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Substituting into a standard series, differentiating one or integrating one term by term produces almost every series anyone needs — and since a power series is unique, each route gives the same answer as the derivatives would.
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 new Maclaurin series from a standard one by substituting for its variable, by differentiating or integrating term by term, or by multiplying through by a power of $x$ — and you can say what each operation does to the radius of convergence. You can also use term-by-term integration on a function with no elementary antiderivative, which is the one situation where a series is not a convenience but the only method there is.
You know four Maclaurin series outright and you can build one from derivatives when you have to. Building from derivatives is the definition, and it is almost never the quickest route. This lesson is the quick routes, and the reason they are not cheating.
Substitution: replacing the variable of a known series by an expression.
Term-by-term differentiation: differentiating each term of a power series separately.
Term-by-term integration: the same for integrals, with a constant to fix afterwards.
Uniqueness: a function has at most one power series about a given centre — so any route to one gives the same series.
The coefficients of a Taylor series are forced: $c_n = f^{(n)}(a)/n!$. So a function has at most one power series about a given centre, and any legitimate way of producing one produces that one. Substitution is not a less rigorous shortcut; it is the same answer, reached sensibly.
Substituting. Replace the variable of a standard series by anything you like. From $e^u$ with $u = -x^2$ comes $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \cdots$ in one line. Two things move at once, and both have to be tracked: the coefficients pick up whatever constant was substituted, and a substitution of $x^m$ spreads the terms out so that only every $m$th power survives. The radius moves too — $|u| < R$ becomes a condition on $x$, and has to be solved as one.
Differentiating and integrating term by term. Inside the radius of convergence both are legal, and both keep the radius. Only the endpoints can change, and they must be retested. This is how the series nobody memorises get built: $\frac{1}{(1-x)^2}$ by differentiating the geometric series, $\ln(1+x)$ and $\arctan x$ by integrating it.
Multiplying by a power of $x$. The cheapest operation of all: it shifts every power up and changes nothing else, not even the radius.
Why this matters beyond convenience. $e^{-x^2}$ has no elementary antiderivative — no combination of the functions you know differentiates to it. Its series does, term by term. That is not a trick for avoiding work; it is the only way such an integral is computed at all.
Another way: picture
Think of the four standard series as four tools on a bench and the operations as ways of adjusting them. Nobody forges a new tool for each job. You pick the nearest one up, change what its variable stands for, and perhaps differentiate or integrate it — and because a power series is unique, the adjusted tool is exactly the one forging from scratch would have produced.
Another way: steps
| Operation | Effect on the terms | Effect on the radius |
|---|---|---|
| Substitute $u = kx$ | coefficients scale by $k^n$ | divided by $\lvert k\rvert$ |
| Substitute $u = kx^m$ | only every $m$th power survives | the $m$th root of the old one, over $\lvert k\rvert$ |
| Differentiate term by term | $x^n \to n x^{n-1}$ | unchanged |
| Integrate term by term | $x^n \to \dfrac{x^{n+1}}{n+1}$ | unchanged |
| Multiply by $x^k$ | every power shifts up by $k$ | unchanged |
Only substitution moves the radius, and it moves it because the radius is a statement about the variable. The other three leave it alone — though integrating can gain an endpoint and differentiating can lose one, which is why an interval has to be rechecked even when the radius has not moved.
Substituting into the powers but not the coefficients. $\sin(2x)$ is $2x - \frac{(2x)^3}{3!} + \cdots$, and the $(2x)^3$ is $8x^3$. Every power of the variable picks up the matching power of the constant.
Losing the radius in a substitution. A series in $x^2$ has its radius in $x$, not in $x^2$: if the condition is $x^2 < \tfrac13$ then the radius is $\tfrac{1}{\sqrt3}$.
Thinking term-by-term calculus is a liberty. Inside the radius it is a theorem, not a liberty — and outside it nothing at all is permitted, not even adding the terms up.
For $\cos(3x)$: start from $\cos u = 1 - \tfrac{u^2}{2!} + \tfrac{u^4}{4!} - \cdots$ and put $u = 3x$.
Name it, then say what $u$ is.
$1 - \dfrac{9x^2}{2!} + \dfrac{81x^4}{4!} - \cdots$ — the $3$ is raised to the same power as the $x$.
Coefficients and powers together.
Still only even powers, and still an infinite radius: $|3x| < \infty$ is no condition at all.
$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$ for $|x| < 1$, by putting $u = -x$ into the geometric series.
Substitute first.
Integrating term by term: $x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots + C$, and $C = 0$ because $\ln 1 = 0$.
Fix the constant at the centre.
So $\ln(1+x) = x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots$, radius still $1$ — and at $x = 1$ it now converges, which the original did not. Integrating gained an endpoint.
It is $\dfrac{1}{1-u}$ with $u = x^2$, multiplied by $x$ — two operations, both cheap.
Name the series and the substitution.
$\dfrac{1}{1-x^2} = 1 + x^2 + x^4 + \cdots$, and multiplying by $x$ shifts every power up by one: $x + x^3 + x^5 + \cdots$.
Substitute, then multiply.
The condition was $|u| < 1$, that is $x^2 < 1$, so the radius in $x$ is $1$; multiplying by $x$ did not change it. Only odd powers survive, which is right — the function is odd.
You want the coefficient of $x^{4}$ in the Maclaurin series of $\cos(6x)$. Put the steps in the order you do them.
Number the steps in order (write the number in the box):
Each of these is a standard series in disguise. Say which series to start from, and what to substitute.
| Start from | Substitute | |
|---|---|---|
| $e^{-x^{2}}$ | ||
| $\dfrac{1}{1 + x}$ | ||
| $\sin(2x)$ |
Find the coefficient of $x^{4}$ in the Maclaurin series of $e^{2x^{2}}$.
Answer:
From $\dfrac{1}{1-x} = 1 + x + x^{2} + \cdots$ for $|x| < 1$, differentiating both sides term by term gives a series for $\dfrac{1}{(1-x)^{2}}$. What is the coefficient of $x^{3}$ in it, and what is its radius?
Match each function to the operation on a standard series that produces it.
| Integrate the geometric series in $-x$ | Differentiate the geometric series | Substitute $u = x^{2}$ into the series for $e^{u}$ | Integrate the geometric series in $-x^{2}$ | |
|---|---|---|---|---|
| $\ln(1 + x)$ | ||||
| $\dfrac{1}{(1 - x)^{2}}$ | ||||
| $e^{x^{2}}$ | ||||
| $\arctan x$ |
$\displaystyle\int_0^1 e^{-x^{4}}\,dx$ has no elementary antiderivative. Using the series, its value is $1 - \dfrac{1}{A} + \dfrac{1}{B} - \cdots$. Give $A$, then $B$.
A = p, B = q
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
You want the coefficient of $x^{4}$ in the Maclaurin series of $\cos(3x)$. Put the steps in the order you do them.
Number the steps in order (write the number in the box):
You can build new series from the standard ones. Without looking: which of the four operations changes the radius of convergence, and why is substituting into a known series not a less rigorous route than differentiating?
9. Your turn: the Maclaurin series of $\dfrac{x}{1 - x^{2}}$, step 3