2.3 - Maclaurin expansions

Maclaurin series

A Maclaurin series is a power-series expansion about $x=0$. Its initial coefficients are determined by derivatives evaluated at zero. Truncating it after a few terms gives a polynomial approximation near zero, wherever the expansion represents the function.

Initial Maclaurin terms

For a function with a Maclaurin expansion,

Here $2!=2$, $3!=6$ and $4!=24$. For a truncated approximation, replace the equality and omitted terms by $\approx$.

The Maclaurin formula is in the formula booklet. Use it to calculate the required initial coefficients.

Worked example 1: an exponential with a scaled argument

Question. Find the Maclaurin expansion of $f(x)=e^{2x}$ up to and including the term in $x^3$.

Plan. Differentiate three times, evaluate at zero, then divide the higher derivatives by their factorials.

The chain rule gives

At zero, $f(0)=1$, $f'(0)=2$, $f''(0)=4$ and $f'''(0)=8$. Hence

Answer. $\displaystyle e^{2x}=1+2x+2x^2+\frac43x^3+\cdots$.

Check. At $x=0.1$, the polynomial gives $1.221333333\ldots$, while direct evaluation gives $e^{0.2}=1.221402758\ldots$. The small difference, about $0.0000694$, is consistent with an approximation rather than an exact cubic identity.

Worked example 2: a negative inner scale

Question. Find the Maclaurin expansion of $f(x)=e^{-2x}$ up to and including the term in $x^3$.

Plan. Retain the factor $-2$ at every differentiation before evaluating at zero.

Substituting $x=0$ gives the values $1,-2,4,-8$. Therefore

Answer. $\displaystyle e^{-2x}=1-2x+2x^2-\frac43x^3+\cdots$.

Check. Direct evaluation at $x=0.1$ gives $e^{-0.2}=0.818730753\ldots$; the cubic gives $0.818666667\ldots$. Also, multiplying the two example polynomials gives constant term $1$ and zero coefficients of $x$, $x^2$ and $x^3$, agreeing with $e^{2x}e^{-2x}=1$ to the calculated degree.

Watch out

The coefficient of $x^3$ is $f'''(0)/6$, not $f'''(0)$. For $e^{-2x}$, forgetting the factorial would incorrectly give $-8x^3$ instead of $-4x^3/3$.

Problem-solving

Simplify a first derivative into a relation involving $x$ and $y$ where possible. Successive differentiation of that relation often gives the initial Maclaurin coefficients with much less algebra.

Worked example 3: tangent through the cubic term

Question. Starting from $y=\tan x$, derive its Maclaurin expansion up to and including the term in $x^3$.

Plan. Use $y'=1+y^2$ and its successive derivatives to evaluate the needed values at zero.

Since $y(0)=0$, the derivative relation gives

The Maclaurin formula now gives

We are working near zero, on the central tangent branch $-\pi/2<x<\pi/2$.

Answer. $\displaystyle \tan x=x+\frac13x^3+\cdots$, with zero constant and quadratic coefficients.

Check. Multiply the proposed polynomial by the supplied initial cosine terms:

These are the initial sine terms, agreeing with $\tan x\cos x=\sin x$.

Worked example 4: tangent with a scaled argument

Question. For $y=\tan(2x)$, use successive implicit differentiation to find the Maclaurin expansion through the term in $x^3$.

Plan. Start with $y'=2(1+y^2)$, then differentiate this relation twice.

The chain and product rules give

At zero, $y=0$. Hence

Substitute into the Maclaurin formula:

The local calculation is on the central branch $-\pi/4<x<\pi/4$.

Answer. $\displaystyle \tan(2x)=2x+\frac83x^3+\cdots$.

Check. Multiplying $2x+8x^3/3$ by $1-2x^2$, the cosine expansion for $\cos(2x)$, gives $2x-4x^3/3$ through degree $3$. This agrees with the sine expansion for $\sin(2x)$, so the quotient is consistent.