Binomial theorem for rational indices (approximate values): Question bank

Before we start, one reminder in plain words so every symbol below is earned:

$(1 + x)^{n}$ means "one plus a small number $x$ , raised to a power $n$ ".
$n$ here may be a fraction (like $\frac{1}{2}$ = square root) or negative (like $- 1$ = reciprocal), not just a counting number.
The expansion $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ runs forever unless $n$ is a non-negative whole number.

First: what does " $∣ x ∣ < 1$ " actually mean, and where does it come from?

Before the traps, let us earn the convergence rule instead of just quoting it. Look at two neighbouring terms of the series, the $r$ -th and the $(r + 1)$ -th, and ask: is each new term bigger or smaller than the one before it? The ratio of consecutive terms is what decides whether the running total settles down or explodes.

Now we can see this shrinkage directly. In the figure below, each bar is the actual size $∣ a_{r} x^{r} ∣$ of a term of $(1 + x)^{1/2}$ at $x = 0.2$ .

True or false — justify

(1 + x)^{n}

has a finite expansion for every value of

n

False. It terminates only when

n

is a non-negative integer, because then a factor

(n - n) = 0

appears and kills every later term; for a fraction or negative

n

no factor is ever zero, so it runs forever.

The series

(1 + x)^{n} = 1 + n x + \dots

is valid for any real

x

when

n

is a fraction.

False. It only converges when the absolute value of

x

is less than

1

, i.e.

∣ x ∣ < 1

; the ratio of consecutive terms tends to

∣ x ∣

, so at

x = 5

each term is bigger than the last and the sum blows up.

(1 - x)^{- 1} = 1 + x + x^{2} + x^{3} + \dots

is just an infinite geometric series in disguise.

True. It is exactly the Infinite Geometric Series with first term

1

and ratio

x

, whose sum

\frac{1}{1 - x}

matches

(1 - x)^{- 1}

— and both need

∣ x ∣ < 1

for the same reason.

If two approximations both use "keep 2 terms", they are equally accurate.

False. Accuracy depends on how small

x

is, not on the term count;

(1 + 0.001)^{50}

with two terms beats

(1 + 0.4)^{1/2}

with two terms, because the leftover is dominated by the first dropped term, whose size is set by the power of

x

The condition

∣ x ∣ < 1

is about

x

, so

(2 + 3)^{1/2}

can never be expanded.

False. The number

5^{1/2}

is fine; you just can't use the

(2 + 3)

split. Factor the larger part:

(4 + 1)^{1/2} = 2 (1 + \frac{1}{4})^{1/2}

, and now

∣ x ∣ = \frac{1}{4} < 1

works.

For

(1 - x)^{1/2}

the coefficient of

x^{2}

becomes positive because

(- x)^{2} = x^{2}

False. Substituting

x \to - x

into

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

gives

- \frac{1}{2} x - \frac{1}{8} x^{2}

: the

x^{2}

term stays negative because its coefficient

- \frac{1}{8}

is unchanged and

(- x)^{2}

is positive. Only odd-power signs flip.

The truncation error of a binomial approximation is about the size of the first term you dropped.

True. Since consecutive terms shrink by a factor near

∣ x ∣ < 1

, the leftover tail is like a geometric chain

t + ∣ x ∣ t + ∣ x ∣^{2} t + \dots

whose sum

\frac{t}{1 - ∣ x ∣}

is dominated by its lead term

t

; so the first omitted term sets the error's size — the idea behind Error & Approximation in Calculus.

The rational-index expansion is the same object as the Maclaurin series of

(1 + x)^{n}

True. Computing the coefficient

a_{r} = \frac{f ^{(r)} ( 0 )}{r !}

for

f (x) = (1 + x)^{n}

(with

r

and

a_{r}

as defined above) reproduces the product-form coefficient exactly — see Maclaurin & Taylor Series.

(1 + x)^{- 2} = 1 - 2 x + 3 x^{2} - 4 x^{3} + \dots

can be got by differentiating

(1 + x)^{- 1}

True (up to a sign). Differentiating

(1 + x)^{- 1} = 1 - x + x^{2} - \dots

gives

- (1 + x)^{- 2} = - 1 + 2 x - 3 x^{2} + \dots

; multiply by

- 1

to recover the stated series.

Spot the error

(a + b)^{n} = a^{n} + b^{n} \cdot (small stuff)

, so I can expand

(3 + 7)^{1/2}

directly with

x = 7

The error is not factoring out the dominant term. You must write

(3 + 7)^{1/2}

as the larger base times

(1 + small)

; here

1 0^{1/2}

needs

10 = 9 (1 + \frac{1}{9})

, giving

3 (1 + \frac{1}{9})^{1/2}

with

∣ x ∣ = \frac{1}{9} < 1

(998)^{1/3} \approx 99 8^{1/3}

, and since

998 \approx 1000

, answer

= 10

exactly."

Rounding the base to

1000

throws away the whole correction. The point of the method is to keep the small deviation:

998 = 1000 (1 - 0.002)

, then

10 (1 - 0.002)^{1/3} \approx 9.9933

, not

10

"To expand

(1 + x)^{1/2}

, use

(r 1/2) = \frac{( 1/2 )!}{r ! ( 1/2 - r )!}

Factorials of fractions don't exist, so the

\frac{n !}{r ! ( n - r )!}

form is illegal here. Use the product form

a_{r} = \frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !}

, which never asks for a fractional factorial.

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

always, so I'll use it for

x = 0.5, n = 10

With

x = 0.5

the second term

\frac{n ( n - 1 )}{2} x^{2} = \frac{90}{2} (0.25) = 11.25

is larger than

1 + n x = 6

; here

x

is not "small" (the ratio-to-1 test fails), so one term is worthless.

(1 - x)^{- 1} = 1 + x + x^{2} + \dots

, so

(1 + x)^{- 1} = 1 + x + x^{2} + \dots

too."

You forgot to substitute

x \to - x

(1 + x)^{- 1} = 1 - x + x^{2} - x^{3} + \dots

— the signs alternate, they don't all stay positive.

"Since

∣ x ∣ < 1

is required,

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

is invalid because

1.02 > 1

The constraint is on

x = 0.02

, the small added part, not on the whole number

1.02

. Here

∣ x ∣ = 0.02 < 1

, so it is perfectly valid.

Why questions

Why does the integer binomial series stop but the rational one doesn't?

For integer

n

, at

r = n + 1

the factor

(n - n)

appears in the numerator, zeroing that and all later coefficients; a fraction or negative

n

never equals any integer

r - 1

, so no factor is ever zero.

Why does the window come out as

∣ x ∣ < 1

and not, say,

∣ x ∣ < 2

Because the ratio of consecutive terms tends to

∣ x ∣

(from

\frac{n - r}{r + 1} \to - 1

times

x

); the tail behaves like a geometric chain of ratio

∣ x ∣

, and geometric chains only settle to a finite sum when that ratio is below

1

Why is "keep only

1 + n x

" often enough for tiny

x

Because each successive term

a_{r} x^{r}

carries an extra power of

x

, and if

x

is tiny (

x = 0.001

), then

x^{2}

is a millionth the size — the first correction

n x

already captures nearly all of the deviation from

1

Why must we always massage a problem into the shape

(1 \pm small)

The series is only defined and convergent in the form

(1 + x)^{n}

with

∣ x ∣ < 1

; any other base has to be factored out first so the leftover deviation is a genuine small number.

Why does the truncation error roughly equal the first term you dropped?

The dropped terms shrink by a factor near

∣ x ∣

each step, so the leftover is a geometric tail

t + ∣ x ∣ t + \dots = \frac{t}{1 - ∣ x ∣}

, close to its lead term

t

when

∣ x ∣

is small — see Figure s01's shrinking bars and Error & Approximation in Calculus.

Why is

0.96

slightly below

0.98

, not above?

Writing

(1 - 0.04)^{1/2} \approx 1 - 0.02 - \frac{1}{8} (0.04)^{2}

, the

x^{2}

correction is negative, pulling the value a hair under the first-order guess

0.98

Why can the reciprocal case

(1 + x)^{- 1}

be viewed two completely different ways?

As a binomial series with

n = - 1

, or as a geometric series with ratio

- x

; both give

1 - x + x^{2} - \dots

and both demand

∣ x ∣ < 1

, showing Convergence of Series and the binomial condition are the same idea.

Edge cases

What happens to the series at exactly

x = + 1

for

n = - 1

(1 - 1)^{- 1}

would be

\frac{1}{0}

, undefined; and the series

1 - 1 + 1 - 1 + \dots

does not settle to a value. The right endpoint

∣ x ∣ = 1

is excluded for good reason.

What happens at

x = + 1

for a positive fractional

n

, say

n = \frac{1}{2}

The value

(1 + 1)^{1/2} = 2

is finite, and the series

1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \dots

has terms that alternate in sign after the first and shrink to

0

, so by the alternating-series idea it converges (indeed to

2

). So

x = + 1

is a genuine borderline: it fails for

n = - 1

but succeeds for

0 < n < 1

— exactly the delicate endpoint studied in Convergence of Series.

What about the other endpoint,

x = - 1

, for

n = - 1

Here

(1 + (- 1))^{- 1} = (0)^{- 1} = \frac{1}{0}

is again undefined, and the series becomes

1 + 1 + 1 + \dots

, which diverges to infinity. So both endpoints fail for

n = - 1

Does the

x = - 1

endpoint ever behave nicely for some

n

Yes — for a positive fractional

n

such as

n = \frac{1}{2}

, the value

(1 + (- 1))^{1/2} = 0

is finite and the series converges there (to

0

). So the endpoint

x = - 1

is a genuine borderline: it can converge or diverge depending on the sign and size of

n

, which is exactly the subtlety studied in Convergence of Series.

What does the expansion give when

n = 0

(1 + x)^{0} = 1

, and every term after the "

1

" carries a factor of

n = 0

, so the whole tail vanishes — the series correctly collapses to just

1

What if

x = 0

regardless of

n

(1 + 0)^{n} = 1

; only the constant term survives since every higher term has a factor of

x = 0

. The approximation is exact and trivial here.

Can

n

be a positive integer and still be written as the infinite series?

Yes — the infinite form still holds, but all terms beyond

x^{n}

are automatically zero, so it silently reduces to the finite Binomial Theorem for Positive Integral Index.

What is the largest

x

for which

(1 + x)^{1/2} \approx 1 + \frac{1}{2} x

is "good"?

There's no sharp cutoff, but the first dropped term is

\frac{1}{8} x^{2}

; for

x = 0.1

that is

\approx 0.00125

(three good digits), while at

x = 0.5

it balloons to

\approx 0.03

— accuracy degrades smoothly as

x

approaches

1

If the small part is negative, say

x = - 0.03

, does convergence still hold?

Yes — convergence depends on

∣ x ∣

, and

∣ - 0.03∣ = 0.03 < 1

, so

(1 - 0.03)^{n}

expands perfectly well; only the signs of odd-power terms change.

Recall Two linked series, seen side by side — and

why truncation works The picture below is the visual argument behind the truncation-error rule. It plots the true reciprocal $\frac{1}{1 - x} = (1 - x)^{- 1}$ (red) against three truncations — keeping $1 + x$ , then $1 + x + x^{2}$ , then one more. Read it as a story:

Near $x = 0$ every truncation hugs the red curve; the gap between a truncation and the truth is essentially the next term you would have added — the first dropped term, exactly as claimed above.
As $∣ x ∣ \to 1$ the curves peel apart and the red one races to infinity; this is the convergence radius $∣ x ∣ < 1$ made visual — outside it, no finite truncation can keep up. This is also why $(1 - x)^{- 1}$ is the same as the Infinite Geometric Series sum, and why truncating any binomial series is the Maclaurin & Taylor Series act of matching a curve near $x = 0$ one term at a time.

Connections

Binomial theorem for rational indices (approximate values) (index 3.3.14) — the parent this bank drills.
Convergence of Series — the $∣ x ∣ < 1$ window and the $x = \pm 1$ endpoint questions.
Infinite Geometric Series — the reciprocal-case duality $(1 - x)^{- 1} = \frac{1}{1 - x}$ .
Maclaurin & Taylor Series — why the coefficients $a_{r}$ are what they are.
Error & Approximation in Calculus — the truncation-error edge cases.

First: what does "∣x∣<1" actually mean, and where does it come from?

True or false — justify

Spot the error

Why questions

Edge cases

Connections

First: what does " $∣ x ∣ < 1$ " actually mean, and where does it come from?