A Maclaurin series is just a Taylor series centered at 00 — that one fact settles the whole comparison. If the center is a=0a = 0, it is Maclaurin; if the center is any other value, it is a Taylor series. The naming change is small, but the center matters because a series is most useful near the point where it is built.

The two side by side

                  Taylor series                 Maclaurin series
Center            any value a                   a = 0 (special case)
General term      f⁽ⁿ⁾(a)/n! · (x − a)ⁿ         f⁽ⁿ⁾(0)/n! · xⁿ
Built to describe the function near x = a        the function near x = 0
Relationship      the general case              one special case of Taylor

In formulas, if a function has enough derivatives at aa, its Taylor series about aa is

n=0f(n)(a)n!(xa)n\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

Set a=0a = 0 and you get the Maclaurin series:

n=0f(n)(0)n!xn\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

The structure does not change. The center changes.

When to use which

Use a Maclaurin series when 00 is the natural reference point or when derivatives at 00 are easy to compute. Use a Taylor series around another value aa when you need a good local approximation near that value — to estimate behavior near x=3x = 3, expanding around a=3a = 3 usually beats expanding around 00.

Why does this matter? The coefficients come from derivatives evaluated at the center, so changing the center usually changes the numbers in the series. Both expansions can be correct, but one may be far more practical for the value you care about. And do not over-claim: a Taylor or Maclaurin series is always a local expansion; whether it equals the function on an interval depends on the function and where the series converges.

Worked example: exe^x at two different centers

Take

f(x)=exf(x) = e^x

a clean comparison because every derivative of exe^x is still exe^x.

Maclaurin series at a=0a = 0. Every derivative value is f(n)(0)=1f^{(n)}(0) = 1, so

ex=n=0xnn!e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}

with first terms

1+x+x22!+x33!+1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

Taylor series at a=1a = 1. Every derivative value at the center is f(n)(1)=ef^{(n)}(1) = e, so

ex=n=0en!(x1)ne^x = \sum_{n=0}^{\infty}\frac{e}{n!}(x-1)^n

with first terms

e+e(x1)+e2!(x1)2+e + e(x-1) + \frac{e}{2!}(x-1)^2 + \cdots

The function stayed the same; only the center changed. That is the entire difference between Taylor and Maclaurin series in one example.

Frequently confused points

  • Treating them as different ideas. They are not separate theories — Maclaurin is one special case of Taylor.
  • Ignoring the center. Two valid series for the same function can differ; the one centered near your target value is the more useful approximation.
  • Assuming the series always equals the function. Not automatic. It depends on the function and the interval. The safe statement is that the series gives a local expansion around its center, then you check convergence if the problem asks.

Apply it yourself

Write the series for sinx\sin x twice: once at a=0a = 0 and once at a=π/4a = \pi/4. Comparing those two expansions is one of the fastest ways to make the difference stick. These series appear whenever you approximate functions, study local behavior, solve differential equations, or replace a complicated expression with an easier polynomial — and the recurring question is simply which point makes the local model most useful.

Frequently Asked Questions

What is the difference between a Taylor series and a Maclaurin series?
A Maclaurin series is the Taylor series centered at $a = 0$. The formula pattern is the same; only the center changes.
Is a Maclaurin series always the best choice?
No. It is convenient when $x = 0$ is a natural center or when derivatives at $0$ are simple. If you care about values near some other point, a Taylor series centered there is usually more useful.

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