Taylor series — derivation from power series: Question bank

True or false — justify

A power series that equals

f

near its centre always converges to

f

for every real

x

False. It converges to

f

only inside the radius of convergence, and even there only if the remainder

R_{n} \to 0

. Outside, terms may blow up entirely.

If two power series (in the same

(x - a)^{n}

) are equal on an interval, their coefficients must be equal term by term.

True. Differentiating

n

times and setting

x = a

isolates

c_{n} = f^{(n)} (a) / n!

uniquely, so a function has at most one power-series representation about

a

The Taylor coefficient is

c_{n} = f^{(n)} (a)

False. You must divide by

n!

c_{n} = f^{(n)} (a) / n!

. Each of the

n

differentiations peels off an exponent, and those factors multiply to

n!

, which the division cancels.

Every infinitely differentiable function equals its Taylor series near the centre.

False. The classic counterexample

f (x) = e^{- 1/ x^{2}}

(with

f (0) = 0

) is smooth with all derivatives zero at

0

, so its Maclaurin series is

0

— yet

f

itself is not zero. The coefficients exist; equality still fails.

For

f

centred at

a = 1

, the series is written in powers of

x

False. The building blocks are

(x - a)^{n} = (x - 1)^{n}

. Writing powers of

x

is a Maclaurin habit that only holds when

a = 0

The Maclaurin series is a different rule from the Taylor series.

False. Maclaurin is just the special case

a = 0

of the same formula; nothing new is being defined.

A geometric series

\sum x^{n} = \frac{1}{1 - x}

is a Taylor series.

True. It is the Taylor (Maclaurin) series of

\frac{1}{1 - x}

about

0

— every coefficient

c_{n} = 1

because

f^{(n)} (0) / n! = n! / n! = 1

f^{'} (a) = 0

, the linear (degree-1) Taylor approximation is just the constant

f (a)

True. The degree-1 term is

f^{'} (a) (x - a)

; when

f^{'} (a) = 0

it vanishes, so the best line is flat — geometrically the tangent is horizontal.

Adding more Taylor terms always improves accuracy at every

x

False. More terms sharpen accuracy near the centre and inside convergence, but outside the radius of convergence adding terms makes the sum diverge — accuracy gets worse, not better.

Spot the error

e^{x} = 1 + x + x^{2} + x^{3} + \dots

, so it's the same as the geometric series."

The

e^{x}

series has

\frac{1}{n !}

dividing each term:

1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots

. Dropping the factorials turns it into

\frac{1}{1 - x}

, a completely different function.

"For

ln x

about

a = 1

ln x = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots

The powers must be

(x - 1)

, not

x

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

. Using

x^{n}

silently re-centres at

0

, where

ln

isn't even defined.

sin x

's series has a

c_{0}

term because

sin

is a nice smooth function."

c_{0} = f (a) = sin 0 = 0

, so there is no constant term. Only odd powers survive because even-order derivatives of

sin

are

0

at the origin.

cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{4} - \dots

The denominators are factorials, not the bare exponents:

cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

. Since

2! = 2

the first mistake hides, but

\frac{x ^{4}}{4}

should be

\frac{x ^{4}}{24}

"The Taylor derivation proves

\sum \frac{f ^{(n)} ( a )}{n !} (x - a)^{n}

equals

f (x)

The derivation only shows what the coefficients must be if

f

equals a power series. Proving actual equality needs a separate remainder argument — see Taylor's theorem with remainder.

"To find

c_{3}

I differentiate three times and read off the coefficient of

(x - a)^{3}

After differentiating three times, the surviving

c_{3}

term is the constant

3! c_{3}

, not a

(x - a)^{3}

term. You then set

x = a

and divide by

3!

Why questions

Why does setting

x = a

isolate a single coefficient?

Every term except the current constant carries a factor

(x - a)^{k}

with

k \geq 1

, and

(x - a) = 0

x = a

. So all those terms die, leaving one survivor.

Why does the factorial

n!

appear rather than, say,

n

Differentiating

(x - a)^{n}

peels exponents one at a time:

n

, then

n - 1

, then

n - 2

, ... down to

1

. Their product is

n \cdot (n - 1) \dots 1 = n!

Why is a power series most accurate near its centre?

Near

x = a

the displacement

(x - a)

is tiny, so high powers

(x - a)^{n}

shrink extremely fast and the first few terms dominate — the tail contributes almost nothing.

Why do we bother centring away from

0

(e.g.

ln x

a = 1

Because a Maclaurin series needs all derivatives at

0

, but

ln x

is undefined there. Choosing a centre where

f

is well-behaved lets the expansion exist at all.

Why is the degree-1 Taylor polynomial the same as linear approximation?

Truncating the series after

n = 1

gives

f (a) + f^{'} (a) (x - a)

— exactly the tangent line, which is the standard linear/differential estimate.

Why can L'Hôpital often be replaced by leading Taylor terms?

Near the centre a function is its leading term (e.g.

sin x \approx x

), so a

0/0

ratio reduces to a ratio of leading powers, giving the limit directly without repeated differentiation.

Edge cases

What is the Taylor series of a constant function

f (x) = 5

Just

5

. Every derivative is

0

, so

c_{0} = 5

and all higher

c_{n} = 0

— the "series" is a single term.

What is the Taylor series of a polynomial

f (x) = 3 x^{2} - x + 4

about

a = 0

The polynomial itself:

4 - x + 3 x^{2}

. Derivatives past order 2 vanish, so the series terminates and reproduces

f

exactly everywhere.

What happens to the series if

f

is not differentiable at the centre (e.g.

∣ x ∣

a = 0

The coefficients aren't defined —

f^{'} (0)

doesn't exist — so no Taylor series about that centre exists at all. Smoothness at

a

is a prerequisite.

What does the "radius of convergence

= 0

" case mean?

The series converges only at

x = a

and nowhere else, so it's useless as an approximation off the centre — the coefficients grow too fast for any nonzero displacement.

If all derivatives at

a

are zero but

f

is not the zero function, what is the Taylor series?

It is identically

0

, yet

f \neq = 0

. This shows the derivation determines coefficients but cannot force the series to equal

f

(the

e^{- 1/ x^{2}}

situation).

What is the degree-

0

Taylor "series" and what does it represent geometrically?

Just

f (a)

— a horizontal line at the function's height at the centre. It matches only the starting value, not slope or curvature.

Recall One-line summary of every trap

The derivation gives coefficients, not equality; the powers are $(x - a)^{n}$ not $x^{n}$ unless $a = 0$ ; and $n!$ is non-negotiable.

Connections

Parent: Taylor series derivation
Power series — radius & interval of convergence (where equality can even hold)
Taylor's theorem with remainder (why coefficients ≠ equality)
Maclaurin series — common expansions (the $a = 0$ case)
Linear approximation & differentials (the $n = 1$ truncation)
Geometric series (all-ones coefficients)
L'Hôpital's rule (leading-term shortcut)