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Calculus BC · Unit 10 · 10.11
AP Calculus BC · Infinite Sequences and Series
Finding Taylor Polynomial Approximations
Match the value and the derivatives at one point: derivatives on top, factorials underneath.
key idea$P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}\,(x-a)^{k}$
The procedure
A Taylor polynomial matches a function's value and derivatives at a single point $a$, up to some order. Each new term corrects the previous approximation using one more derivative, and the whole construction is book-keeping: derivatives on top, factorials underneath.
1
Identify the centre $a$
‘About $x=a$’ or ‘centred at $a$’. A Maclaurin polynomial is simply the case $a=0$.
in $f$ and a centre out the value of $a$
2
Differentiate repeatedly
Compute $f,f',f'',\dots$ up to the required order. Do not simplify aggressively — you only need each one evaluated at $a$.
out a list of derivatives
3
Evaluate every derivative at $a$
These numbers are the only thing that survives. Tabulating them keeps the next step mechanical.
out $f(a),f'(a),f''(a),\dots$
4
Assemble the terms
Term $k$ is $\dfrac{f^{(k)}(a)}{k!}(x-a)^{k}$. The factorial is $k!$, matching the power — not $n!$.
out the polynomial
5
Check the order
$P_n$ has degree at most $n$, so it has $n+1$ terms including the constant. A vanishing derivative can make the actual degree lower.
out a verified $P_n$
| Function | Centre | $P_3(x)$ |
|---|
| $e^{x}$ | $0$ | $1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}$ |
| $\sin x$ | $0$ | $x-\dfrac{x^{3}}{6}$ |
| $\cos x$ | $0$ | $1-\dfrac{x^{2}}{2}$ |
| $\ln x$ | $1$ | $(x-1)-\dfrac{(x-1)^{2}}{2}+\dfrac{(x-1)^{3}}{3}$ |
| $\dfrac{1}{1-x}$ | $0$ | $1+x+x^{2}+x^{3}$ |
Rows 2 and 3 have missing terms, because the corresponding derivatives vanish at $0$. $P_3$ for $\sin x$ genuinely has degree $3$ but only two terms — degree at most $n$, not exactly $n$.
| $k$ | $f^{(k)}(x)$ for $f=e^{x}\sin x$ | $f^{(k)}(0)$ | Term |
|---|
| $0$ | $e^{x}\sin x$ | $0$ | $0$ |
| $1$ | $e^{x}\left(\sin x+\cos x\right)$ | $1$ | $x$ |
| $2$ | $2e^{x}\cos x$ | $2$ | $\dfrac{2}{2!}x^{2}=x^{2}$ |
| $3$ | $2e^{x}\left(\cos x-\sin x\right)$ | $2$ | $\dfrac{2}{3!}x^{3}=\dfrac{x^{3}}{3}$ |
Tabulating the derivatives keeps the assembly mechanical and makes the factorials impossible to misplace. Every row contributes exactly one term.
Coefficient of $(x-a)^{k}$$\dfrac{f^{(k)}(a)}{k!}$
Number of terms in $P_n$$n+1$, including the constant
Maclaurinthe case $a=0$
Worked examples
EX 1
A Maclaurin polynomial
centre $a=0$Find $P_4(x)$ for $f(x)=e^{x}$ about $x=0$.
1
Differentiate
Every derivative of $e^{x}$ is $e^{x}$.
2
Evaluate at the centre
$f^{(k)}(0)=1$ for every $k$.
3
Assemble the terms
$\dfrac{1}{k!}x^{k}$ for $k=0,1,2,3,4$.
4
Write the polynomial
$P_4(x)=1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+\dfrac{x^{4}}{24}$.
5
Check the count
Five terms for degree $4$ — $n+1$, as expected.
EX 2
A centre away from zero
powers of $(x-a)$Find $P_3(x)$ for $f(x)=\ln x$ about $x=1$.
1
Differentiate
$f'=x^{-1}$, $f''=-x^{-2}$, $f'''=2x^{-3}$.
2
Evaluate at $a=1$
$f(1)=0$, $f'(1)=1$, $f''(1)=-1$, $f'''(1)=2$.
3
Divide by the factorials
$\dfrac{0}{0!},\ \dfrac{1}{1!},\ \dfrac{-1}{2!},\ \dfrac{2}{3!}$.
4
Attach the powers
Each term carries $(x-1)^{k}$, not $x^{k}$.
5
Write it
$P_3(x)=(x-1)-\dfrac{(x-1)^{2}}{2}+\dfrac{(x-1)^{3}}{3}$.
EX 3
Missing terms
vanishing derivativesFind $P_5(x)$ for $f(x)=\sin x$ about $x=0$.
1
List the derivative values
$f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=-1$, $f^{(4)}(0)=0$, $f^{(5)}(0)=1$.
2
Note the pattern
Every even derivative vanishes at $0$, because $\sin0=0$.
3
Assemble the surviving terms
$x$, $\dfrac{-1}{3!}x^{3}$, $\dfrac{1}{5!}x^{5}$.
4
Write it
$P_5(x)=x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120}$.
5
Read the structure
Only odd powers appear, which is exactly what an odd function requires of its approximation.
EX 4
Working backwards from a polynomial
coefficients carry derivativesThe Maclaurin polynomial of $f$ begins $3-2x+5x^{2}-\dfrac{x^{3}}{2}$. Find $f''(0)$ and $f'''(0)$.
1
Match the general form
The coefficient of $x^{k}$ is $\dfrac{f^{(k)}(0)}{k!}$.
2
Use the $x^{2}$ coefficient
$\dfrac{f''(0)}{2!}=5$, so $f''(0)=10$.
3
Use the $x^{3}$ coefficient
$\dfrac{f'''(0)}{3!}=-\dfrac12$, so $f'''(0)=-3$.
4
Sanity-check the easy ones
$f(0)=3$ and $f'(0)=-2$, read straight off. ✓
5
State the principle
A Taylor coefficient is a derivative divided by a factorial. Reading one off means multiplying the factorial back.
Where it goes wrong
The factorial matches the power
Term $k$ carries $k!$, not $n!$. Writing every term over $n!$ is the most common structural error in this topic.
Powers are $(x-a)^{k}$
For a centre away from zero the variable is the displacement from $a$. Using $x^{k}$ silently recentres the polynomial at the origin.
Degree at most $n$, not exactly $n$
$P_4$ for $\cos x$ has no $x^{3}$ term. Missing terms mean a vanishing derivative, not a mistake.
Derivatives are evaluated at $a$
The polynomial's coefficients are numbers. Leaving an $x$ inside a coefficient means the evaluation step was skipped.