The Main Idea 

Taylor series provide a systematic method for finding power series representations of functions using derivatives. Instead of clever algebraic tricks, you can now represent almost any “nice” function as a power series by following a straightforward formula.

If a function [latex]f[/latex] has a power series representation at [latex]x = a[/latex], then the coefficients must be determined by the function’s derivatives at that point. By matching the function value and all its derivatives at [latex]x = a[/latex], we get:

[latex]c_n = \frac{f^{(n)}(a)}{n!}[/latex]

The Taylor series formula: For a function with derivatives of all orders at [latex]x = a[/latex]:

[latex]\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots[/latex]

Maclaurin series: When [latex]a = 0[/latex], this becomes the Maclaurin series:

[latex]\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots[/latex]

Since power series representations are unique, if a function has any power series representation at [latex]x = a[/latex], it must be the Taylor series. This means Taylor series are not just one way to find power series – they’re the only way for functions that can be represented this way.

The key requirement: The function must have derivatives of all orders at the point of expansion. Functions like [latex]e^x[/latex], [latex]\sin x[/latex], [latex]\cos x[/latex], and [latex]\ln(1+x)[/latex] all satisfy this condition and have Taylor series representations.

The Main Idea 

Taylor polynomials are the finite “building blocks” of Taylor series. Instead of working with an infinite sum, you can use just the first few terms to create polynomial approximations of your function that become increasingly accurate as you add more terms.

The [latex]n[/latex]th Taylor polynomial [latex]p_n(x)[/latex] is simply the first [latex]n+1[/latex] terms of the Taylor series:

[latex]p_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n[/latex]

What each polynomial captures:

• [latex]p_0(x) = f(a)[/latex]: Matches the function value at [latex]x = a[/latex]
• [latex]p_1(x)[/latex]: Matches function value and slope
• [latex]p_2(x)[/latex]: Matches function value, slope, and curvature
• [latex]p_3(x)[/latex]: Matches function value, slope, curvature, and rate of curvature change

Key patterns for common functions:

• [latex]e^x[/latex]: All derivatives equal 1 at [latex]x = 0[/latex], giving [latex]p_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}[/latex]
• [latex]\sin x[/latex]: Only odd powers appear, with alternating signs
• [latex]\cos x[/latex]: Only even powers appear, with alternating signs

As [latex]n[/latex] increases, [latex]p_n(x)[/latex] provides a better approximation of [latex]f(x)[/latex] in a neighborhood around [latex]x = a[/latex]. The graphs show how higher-degree polynomials “hug” the original function more closely over larger intervals.

The Main Idea 

When you use a Taylor polynomial to approximate a function, you need to know how accurate your approximation is. Taylor’s theorem with remainder gives you both the exact form of the error and a way to bound it.

The remainder formula: The error when using the [latex]n[/latex]th Taylor polynomial is:

[latex]R_n(x) = f(x) - p_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}[/latex]

where [latex]c[/latex] is some unknown point between [latex]a[/latex] and [latex]x[/latex].

Even though you can’t find the exact value of [latex]c[/latex], you can still bound your error. If you know that [latex]|f^{(n+1)}(x)| \leq M[/latex] on your interval, then:

[latex]|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}[/latex]

Practical applications:

• Proving convergence: If [latex]\lim_{n \to \infty} R_n(x) = 0[/latex], then the Taylor series converges to [latex]f(x)[/latex]
• Error bounds: You can determine how many terms you need for a desired accuracy
• Quality control: You know the maximum possible error in your approximation

Strategy for finding bounds: Look for the maximum value of [latex]|f^{(n+1)}(x)|[/latex] on your interval. For functions like [latex]\sin x[/latex] and [latex]\cos x[/latex], all derivatives have absolute value at most 1, making the bound especially clean.

As you move further from the center [latex]a[/latex], the term [latex]|x-a|^{n+1}[/latex] grows larger, requiring more terms for the same accuracy. Taylor polynomials work best near their center point.

The Main Idea 

Finding a Taylor series is only half the battle – you also need to prove it actually converges to the function you want. Just because you can write down the series doesn’t guarantee it represents your function.

The key question: Does [latex]\lim_{n \to \infty} p_n(x) = f(x)[/latex]? In other words, do the Taylor polynomials actually approach the function as you add more terms?

The convergence test: A Taylor series converges to [latex]f(x)[/latex] if and only if the remainder approaches zero: [latex]\lim_{n \to \infty} R_n(x) = 0[/latex].

 Use Taylor’s theorem to bound the remainder:

[latex]|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}[/latex]

where [latex]M[/latex] bounds all [latex](n+1)[/latex]th derivatives on your interval.

Finding intervals of convergence: Use standard tests (ratio test, root test) just like with power series. The Taylor series will have some radius of convergence, but proving it converges to the actual function requires the remainder analysis.

A Taylor series might converge to some function, but that doesn’t mean it converges to the function you started with. Only by showing [latex]R_n(x) \to 0[/latex] can you guarantee the series represents your original function.