The Main Idea 

Once you have power series representations for basic functions, you can combine and manipulate them to create power series for more complex functions. This is like having a toolkit where each known power series becomes a building block for constructing new ones.

The three key operations: For power series that converge on a common interval, you can:

• Addition/Subtraction: [latex]\sum c_n x^n \pm \sum d_n x^n = \sum (c_n \pm d_n) x^n[/latex]
• Scalar and Power Multiplication: [latex]bx^m \sum c_n x^n = \sum bx^m c_n x^n[/latex]
• Composition: Replace [latex]x[/latex] with [latex]bx^m[/latex] in your series: [latex]\sum c_n (bx^m)^n[/latex]

Since [latex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n[/latex] for [latex]|x| < 1[/latex], you can use substitution to create new series. For example:

• Replace [latex]x[/latex] with [latex]-x^2[/latex]: [latex]\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n[/latex]
• Multiply by [latex]3x[/latex]: [latex]\frac{3x}{1+x^2} = 3x \sum_{n=0}^{\infty} (-x^2)^n[/latex]

When combining series, your new series converges on the intersection of the original convergence intervals. If one series works on [latex](-1,1)[/latex] and another on [latex](-2,2)[/latex] , their combination works on [latex](-1,1)[/latex] .

Strategy for compositions: When you substitute [latex]bx^m[/latex] for [latex]x[/latex] in a series that converges for [latex]|x| < R[/latex], the new series converges when [latex]|bx^m| < R[/latex]. Solve this inequality to find your new interval. You can also work backwards. If you see [latex]\sum 2^n x^n = \sum (2x)^n[/latex], recognize this as [latex]\frac{1}{1-2x}[/latex] with convergence when [latex]|2x| < 1[/latex], or [latex]|x| < \frac{1}{2}[/latex].

The Main Idea 

When you multiply two power series together, you’re essentially multiplying two “infinite polynomials.” Just like with regular polynomial multiplication, you distribute each term and collect like powers of [latex]x[/latex] to form new coefficients.

The multiplication process: To multiply [latex]\sum c_n x^n[/latex] and [latex]\sum d_n x^n[/latex], you need to find all the ways to get each power [latex]x^n[/latex] in the product. For the coefficient of [latex]x^n[/latex], collect all combinations where the powers add to [latex]n[/latex]:

[latex]e_n = c_0 d_n + c_1 d_{n-1} + c_2 d_{n-2} + \cdots + c_n d_0[/latex]

This is called the Cauchy product. Each new coefficient [latex]e_n[/latex] comes from all possible ways to pair terms from the two original series that multiply to give [latex]x^n[/latex].

If both original series converge on interval [latex]I[/latex], then their product series also converges on [latex]I[/latex] and represents the product of the two functions.

Practical strategy: Instead of using the general formula, you can often work out the first several terms by hand:

• Multiply each term in the first series by each term in the second series
• Group terms with the same power of [latex]x[/latex]
• Add up the coefficients for each power

When you see a rational function that factors into simpler pieces, consider whether you can find power series for the individual factors and then multiply them together.

The Main Idea 

One of the most powerful features of power series is that you can differentiate and integrate them term-by-term, just like polynomials. This transforms complex calculus problems into straightforward arithmetic operations.

The process is simple:

• To differentiate: [latex]\frac{d}{dx}\sum c_n x^n = \sum nc_n x^{n-1}[/latex]
• To integrate: [latex]\int \sum c_n x^n dx = C + \sum c_n \frac{x^{n+1}}{n+1}[/latex]

You apply the power rule to each term individually. For [latex]c_n x^n[/latex], the derivative becomes [latex]nc_n x^{n-1}[/latex] and the integral becomes [latex]c_n \frac{x^{n+1}}{n+1}[/latex].

Both the differentiated and integrated series have the same radius of convergence as the original series. However, endpoint behavior might change – always check endpoints separately if needed.

This technique lets you find power series for functions that would be difficult to obtain directly:

• Start with [latex]\frac{1}{1-x} = \sum x^n[/latex]
• Differentiate to get [latex]\frac{1}{(1-x)^2} = \sum (n+1)x^n[/latex]
• Integrate [latex]\frac{1}{1+x} = \sum (-1)^n x^n[/latex] to get [latex]\ln(1+x) = \sum (-1)^{n+1}\frac{x^n}{n}[/latex]

When integrating, use initial conditions to find [latex]C[/latex]. For example, if [latex]f(0) = 0[/latex], then [latex]C = 0[/latex].

The Main Idea 

When you find a power series representation for a function, you might wonder: “Could there be another, completely different power series for the same function?” The answer is no – power series representations are unique.

The uniqueness principle: If two power series [latex]\sum c_n (x-a)^n[/latex] and [latex]\sum d_n (x-a)^n[/latex] both converge and represent the same function on an open interval containing [latex]a[/latex], then their coefficients must be identical: [latex]c_n = d_n[/latex] for all [latex]n \geq 0[/latex].

Think of power series as “infinite polynomials.” Just as two polynomials can only be equal if they have identical coefficients, the same principle applies to power series. If [latex]c_0 + c_1 x + c_2 x^2 + \cdots = d_0 + d_1 x + d_2 x^2 + \cdots[/latex] for all [latex]x[/latex] in some interval, then [latex]c_0 = d_0[/latex], [latex]c_1 = d_1[/latex], [latex]c_2 = d_2[/latex], and so on.

This uniqueness means you can use any valid method to find a power series – algebraic manipulation, differentiation, integration, or others we’ll learn later. No matter which approach you use, you’ll always get the same series for a given function centered at a specific point.