Linear/Quadratic Approximation Cheatsheet

1. Definitions

1.1 Linear approximation

$$f(x) \approx f(0) + f'(0)\,x$$

Replace the curve with the straight line that matches it at $x=0$ in height and slope.

1.1.1 Slope intuition

A line through $(0, f(0))$ needs one more number: its slope. Pick the slope wrong and the line peels away from the curve immediately. Pick $f'(0)$ and it stays glued for as long as possible.

1.1.2 Reconstructing a and b

The other way to read the formula: linear approximation is a search for the best $a + bx$.

Write $P(x) = a + bx$ and demand it agree with $f$ at $0$:

$$\begin{aligned} P(0) &= a &&\Rightarrow\ a = f(0) \\ P'(0) &= b &&\Rightarrow\ b = f'(0) \end{aligned}$$

There is no choice left — matching height and slope forces both coefficients. So if you hand the formula something that already is a line, it must hand the same line back:

1.2 Quadratic approximation

$$f(x) \approx f(0) + f'(0)\,x + \frac{f''(0)}{2}x^2$$

Same idea, one more term: match the curve at $x=0$ in height, slope, and curvature.

1.2.1 Parabola intuition

A line can only match height and slope; it is always straight while the curve bends. Adding an $x^2$ term lets the approximation bend too, and $f''(0)$ says how much.

1.2.2 Reconstructing a, b and c

Now the search is for the best $a + bx + cx^2$. Write $P(x) = a + bx + cx^2$:

$$\begin{aligned} P(0) &= a &&\Rightarrow\ a = f(0) \\ P'(0) &= b &&\Rightarrow\ b = f'(0) \\ P''(0) &= 2c &&\Rightarrow\ c = \tfrac{1}{2}f''(0) \end{aligned}$$

That $\tfrac{1}{2}$ is not a fudge factor — it is there because $P''(0)$ comes out as $2c$ rather than $c$.

Set $c=0$ and the quadratic formula collapses into the linear one. Linear approximation is the quadratic with curvature switched off, which is why a function with $f''(0)=0$ has the same linear and quadratic approximation.

2. Cheatsheet

2.1 Near 0

Each entry lists $f(0)$, $f'(0)$ and $f''(0)$, then both approximations, then the values against a direct calculation, then the graphs.

2.1.1 Trigonometric

2.1.1.1$\sin x$

$f(x)=\sin x$	$f(0)=0$
$f^{\prime }(x)=\cos x$	$f^{\prime }(0)=1$
$f^{\prime \prime }(x)=-\sin x$	$f^{\prime \prime }(0)=0$

linear

$\sin x\approx x$

quadratic

$\sin x\approx x$

x	direct	linear	quadratic	quad. error
0	0.000000	0.000000	0.000000	0.000000
0.01	0.010000	0.010000	0.010000	1.67e-7
0.1	0.099833	0.100000	0.100000	0.000167
0.5	0.479426	0.500000	0.500000	0.020574
1	0.841471	1.000000	1.000000	0.158529

sin xlinearquadratic

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Odd function, so f″(0) = 0 and the quadratic term vanishes — the quadratic approximation is the linear one.

2.1.1.2$\cos x$

$f(x)=\cos x$	$f(0)=1$
$f^{\prime }(x)=-\sin x$	$f^{\prime }(0)=0$
$f^{\prime \prime }(x)=-\cos x$	$f^{\prime \prime }(0)=-1$

linear

$\cos x\approx 1$

quadratic

$\cos x\approx 1-\frac{x^2}{2}$

x	direct	linear	quadratic	quad. error
0	1.000000	1.000000	1.000000	0.000000
0.01	0.999950	1.000000	0.999950	4.17e-10
0.1	0.995004	1.000000	0.995000	4.17e-6
0.5	0.877583	1.000000	0.875000	0.002583
1	0.540302	1.000000	0.500000	0.040302

cos xlinearquadratic

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Flat at 0, so the linear approximation is just the constant 1. All the information is in the quadratic term.

2.1.1.3$\tan x$

$f(x)=\tan x$	$f(0)=0$
$f^{\prime }(x)={\sec }^{2}x$	$f^{\prime }(0)=1$
$f^{\prime \prime }(x)=2{\sec }^{2}x\tan x$	$f^{\prime \prime }(0)=0$

linear

$\tan x\approx x$

quadratic

$\tan x\approx x$

x	direct	linear	quadratic	quad. error
0	0.000000	0.000000	0.000000	0.000000
0.01	0.010000	0.010000	0.010000	3.33e-7
0.1	0.100335	0.100000	0.100000	0.000335
0.5	0.546302	0.500000	0.500000	0.046302
1	1.557408	1.000000	1.000000	0.557408

tan xlinearquadratic

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Also odd, so again the quadratic term vanishes.

2.1.1.4$\sec x$

$f(x)=\sec x$	$f(0)=1$
$f^{\prime }(x)=\sec x\tan x$	$f^{\prime }(0)=0$
$f^{\prime \prime }(x)=\sec x{\tan }^{2}x+{\sec }^{3}x$	$f^{\prime \prime }(0)=1$

linear

$\sec x\approx 1$

quadratic

$\sec x\approx 1+\frac{x^2}{2}$

x	direct	linear	quadratic	quad. error
0	1.000000	1.000000	1.000000	0.000000
0.01	1.000050	1.000000	1.000050	2.08e-9
0.1	1.005021	1.000000	1.005000	2.09e-5
0.5	1.139494	1.000000	1.125000	0.014494
1	1.850816	1.000000	1.500000	0.350816

sec xlinearquadratic

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The mirror of cos x — same flat linear part, opposite sign on the quadratic term.

2.1.1.5$\cot x$

$$\cot x=\frac{\cos x}{\sin x},\sin 0=0$$

No approximation at 0. cot x blows up there, so there is no finite value, slope, or curvature to match.

2.1.1.6$\csc x$

$$\csc x=\frac{1}{\sin x},\sin 0=0$$

Same problem as cot x — undefined at 0, so no expansion exists there.

2.1.2 Exponents and logarithms

2.1.2.1$e^x$

$f(x)=e^x$	$f(0)=1$
$f^{\prime }(x)=e^x$	$f^{\prime }(0)=1$
$f^{\prime \prime }(x)=e^x$	$f^{\prime \prime }(0)=1$

linear

$e^x\approx 1+x$

quadratic

$e^x\approx 1+x+\frac{x^2}{2}$

x	direct	linear	quadratic	quad. error
0	1.000000	1.000000	1.000000	0.000000
0.01	1.010050	1.010000	1.010050	1.67e-7
0.1	1.105171	1.100000	1.105000	0.000171
0.5	1.648721	1.500000	1.625000	0.023721
1	2.718282	2.000000	2.500000	0.218282

e^xlinearquadratic

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Every derivative is e^x, so every coefficient at 0 is 1. This is the cleanest one to remember.

2.1.2.2$\ln (1+x)$

$f(x)=\ln (1+x)$	$f(0)=0$
$f^{\prime }(x)=\frac{1}{1+x}$	$f^{\prime }(0)=1$
$f^{\prime \prime }(x)=-\frac{1}{(1+x{)}^2}$	$f^{\prime \prime }(0)=-1$

linear

$\ln (1+x)\approx x$

quadratic

$\ln (1+x)\approx x-\frac{x^2}{2}$

x	direct	linear	quadratic	quad. error
0	0.000000	0.000000	0.000000	0.000000
0.01	0.009950	0.010000	0.009950	3.31e-7
0.1	0.095310	0.100000	0.095000	0.000310
0.5	0.405465	0.500000	0.375000	0.030465
1	0.693147	1.000000	0.500000	0.193147

ln(1 + x)linearquadratic

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Written as ln(1 + x) rather than ln x precisely so that the expansion point sits at 0 — ln x has no expansion there.

2.1.2.3$(1+x{)}^r$

$f(x)=(1+x{)}^r$	$f(0)=1$
$f^{\prime }(x)=r(1+x{)}^{r-1}$	$f^{\prime }(0)=r$
$f^{\prime \prime }(x)=r(r-1)(1+x{)}^{r-2}$	$f^{\prime \prime }(0)=r(r-1)$

linear

$(1+x{)}^r\approx 1+rx$

quadratic

$(1+x{)}^r\approx 1+rx+\frac{r(r-1)}{2}{x}^2$

r =1/2−1/2−123

x	direct	linear	quadratic	quad. error
0	1.000000	1.000000	1.000000	0.000000
0.01	1.004988	1.005000	1.004987	6.21e-8
0.1	1.048809	1.050000	1.048750	5.88e-5
0.5	1.224745	1.250000	1.218750	0.005995
1	1.414214	1.500000	1.375000	0.039214

(1 + x)^rlinearquadratic

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The workhorse. r = 1/2 gives √(1+x), r = −1 gives 1/(1+x). At r = 2 the quadratic approximation is exact — (1+x)² already is a quadratic, so there is nothing to discard.

2.1.2.4$a^x$

$f(x)=a^x$	$f(0)=1$
$f^{\prime }(x)=a^x\ln a$	$f^{\prime }(0)=\ln a$
$f^{\prime \prime }(x)=a^x{\ln }^{2}a$	$f^{\prime \prime }(0)={\ln }^{2}a$

linear

$a^x\approx 1+x\ln a$

quadratic

$a^x\approx 1+x\ln a+\frac{{\ln }^{2}a}{2}{x}^2$

a =2310

x	direct	linear	quadratic	quad. error
0	1.000000	1.000000	1.000000	0.000000
0.01	1.006956	1.006931	1.006955	5.56e-8
0.1	1.071773	1.069315	1.071717	5.65e-5
0.5	1.414214	1.346574	1.406630	0.007583
1	2.000000	1.693147	1.933374	0.066626

a^xlinearquadratic

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Generalises e^x — set a = e and ln a = 1, recovering 1 + x + x²/2.

3. Combining approximations

Rather than differentiating a complicated function twice, build its approximation out of the entries above. The one move that covers most cases: substitute, then truncate.

operation	rule
$f + g$	add the approximations
$k \cdot f$	scale the approximation by $k$
$f \cdot g$	multiply the approximations, then truncate
$f \circ g$, where $g(0) = 0$	substitute $g$'s approximation into $f$'s, then truncate

Truncate means: discard every power above the order you are working to — $x^2$ and up for linear, $x^3$ and up for quadratic.

For composition there is also the direct form, which is just the chain rule:

$$f(g(x)) \approx f(g(0)) + f'(g(0))\,g'(0)\,x$$

Two conditions worth keeping in mind:

• The inside function must land near 0. Every entry above is an expansion around $0$, so substitution is only legal when $g(0) = 0$.
• Truncating is not optional. The product of two linear approximations contains an $x^2$ term that is not trustworthy — the inputs were only accurate to first order, so the second-order information in the product is noise.

That second point is worth seeing. Multiplying two linear approximations gives

$$\big(f(0) + f'(0)x\big)\big(g(0) + g'(0)x\big) = f(0)g(0) + \big[f'(0)g(0) + f(0)g'(0)\big]x + f'(0)g'(0)x^2$$

Drop the $x^2$ and the coefficient of $x$ is $f'(0)g(0) + f(0)g'(0)$ — exactly $(fg)'(0)$ from the product rule. The shortcut agrees with the definition, which is why it is safe.