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# Taylor and MacLaurin series

## Higher derivative

For an arbitrary integer $k$, the derivative of order $k$ of a function $f$ is computed by differentiating it $k$ times. It is defined by induction $$f^{(0)}(x_0) =f(x_0),\qquad f^{(k+1)}(x_0) = \frac{d}{dx}f^{(k)}(x_0)$$

The usual derivative is of order 1; the second derivative is of order 2, etc. The derivative of order $k$ is also denoted by $\displaystyle \frac{d^k}{dx^k}f(x)$.

The derivative of $f(x) = e^x$ is $f'(x)=e^{x}$; so its second derivative is $e^{x}$ as well and so on. Therefore $f^{(k)}(x) = e^{x}$ for all $k$.

The derivatives of order $k$ of some common functions are summarised below. \begin{gather*} (e^x)^{(k)} = e^x\\ (\sin x)^{(2k)} = (-1)^k\sin x,\quad (\sin x)^{(2k+1)} = (-1)^k\cos x\\ (\cos x)^{(2k)} = (-1)^k\cos x,\quad (\cos x)^{(2k+1)} = (-1)^{k+1}\sin x\\ ( x^\alpha)^{(k)} = \alpha (\alpha -1)\dots(\alpha -k+1) x^{\alpha -k} \end{gather*}

All these formulae an be proved by induction.

$x^4$ and its multiple derivatives

## Asymptotic expansions

The Taylor series approximates a function by a polynomial near a point \begin{align*} f(x) &= f(x_0) + f'(x_0) (x-x_0)+ \frac{f'(x_0)}{2!}(x-x_0)^2 +\dots\\ &= \sum_{k=0}^{\infty} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k \end{align*} It refines the approximation of a function by its tangent.

The series is also called Taylor expansion or asymptotic expansion. The polynomial up to order $k$ is the Taylor expansion of order $k$.

When $x_0=0$, the Taylor series is called a MacLaurin series $$f(x) = f(0) + f'(0) x+ \frac{f'(0)}{2!}x^2 +\dots + \frac{f^{(k)}(0)}{k!}x^k +\dots.$$

For instance, the MacLaurin series of the function $f(x) = e^x$ is $$e^x = 1+x+\frac{x^2}{2!}+\dots+\frac{x^k}{k!}+\dots$$ We calculate this using $f^{(k)}(x) = e^{x}$, and so $f^{(k)}(0) = 1$ for all $k$.

Partial sums of the MacLaurin series locally approximate $e^x$

## Justification of MacLaurin series

The MacLaurin series of an indefinitely differentiable function $f$ is $$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$$

We explain how to find the coefficients of the polynomial. Consider an infinite polynomial that approximates $f$ $$g(x) = a_0+a_1x+a_2x^2+\dots +a_kx^k+\dots$$ We must have $f(0) = g(0) = a_0$. By differentiation, $$g'(x) = a_1+2 a_2x^2+\dots+k a_kx^{k-1}+\dots$$ Therefore $f'(0) = g'(0) = a_1$. Repeating the argument, we find $$a_k = \frac{f^{(k)}(0)}{k!} .$$

We clarify the identification of a function and its MacLaurin series

• Depending on the function, the series may converge for all $x$ or for certain values of $x$ only. When the series converges, the function generally equals the series (though there are some exceptions).
• This said, the function can always be approximated by a polynomial, in the sense that, for some constant $M$, $$\vert f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^k\vert \le M \vert x\vert ^{n+1}$$

## Asymptotic expansion of common functions

The MacLaurin series of common functions are obtained by computing their successive derivatives. Some are given below.

\begin{align*} e^x & = 1+x + \dots+ \frac{x^k}{k!}+\dots\\ \sin x &= x - \frac{x^3}{3!} + \dots+ (-1)^k \frac{x^{2k+1}}{(2k+1)!}+\dots\\ \cos x &= 1- \frac{x^2}{2!} + \dots+ (-1)^k \frac{ x^{2k}}{(2k)!}+\dots\\ (1+x)^n &= 1+x + \dots+ \frac{n(n-1)\dots(n-k+1) }{k!}x^k +\dots\\ \frac{1}{1+x} &= 1- x + \dots+ (-1)^k x^k+\dots\\ \ln(1+x) &= x - \frac{x^2}{2}+ \dots+ (-1)^{k+1}\frac{x^k}{k}+\dots\\ \end{align*}

For instance, for $f(x) = \sin x$, we have that $$f^{(2k)}(x) = (-1)^k\sin x,\qquad f^{(2k+1)}(x) = (-1)^k\cos x.$$ Therefore, $f^{(2k)}(0) = 0$ and $f^{(2k+1)}(0) = (-1)^k$. We deduce that $$\sin x = \sum_{k'=0}^{\infty} \frac{f^{(k')}(0)}{k'!}x^{k'} = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1} }{(2k+1)!}$$