Electricity-And-Magnetism · Unit 1 · Video 1 · Interactive Practice

Series Approximations: Keeping One or Two Terms

IKey Formulas

Formula	Name	Valid when
$f(x) \approx f(0) + f'(0)\,x + \dfrac{f''(0)}{2!}\,x^2 + \cdots$	Series about $x = 0$	$f$ well behaved near $0$
$(1+x)^m \approx 1 + mx$	Power expansion ( $m$ any number)	$\|x\| \ll 1$
$e^x \approx 1 + x, \qquad \ln(1+x) \approx x$	Exponential and logarithm	$\|x\| \ll 1$
$\sin x \approx x, \quad \tan x \approx x, \quad \cos x \approx 1 - \dfrac{x^2}{2}$	Small angles	$\|x\| \ll 1$ , $x$ in radians

Key Insight: Keep the constant and the first term that varies with $x$ . The leading term is the limiting value, the next term is the first correction, and the error is roughly the size of the first term you drop.

IIThe Truncated Series and Its Error

Near $x = 0$ a function hugs its truncated series, and the error is roughly the first dropped term.

Function Terms kept x 0.50

IIISmall Angles on the Unit Circle

For a small angle in radians, the arc $x$ , the height $\sin x$ and $\tan x$ nearly coincide.

Angle (degrees) 20°

💡 An angle in degrees must be converted first, $x = \theta\,\pi/180$ : five degrees is $0.0873$ rad and $\sin 5\degree = 0.0872$ , not $5$ .

IVCasting into Standard Form

Factoring out $b$ gives $(b-y)^{-1/3} = b^{-1/3}(1 - y/b)^{-1/3}$ , a power expansion with $x = -y/b$ , $m = -\tfrac13$ .

y (with b = 8) 0.08

VQuiz Questions

Problem 1 · The Power Expansion

Given: $(1+x)^m \approx 1 + mx$ — estimate $\sqrt{1.02}$ .

✅ Correct! With

x = 0.02

and

m = \tfrac12

1 + mx = 1.01

; the exact value is

1.00995

❌ Not quite. That is

1 + x

: the power

m = \tfrac12

must multiply

x

❌ Not quite.

x = +0.02

and

m = +\tfrac12

, so the correction

mx

is positive.

❌ Not quite. A square root is the power

m = \tfrac12

, not

m = 2

❌ Not quite. Write

\sqrt{1.02} = (1 + 0.02)^{1/2}

and read off

x

and

m

Show solution

Write $\sqrt{1.02} = (1 + 0.02)^{1/2}$ , so $x = 0.02$ and $m = \tfrac12$ . Since $|x| \ll 1$ :

\sqrt{1.02} \approx 1 + \tfrac12(0.02) = 1.01

The exact value is $1.0099505\ldots$ The error, $-0.00005$ , matches the first dropped term $\tfrac{m(m-1)}{2}x^2 = -\tfrac18(0.02)^2 = -0.00005$ .

Problem 2 · Cosine Needs the Square Term

Given: $x = 0.2$ rad — estimate $\cos(0.2)$ by keeping the constant and the first term that varies with $x$ .

✅ Correct!

1 - \tfrac{(0.2)^2}{2} = 0.98

, against the exact

0.98007

❌ Not quite.

1

is only the constant term; it carries no information about

x

. The first term that varies is

-x^2/2

❌ Not quite. Cosine has slope

0

x = 0

, so there is no

-x

term; the first varying term is

-x^2/2

❌ Not quite. The square term carries

1/2!

: it is

-x^2/2

, not

-x^2

❌ Not quite. Use

\cos x \approx 1 - x^2/2

Show solution

For cosine, $f(0) = 1$ , $f'(0) = -\sin 0 = 0$ and $f''(0) = -\cos 0 = -1$ , so the first term that varies is $\tfrac{f''(0)}{2!}x^2 = -\tfrac{x^2}{2}$ :

\cos(0.2) \approx 1 - \frac{(0.2)^2}{2} = 1 - 0.02 = 0.98

The exact value is $0.980067\ldots$ ; the error $0.000067$ is about the next term $x^4/24 = 0.0000667$ .

Problem 3 · Cast into Standard Form

Given: $(27 + y)^{1/3}$ with $|y| \ll 27$ — find a two-term approximation.

After factoring out $27$ , what are $x$ and $m$ ?

The approximation is

✅ Correct!

(27+y)^{1/3} \approx 3 + \tfrac{y}{27}

: the limiting value

3

plus a small correction.

❌ Not quite.

27 + y

does not start with

1

; factor out

27

first, which makes

x = y/27

❌ Check $x$ and $m$ .

27 + y = 27\left(1 + \tfrac{y}{27}\right)

and the power is the cube root,

m = \tfrac13

❌ Close, but… The factor

27^{1/3} = 3

multiplies the whole bracket, correction included:

3\left(1 + \tfrac{y}{81}\right) = 3 + \tfrac{y}{27}

❌ Check the result. Expand

3\left(1 + \tfrac{y}{27}\right)^{1/3} \approx 3\left(1 + \tfrac13 \cdot \tfrac{y}{27}\right)

and multiply out.

Show solution

Cast: $27 + y = 27\left(1 + \tfrac{y}{27}\right)$ , so

(27 + y)^{1/3} = 27^{1/3}\left(1 + \frac{y}{27}\right)^{1/3} = 3\left(1 + \frac{y}{27}\right)^{1/3}

Identify: $x = \tfrac{y}{27}$ , $m = \tfrac13$ , and $|x| \ll 1$ because $|y| \ll 27$ .

Expand:

3\left(1 + \frac{y}{27}\right)^{1/3} \approx 3\left(1 + \frac{y}{81}\right) = 3 + \frac{y}{27}

Check with $y = 0.54$ : $3 + 0.02 = 3.02$ , while $27.54^{1/3} = 3.01987\ldots$

Problem 4 · Difference of Nearly Equal Terms

Given: $\dfrac{1}{d-a} - \dfrac{1}{d+a}$ with $a \ll d$ — find its leading nonzero approximation.

✅ Correct! The leading terms

\tfrac1d

cancel, so the first terms that vary carry the whole answer:

\tfrac{2a}{d^2}

❌ Not quite. Keeping only the constant

\tfrac1d

in each term makes them cancel; keep the first term that varies with

a/d

as well.

❌ Not quite.

\tfrac{1}{d-a}

is the larger term, so the difference is positive:

(1 - a/d)^{-1} \approx 1 + a/d

❌ Not quite. The result must vanish when

a = 0

, since the two terms are then equal;

\tfrac2d

does not.

❌ Not quite. Factor

d

out of each denominator and use

(1+x)^{-1} \approx 1 - x

Show solution

Factor $d$ out of each denominator and use $m = -1$ :

\frac{1}{d \mp a} = \frac1d\left(1 \mp \frac{a}{d}\right)^{-1} \approx \frac1d\left(1 \pm \frac{a}{d}\right)

Subtract:

\frac{1}{d-a} - \frac{1}{d+a} \approx \frac1d\left(1 + \frac ad\right) - \frac1d\left(1 - \frac ad\right) = \frac{2a}{d^2}

Check with $d = 10$ , $a = 0.1$ : exact $\tfrac{1}{9.9} - \tfrac{1}{10.1} = 0.0020002$ , approximation $0.002$ .

Solved: 0 / 4