Back to the on-screen lesson ·
Differentiating term by term forces the coefficients to be $f^{(n)}(a)/n!$. The series for $e^x$, $\sin x$, $\cos x$ and $1/(1-x)$ are worth knowing outright, because almost every other series is one of them in disguise.
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 can build a Taylor series from a function's derivatives, dividing the $n$th derivative at the centre by $n$ factorial, and you can write down the Maclaurin series of $e^x$, $\sin x$, $\cos x$ and $\dfrac{1}{1-x}$ without deriving them. You can say what the factorial denominators are for, why two of these series converge everywhere and two have radius $1$, and why an odd function's series contains only odd powers.
You can find where a power series converges. That answered where, and left what open: a power series converges to some number at each point inside its radius, and nothing so far said which number. This lesson runs the question the other way — start from a function and build the series that represents it.
Taylor series of $f$ about $a$: $\sum \dfrac{f^{(n)}(a)}{n!}(x - a)^n$.
Maclaurin series: a Taylor series about $0$.
Taylor polynomial of degree $n$: the series cut off after the $x^n$ term — a polynomial you can actually compute with.
$f^{(n)}(a)$: the $n$th derivative, evaluated at the centre.
Suppose $f(x) = c_0 + c_1(x - a) + c_2(x - a)^2 + \cdots$ near $a$. Put $x = a$: every term but the first dies, so $c_0 = f(a)$. Differentiate and put $x = a$ again: $c_1 = f'(a)$. Differentiate twice: the $x^2$ term brings down a $2$, so $c_2 = \frac{f''(a)}{2}$. Continuing forces every coefficient:
$$c_n = \frac{f^{(n)}(a)}{n!}, \qquad f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n.$$
The coefficients were not chosen; they were forced. If a function has a power series at all, that is the one it has — which is why any legitimate route to a series gives the same answer.
The factorial is structural. It is what makes the terms shrink faster than any power grows, and it is the whole reason $e^x$ and $\sin x$ have series valid for every $x$. A series written without it converges nowhere useful.
Four to know outright, all about $0$:
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \quad (R = \infty)$$ $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \quad (R = \infty)$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \quad (R = \infty)$$ $$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \quad (R = 1)$$
Two more follow from the last by integrating term by term and are worth knowing too: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ and $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$, both of radius $1$.
A radius of $1$ is not a defect of the method. It is the distance from the centre to the nearest place the function breaks down: $\frac{1}{1-x}$ has an asymptote at $x = 1$, and no series about $0$ could ever reach past it.
Another way: picture
Picture the graph of $\sin x$ with polynomials drawn over it: first $x$, then $x - \tfrac{x^3}{6}$, then five terms, then seven. Each hugs the curve for longer before peeling away. The Taylor series is the limit of that sequence of ever-better local fits, and the radius of convergence is how far the hugging eventually reaches.
Another way: steps
| Function | Series | Powers present | Radius |
|---|---|---|---|
| $e^{x}$ | $\sum \dfrac{x^n}{n!}$ | all | infinite |
| $\sin x$ | $\sum \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ | odd only | infinite |
| $\cos x$ | $\sum \dfrac{(-1)^n x^{2n}}{(2n)!}$ | even only | infinite |
| $\dfrac{1}{1-x}$ | $\sum x^n$ | all | $1$ |
| $\ln(1+x)$ | $\sum \dfrac{(-1)^{n+1}x^n}{n}$ | all, from $n = 1$ | $1$ |
| $\arctan x$ | $\sum \dfrac{(-1)^n x^{2n+1}}{2n+1}$ | odd only | $1$ |
Two patterns are worth naming. An odd function has only odd powers and an even function only even ones, which halves the work and catches errors: a cosine series with an $x^3$ in it is wrong on symmetry grounds alone. And factorials in the denominator go with an infinite radius, plain integers with a radius of $1$ — the bottom three are all the geometric series in disguise, and they inherit its limit.
Dropping the factorials. $c_n = \dfrac{f^{(n)}(a)}{n!}$, and writing $\dfrac{1}{n}$ instead is the quiet version of the same error — it gives a series that converges, to the wrong function.
"A Taylor series equals its function everywhere." Only inside the radius. $\dfrac{1}{1-x}$ has a series valid on $(-1, 1)$ and useless at $x = 2$, where the function itself is perfectly well defined.
Forgetting that the centre shifts the powers. A series about $a$ is built from $(x - a)^n$, not from $x^n$. Expanding those brackets out is almost always the wrong move: it destroys the one feature the form was chosen for.
Derivatives at $0$: $\cos 0 = 1$, $-\sin 0 = 0$, $-\cos 0 = -1$, $\sin 0 = 0$, then it repeats.
The odd ones all vanish.
Divide the $n$th by $n!$: $c_0 = 1$, $c_1 = 0$, $c_2 = -\tfrac{1}{2!}$, $c_3 = 0$, $c_4 = \tfrac{1}{4!}$.
Definition, term by term.
So $\cos x = 1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots$: only even powers, and the signs alternate. The cosine is an even function, and its series shows it.
To estimate $\sin(0.2)$, take the first two terms: $x - \tfrac{x^3}{6}$.
Degree three.
$0.2 - \dfrac{0.008}{6} = 0.2 - 0.00133 = 0.19867$.
Arithmetic only.
The true value is $0.198669\ldots$, so two terms gave five correct decimal places. This is how a calculator does it.
Rather than differentiating, start from $e^{u} = 1 + u + \tfrac{u^2}{2!} + \cdots$, which is one of the four.
A standard series in disguise.
Put $u = -x$: $1 - x + \tfrac{x^2}{2!} - \tfrac{x^3}{3!} + \cdots$.
Substitute.
The signs now alternate and the radius is still infinite, since $|-x|$ and $|x|$ are the same size. Every derivative of $e^{-x}$ at zero is $\pm 1$, which is the same answer reached the long way.
Match each function to its Maclaurin series.
| $1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \cdots$ | $x - \dfrac{x^{3}}{3!} + \dfrac{x^{5}}{5!} - \cdots$ | $1 - \dfrac{x^{2}}{2!} + \dfrac{x^{4}}{4!} - \cdots$ | $1 + x + x^{2} + x^{3} + \cdots$ | |
|---|---|---|---|---|
| $e^{x}$ | ||||
| $\sin x$ | ||||
| $\cos x$ | ||||
| $\dfrac{1}{1 - x}$ |
Give the radius of convergence of each standard Maclaurin series.
| Radius | |
|---|---|
| $e^{x}$ | |
| $\cos x$ | |
| $\dfrac{1}{1 - x}$ | |
| $\arctan x$ |
In the Maclaurin series of $\dfrac{1}{1 - x}$, what is the coefficient of $x^{3}$? Give it to four decimal places.
Answer:
Which of these is the Maclaurin series of $\ln(1 + x)$?
A function $f$ is to be expanded as a Maclaurin series up to $x^{4}$, from the definition. Put the steps in the order you do them.
Number the steps in order (write the number in the box):
Use the standard series to evaluate $\displaystyle\lim_{x\to 0}\frac{1 - \cos x}{x^{2}}$, then $\displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^{3}}$. Give the two limits.
first limit = p, second limit = q
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In the Maclaurin series of $e^x$, what is the coefficient of $x^{3}$? Give it to four decimal places.
Answer:
You can build and recognise the standard Maclaurin series. Without looking: what is the coefficient of $x^n$ in a series about zero, and which two of the four standard series have a finite radius?
9. Your turn: the Maclaurin series of $e^{-x}$, step 3