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Inside its radius a power series behaves like a polynomial, which turns one geometric series into the whole catalogue.
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 differentiate and integrate a power series term by term inside its radius and say what makes that a theorem rather than an obvious rearrangement, keep the radius unchanged under both operations while retesting both endpoints afterwards, build a new series from a standard one by substituting and recompute the radius the substitution produces, divide by the new exponent rather than the old when integrating, and evaluate an integral with no elementary antiderivative by expanding and integrating term by term.
$\dfrac{d}{dx}(f + g) = f' + g'$ extends to any finite sum by induction. An infinite sum is a limit, and induction says nothing about limits — so whether $\dfrac{d}{dx}\sum f_n = \sum f_n'$ is a genuine question, and for general series of functions the answer is no.
For power series inside their radius the answer is yes, and that is a theorem worth naming. It is what makes a power series behave like a polynomial of infinite degree rather than like an arbitrary limit of functions.
To operate term by term is to apply an operation to each term and sum the results, asserting that the result equals the operation applied to the sum.
The standard series are the four worth knowing outright: the geometric series, $e^x$, $\ln(1+x)$ and $\arctan x$, together with $\sin x$ and $\cos x$. Every other series in this course is one of those substituted into, differentiated, integrated or multiplied.
The radius is unchanged by differentiation and integration, but the endpoints are not: integrating can gain one and differentiating can lose one.
Theorem. Let $f(x) = \sum_{n\ge0}c_n(x-a)^n$ have radius $R > 0$. Then on $(a - R, a + R)$, $f$ is infinitely differentiable, and $$f'(x) = \sum_{n\ge1}nc_n(x-a)^{n-1}, \qquad \int_a^x f = \sum_{n\ge0}\frac{c_n(x-a)^{n+1}}{n+1},$$ both with the same radius $R$.
The radius is unchanged because dividing or multiplying $c_n$ by $n$ does not change $\lim\sqrt[n]{|c_n|}$ — the $n$th root of $n$ tends to $1$.
The endpoints are not covered. They must be retested after every operation. Integrating divides the coefficients by $n+1$, making the terms smaller, so an endpoint can be gained; differentiating multiplies by $n$, so one can be lost.
Building new series. Four operations, each cheap:
Why this matters. Almost nothing is expanded from the definition. One series plus these four moves generates the whole catalogue — and term-by-term integration evaluates integrals that have no elementary antiderivative at all.
Uniqueness. If two power series about the same centre agree as functions near it, their coefficients agree. So however a series was obtained, it is the series, and $c_n = f^{(n)}(a)/n!$ regardless.
Another way: picture
The partial sums of $\sum x^n$ drawn against $\dfrac{1}{1-x}$ on $(-1,1)$: each is a polynomial, each hugs the curve more closely and over a wider stretch, and all of them fly away near $x = 1$. Differentiating any one of them is ordinary calculus; the theorem says the limit of the derivatives is the derivative of the limit, which the picture makes plausible and does not prove.
Another way: steps
Start with $\dfrac{1}{1-x} = \sum_{n\ge0}x^n$ on $(-1,1)$.
Substitute $-x$: $\dfrac{1}{1+x} = \sum(-1)^n x^n$, same radius.
Integrate that from $0$: $\ln(1+x) = \sum_{n\ge1}\dfrac{(-1)^{n+1}x^n}{n}$, radius still $1$ — and now $x = 1$ has been gained, giving $\ln 2 = 1 - \tfrac12 + \tfrac13 - \cdots$, the alternating harmonic series.
Substitute $-x^2$ into the geometric series: $\dfrac{1}{1+x^2} = \sum(-1)^n x^{2n}$, radius $1$ (the square root of $1$).
Integrate that: $\arctan x = \sum_{n\ge0}\dfrac{(-1)^n x^{2n+1}}{2n+1}$, and again $x = 1$ is gained: $\dfrac{\pi}{4} = 1 - \tfrac13 + \tfrac15 - \cdots$, Leibniz's formula for $\pi$.
Differentiate the geometric series: $\dfrac{1}{(1-x)^2} = \sum_{n\ge1}nx^{n-1}$.
Five well-known series, none computed from the definition, all obtained in a line each from the first. That is the working method, and it is also why the standard series are worth memorising: they are the seeds.
Operating outside the radius. The theorem is about the open interval. Nothing permits term-by-term anything at or beyond the boundary.
Assuming the endpoints carry over. They must be retested. Integration can gain one; differentiation can lose one.
Dividing by $n$ instead of $n+1$ when integrating. $\int x^n = \frac{x^{n+1}}{n+1}$; the denominator is the new exponent.
Forgetting the constant of integration. $\int_a^x$ is the clean way to avoid it; an indefinite integral needs the constant fixed by a known value.
Losing a substitution's effect on the radius. $u = kx$ divides it by $|k|$; $u = x^2$ takes the square root.
Shifting the index carelessly when differentiating. $\sum_{n\ge0}c_nx^n$ differentiates to $\sum_{n\ge1}nc_nx^{n-1}$: the $n=0$ term is a constant and disappears.
Assuming it works for general series of functions. It does not. Power series inside the radius are a very special case, and the reason is that the convergence there is uniform on compact subsets — a fact this course states and real analysis proves.
Differentiating a sum termwise feels like something that must be true, because it is true of every sum anyone has differentiated before. That intuition comes from finite sums, where it is a consequence of the sum rule applied finitely often, and it does not extend to limits.
For general series of functions it is false. There are series of perfectly smooth functions that converge to a function which is nowhere differentiable, and others where $\sum f_n'$ converges to something that is not the derivative of $\sum f_n$. Nothing about "add up the derivatives" is automatic.
What makes power series different is that inside the radius they converge extremely well — uniformly on every closed subinterval, a notion this course names and real analysis develops — and that is exactly the hypothesis under which limits and derivatives may be exchanged. The radius is not decoration on the theorem; it is where the theorem lives.
The practical consequences are two. First, never operate at or beyond the boundary: the theorem does not reach there, and the endpoints of the new series must be retested by hand. Second, inside the radius you may be as free as you like — substitute, differentiate, integrate, multiply — and that freedom is what turns one geometric series into the whole catalogue, including a value for $\pi$ and an integral that no antiderivative can reach.
Find $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^{n}}$. It is not geometric — the $n$ on top rules that out.
A numerical series, approached through a function.
Start from $\sum_{n\ge0}x^n = \dfrac{1}{1-x}$ and differentiate: $\sum_{n\ge1}nx^{n-1} = \dfrac{1}{(1-x)^2}$. Multiply by $x$: $\sum_{n\ge1}nx^{n} = \dfrac{x}{(1-x)^2}$.
Differentiating brings the $n$ down.
Put $x = \tfrac12$, which is inside the radius: the sum is $\dfrac{1/2}{1/4} = 2$. A numerical series evaluated by turning it into a function first.
$\displaystyle\int_0^{1}\frac{\sin x}{x}dx$: the integrand has no elementary antiderivative, and the integral is a perfectly definite number.
The situation the theorem exists for.
$\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$ with infinite radius, so $\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots$ — and the singularity at $0$ was never real.
Divide the series, not the functions.
Integrate term by term: $1 - \dfrac{1}{18} + \dfrac{1}{600} - \cdots \approx 0.9461$. The series alternates with decreasing terms, so three terms are right to within $\frac{1}{35280}$.
A value and an error bound together.
Start from $\dfrac{1}{1+x} = \sum_{n\ge0}(-1)^{n}x^{n}$, radius $1$.
Substitute $-x$ into the geometric series.
Differentiate both sides: the left gives $-\dfrac{1}{(1+x)^{2}}$, and the right gives $\sum_{n\ge1}(-1)^{n}nx^{n-1}$.
Term by term, permitted inside the radius.
Multiply by $-1$: $\dfrac{1}{(1+x)^{2}} = \sum_{n\ge1}(-1)^{n+1}nx^{n-1} = \sum_{m\ge0}(-1)^{m}(m+1)x^{m}$, radius still $1$ — and at $x = \pm1$ the terms no longer tend to zero, so both endpoints have been lost.
Find the coefficient of $x^{4}$ in the Maclaurin series of $\dfrac{1}{1 - 4x}$.
Answer:
Match each function to its Maclaurin series. (One of them, $e^{x}$, has radius infinite.)
| $\sum_{n\ge0}x^{n}$ | $\sum_{n\ge0}\dfrac{x^{n}}{n!}$ | $\sum_{n\ge1}\dfrac{(-1)^{n+1}x^{n}}{n}$ | $\sum_{n\ge0}\dfrac{(-1)^{n}x^{2n+1}}{2n+1}$ | |
|---|---|---|---|---|
| $\dfrac{1}{1 - x}$ | ||||
| $e^{x}$ | ||||
| $\ln(1 + x)$ | ||||
| $\arctan x$ |
A series in $u$ has radius $1$. Give the radius in $x$ after each operation.
| New radius | |
|---|---|
| Differentiate term by term | |
| Integrate term by term | |
| Substitute $u = 3x$ |
$\displaystyle\sum_{n\ge0}x^{n}$ converges on $(-1, 1)$ and at neither endpoint. Integrating term by term gives $\displaystyle\sum_{n\ge1}\frac{x^{n}}{n}$. What has changed?
Integrating $\dfrac{1}{1 - 5x} = \sum_{n\ge0}5^{\,n}x^{n}$ term by term from $0$ gives a series $\sum_{n \ge 0} d_n x^{n+1}$. Find $d_{3}$.
Answer:
$\displaystyle\int_0^{1/2} e^{-x^{2}}dx$ has no elementary antiderivative. Expanding and integrating term by term gives $\dfrac{1}{2} - c + \cdots$. Find the second term $c$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A power series is integrated one term at a time. Practise the move on a finite one: find the antiderivative $F$ of $f(x) = 6x^{2} + 2x + 2$ with $F(0) = 0$.
Answer:
You can produce a series for almost any function by bending a standard one, and say where the operations are permitted. Say in your own words why integrating can gain an endpoint and differentiating can lose one. Next: where the coefficients come from in the first place.
10. Your turn: a series for $\dfrac{1}{(1 + x)^{2}}$, step 3
Retest the ends after differentiating.