The Main Idea 

The binomial series generalizes the familiar binomial theorem to handle any real exponent [latex]r[/latex], not just positive integers. This powerful tool lets you find power series for functions like [latex]\sqrt{1+x}[/latex], [latex]\frac{1}{1+x}[/latex], and [latex](1+x)^{-2}[/latex].

The generalized binomial coefficient: For any real number [latex]r[/latex] and non-negative integer [latex]n[/latex]:

[latex](\begin{array}{c}r \ n\end{array}) = \frac{r(r-1)(r-2)\cdots(r-n+1)}{n!}[/latex]

The binomial series formula: [latex](1+x)^r = \sum_{n=0}^{\infty} (\begin{array}{c}r \ n\end{array}) x^n = 1 + rx + \frac{r(r-1)}{2!}x^2 + \frac{r(r-1)(r-2)}{3!}x^3 + \cdots[/latex]

When [latex]r[/latex] is a non-negative integer, the series terminates naturally (since higher-order derivatives become zero), giving you the familiar finite binomial expansion. When [latex]r[/latex] is any other real number, you get an infinite series.

Interval of convergence: The binomial series converges for [latex]|x| < 1[/latex]. Endpoint behavior depends on [latex]r[/latex]:

• [latex]r \geq 0[/latex]: converges at both [latex]x = \pm 1[/latex]
• [latex]-1 < r < 0[/latex]: converges at [latex]x = 1[/latex], diverges at [latex]x = -1[/latex]
• [latex]r < -1[/latex]: diverges at both endpoints

Common applications:

• [latex]\sqrt{1+x} = (1+x)^{1/2}[/latex] with [latex]r = \frac{1}{2}[/latex]
• [latex]\frac{1}{\sqrt{1+x}} = (1+x)^{-1/2}[/latex] with [latex]r = -\frac{1}{2}[/latex]
• [latex]\frac{1}{1+x} = (1+x)^{-1}[/latex] with [latex]r = -1[/latex]

The binomial series is particularly valuable for approximating roots and reciprocals when [latex]x[/latex] is small. You can use just the first few terms to get good approximations, with Taylor’s theorem helping you bound the error.

The Main Idea 

Instead of deriving Taylor series from scratch every time, you can build a powerful toolkit of standard series and use them to find series for more complex functions. This reference approach makes Taylor series much more practical and efficient.

Your essential reference table:

• [latex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n[/latex] for [latex]|x| < 1[/latex]
• [latex]e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}[/latex] for all [latex]x[/latex]
• [latex]\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}[/latex] for all [latex]x[/latex]
• [latex]\cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}[/latex] for all [latex]x[/latex]
• [latex]\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}[/latex] for [latex]-1 < x \leq 1[/latex]
• [latex]\tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}[/latex] for [latex]|x| \leq 1[/latex]
• [latex](1+x)^r = \sum_{n=0}^{\infty} \binom{r}{n} x^n[/latex] for [latex]|x| < 1[/latex]

Building new series from old ones: Use substitution, differentiation, integration, and algebraic manipulation:

• For [latex]\cos \sqrt{x}[/latex]: substitute [latex]\sqrt{x}[/latex] into the cosine series
• For [latex]\sinh x = \frac{e^x - e^{-x}}{2}[/latex]: combine exponential series
• For [latex]\frac{1}{\sqrt{1+x}}[/latex]: differentiate the series for [latex]\sqrt{1+x}[/latex]

Key patterns to recognize:

• Exponential-related functions: Use [latex]e^x[/latex] series with appropriate substitutions
• Trigonometric variants: Modify [latex]\sin x[/latex] and [latex]\cos x[/latex] series
• Rational functions: Often relate to geometric series [latex]\frac{1}{1-x}[/latex]
• Root and power functions: Use binomial series latex^r[/latex]

Instead of computing derivatives repeatedly, ask “Which standard series can I modify to get this function?” This approach is faster, less error-prone, and builds your intuition about how different functions relate to each other.

The Main Idea 

Power series offer a systematic approach to solving differential equations that can’t be handled by separation of variables, integrating factors, or other elementary techniques. The method transforms a differential equation into a problem about finding coefficients in a power series.

Assume your solution has the form [latex]y = \sum_{n=0}^{\infty} c_n x^n[/latex], then use the differential equation to determine what the coefficients [latex]c_n[/latex] must be.

Problem-Solving Strategy:

1. Assume a power series solution: [latex]y = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots[/latex]
2. Differentiate term-by-term: [latex]y' = c_1 + 2c_2 x + 3c_3 x^2 + \cdots[/latex]
3. Substitute into the differential equation: Replace [latex]y[/latex], [latex]y'[/latex], etc. with their series
4. Match coefficients: Use the uniqueness of power series to equate coefficients of like powers
5. Apply initial conditions: Use [latex]y(0)[/latex], [latex]y'(0)[/latex], etc. to find specific coefficient values

The uniqueness of power series representations means that if two power series are equal, their corresponding coefficients must be equal. This gives you a system of equations to solve for the coefficients.

Often you’ll find that [latex]c_{n+k} = f(c_n)[/latex] for some function [latex]f[/latex]. This creates patterns where coefficients can be expressed in terms of just a few initial coefficients.

Power series solutions are especially valuable for:

• Linear differential equations with polynomial coefficients
• Equations where elementary methods fail (like Airy’s equation [latex]y'' - xy = 0[/latex])
• Problems requiring solutions near specific points

Sometimes your coefficient pattern will match a known Taylor series (like [latex]e^x[/latex]), giving you a closed-form solution. Other times, the power series itself is the most useful form of the solution.

The Main Idea 

Some integrals simply can’t be solved using the integration techniques you’ve learned. Functions like [latex]e^{-x^2}[/latex], [latex]\frac{\sin x}{x}[/latex], and [latex]\frac{1}{\sqrt{1-k^2\sin^2\theta}}[/latex] have antiderivatives that aren’t expressible as elementary functions. Power series provide the key to unlocking these “impossible” integrals.

When you can’t integrate directly, represent the integrand as a power series, then integrate term by term. This transforms an unsolvable integral into a manageable infinite series.

Problem-Solving Strategy:

1. Express the integrand as a power series using known series or substitutions
2. Integrate term by term – power series can be integrated just like polynomials
3. Evaluate definite integrals by substituting limits into your series
4. Estimate accuracy using alternating series test or other convergence tools

Power series don’t just solve differential equations – they extend your integration capabilities to a vast class of functions that would otherwise be inaccessible. When traditional integration fails, power series succeed by transforming the problem into polynomial arithmetic.