Taylor Expansion


Taylor expansion around x = a is to express an arbitrary function f (x) in a form like the expression below.


When a = 0, it can be written as follows. This case is called “Maclaurin’s expansion” specially.


\(f^{(k)}(x)\) means the \(k\)th order derivative of the function \(f(x)\).

A Process of Taylor expansion

In this example, let us talk about from the first term of the Taylor expansion around x=a to the \(n\)th.

①Find the n th derivative from the first derivative of the function f (x). For example, in the case of n = 3, it is necessary to obtain the first derivative, the second derivative, and the third derivative.

②Substitute x = a into the every derivative obtained by f (x) and ①.

③Calculate the sum of derivatives obtained in ②.


f(x)=sinx, n=3 Maclaurin’s expansion

①First-order, second-order and third-order derivatives of sinx are cosx, -sinx, and -cosx, respectively.

②We get a=0 because we are thinking about McLoughlin deployment now. Therefore, substitute x = 0, and we get sin 0 = 0, cos 0 = 1, – sin 0 = 0 and – cos 0 = – 1.

③Calculate the sum of derivatives.

\begin{eqnarray} f(x)&=&\frac{f(0)}{0!}(x-0)^0+\frac{f^{(1)}(0)}{1!}(x-0)^1+\frac{f^{(2)}(0)}{2!}(x-0)^2+\frac{f^{(3)}(0)}{3!}(x-0)^3 \\&=&sin0+(cos0)x+\frac{-sin0}{2}x^2+\frac{-cos0}{6}x^3\\&=&x-\frac{1}{6}x^3 \end{eqnarray}


f(x)=cosx, n=3 Maclaurin’s expansion

Calculate in exactly the same way as for sinx.

\begin{eqnarray} cosx&=&\frac{f(0)}{0!}(x-0)^0+\frac{f^{(1)}(0)}{1!}(x-0)^1+\frac{f^{(2)}(0)}{2!}(x-0)^2+\frac{f^{(3)}(0)}{3!}(x-0)^3 \\&=&cos0+(-sin0)x+\frac{-cos0}{2}x^2+\frac{sin0}{6}x^3\\&=&1-0・x-\frac{1}{2}x^2+\frac{0}{6}x^3\\&=&x-\frac{1}{2}x^2 \end{eqnarray}

Odd and even functions

Since sinx is an odd function, it includes odd functions \(x\) and \(x^3\). Similarly, since cosx is an even function, it includes the even function \(x^2\).


We confirmed the definition of Taylor expansion and its examples.

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