FP2.6 - Maclaurin and Taylor series

The $n$th derivative of $y=f(x)$ is written $f^{(n)}(x)$ or $\dfrac{d^ny}{dx^n}$. Higher derivatives are found by differentiating successively, but patterns, recurrences and a derivative table prevent repeated algebra from becoming unmanageable.

For $y=\ln(2+x)$,$$y'=\frac1{2+x},\qquad y''=-\frac1{(2+x)^2},\qquad y'''=\frac2{(2+x)^3}.$$The powers, alternating signs and factorial factors give $$y^{(n)}=\frac{(-1)^{n-1}(n-1)!}{(2+x)^n},\qquad n\ge1.$$ Substituting $n=1,2,3$ checks the general expression.

When a function satisfies a differential relation, differentiating that relation can be shorter than starting again from the original function. Keep every product-rule term: differentiating $x,y''$ gives $y''+x,y'''$. Values of lower derivatives at a point can then determine a required higher derivative there.

A Maclaurin series is a Taylor series centred at $0$. Its coefficients are chosen so that the polynomial and the function have the same successive derivatives at $x=0$:$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+\cdots.$$

If $f(x)=a_0+a_1x+a_2x^2+\cdots$, then setting $x=0$ gives $a_0=f(0)$. Differentiate once and set $x=0$ to get $a_1=f'(0)$; differentiating $r$ times gives $f^{(r)}(0)=r!a_r$. Hence $a_r=f^{(r)}(0)/r!$.

The logarithmic expansion follows because for $r\ge1$, $$\left.\frac{d^r}{dx^r}\ln(1+x)\right|_{x=0}=(-1)^{r-1}(r-1)!.$$ Standard series may then be adapted and combined. For example,$$\ln\frac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=2x+\frac{2x^3}{3}+O(x^5).$$ Even powers cancel, so only the needed terms should be retained.

A Taylor series expands $f(x)$ about a chosen centre $x=a$ in powers of the displacement $h=x-a$:$$f(x)=f(a)+f'(a)h+\frac{f''(a)}{2!}h^2+\frac{f'''(a)}{3!}h^3+\cdots.$$The coefficients therefore describe the function's value and successive local changes at $a$, not at $0$.

For the official example $f(x)=\sin x$ about $a=\pi$, let $h=x-\pi$. The derivative values are$$f(\pi)=0,\quad f'(\pi)=-1,\quad f''(\pi)=0,\quad f'''(\pi)=1.$$Therefore, up to and including $h^3$,$$\sin x=-(x-\pi)+\frac{(x-\pi)^3}{6}+O\!\left((x-\pi)^4\right).$$ Directly using $\sin(\pi+h)=-\sin h$ confirms the signs.

The Taylor-series method solves a differential equation locally by using the equation and its derivatives to generate $y(a),y'(a),y''(a),\ldots$ at the expansion centre. A term through $(x-a)^n$ requires derivative values through $y^{(n)}(a)$.

For the official model$$y''+xy'+y=0,\qquad y(0)=1,\quad y'(0)=0,$$the original equation at $0$ gives $y''(0)=-1$. Differentiating once gives$$y'''+2y'+xy''=0,$$so $y'''(0)=0$. Differentiating again gives$$y''''+3y''+xy'''=0,$$so $y''''(0)=3$. Hence, through $x^4$,y=1-rac{x^2}{2!}+\frac{3x^4}{4!}+O(x^5)=1-\frac{x^2}{2}+\frac{x^4}{8}+O(x^5).

Equivalently, substitute $y=a_0+a_1x+a_2x^2+\cdots$ and its differentiated series into the equation, then equate coefficients. The derivative-at-the-centre method is usually shorter when initial values are supplied explicitly.