Single-Variable-Calculus · Unit 8 · Video 2 · Interactive Practice

Why the Quadratic Term Carries a One Half, and Where Fifty-One Eighths Comes From

IKey Formulas

Formula	Name	Where it comes from
$f(x) \approx f(0) + f'(0)\,x + \dfrac{f''(0)}{2}\,x^2$	Quadratic approximation ( $x$ near $0$ )	Must be exact when $f$ is a parabola
$f(x) = a + bx + cx^2 \;\Rightarrow\; f''(0) = 2c$	Why the $\tfrac{1}{2}$	$c = \tfrac{f''(0)}{2}$ is the only choice that recovers $c$
$e^x \approx 1 + x + \tfrac{1}{2}x^2$ , $\ln(1+x) \approx x - \tfrac{1}{2}x^2$ , $(1+x)^r \approx 1 + rx + \tfrac{r(r-1)}{2}x^2$	Catalogue near $0$	One $f''(0)$ each
$\ln a_k - 1 \approx -\dfrac{1}{2k}$ , where $a_k = \left(1 + \tfrac{1}{k}\right)^k$	Rate of convergence to $e$	Keep the quadratic term of $\ln(1 + \tfrac{1}{k})$

Key Insight: For a product, skip the product rule: multiply catalogue rows and keep terms up to $x^2$ , which gives $e^{-3x}(1+x)^{-1/2} \approx 1 - \tfrac{7}{2}x + \tfrac{51}{8}x^2$ .

IIVisualization 1 — Why the Factor Is One Half

The formula must reproduce every parabola exactly, and only the factor $\tfrac{1}{2}$ on $f''(0)$ does.

x² term of the formula

IIIVisualization 2 — How Fast $(1 + 1/k)^k$ Reaches $e$

The quadratic term predicts the gap $\ln a_k - 1 \approx -\tfrac{1}{2k}$ : ten times larger $k$ , ten times smaller gap.

k 100

IVVisualization 3 — Where Fifty-One Eighths Comes From

Of the nine products of catalogue terms, three land on $x^2$ , and together they make $\tfrac{51}{8}$ .

Step 1

VQuiz Questions

Problem 1 · A Catalogue Row with a Substitution

Find the quadratic approximation of $e^{2x}$ for $x$ near $0$ .

✅ Correct!

f''(0) = 4

, and the formula halves it:

\tfrac{4}{2}x^2 = 2x^2

❌ The factor $\tfrac{1}{2}$ is missing.

f''(0) = 4

is twice the

x^2

coefficient, so the term is

\tfrac{4}{2}x^2

, not

4x^2

❌ The substitution was skipped. Replace

x

2x

\tfrac{1}{2}x^2

: the term becomes

\tfrac{1}{2}(2x)^2

❌ Square the whole of $2x$ .

(2x)^2 = 4x^2

, so

\tfrac{1}{2}(2x)^2 = 2x^2

, not

\tfrac{2x^2}{2}

❌ Not quite. Substitute

u = 2x

into

e^u \approx 1 + u + \tfrac{1}{2}u^2

Show solution

By the catalogue: put $u = 2x$ in $e^u \approx 1 + u + \tfrac{1}{2}u^2$ :

e^{2x} \approx 1 + 2x + \tfrac{1}{2}(2x)^2 = 1 + 2x + 2x^2.

By the formula: $f(x) = e^{2x}$ gives $f(0) = 1$ , $f'(0) = 2$ , $f''(0) = 4$ , so

e^{2x} \approx 1 + 2x + \frac{4}{2}x^2 = 1 + 2x + 2x^2.

Problem 2 · The Power Formula with a Negative Exponent

Find the quadratic approximation of $\dfrac{1}{(1+x)^2}$ for $x$ near $0$ . Watch the sign of the $x^2$ term.

✅ Correct! With

r = -2

\tfrac{r(r-1)}{2} = \tfrac{(-2)(-3)}{2} = 3

, and two minus signs multiply to a plus.

❌ Check the sign.

r = -2

and

r - 1 = -3

are both negative, so their product

(-2)(-3) = 6

is positive.

❌ The factor $\tfrac{1}{2}$ is missing.

r(r-1) = 6

f''(0)

; the

x^2

coefficient is half of it.

❌ That used $r + 1$ . The formula has

r(r-1)

: with

r = -2

the second factor is

-3

, not

-1

❌ Not quite. Use

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

with

r = -2

Show solution

Write $\dfrac{1}{(1+x)^2} = (1+x)^{-2}$ , so $r = -2$ and $r - 1 = -3$ .

\text{linear coefficient: } r = -2, \qquad \text{quadratic coefficient: } \frac{r(r-1)}{2} = \frac{(-2)(-3)}{2} = 3.

\frac{1}{(1+x)^2} \approx 1 - 2x + 3x^2.

Check by differentiating: $f'(x) = -2(1+x)^{-3}$ , $f''(x) = 6(1+x)^{-4}$ , so $f''(0) = 6$ and $\tfrac{6}{2} = 3$ .

Problem 3 · A Product of Two Catalogue Rows

Find the quadratic approximation of $e^{x}\sqrt{1+x}$ for $x$ near $0$ , without the product rule.

✅ Correct! Three products land on

x^2

\tfrac{1}{2} + \tfrac{1}{2} - \tfrac{1}{8} = \tfrac{7}{8}

❌ The cross term is missing. The two linear terms multiply too:

x \cdot \tfrac{1}{2}x = \tfrac{1}{2}x^2

❌ Check the sign of the $\sqrt{1+x}$ term. With

r = \tfrac{1}{2}

r - 1 = -\tfrac{1}{2}

, so

\tfrac{r(r-1)}{2} = -\tfrac{1}{8}

❌ The $x^2$ coefficient of $e^x$ is $\tfrac{1}{2}$ , not $1$ .

f''(0) = 1

for

e^x

, and the formula halves it.

❌ Not quite. Multiply

1 + x + \tfrac{1}{2}x^2

1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2

and keep terms up to

x^2

Show solution

Catalogue rows: $e^x \approx 1 + x + \tfrac{1}{2}x^2$ . For $\sqrt{1+x} = (1+x)^{1/2}$ , $r = \tfrac{1}{2}$ and $r - 1 = -\tfrac{1}{2}$ :

\sqrt{1+x} \approx 1 + \tfrac{1}{2}x + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}x^2 = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2.

Multiply, keeping terms up to $x^2$ :

constant: $1 \cdot 1 = 1$
$x$ terms: $x + \tfrac{1}{2}x = \tfrac{3}{2}x$
$x^2$ terms: $\tfrac{1}{2}x^2 \cdot 1 + x \cdot \tfrac{1}{2}x + 1 \cdot \left(-\tfrac{1}{8}x^2\right) = \left(\tfrac{4}{8} + \tfrac{4}{8} - \tfrac{1}{8}\right)x^2 = \tfrac{7}{8}x^2$

e^{x}\sqrt{1+x} \approx 1 + \tfrac{3}{2}x + \tfrac{7}{8}x^2.

Every other product is $x^3$ or higher and is dropped.

Problem 4 · The Rate of Convergence to $e^2$

Given: $b_k = \left(1 + \tfrac{2}{k}\right)^k \to e^2$ , so $\ln b_k \to 2$ . Find what the quadratic term of $\ln(1+x) \approx x - \tfrac{1}{2}x^2$ predicts for the gap $\ln b_k - 2$ .

✅ Correct!

k\left(\tfrac{2}{k} - \tfrac{1}{2}\cdot\tfrac{4}{k^2}\right) = 2 - \tfrac{2}{k}

, so

b_k

approaches

e^2

from below, with a gap shrinking like

\tfrac{1}{k}

❌ That is the gap for $a_k = (1 + \tfrac{1}{k})^k$ . Here

x = \tfrac{2}{k}

, and

x^2 = \tfrac{4}{k^2}

❌ Square the $2$ as well.

\left(\tfrac{2}{k}\right)^2 = \tfrac{4}{k^2}

, and half of it is

\tfrac{2}{k^2}

❌ The factor $\tfrac{1}{2}$ is missing. The quadratic term of

\ln(1+x)

-\tfrac{1}{2}x^2

, not

-x^2

❌ Not quite. Put

x = \tfrac{2}{k}

into

x - \tfrac{1}{2}x^2

, then multiply by

k

Show solution

Take logs: $\ln b_k = k \ln\left(1 + \tfrac{2}{k}\right)$ . For large $k$ , $x = \tfrac{2}{k}$ is near $0$ , so keep the quadratic term:

\ln\left(1 + \tfrac{2}{k}\right) \approx \frac{2}{k} - \frac{1}{2}\cdot\frac{4}{k^2} = \frac{2}{k} - \frac{2}{k^2}.

Multiply by $k$ :

\ln b_k \approx 2 - \frac{2}{k}, \qquad \ln b_k - 2 \approx -\frac{2}{k}.

The linear term alone gives only the limit $2$ . Check at $k = 100$ : predicted $-0.02$ ; true $100\ln(1.02) - 2 \approx -0.01974$ . The gap is negative, so $b_k < e^2$ .

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Why the Quadratic Term Carries a One Half, and Where Fifty-One Eighths Comes From

IKey Formulas

IIVisualization 1 — Why the Factor Is One Half

IIIVisualization 2 — How Fast (1+1/k)k(1 + 1/k)^k(1+1/k)k Reaches eee

IVVisualization 3 — Where Fifty-One Eighths Comes From

VQuiz Questions

IIIVisualization 2 — How Fast $(1 + 1/k)^k$ Reaches $e$