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The coefficients of a power series are forced to be $f^{(n)}(a)/n!$, so a function has at most one — and need not equal it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to derive the Taylor coefficient formula by differentiating a series term by term and setting $x = a$, build a Taylor series about any centre by evaluating derivatives there and dividing by the factorial, write the result in powers of $(x - a)$, read a high derivative off a series obtained by any route because the series is unique, recognise a Taylor polynomial as the best local fit of its degree, and give an example of a function that is infinitely differentiable and not equal to its Taylor series.
The last lesson bent known series into new ones, which is the practical method. But it leaves an obvious question unanswered: where did $\sum \frac{x^n}{n!}$ come from in the first place?
The answer is a formula. If a function has a power series about $a$, then differentiating it $n$ times and setting $x = a$ kills every term but one and recovers the coefficient. So the coefficients are forced: $c_n = f^{(n)}(a)/n!$. There is at most one power series a function can have about a point, and this is it.
The Taylor series of $f$ about $a$ is $\sum_{n\ge0}\dfrac{f^{(n)}(a)}{n!}(x-a)^n$. When $a = 0$ it is called the Maclaurin series, which is a name and not a different idea.
The Taylor polynomial $T_N$ is the series cut off after the $(x-a)^N$ term. It is the polynomial of degree $N$ that matches $f$ and its first $N$ derivatives at $a$ — the best possible local fit of that degree.
A function is analytic at $a$ if its Taylor series actually converges to it near $a$. Not every infinitely differentiable function is, which is the surprise of this lesson.
The formula. Suppose $f(x) = \sum_{n\ge0}c_n(x-a)^n$ near $a$. Differentiate $n$ times and set $x = a$: every term below $n$ has been differentiated to nothing, every term above still has a factor $(x-a)$, and the $n$th contributes $n!\,c_n$. So $$c_n = \frac{f^{(n)}(a)}{n!}.$$
Uniqueness. The coefficients are determined by $f$, so a function has at most one power series about a point. However you obtained a series — by substituting, integrating, multiplying — it is the Taylor series, and $f^{(n)}(a) = n!\,c_n$ can be read straight off it. That is often the cheapest way to find a high derivative.
Taylor polynomials. $T_N(x) = \sum_{n=0}^{N}\dfrac{f^{(n)}(a)}{n!}(x-a)^n$ agrees with $f$ to order $N$ at $a$. $T_1$ is the tangent line from Calculus I; $T_2$ adds curvature; each extra term buys one more order of agreement.
The catch. Having a Taylor series is not the same as being equal to it. The series may diverge away from $a$, or converge to the wrong function. The standard example is $$f(x) = \begin{cases}e^{-1/x^2} & x \ne 0\\ 0 & x = 0\end{cases},$$ which is infinitely differentiable with every derivative zero at $0$. Its Maclaurin series is identically $0$, and converges everywhere — to the wrong function, at every point but one.
So "$f$ equals its Taylor series" is a further claim, settled by showing the remainder tends to zero. That is the next lesson.
The six standard series are worth knowing outright, because in practice they are how almost every Taylor series is obtained.
Another way: picture
$\sin x$ drawn with $T_1 = x$, $T_3 = x - \frac{x^3}{6}$ and $T_5$ on the same axes. Each hugs the curve over a wider stretch before peeling away, and each peels away upwards or downwards according to the sign of the first term it omitted. The picture of a Taylor polynomial is a local fit that is excellent near the centre and eventually worthless, and the remainder measures where 'eventually' begins.
Another way: steps
$T_N$ is defined by matching derivatives, and it is worth seeing why that is the right thing to match.
The tangent line $T_1(x) = f(a) + f'(a)(x-a)$ is the unique line with $T_1(a) = f(a)$ and $T_1'(a) = f'(a)$, and Calculus I showed it is the best linear approximation near $a$: the error is $o(x - a)$. $T_2$ adds the condition $T_2''(a) = f''(a)$, and the error becomes $o((x-a)^2)$.
In general $T_N$ is the unique polynomial of degree $N$ agreeing with $f$ in value and first $N$ derivatives at $a$, and its error is $o((x-a)^N)$. Each extra derivative matched buys one more power of smallness in the error — near the centre.
Away from the centre it buys nothing in particular, and for a series with finite radius it eventually buys damage. $T_N$ for $\frac{1}{1-x}$ is a polynomial, which is finite everywhere; the function has a pole at $1$. No polynomial can imitate that, and the higher $N$ goes the more violently $T_N$ misbehaves past $x = 1$.
So a Taylor polynomial is a local instrument with a controlled error, and the control is the subject of the next lesson.
Forgetting to evaluate at the centre. The coefficients are numbers. A series with $f'(x)$ in it is not a series.
Writing $x^n$ instead of $(x-a)^n$. A Taylor series about $3$ is in powers of $(x - 3)$; anything else describes a different function.
Multiplying by $n!$ instead of dividing. The coefficient is $f^{(n)}(a)/n!$. The multiplication is for going the other way, from coefficient to derivative.
Guessing the pattern from two derivatives. Three or four is the minimum, and for a product or quotient the pattern often has no simple form at all — which is a reason to use the standard series instead.
Assuming convergence means equality. $e^{-1/x^2}$ is the standing counterexample: series convergent everywhere, equal to the function nowhere but the centre.
Assuming infinite differentiability is enough. It is necessary and not sufficient; that is the same example.
Computing a Taylor series from the definition when a standard one would do. $e^{x^2}$ by repeated differentiation is unpleasant; by substitution it is one line.
The formula $c_n = f^{(n)}(a)/n!$ works for any infinitely differentiable function, and it always produces a series. It is very easy to slide from "$f$ has a Taylor series" to "$f$ equals its Taylor series", and those are different statements.
The series can fail in two ways. It can diverge away from the centre, which is ordinary — $\frac{1}{1-x}$ has radius $1$ and the function exists far beyond that. Or it can converge and converge to the wrong thing, which is not ordinary and is why the point needs making.
The example is $f(x) = e^{-1/x^2}$ for $x \ne 0$, with $f(0) = 0$. It is infinitely differentiable, and every derivative at $0$ is zero — the exponential flattens faster than any polynomial can notice. So its Maclaurin series is $0 + 0x + 0x^2 + \cdots$, which converges beautifully, everywhere, to the zero function. And $f$ is not the zero function anywhere except at the origin.
So the coefficients carry only what the derivatives at one point know, and for this function they know nothing. Establishing that $f$ equals its series requires showing that the remainder $f - T_N$ tends to zero, which is a separate piece of work with its own theorem — and it is what the next lesson is about.
Until that work is done, the right phrasing is "the Taylor series of $f$", never "$f$ is its Taylor series".
$f(x) = \ln x$ about $a = 1$: $f(1) = 0$, $f'(x) = x^{-1}$ so $f'(1) = 1$, $f''(x) = -x^{-2}$ so $f''(1) = -1$, $f'''(x) = 2x^{-3}$ so $f'''(1) = 2$.
Differentiate, then evaluate at the centre.
In general $f^{(n)}(1) = (-1)^{n+1}(n-1)!$, so $c_n = \dfrac{(-1)^{n+1}}{n}$.
The factorial in the derivative and the one in the formula nearly cancel.
$\ln x = \sum_{n\ge1}\dfrac{(-1)^{n+1}(x-1)^{n}}{n}$, radius $1$ about $1$ — which is the $\ln(1+u)$ series with $u = x - 1$, as it must be. The radius is $1$ because the function fails to exist at $0$, a distance $1$ away.
Find $f^{(10)}(0)$ for $f(x) = x^{2}e^{x^{2}}$. Differentiating ten times is out of the question.
The question is about a coefficient in disguise.
$e^{u} = \sum \dfrac{u^{n}}{n!}$, so $e^{x^2} = \sum \dfrac{x^{2n}}{n!}$ and $x^2 e^{x^2} = \sum_{n\ge0}\dfrac{x^{2n+2}}{n!}$.
Two substitutions, no differentiation.
The $x^{10}$ term is $n = 4$: coefficient $\dfrac{1}{4!} = \dfrac{1}{24}$. So $f^{(10)}(0) = 10!\cdot\dfrac{1}{24} = 151200$. Uniqueness is what permits this: the series obtained by substitution is the Taylor series.
Substitute $u = -x^{2}$ into the geometric series: $\dfrac{1}{1 + x^{2}} = \sum_{n\ge0}(-1)^{n}x^{2n}$, radius $1$.
Cheaper than four differentiations.
The $x^{4}$ term is $n = 2$, with coefficient $(-1)^{2} = 1$.
Only even powers appear.
So $f^{(4)}(0) = 4!\times 1 = 24$. And $f^{(3)}(0) = 0$, because there is no $x^{3}$ term — every odd derivative of an even function vanishes at $0$.
For $f(x) = x^{3}$, use the second-order Taylor polynomial at $x = 4$ to estimate $f\!\left(4 + \dfrac{1}{6}\right)$.
Answer:
A function's Maclaurin series has $5x^{5}$ as its $x^{5}$ term. What is $f^{(5)}(0)$?
Answer:
Match each function to its coefficient of $x^{3}$. (For reference, $\cos x$ has coefficient $0$.)
| $\dfrac{1}{6}$ | $-\dfrac{1}{6}$ | $0$ | $1$ | |
|---|---|---|---|---|
| $e^{x}$ | ||||
| $\sin x$ | ||||
| $\cos x$ | ||||
| $\dfrac{1}{1 - x}$ |
Put the steps of building the Taylor series of a function about $x = 8$ into the order you do them.
Number the steps in order (write the number in the box):
For $f(x) = x^{3}$ about $x = 3$, fill in the derivatives at the centre and the first three coefficients.
| Value | |
|---|---|
| $f(3)$, and so $c_0$ | |
| $f'(3)$, and so $c_1$ | |
| $f''(3)$ | |
| $c_2 = f''(3)/2!$ |
Find $\displaystyle\lim_{x\to0}\frac{\sin(8x) - 8x}{x^{3}}$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = x^{3}$, use the second-order Taylor polynomial at $x = 5$ to estimate $f\!\left(5 + \dfrac{1}{2}\right)$.
Answer:
You can build a Taylor series about any centre and read derivatives back off one. Say in your own words why a convergent Taylor series need not converge to the function it came from. Next: the theorem that settles when it does.
10. Your turn: the Maclaurin series of $\dfrac{1}{1 + x^{2}}$, and $f^{(4)}(0)$, step 3