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Taylor Series

Taylor Series

The power series in x – a for a function f . Note: If a = 0 the series is called a Maclaurin series.

 

Taylor series formula: sum from k=0 to n of f^(k)(a)/k! times (x-a)^k, expanded showing f(a)+f'(a)(x-a)+f''(a)/2!(x-a)²+...

 

 

See also

Taylor series remainder, convergence tests, power series convergence, sigma notation, factorial, nth derivative

Key Formula

f(x)=n=0f(n)(a)n!(xa)n=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+f(x) = \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
Where:
  • f(x)f(x) = The function being represented as a series
  • aa = The center point around which the series is expanded
  • f(n)(a)f^{(n)}(a) = The nth derivative of f evaluated at x = a
  • n!n! = n factorial (the product 1 · 2 · 3 · … · n)
  • (xa)n(x - a)^n = The nth power of (x − a)

Worked Example

Problem: Find the Taylor series for f(x) = eˣ centered at a = 0 and write out the first five terms.
Step 1: Find the derivatives of f(x) = eˣ. Every derivative of eˣ is eˣ.
f(x)=ex,f(x)=ex,f(x)=ex,f(x)=ex,f(4)(x)=exf(x) = e^x,\quad f'(x) = e^x,\quad f''(x) = e^x,\quad f'''(x) = e^x,\quad f^{(4)}(x) = e^x
Step 2: Evaluate each derivative at a = 0. Since e⁰ = 1, every derivative at 0 equals 1.
f(0)=1,f(0)=1,f(0)=1,f(0)=1,f(4)(0)=1f(0) = 1,\quad f'(0) = 1,\quad f''(0) = 1,\quad f'''(0) = 1,\quad f^{(4)}(0) = 1
Step 3: Substitute into the Taylor series formula with a = 0.
ex=n=01n!xn=1+x+x22!+x33!+x44!+e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
Step 4: Write out the first five terms with the factorials computed.
ex=1+x+x22+x36+x424+e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots
Answer: The Taylor series (Maclaurin series) for eˣ is ex=n=0xnn!=1+x+x22+x36+x424+e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots, and it converges for all real xx.

Another Example

This example uses a non-zero center (a = 1) and a function whose derivatives follow a more complex pattern, showing how to handle alternating signs and varying derivative values.

Problem: Find the Taylor series for f(x) = ln(x) centered at a = 1 up to the term containing (x − 1)⁴.
Step 1: Compute the first four derivatives of f(x) = ln(x).
f(x)=lnx,f(x)=1x,f(x)=1x2,f(x)=2x3,f(4)(x)=6x4f(x) = \ln x,\quad f'(x) = \frac{1}{x},\quad f''(x) = -\frac{1}{x^2},\quad f'''(x) = \frac{2}{x^3},\quad f^{(4)}(x) = -\frac{6}{x^4}
Step 2: Evaluate each derivative at a = 1.
f(1)=0,f(1)=1,f(1)=1,f(1)=2,f(4)(1)=6f(1) = 0,\quad f'(1) = 1,\quad f''(1) = -1,\quad f'''(1) = 2,\quad f^{(4)}(1) = -6
Step 3: Substitute into the Taylor series formula, dividing each derivative value by its corresponding factorial.
lnx0+11!(x1)+12!(x1)2+23!(x1)3+64!(x1)4\ln x \approx 0 + \frac{1}{1!}(x-1) + \frac{-1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 + \frac{-6}{4!}(x-1)^4
Step 4: Simplify each coefficient.
lnx(x1)(x1)22+(x1)33(x1)44\ln x \approx (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4}
Step 5: Recognize the general pattern for the full series.
lnx=n=1(1)n+1n(x1)n,0<x2\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n, \quad 0 < x \le 2
Answer: The Taylor series for ln(x) centered at a = 1 is lnx=(x1)(x1)22+(x1)33(x1)44+\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots, valid for 0<x20 < x \le 2.

Frequently Asked Questions

What is the difference between a Taylor series and a Maclaurin series?
A Maclaurin series is simply a Taylor series centered at a = 0. Every Maclaurin series is a Taylor series, but not every Taylor series is a Maclaurin series. When someone says 'Maclaurin series,' they mean the special case where the expansion point is the origin.
When does a Taylor series converge to the original function?
A Taylor series converges to f(x) within its radius of convergence, provided the Taylor series remainder (the error from truncating the series) approaches zero as you include more terms. Some functions, like eˣ, sin(x), and cos(x), have Taylor series that converge for all real x. Others, like ln(x) centered at 1, converge only on a limited interval. You can use convergence tests or the remainder theorem to determine the interval.
Why are Taylor series useful?
Taylor series let you approximate complicated functions with polynomials, which are easy to compute, differentiate, and integrate. They are essential in physics and engineering for linearization, in numerical methods for computing values of functions like sin(x) and eˣ, and in calculus for evaluating limits and integrals that have no closed-form solution.

Taylor Series vs. Maclaurin Series

Taylor SeriesMaclaurin Series
DefinitionPower series expansion of f(x) about any point aPower series expansion of f(x) about a = 0
Formulan=0f(n)(a)n!(xa)n\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^nn=0f(n)(0)n!xn\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n
Center pointAny real number aAlways a = 0
When to useWhen you need an expansion near a specific point a ≠ 0When expanding near the origin is natural or convenient
Exampleln(x) about a = 1eˣ about a = 0

Why It Matters

Taylor series appear throughout AP Calculus BC, college calculus, physics, and engineering courses. They are one of the primary tools for approximating functions—calculators and computers use truncated Taylor polynomials to evaluate trigonometric, exponential, and logarithmic functions. Understanding Taylor series also builds the foundation for Fourier series and other advanced topics in analysis and differential equations.

Common Mistakes

Mistake: Forgetting to divide by n! in each term.
Correction: Every term in the Taylor series has the factorial n! in the denominator. When you compute the nth derivative and plug it in, you must divide by n!. For example, if f'''(a) = 6, the third-degree term is 6/3! · (x−a)³ = (x−a)³, not 6(x−a)³.
Mistake: Using the wrong center point when evaluating derivatives.
Correction: The derivatives must all be evaluated at x = a, not at x or at some other value. If your Taylor series is centered at a = 1, you compute f(1), f'(1), f''(1), etc.—not f(0).

Related Terms