Taylor series and approximation

The Taylor polynomial

Idea. Near a point $a$, a smooth function is close to the polynomial that matches its value and first $n$ derivatives at $a$:

$T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x - a)^n .$

With $a = 0$ this is the Maclaurin polynomial. If the remainder tends to 0 as $n \to \infty$, the infinite sum is the Taylor series, and it equals $f$.

Why the factorials. Differentiating $(x - a)^k$ exactly $k$ times gives $k!$. Dividing by $k!$ makes the $k$-th derivative of $T_n$ at $a$ equal to $f^{(k)}(a)$.

Remainder (Lagrange form).

$f(x) - T_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1} \quad \text{for some } c \text{ between } a \text{ and } x.$

So if $\lvert f^{(n+1)}\rvert \le M$ on the interval, the error is at most $\dfrac{M\lvert x - a\rvert^{n+1}}{(n+1)!}$.

Series to know by heart

Function	Maclaurin series	Valid for
$e^x$	$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$	all $x$
$\sin x$	$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$	all $x$
$\cos x$	$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$	all $x$
$\frac{1}{1 - x}$	$1 + x + x^2 + x^3 + \dots$	$\lvert x\rvert < 1$
$\ln(1 + x)$	$x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$	$-1 < x \le 1$
$(1 + x)^\alpha$	$1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \dots$	$\lvert x\rvert < 1$
$\arctan x$	$x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$	$\lvert x\rvert \le 1$
$\tan x$	$x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$	$\lvert x\rvert < \frac{\pi}{2}$
$\sinh x$, $\cosh x$	like $\sin$ and $\cos$, but with all signs $+$	all $x$

Patterns.

• $\sin$ is odd and contains only odd powers. $\cos$ is even and contains only even powers.
• The derivative of the $e^x$ series is itself. The derivative of the $\sin$ series is the $\cos$ series.
• Substituting $x \to -x$ in $\frac{1}{1-x}$ gives $\frac{1}{1 + x} = 1 - x + x^2 - \dots$. Integrating that gives $\ln(1 + x)$.
• Substituting $x \to x^2$ in $\frac{1}{1+x}$ and integrating gives $\arctan x$.

Computing Taylor expansions

Example 1: $e^x$ about $a = 1$. Every derivative equals $e$ at 1, so $e^x = e\left[1 + (x - 1) + \frac{(x - 1)^2}{2!} + \dots\right].$

Example 2: $\ln x$ about $a = 1$. $f(1) = 0$, $f'(1) = 1$, $f''(1) = -1$ and $f'''(1) = 2$. So $\ln x = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \dots$ This is the $\ln(1 + t)$ series with $t = x - 1$.

Example 3: approximate $\sqrt{4.1}$. Take $f(x) = \sqrt x$ and $a = 4$.

• $f(4) = 2$.
• $f'(x) = \frac{1}{2\sqrt x}$, so $f'(4) = \frac14$.
• $f''(x) = -\frac{1}{4x^{3/2}}$, so $f''(4) = -\frac{1}{32}$.

$\sqrt{4.1} \approx 2 + \frac{0.1}{4} - \frac{1}{32}\cdot\frac{0.01}{2} = 2 + 0.025 - 0.00015625 = 2.02484375 .$

The true value is $2.0248457\ldots$. The first-order (linear) approximation 2.025 is already within $1.6 \times 10^{-4}$.

Example 4: an error bound. Approximate $e^{0.5}$ by $T_3 = 1 + 0.5 + 0.125 + 0.0208\overline{3} = 1.6458\overline{3}$.

• Every derivative of $e^x$ on $[0, 0.5]$ is at most $e^{0.5} < 1.65$.
• So the error is at most $\frac{1.65 \times 0.5^4}{4!} = \frac{1.65 \times 0.0625}{24} \approx 0.0043$.

The true error is $1.64872 - 1.64583 \approx 0.0029$, within the bound.