Unit 10 Notes: Taylor & Maclaurin Series (AP Calculus BC)

Taylor Polynomial Approximations

What a Taylor polynomial is

A Taylor polynomial is a polynomial that imitates a function near a specific point. The key idea is: if you match enough derivative information at a point, a polynomial can “hug” the function extremely well in a neighborhood of that point.

Concretely, suppose you care about a function $f(x)$ near $x=a$. You build a polynomial $P_n(x)$ of degree $n$ so that it matches:

• the function value: $P_n(a)=f(a)$
• the first derivative: $P_n'(a)=f'(a)$
• the second derivative: $P_n''(a)=f''(a)$
• and so on, up through the $n$th derivative.

This creates a polynomial approximation that is “locally tuned” to the function’s behavior at $a$.

Why Taylor polynomials matter

Taylor polynomials matter because they let you:

• Approximate values of complicated functions using simple arithmetic.
• Estimate errors (how accurate your approximation is) using rigorous bounds.
• Set up Taylor series (infinite versions of these polynomials), which power many convergence and representation questions in Unit 10.

A good mental model: a Taylor polynomial is like taking a “snapshot” of the function at $x=a$, including slope, concavity, and higher-order shape, then reconstructing the nearby curve using a polynomial.

How the Taylor polynomial formula works

The $n$th-degree Taylor polynomial for $f(x)$ about $x=a$ is

$P_n(x)=\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$

Here:

• $f^{(k)}(a)$ means the $k$th derivative evaluated at $a$.
• $k!$ (factorial) normalizes the coefficients.
• $(x-a)^k$ shifts the approximation to be centered at $a$.

When $a=0$, it’s called a Maclaurin polynomial (a special case of Taylor).

Notation you’ll see (and how to translate it)

Example 1: Build a Maclaurin polynomial for $\sin(x)$

You often know derivative cycles for trig functions, so $\sin(x)$ is a classic.

We compute derivatives and evaluate at $0$:

• $f(x)=\sin(x)$ so $f(0)=0$
• $f'(x)=\cos(x)$ so $f'(0)=1$
• $f''(x)=-\sin(x)$ so $f''(0)=0$
• $f'''(x)=-\cos(x)$ so $f'''(0)=-1$
• $f^{(4)}(x)=\sin(x)$ so $f^{(4)}(0)=0$
• $f^{(5)}(x)=\cos(x)$ so $f^{(5)}(0)=1$

The degree-5 Maclaurin polynomial is

$P_5(x)=\sum_{k=0}^{5} \frac{f^{(k)}(0)}{k!}x^k$

Only the nonzero derivative values contribute:

$P_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}$

So near $0$,

$\sin(x)\approx x-\frac{x^3}{6}+\frac{x^5}{120}$

A common mistake here is forgetting factorials (for example, writing $x-\frac{x^3}{3}$ instead of $x-\frac{x^3}{6}$).

Example 2: Use a Taylor polynomial to approximate a value

Approximate $e^{0.1}$ using the degree-3 Maclaurin polynomial for $e^x$.

For $f(x)=e^x$, every derivative is $e^x$, so $f^{(k)}(0)=1$ for all $k$. Thus

$P_3(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$

Plug in $x=0.1$:

$e^{0.1}\approx 1+0.1+\frac{0.1^2}{2}+\frac{0.1^3}{6}$

Compute:

• $0.1^2=0.01$ so $\frac{0.01}{2}=0.005$
• $0.1^3=0.001$ so $\frac{0.001}{6}\approx 0.0001667$

So

$e^{0.1}\approx 1.1051667$

This is the basic move behind many AP questions: “Use a Taylor polynomial of degree $n$ about $a$ to approximate $f(b)$.”

Exam Focus

• Typical question patterns:
  • “Find the Taylor polynomial of degree $n$ for $f(x)$ about $x=a$.”
  • “Use a Maclaurin polynomial to approximate $f(c)$ and show work.”
  • “Given values of derivatives at $a$, write $P_n(x)$ (no re-differentiation needed).”
• Common mistakes:
  • Mixing up $a$ and $x$ (writing powers of $x$ when the center is $a\neq 0$; you need $(x-a)^k$).
  • Forgetting factorials in coefficients.
  • Using a polynomial beyond its “safe zone” (approximating far from $a$ without any error reasoning).

Lagrange Error Bound

What the error term is

A Taylor polynomial is an approximation, so you need a way to quantify the gap:

$R_n(x)=f(x)-P_n(x)$

This difference $R_n(x)$ is called the **remainder** (or error) after using the degree-$n$ Taylor polynomial.

Why the Lagrange error bound matters

The AP exam often asks for a guaranteed bound on the error: not “what the exact error is” (usually hard), but “how big it could be at most.” The Lagrange Error Bound gives a clean way to do that if you can bound the next derivative.

This is especially important when you approximate numerical values (like $\sin(0.2)$ or $e^{0.1}$) and must justify the number of correct decimal places.

The Lagrange Error Bound formula (how it works)

If you use the degree-$n$ Taylor polynomial about $a$, and if $f^{(n+1)}(t)$ exists on the interval between $a$ and $x$, then

$|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}$

where $M$ is any number such that

$|f^{(n+1)}(t)|\le M$

for all $t$ between $a$ and $x$.

Interpretation:

• The error depends on the size of the next derivative (captured by $M$).
• The error shrinks quickly when $|x-a|$ is small and $n$ is larger, because factorials grow fast.

A subtle but crucial point: the bound uses the maximum possible size of the $n+1$ derivative on the interval. If you underestimate $M$, your “bound” may be false.

Example 1: Bound the error for approximating $e^{0.1}$ with $P_3(x)$

We previously used $P_3(x)$ about $a=0$ to approximate $e^{0.1}$. Now bound the error.

Here $f(x)=e^x$ and $n=3$, so we need $f^{(4)}(x)=e^x$.

On the interval between $0$ and $0.1$, the maximum of $e^t$ occurs at $t=0.1$ (because $e^t$ increases). So we can take

$M=e^{0.1}$

Then

$|R_3(0.1)|\le \frac{e^{0.1}}{4!}|0.1|^4$

Compute pieces:

• $4!=24$
• $0.1^4=0.0001$

So

$|R_3(0.1)|\le \frac{e^{0.1}}{24}\cdot 0.0001$

Since $e^{0.1}\approx 1.1052$, the bound is approximately

$|R_3(0.1)|\le \frac{1.1052}{24}\cdot 0.0001\approx 0.0000046$

So the approximation is accurate to about the millionths place (and certainly to 4 decimal places).

Example 2: Choosing $M$ correctly

Suppose you approximate $\sin(x)$ near $0$ using $P_5(x)$ and want an error bound at $x=0.3$.

Here $n=5$, so we need the 6th derivative. The derivatives of $\sin(x)$ cycle, and the 6th derivative is $-\sin(x)$. So

$|f^{(6)}(t)|=|\sin(t)|\le 1$

for all real $t$. You can safely choose $M=1$.

Then

$|R_5(0.3)|\le \frac{1}{6!}|0.3|^6$

This is usually enough for AP justification; you don’t need to fully decimal-evaluate unless asked.

Exam Focus

• Typical question patterns:
  • “Use the Lagrange Error Bound to find an upper bound on the error when approximating $f(x)$ by $P_n(x)$ at $x=b$.”
  • “Find the degree $n$ needed so the approximation error is less than a given tolerance.”
  • “Justify that the approximation is within $0.001$ (or has a certain number of correct decimals).”
• Common mistakes:
  • Using $f^{(n)}$ instead of $f^{(n+1)}$ in the bound.
  • Picking $M$ from the wrong interval (you must use the interval between $a$ and the target $x$).
  • Treating the bound as the exact error; it’s only a guaranteed maximum.

Radius and Interval of Convergence of Power Series

What a power series is

A power series is an infinite polynomial-like expression, typically centered at some point $a$:

$\sum_{n=0}^{\infty} c_n(x-a)^n$

For each fixed $x$, this becomes an infinite series of numbers. The big question is: for which $x$ values does it converge?

Why convergence matters

If a power series converges, it defines a function (you can plug in $x$ and get a finite value). If it diverges, it doesn’t represent a meaningful value there.

On the AP exam, you’re often asked to:

• find the radius of convergence (how far from the center it converges)
• find the interval of convergence (the exact set of $x$ values, including endpoint checks)

The radius vs. the interval (what’s the difference?)

For many power series centered at $a$, there exists a number $R$ such that:

• the series converges for $|x-a|<R$
• the series diverges for $|x-a|>R$
• at $|x-a|=R$ (the endpoints), convergence must be checked separately

That $R$ is the radius of convergence. The interval of convergence is typically

$(a-R,a+R)$

plus whatever endpoints actually converge.

How to find the radius (Ratio Test is the workhorse)

The most common method is the Ratio Test applied to the general term.

If your series is

$\sum_{n=0}^{\infty} u_n$

and

$L=\lim_{n\to\infty} \left|\frac{u_{n+1}}{u_n}\right|$

then:

• if $L<1$, the series converges
• if $L>1$ (or infinite), the series diverges
• if $L=1$, the test is inconclusive

In a power series, $u_n$ depends on $x$, so the Ratio Test usually produces an inequality in $x$ that gives $|x-a|<R$.

Example 1: Find the interval of convergence

Find the interval of convergence of

$\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}$

Let

$u_n=\frac{(x-2)^n}{n}$

Compute the ratio:

$\left|\frac{u_{n+1}}{u_n}\right|=\left|\frac{(x-2)^{n+1}}{n+1}\cdot \frac{n}{(x-2)^n}\right|=\left|x-2\right|\cdot \frac{n}{n+1}$

Take the limit:

$L=\lim_{n\to\infty} \left|x-2\right|\cdot \frac{n}{n+1}=|x-2|$

Convergence requires $|x-2|<1$, so the radius is $R=1$ and the “open interval” is $1<x<3$.

Now check endpoints:

• At $x=3$: series becomes $\sum_{n=1}^{\infty} \frac{1}{n}$, which diverges (harmonic series).
• At $x=1$: series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, which converges (alternating harmonic).

So the interval of convergence is

$[1,3)$

Example 2: A power series with factorials

Find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{n!}{3^n}(x+1)^n$

Let

$u_n=\frac{n!}{3^n}(x+1)^n$

Ratio:

$\left|\frac{u_{n+1}}{u_n}\right|=\left|\frac{(n+1)!}{3^{n+1}}(x+1)^{n+1}\cdot \frac{3^n}{n!}\cdot \frac{1}{(x+1)^n}\right|=\left|\frac{n+1}{3}(x+1)\right|$

As $n\to\infty$, this ratio grows without bound unless $x=-1$ makes the factor $(x+1)$ equal to zero. So:

• If $x\neq -1$, the ratio tends to infinity, so the series diverges.
• If $x=-1$, every term after the first is zero, so it converges.

This means the radius of convergence is $R=0$ (only converges at the center).

That’s a good reminder: not every power series converges on an interval; sometimes it collapses to a single point.

Exam Focus

• Typical question patterns:
  • “Find the radius of convergence and interval of convergence of a power series.”
  • “Use the Ratio Test (or Root Test) to determine convergence.”
  • “Check endpoints separately and classify convergence (absolute vs conditional).”
• Common mistakes:
  • Forgetting to check endpoints after finding $|x-a|<R$.
  • Algebra errors when simplifying the ratio—especially with factorials and powers.
  • Assuming endpoints behave the same; they often differ (one converges, the other diverges).

Finding Taylor and Maclaurin Series

What a Taylor series is

A Taylor series is what you get if you keep increasing the Taylor polynomial degree forever. Formally, the Taylor series of $f(x)$ about $x=a$ is

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

The corresponding Maclaurin series is the special case $a=0$:

$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$

Important nuance: Writing down the Taylor series is one step; asserting it actually equals $f(x)$ is another. On the AP exam, you often use known expansions or are implicitly working where equality is valid.

Why learning to find series matters

You’re building a toolkit to:

• generate power series representations for functions,
• approximate values efficiently,
• analyze limits and integrals using series,
• and solve differential equation–style modeling problems (in broader calculus contexts).

Two main methods for finding series

1. Derivative method (from the definition): compute $f^{(n)}(a)$ patterns.
2. Known series + algebra: start from a “library” of standard series (like geometric) and transform.

The second method is usually faster and more common in AP problems, but the derivative method is essential when a function isn’t easily linked to a known series.

A small “library” of standard Maclaurin series

These are commonly used in AP Calculus BC:

1. Geometric series

$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$

(valid for $|x|<1$)

2. Exponential

$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

(converges for all real $x$)

3. Sine

$\sin(x)=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$

(converges for all real $x$)

4. Cosine

$\cos(x)=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!}$

(converges for all real $x$)

5. Natural log (often derived via integration)

$\ln(1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$

(valid for $-1<x\le 1$)

6. Arctangent (often derived from geometric)

$\arctan(x)=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$

(valid for $|x|\le 1$)

A frequent mistake is memorizing these without the conditions (intervals) where they converge.

Example 1: Find the Taylor series for $\ln(x)$ about $a=1$

This is a classic BC skill because it produces a power series centered at $1$.

Let $f(x)=\ln(x)$. Compute derivatives:

• $f(1)=\ln(1)=0$
• $f'(x)=\frac{1}{x}$ so $f'(1)=1$
• $f''(x)=-\frac{1}{x^2}$ so $f''(1)=-1$
• $f'''(x)=\frac{2}{x^3}$ so $f'''(1)=2$
• $f^{(4)}(x)=-\frac{6}{x^4}$ so $f^{(4)}(1)=-6$

A pattern emerges: for $n\ge 1$,

$f^{(n)}(1)=(-1)^{n-1}(n-1)!$

Plug into the Taylor series formula:

$\ln(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n$

Since $\frac{(n-1)!}{n!}=\frac{1}{n}$, this simplifies to

$\ln(x)=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{(x-1)^n}{n}$

This is equivalent to the well-known series for $\ln(1+u)$ if you set $u=x-1$.

Example 2: Find a Maclaurin series using known series (no derivatives)

Find a power series for

$\frac{1}{1+3x}$

Start from the geometric series:

$\frac{1}{1-r}=\sum_{n=0}^{\infty} r^n$

Here you want $1+3x$ in the denominator, so rewrite:

$\frac{1}{1+3x}=\frac{1}{1-(-3x)}$

So $r=-3x$. Substitute:

$\frac{1}{1+3x}=\sum_{n=0}^{\infty} (-3x)^n=\sum_{n=0}^{\infty} (-1)^n 3^n x^n$

Convergence requires $|-3x|<1$, so $|x|<\frac{1}{3}$.

Exam Focus

• Typical question patterns:
  • “Find the Maclaurin series for $f(x)$” (often trig, exponential, log).
  • “Find the Taylor series centered at $a$ and state its interval of convergence.”
  • “Use a known series (often geometric) to produce a new one by substitution.”
• Common mistakes:
  • Dropping the center shift: writing powers of $x$ instead of $(x-a)$.
  • Not simplifying factorial patterns correctly (like turning $\frac{(n-1)!}{n!}$ into $\frac{1}{n}$).
  • Forgetting to transform the convergence condition when substituting (for example, $|r|<1$ becomes a condition on $x$).

Representing Functions as Power Series

What “represent as a power series” means

To represent a function as a power series means to rewrite it in the form

$f(x)=\sum_{n=0}^{\infty} c_n(x-a)^n$

on some interval of convergence.

This is more than an algebra trick: once you have a power series, you can often differentiate, integrate, and approximate in ways that are difficult with the original function.

Why this is powerful (big ideas)

Power series are like “function LEGO bricks.” Once you know a few standard series, you can build many others by:

• Substitution (replace $x$ with another expression)
• Multiplying by a power (shift indices)
• Differentiating/integrating term-by-term (within the interval of convergence)

On the AP exam, this shows up in problems that ask for a series for a new function, and often also ask for the radius/interval of convergence after the transformation.

Core tools for transforming series

Suppose

$f(x)=\sum_{n=0}^{\infty} c_n x^n$

Then (within the interval of convergence):

1. Multiply by $x^m$:

$x^m f(x)=\sum_{n=0}^{\infty} c_n x^{n+m}$

2. Differentiate:

$f'(x)=\sum_{n=1}^{\infty} n c_n x^{n-1}$

3. Integrate:

$\int f(x)\,dx=C+\sum_{n=0}^{\infty} \frac{c_n}{n+1}x^{n+1}$

A common misconception is that you can always differentiate/integrate any series anywhere. For AP purposes, the safe statement is: you can do these operations on the interval of convergence, and the resulting series has the same radius of convergence.

Example 1: Derive the series for $\ln(1+x)$ by integrating a geometric series

Start with

$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$

Replace $x$ with $-x$:

$\frac{1}{1+x}=\sum_{n=0}^{\infty} (-1)^n x^n$

This holds for $|x|<1$.

Now integrate both sides from $0$ to $x$. The left side:

$\int_0^x \frac{1}{1+t}\,dt=\ln(1+x)$

The right side integrates term-by-term:

$\int_0^x \sum_{n=0}^{\infty} (-1)^n t^n\,dt=\sum_{n=0}^{\infty} (-1)^n\frac{x^{n+1}}{n+1}$

Rewrite the index with $k=n+1$:

$\ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{x^k}{k}$

This is the standard Maclaurin series for $\ln(1+x)$.

Convergence: the original geometric series required $|x|<1$. At endpoints, you check separately; this series converges at $x=1$ (alternating harmonic) and diverges at $x=-1$ (harmonic-like), giving $-1<x\le 1$.

Example 2: Create a power series for $\arctan(x)$

A common route is to start from

$\frac{1}{1+x^2}$

Use geometric series with $r=-x^2$:

$\frac{1}{1-(-x^2)}=\sum_{n=0}^{\infty} (-x^2)^n=\sum_{n=0}^{\infty} (-1)^n x^{2n}$

(valid for $|x|<1$).

Now integrate term-by-term:

$\int_0^x \frac{1}{1+t^2}\,dt=\arctan(x)$

and

$\int_0^x \sum_{n=0}^{\infty} (-1)^n t^{2n}\,dt=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$

So

$\arctan(x)=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$

This representation is useful for approximating $\arctan(1)$ (which relates to $\pi$) and for many series-based approximation questions.

Example 3: Represent a nonstandard function by substitution and scaling

Find a power series for

$\frac{x^2}{4-x}$

First rewrite it to match geometric form. Factor the denominator:

$\frac{x^2}{4-x}=\frac{x^2}{4\left(1-\frac{x}{4}\right)}=\frac{x^2}{4}\cdot \frac{1}{1-\frac{x}{4}}$

Now apply geometric series with $r=\frac{x}{4}$:

$\frac{1}{1-\frac{x}{4}}=\sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n$

So

$\frac{x^2}{4-x}=\frac{x^2}{4}\sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n=\sum_{n=0}^{\infty} \frac{x^{n+2}}{4^{n+1}}$

Convergence comes from $\left|\frac{x}{4}\right|<1$, so $|x|<4$.

Common pitfall: forgetting to carry the prefactor $\frac{x^2}{4}$ into the series.

How these representations connect back to Taylor polynomials

Once you have a power series, a Taylor polynomial is just a truncation (cutting off after finitely many terms). For example, if

$f(x)=\sum_{n=0}^{\infty} c_n x^n$

then a degree-$N$ Maclaurin polynomial is

$P_N(x)=\sum_{n=0}^{N} c_n x^n$

That’s why “represent as a power series” and “approximate with a polynomial” are two sides of the same idea.

Exam Focus

• Typical question patterns:
  • “Find a power series representation for $g(x)$ using a known series for $f(x)$.”
  • “Differentiate/integrate a power series to obtain a series for a related function, and state the interval of convergence.”
  • “Use a power series to approximate a definite integral or function value.”
• Common mistakes:
  • Not updating the convergence condition after substitution (for example, $|r|<1$ becomes an inequality in $x$).
  • Differentiating/integrating incorrectly (especially index shifts and new starting indices).
  • Forgetting the constant of integration when integrating an indefinite series (when appropriate); using a definite integral from $0$ often avoids this.