Finding Taylor or Maclaurin Series for a Function

AP Calculus BC · Infinite Sequences and Series

Finding Taylor or Maclaurin Series

Almost never from scratch: substitute into a series you already know.

key idea$f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^{k}$

The procedure

A Taylor series is the Taylor polynomial without a stopping point. Deriving one from scratch means finding a pattern in the derivatives; in practice you should almost always manipulate a known series instead, which is faster and far less error-prone.

Check the four series you know

$e^{x}$, $\sin x$, $\cos x$ and $\dfrac{1}{1-x}$. If the function is any of these composed with something simple, substitute rather than differentiate.

in a function out a strategy

Substitute into a known series

Replacing $x$ by $-x^{2}$, $3x$ or similar is legal term by term, and the radius scales accordingly.

out a new series, almost free

Differentiate or integrate term by term

Both are legal inside the interval of convergence and preserve the radius. Integration introduces a constant, fixed by evaluating at the centre.

out a related series

Only then derive from the definition

If no manipulation applies, compute $f^{(k)}(a)$ for $k=0,1,2,\dots$, find the pattern, and write the general term.

out a general term in $k$

State the general term and the centre

A series answer is a formula for the $k$th term, not three terms and an ellipsis — unless the question explicitly asks for the first few.

out the finished series

Function	Maclaurin series	Radius
$e^{x}$	$\displaystyle\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$	$\infty$
$\sin x$	$\displaystyle\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$	$\infty$
$\cos x$	$\displaystyle\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(2k)!}$	$\infty$
$\dfrac{1}{1-x}$	$\displaystyle\sum_{k=0}^{\infty}x^{k}$	$1$
$\ln(1+x)$	$\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{k}}{k}$	$1$
$\arctan x$	$\displaystyle\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}$	$1$

The first four are worth memorising outright; the last two follow from the fourth by integrating term by term. Nearly every exam question is one substitution away from this table.

four shortcuts

Substitution into a known series is legal and preserves convergence — scaled by the substitution.
Term-by-term differentiation and integration are legal inside the interval of convergence.
$\sin$ has only odd powers; $\cos$ has only even ones. Their parity is a free error check.
A complete answer states the general term, not just the first few terms.

$\sin x$only odd powers

$\cos x$only even powers

$e^{x}$every power, all positive

Worked examples

EX 1

Substitution

$e^{-x^{2}}$ from $e^{x}$

Find the Maclaurin series for $e^{-x^{2}}$.

Start from the known series

$e^{u}=\displaystyle\sum_{k=0}^{\infty}\dfrac{u^{k}}{k!}$.

Substitute

Let $u=-x^{2}$: $\displaystyle\sum_{k=0}^{\infty}\dfrac{\left(-x^{2}\right)^{k}}{k!}$.

Simplify the powers

$\left(-x^{2}\right)^{k}=(-1)^{k}x^{2k}$.

Write the series

$\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}x^{2k}}{k!}=1-x^{2}+\dfrac{x^{4}}{2}-\dfrac{x^{6}}{6}+\cdots$

Note the alternative

Differentiating $e^{-x^{2}}$ repeatedly is punishing; the substitution took one line and no derivatives at all.

EX 2

Term-by-term integration

$\ln(1+x)$

Derive the Maclaurin series for $\ln(1+x)$ from a geometric series.

Start from the geometric series

$\dfrac{1}{1-u}=\displaystyle\sum_{k=0}^{\infty}u^{k}$ for $\lvert u\rvert\lt1$.

Substitute $u=-t$

$\dfrac{1}{1+t}=\displaystyle\sum_{k=0}^{\infty}(-1)^{k}t^{k}$.

Integrate from $0$ to $x$

$\ln(1+x)=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}x^{k+1}}{k+1}$.

Re-index for tidiness

$=\displaystyle\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}x^{k}}{k}=x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\cdots$

Check the constant

At $x=0$ both sides give $0$, so no constant of integration is needed. The radius is still $1$.

EX 3

Deriving from the definition

when nothing else applies

Find the Taylor series for $f(x)=\dfrac{1}{x}$ about $x=1$.

Differentiate repeatedly

$f'=-x^{-2}$, $f''=2x^{-3}$, $f'''=-6x^{-4}$, so $f^{(k)}(x)=(-1)^{k}k!\,x^{-(k+1)}$.

Evaluate at $a=1$

$f^{(k)}(1)=(-1)^{k}k!$.

Divide by $k!$

The coefficient is $\dfrac{(-1)^{k}k!}{k!}=(-1)^{k}$.

Write the series

$\displaystyle\sum_{k=0}^{\infty}(-1)^{k}(x-1)^{k}$.

Recognise it

A geometric series with ratio $-(x-1)$, so it converges for $\lvert x-1\rvert\lt1$ — a check that the pattern was found correctly.

EX 4

Combining a known series with a factor

$x\cos(2x)$

Find the Maclaurin series for $x\cos(2x)$.

Start from the cosine series

$\cos u=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}u^{2k}}{(2k)!}$.

Substitute $u=2x$

$\cos(2x)=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}2^{2k}x^{2k}}{(2k)!}$.

Multiply by $x$

$x\cos(2x)=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}4^{k}x^{2k+1}}{(2k)!}$.

Write out a few terms

$x-2x^{3}+\dfrac{2x^{5}}{3}-\cdots$

Check the parity

$x\cos(2x)$ is an odd function, and every exponent in the answer is odd. ✓

Where it goes wrong

Reach for a known series first

Differentiating $e^{-x^{2}}$ four times to find a pattern is a serious time sink. Substitution gets there in one line.

Substitute into the whole term

Replacing $x$ by $2x$ affects the power and its coefficient: $(2x)^{2k}=4^{k}x^{2k}$. Dropping the $4^{k}$ is the standard slip.

Give the general term

‘Find the Maclaurin series’ asks for a formula in $k$. Three terms and an ellipsis is an incomplete answer unless the question asks only for those.

Parity is a free check

An even function's series has only even powers; an odd function's has only odd ones. A stray term of the wrong parity means an error upstream.

Course map. Topic 10.14 · LO LIM-8.E, LIM-8.F — represent a function as a Taylor series and interpret Taylor series. BC-only.