot x+\dfrac{1}{2!}x^2+\dfrac{0}{3!}x^3+\dfrac{1}{4!}x^4+\dfrac{0}{5!}x^5+\dfrac{1}{6!}x^6+\cdots \]
\[ f(x) = 1 +\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\cdots \]
\[ f(x) = \sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!}. \]
Now let's see how can we solve this using the hyperbolic cosine definition:
\[ \cosh(x)=\dfrac{e^x+e^{-x}}{2} \]
\[ e^x = \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \]
\[ \begin{align} e^{-x} &= \sum_{n=0}^{\infty}\dfrac{(-x)^n}{n!} \\ e^{-x} &= \sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{n!} \end{align}\]
\[ \begin{align} e^{x} &= 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\cdots \\ \\ e^{-x} &= 1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\dfrac{x^5}{5!}+\cdots \\ \\ e^x+e^{-x} &= 2+0+2\dfrac{x^2}{2!}+0+2\dfrac{x^4}{4!}+0+\cdots \\ \\ e^x+e^{-x} &= 2+2\dfrac{x^2}{2!}+2\dfrac{x^4}{4!}+\cdots \end{align}\]
\[ \begin{align} \dfrac{e^x+e^{-x}}{2} &= \dfrac{1}{2}\left(2+2\dfrac{x^2}{2!}+2\dfrac{x^4}{4!}+\cdots\right) \\ \\ \dfrac{e^x+e^{-x}}{2} &= 1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots \end{align}\]
\[ f(x) = \sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!}, \]
Which is the same as the first part.
\[ M_f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n \]
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:
\[ \begin{align} e^x &= \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \\ \sin(x) &= \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!} \\ \cos(x) &= \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!} \\ \ln(1+x) &= \sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n} \\ \sinh(x) &= \sum_{n=0}^{\infty}\dfrac{x^{2n+1}}{(2n+1)!} \\ \cosh(x) &= \sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!}\end{align}\]
\[ \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| <1\]
A Maclaurin series is just a Taylor series centered at \(x=0\).
To find a Maclaurin series, you need first to calculate the derivatives of the given function and evaluate it at \( x=0\), then apply the Maclaurin series formula.
No, a Maclaurin series is a special case of a Taylor series centered at \( x=0 \).
It is named after Colin Maclaurin because he studies this particular case of the Taylor series in depth.
The formula for the Maclaurin series is given by the derivatives of the given function evaluated at \( x=0\). To see the precise formula take a look at our Maclaurin series article.
What is the main difference between the Taylor series and the Maclaurin series?
The Maclaurin series is a particular case of the Taylor series centered at \( x=0 \).
The Maclaurin series expansion for a function \( f \) is given by \[ M_f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n\]
True
If \( M_f(x) \) is the Maclaurin series expansion of \( f(x) \) then the following is true: \[ M^{(n)}_f(0) = f^{(n)}(0) \]
True
Can any function \( f(x) \) be written as a Maclaurin series expansion \[ f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n?\]
No, this is only possible if \(f\) has the derivatives of all orders at \( x=0\). Also, the equality is only true inside the convergence interval.
Given a function \( f\) with derivatives of all orders at \( x=0 \), what are the steps to write the Maclaurin series of \( f\)?
First, find all the derivatives of \( f\);
Second, evaluate the derivatives at \( x=0 \);
Third, apply the results in the Maclaurin series formula;
Forth, find the interval of convergence.
Given the Maclaurin series \( e^x = \sum_{n=0}^{\infty}\dfrac{x^n}{n!} \) how can you find a series expansion of \( x^2e^x\)?
You can multiply the whole series by \( x^2\) to get \[x^2e^x = \sum_{n=0}^{\infty}\dfrac{x^{n+2}}{n!} \]
Already have an account? Log in
Open in AppThe first learning app that truly has everything you need to ace your exams in one place
Sign up to highlight and take notes. It’s 100% free.
Save explanations to your personalised space and access them anytime, anywhere!
Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of StudySmarter.
Already have an account? Log in
Already have an account? Log in
The first learning app that truly has everything you need to ace your exams in one place
Already have an account? Log in