A Taylor series builds a polynomial that matches a function closely near one point. Then it keeps adding more terms. Each new term brings it closer to the function. Within some fixed distance from that point, called the radius, this process reaches convergence: the polynomial matches the function exactly.
Later in this course, many ideas depend on convergence in exactly this sense. Two examples are taking a derivative under an integral sign, and writing a solution as a series. Both need convergence to be something you can actually check, not just something you hope is true.
A Taylor series is like a picture taken with a longer and longer lens. The first couple of terms capture only the area close to the center. They capture it sharply. Each new term brings a little more of the area around it into focus.
For a well-behaved function, the whole picture eventually matches the real function. This holds out to some fixed distance from where the picture was centered. That distance is the radius of convergence. Past it, the picture can blur, or even lie outright.
The Taylor series of f centered at a is f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots. It has one term per derivative of f, evaluated at a. Each term is weighted by a factorial. That weight makes higher-order corrections shrink fast enough for the sum to make sense.
Convergence isn't automatic just because a series is built from derivatives that all exist. A function can be infinitely differentiable at a point and STILL have a Taylor series that converges to the wrong thing, or to nothing, away from that point.
The radius of convergence is the distance from a within which the series is guaranteed to actually equal f. A ratio-test computation, comparing consecutive terms, pins that radius down concretely. Chapter 08 leans on convergence arguments of exactly this kind. Differentiating under the integral sign, and the Leibniz rule, both depend on convergence checked rigorously, rather than just assumed.
Formula
The Taylor Series of f at c
also known as: the Maclaurin series when c=0
Recalled from Calculus II — Integral Calculus & Infinite Series · Taylor and Maclaurin series — first taught there.
If f has derivatives of every order at x=c, its Taylor series centered at c is \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n. When c=0 this is called the Maclaurin series of f. The Taylor series is a power series built entirely from the values of f and its derivatives AT the single point c. Nothing about f away from c is used to generate the coefficients, even though the series may converge to f(x) over a whole interval.
More terms track the curve further
Drag the degree slider up -- the polynomial hugs the true curve over a wider and wider window, but only up to the function's radius of convergence.
Loading interactive…
Worked example
A Maclaurin series's radius of convergence
-
1
State the radius of convergence for the Maclaurin series of f(x)=\dfrac{1}{1-x}.
-
2
f(x) = \sum_{n=0}^{\infty} x^n — a geometric series with ratio x.
-
3
A geometric series converges exactly when |x|<1.
-
4
\boxed{R=1} — the series matches f on (-1,1) and diverges outside it.
Loading interactive…
Where this goes next
Chapter 8 of this course
Chapter 08 defines the Gaussian integral and the Gamma function. Both use exactly this kind of convergence check.
Here is the rule. You can take the derivative of a series, or of an integral, one term at a time. But that move is only legal once you have actually checked uniform convergence, or something equivalent. You cannot just assume it, even if each term on its own looks well-behaved.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
What distinguishes a Maclaurin series from a general Taylor series?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Maclaurin series don't have a special convergence guarantee beyond ordinary Taylor series — that isn't the distinguishing feature.
Both Taylor and Maclaurin series are, in general, infinite — neither is automatically finite.
Maclaurin series can be built for any function with derivatives at 0, not only trig functions.
Why
A Maclaurin series is simply the special case of a Taylor series where the center c is chosen to be 0.
From Calculus II — Integral Calculus & Infinite Series
For the Maclaurin series of \cos(x), what is the coefficient of x^4?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
The k=2 term of \sum\frac{(-1)^k}{(2k)!}x^{2k} gives coefficient \frac{1}{4!}=\frac1{24}\approx0.041667.
Correct answer
Also accepted:
Why
The coefficient of x^4 is \frac{f^{(4)}(0)}{4!}, and f^{(4)}(0)=1 for cosine.
From Calculus II — Integral Calculus & Infinite Series
Find the coefficient of x^5 in the Maclaurin series for f(x)=e^{x}. Round to 6 decimal places.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Every derivative of e^x equals e^x, so f^{(n)}(0)=1 for all n, giving coefficient \frac{1}{n!}; for n=5, \frac{1}{120}\approx0.008333.
Correct answer
Also accepted:
Why
Every derivative of e^x equals e^x itself -- what does that make f^{(5)}(0), and what coefficient does that produce?
From Calculus II — Integral Calculus & Infinite Series
For f(x)=\sin(x), which powers of x appear in its Maclaurin series?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Even powers are exactly the ones that vanish for sine — its ODD derivatives at 0 are the nonzero ones.
Half the powers (the even ones) have coefficient exactly 0, not merely alternating sign.
This describes neither sine nor cosine's actual pattern.
Why
\sin(0)=0 and every even derivative of \sin x vanishes at 0, leaving only odd-power terms in its Maclaurin series.
Check which derivatives of \sin x vanish at 0 -- the even ones or the odd ones?