6.4 Working with Taylor Series

Binomial Series and Taylor Series

Taylor and binomial series let you represent functions as infinite sums of polynomial terms. This is useful when you need to approximate values, evaluate integrals that have no closed-form antiderivative, or solve differential equations that resist standard techniques. This section covers how to build these series, recognize common ones, and put them to work.

Terms of the Binomial Series

The binomial series expands $(1+x)^n$ for any real exponent $n$ (not just positive integers) when $|x| < 1$:

$(1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$

The generalized binomial coefficient for any real $n$ is:

$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$

Note: this is not the same as $\frac{n!}{k!(n-k)!}$, which only works when $n$ is a non-negative integer. The generalized form works for any real $n$, including fractions and negatives. For example, you can expand $(1+x)^{1/2}$ or $(1+x)^{-3}$ this way.

Taylor Series for Common Functions

A Taylor series represents a function as an infinite sum built from its derivatives at a single point. You should have these common Maclaurin series (centered at $a = 0$) memorized:

• $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$   (converges for all $x$)
• $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$   (converges for all $x$)
• $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$   (converges for all $x$)
• $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$   (converges for $|x| < 1$)
• $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$   (converges for $-1 < x \leq 1$)
• $(1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k$   (converges for $|x| < 1$, any real $n$)

These are called analytic functions within their radius of convergence: the Taylor series actually equals the function, not just approximates it.

Techniques for Working with Taylor Series

Building a Taylor series from scratch centered at $x = a$:

1. Compute $f(a),\; f'(a),\; f''(a),\; f'''(a), \ldots$ up to the desired order.
2. Plug into the formula: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$

A Maclaurin series is just a Taylor series with $a = 0$.

Often, though, you don't need to compute derivatives from scratch. Instead, you can manipulate known series. These techniques save significant time:

• Substitution: Replace $x$ with an expression. For instance, the series for $e^{-x^2}$ comes from substituting $-x^2$ for $x$ in the $e^x$ series: $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$
• Multiplication: Multiply a known series by $x^k$ or by another series.
• Differentiation: Differentiate a known series term by term. For example, differentiating the geometric series $\frac{1}{1-x}$ gives the series for $\frac{1}{(1-x)^2}$.
• Integration: Integrate a known series term by term. The series for $\ln(1+x)$ can be obtained by integrating $\frac{1}{1+x} = 1 - x + x^2 - \cdots$

The Taylor Remainder Theorem gives you a bound on the error when you truncate a Taylor series after $n$ terms. The remainder $R_n(x)$ satisfies:

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

where $M$ is an upper bound on $|f^{(n+1)}(c)|$ for $c$ between $a$ and $x$. This tells you how many terms you need for a given accuracy.

Taylor Series in Differential Equations

You can approximate solutions to differential equations by assuming the solution is a power series and then finding its coefficients.

General method:

1. Assume $y = \sum_{n=0}^{\infty} a_n x^n$.
2. Compute $y'$, $y''$, etc. as needed by differentiating term by term.
3. Substitute into the differential equation.
4. Equate coefficients of like powers of $x$ to get a system of equations for the $a_n$.
5. Use initial conditions to pin down specific values, then solve for the remaining coefficients.
6. Write out the series solution.

Example: Solve $y' = y$ with $y(0) = 1$.

1. Assume $y = \sum_{n=0}^{\infty} a_n x^n$.
2. Then $y' = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} (n+1)a_{n+1} x^n$ (re-index to match powers).
3. Setting $y' = y$: $(n+1)a_{n+1} = a_n$ for all $n \geq 0$.
4. From the initial condition: $a_0 = y(0) = 1$.
5. The recurrence gives $a_1 = 1$, $a_2 = \frac{1}{2}$, $a_3 = \frac{1}{6}$, and in general $a_n = \frac{1}{n!}$.
6. The solution is $y = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x$, which you can verify directly.

Complex Integrals via Taylor Series

Some integrals have no elementary antiderivative (like $\int e^{-x^2}\,dx$), but you can still evaluate them by expanding the integrand as a Taylor series and integrating term by term.

Method:

1. Replace the integrand with its Taylor series.
2. Integrate each term of the series individually.
3. Check that the interval of convergence includes your limits of integration.
4. Sum enough terms to reach the desired accuracy.

Example: Evaluate $\int_{0}^{1} \frac{1}{1+x^2}\,dx$ using Taylor series.

1. Recognize that $\frac{1}{1+x^2}$ is a geometric series with $x^2$ substituted for $x$ in $\frac{1}{1-(-x^2)}$: $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots \quad \text{for } |x| < 1$

2. Integrate term by term: $\int_{0}^{1} \left(1 - x^2 + x^4 - x^6 + \cdots\right) dx = \left[x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\right]_{0}^{1}$

3. Evaluate: $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$

4. This is the Leibniz formula, and it converges to $\frac{\pi}{4}$. (The series converges at $x = 1$ even though the radius of convergence is 1, because this is an alternating series whose terms decrease to 0.)

A subtlety worth noting: the geometric series converges for $|x| < 1$, but the integrated series actually converges at $x = 1$ as well. This is justified by Abel's theorem, which guarantees that if a power series converges at an endpoint, its sum equals the limit of the function as you approach that endpoint.

Applications and Convergence

Convergence Analysis

To find the radius of convergence $R$ of a Taylor series, use the ratio test:

$R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$

or the root test:

$\frac{1}{R} = \lim_{n \to \infty} |a_n|^{1/n}$

The series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$. You must check the endpoints $x = a \pm R$ separately.

As you include more terms in a Taylor polynomial, the approximation improves within the interval of convergence. Outside that interval, adding more terms actually makes the approximation worse, which is a common source of confusion.