A Taylor polynomial matches a function's derivatives up to some degree and then stops. If the matching never stops, which series come out for , and , and how do a handful of such series turn into series for almost any function built from them?

Calculus 2

The common Maclaurin series

A Taylor polynomial matches a function’s derivatives up to some degree and then stops. If the matching never stops, which series come out for e^x, \sin x and \cos x, and how do a handful of such series turn into series for almost any function built from them?

You need

Representing functions as power seriesCalculus 2 Second and higher derivativesCalculus 1 Derivatives of exponential functionsCalculus 1

Unlocks

Using series to evaluate limits and integralsCalculus 2

Where you are 5 in Taylor and Maclaurin series

Calculus8

Sequences and series in calculus8

Taylor and Maclaurin series5

Taylor polynomials The Taylor remainder and the Lagrange error bound The binomial series

The common Maclaurin series

Using series to evaluate limits and integrals

Why this exists

Taylor polynomials built Taylor polynomials: the degree n polynomial whose value and first n derivatives at the center a match those of f, with coefficients \displaystyle \frac{f^{(k)}(a)}{k!}. A polynomial of fixed degree is good only near its center. Keeping every term gives a power series, and power series are functions in their own right, which can be added, multiplied, differentiated and integrated term by term (Representing functions as power series).

That lesson built series from the geometric series alone, which reaches rational functions, logarithms and the inverse tangent but not e^x, \sin x or \cos x: no algebra on \displaystyle \frac{1}{1-x} produces them. Their derivatives at 0 repeat in a short cycle, though, so their coefficients can be written down at once. Day 1 does that. Day 2 collects six standard series and shows how substitution, multiplication, addition and calculus turn them into series for functions like x^2\sin(3x) and \displaystyle \cos\sqrt{x} without computing a single derivative.

Definition

Day 1: Taylor and Maclaurin series

Let f have derivatives of every order at a. The Taylor series of f centered at a is

\sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k

= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots

Its partial sums are the Taylor polynomials P_n(x) of Taylor polynomials. A Taylor series centered at 0 is a Maclaurin series. The definition makes no promise that the series converges at any x \ne a, or that where it converges its sum is f(x); both have to be found.

Day 2: building series from known ones

A series is built from a known one when it is obtained from a series already known to equal its function by the moves of Representing functions as power series: substituting an expression for the variable, multiplying by a constant or a power of x, adding, multiplying two series, and differentiating or integrating term by term. Each move keeps the equality on a stated set of x.

Intuition

A Taylor polynomial of degree n agrees with f to order n at the center and drifts away farther out. Raising the degree pushes the drift farther out. For \sin x (Figure 1) the partial sums P_1, P_5, P_9 follow the wave over wider and wider stretches. The radius of the series is infinite, so the stretch grows without end, and each fixed x is eventually inside it.

Figure 1y = \sin x with the partial sums P_1, P_5 and P_9 of its Maclaurin series. Each extra pair of terms keeps the partial sum on the wave over a wider stretch; the series has infinite radius, so every x is eventually reached.

Whether a series has this property is a question about the derivatives. For e^x, \sin x and \cos x the derivatives at 0 stay bounded, while k! grows faster than any power r^k, so the terms \displaystyle \frac{f^{(k)}(0)}{k!}x^k shrink to 0 for every x. For \displaystyle \frac{1}{1-x} the derivatives at 0 are k! exactly, the factorials cancel, and the series converges only for \lvert x \rvert < 1.

Results

Uniqueness. If f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k for every x in an open interval around a, then

c_k = \frac{f^{(k)}(a)}{k!} \quad \text{for every } k

So a function has at most one power series at a given center, and any power series for f, however it was found, is its Taylor series. Read backward, f^{(k)}(a) = k!\,c_k: a high derivative at the center can be read off a coefficient.

The common Maclaurin series.

Function	Series	Converges to the function on
\dfrac{1}{1-x}	\displaystyle\sum_{n=0}^{\infty} x^n	(-1, 1)
e^x	\displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}	all real x
\sin x	\displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}	all real x
\cos x	\displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}	all real x
\ln(1+x)	\displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}	(-1, 1]
\arctan x	\displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}	[-1, 1]

The first is the geometric series. The last two were found by integrating geometric series in Representing functions as power series. The equality for e^x is proved below; the equality for \sin x and \cos x at every x is proved in The Taylor remainder and the Lagrange error bound, which also proves the e^x case a second way and measures how far a partial sum is from the function. Powers (1+x)^k have a series of their own, The binomial series.

Products. If \sum a_n x^n and \sum b_n x^n converge to f(x) and g(x) at a point x inside both open intervals of convergence, then

\begin{aligned} &f(x)\,g(x) \\ &\quad = \sum_{n=0}^{\infty} \Big(\sum_{j=0}^{n} a_j b_{n-j}\Big) x^n \end{aligned}

the coefficient of x^n collecting every product of a term of one series with a term of the other whose powers add to n. This is stated without proof; it rests on absolute convergence inside the interval.

Why it is true

Uniqueness. By term by term differentiation, f can be differentiated k times inside the interval, and

\begin{aligned} f^{(k)}(x) = k!\,c_k &+ \sum_{j > k} j(j-1)\cdots \\ &\quad \cdots(j-k+1)\,c_j (x-a)^{j-k} \end{aligned}

At x = a every term of the sum is 0, leaving f^{(k)}(a) = k!\,c_k.

The series for e^x equals e^x. Let \displaystyle S(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}, which converges for every x (Example 1). Differentiating term by term,

\begin{aligned} S'(x) &= \sum_{n=1}^{\infty} \frac{n\,x^{n-1}}{n!} \\ &= \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = S(x) \end{aligned}

so \displaystyle \frac{d}{dx}\big[S(x)e^{-x}\big] = S'(x)e^{-x} - S(x)e^{-x} = 0 for every x. A function with derivative 0 on the whole line is constant, and at x = 0 the constant is S(0)e^0 = 1. So S(x) = e^x for every real x.

A similar argument works for sine and cosine, since each series is the derivative of the other up to sign (Example 2); this course proves that case in The Taylor remainder and the Lagrange error bound.

Worked examples

Example 1

Day 1: the series for e to the x

Branka wants the Maclaurin series of e^x, its radius of convergence, and the value its partial sum through x^5 gives at \displaystyle x = -\frac12.

Solution

Every derivative of e^x is e^x (Derivatives of exponential functions), so f^{(k)}(0) = e^0 = 1 for every k and \displaystyle c_k = \frac{1}{k!}:

e^x \;\to\; \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

For the radius, apply the The ratio and root tests to the absolute values of the terms. For x \ne 0,

\left\lvert \frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n} \right\rvert = \frac{\lvert x \rvert}{n+1} \to 0

The limit is 0 < 1 for every x, so the series converges absolutely for every real x: the radius is infinite. With the proof above, the sum is e^x everywhere.

At \displaystyle x = -\frac12 the terms through x^5 are \displaystyle 1, -\frac12, \frac18, -\frac1{48}, \frac1{384}, -\frac1{3840}. Over the common denominator 3840:

\frac{3840 - 1920 + 480 - 80 + 10 - 1}{3840} = \frac{2329}{3840}

which is 0.606510 to six places, against e^{-1/2} = 0.606531. Six terms already agree to four decimal places, because the factorials in the denominators grow so fast.

Example 2

Day 1: sine and cosine

Zorana wants the Maclaurin series of \sin x and its radius, and then the series of \cos x.

Solution

The derivatives of \sin x run \sin x, \cos x, -\sin x, -\cos x and then repeat, so at 0 they run 0, 1, 0, -1 and repeat. Every even coefficient is 0, and the odd ones alternate in sign: \displaystyle c_{2n+1} = \frac{(-1)^n}{(2n+1)!}. The Maclaurin series of \sin x is

\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots

The zero terms are skipped, so the ratio test is applied to the series of nonzero terms. For x \ne 0,

\left\lvert \frac{x^{2n+3}}{(2n+3)!} \cdot \frac{(2n+1)!}{x^{2n+1}} \right\rvert = \frac{x^2}{(2n+2)(2n+3)} \to 0

so the series converges for every real x, and the radius is infinite.

For cosine, differentiate the sine series term by term, which keeps the infinite radius:

\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)\,x^{2n}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

since \displaystyle \frac{2n+1}{(2n+1)!} = \frac{1}{(2n)!}. This is the series \displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots, and its coefficients match the derivatives of \cos x at 0, which run 1, 0, -1, 0. Sine is an odd function and its series has only odd powers; cosine is even and has only even powers (Exercise 10).

Example 3

Day 1: cosine centered at pi over 6

Velimir wants the Taylor series of \cos x centered at \displaystyle \frac{\pi}{6} through the \displaystyle \left(x - \frac{\pi}{6}\right)^4 term, its radius, and a second derivation of the same terms.

Solution

The derivatives run \cos x, -\sin x, -\cos x, \sin x and repeat. At \displaystyle \frac{\pi}{6}, where \displaystyle \cos = \frac{\sqrt3}{2} and \displaystyle \sin = \frac12, they run \displaystyle \frac{\sqrt3}{2}, -\frac12, -\frac{\sqrt3}{2}, \frac12, then \displaystyle \frac{\sqrt3}{2} again. Dividing by k!, with \displaystyle h = x - \frac{\pi}{6}:

\begin{aligned} \cos x \;\to\; &\frac{\sqrt3}{2} - \frac12 h - \frac{\sqrt3}{4} h^2 \\ &+ \frac{1}{12}h^3 + \frac{\sqrt3}{48}h^4 - \cdots \end{aligned}

No coefficient here is 0, and the sizes alternate between \displaystyle \frac{\sqrt3}{2\,k!} and \displaystyle \frac{1}{2\,k!}. The ratio of consecutive sizes is \displaystyle \frac{1}{\sqrt3\,(k+1)} or \displaystyle \frac{\sqrt3}{k+1}, so the ratio of consecutive terms is at most \displaystyle \frac{\sqrt3\,\lvert h \rvert}{k+1}, which tends to 0. The radius is infinite.

A second route: by the angle sum formula, \displaystyle \cos x = \cos\left(\frac{\pi}{6} + h\right) = \frac{\sqrt3}{2}\cos h - \frac12\sin h. Substituting the series of Example 2 in h:

\begin{aligned} &\tfrac{\sqrt3}{2}\left(1 - \tfrac{h^2}{2} + \tfrac{h^4}{24}\right) - \tfrac12\left(h - \tfrac{h^3}{6}\right) \\ &\quad = \tfrac{\sqrt3}{2} - \tfrac12 h - \tfrac{\sqrt3}{4} h^2 + \tfrac{1}{12}h^3 + \tfrac{\sqrt3}{48}h^4 \end{aligned}

through h^4. The two routes agree, as uniqueness says they must: once the series of \sin and \cos equal their functions, the angle sum gives a power series for \cos x in powers of h, and there is only one.

Example 4

Day 2: substituting and multiplying by a power

Branka wants the Maclaurin series of f(x) = x^2\sin(3x), where it is valid, and f^{(9)}(0) and f^{(10)}(0).

Solution

The sine series holds for every real u, so it holds with u = 3x:

\sin(3x) = \sum_{n=0}^{\infty} \frac{(-1)^n 3^{2n+1} x^{2n+1}}{(2n+1)!}

Multiplying by x^2 multiplies every partial sum by the same number, so for every real x

\begin{aligned} x^2\sin(3x) &= \sum_{n=0}^{\infty} \frac{(-1)^n 3^{2n+1} x^{2n+3}}{(2n+1)!} \\ &= 3x^3 - \frac92 x^5 + \frac{81}{40} x^7 - \cdots \end{aligned}

By uniqueness this is the Taylor series of f, so f^{(9)}(0) = 9!\,c_9. The power x^9 is x^{2n+3} with n = 3, whose coefficient is \displaystyle \frac{(-1)^3\,3^7}{7!} = -\frac{2187}{5040}. So

\begin{aligned} f^{(9)}(0) &= 9! \cdot \left(-\frac{2187}{5040}\right) \\ &= -72 \cdot 2187 = -157{,}464 \end{aligned}

using \displaystyle \frac{9!}{7!} = 72. Every power in the series is odd, so c_{10} = 0 and f^{(10)}(0) = 0. Nine rounds of the product rule would reach the same numbers with far more work.

Example 5

Day 2: a logarithm and its endpoints

Zorana wants the Maclaurin series of \ln(8 + x^3) and the exact set of x where it equals the function.

Solution

Factor out the value at the center: \displaystyle \ln(8 + x^3) = \ln 8 + \ln\left(1 + \frac{x^3}{8}\right) for x > -2. The series for \ln(1 + u) holds exactly for -1 < u \le 1, so with \displaystyle u = \frac{x^3}{8},

\begin{aligned} \ln(8 + x^3) &= \ln 8 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(\frac{x^3}{8}\right)^n \\ &= \ln 8 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{3n}}{n\,8^n} \end{aligned}

whenever \displaystyle -1 < \frac{x^3}{8} \le 1, that is -8 < x^3 \le 8, that is -2 < x \le 2. The substitution carries the interval (-1, 1] over whole, endpoints included. At x = 2 the series is \displaystyle \ln 8 + \left(1 - \frac12 + \frac13 - \cdots\right) = \ln 8 + \ln 2 = \ln 16, which is the function’s value. At x = -2 every term is \displaystyle -\frac1n, the series diverges, and the function is undefined. The interval is (-2, 2].

Example 6

Day 2: multiplying two series

Velimir wants the Maclaurin series of e^{2x}\cos x through x^4, and a check of one coefficient by differentiating.

Solution

Substituting u = 2x into the series of e^u, and writing out \cos x:

\begin{aligned} e^{2x} &= 1 + 2x + 2x^2 + \tfrac43 x^3 + \tfrac23 x^4 + \cdots \\ \cos x &= 1 - \tfrac12 x^2 + \tfrac1{24} x^4 - \cdots \end{aligned}

Both hold for every real x, so the product rule for series applies everywhere. The coefficient of x^n collects the products whose powers add to n; cosine has no odd terms, which shortens the work:

Power	Products	Coefficient
x^0	1 \cdot 1	1
x^1	2 \cdot 1	2
x^2	\displaystyle 2 \cdot 1 + 1 \cdot \left(-\frac12\right)	\displaystyle \frac32
x^3	\displaystyle \frac43 \cdot 1 + 2 \cdot \left(-\frac12\right)	\displaystyle \frac13
x^4	\displaystyle \frac23 - 1 + \frac1{24}	\displaystyle -\frac{7}{24}

So \displaystyle e^{2x}\cos x = 1 + 2x + \frac32 x^2 + \frac13 x^3 - \frac{7}{24}x^4 + \cdots for every real x.

Check c_2: with f(x) = e^{2x}\cos x, the product rule gives f'(x) = e^{2x}(2\cos x - \sin x) and f''(x) = e^{2x}(3\cos x - 4\sin x). So f''(0) = 3 and \displaystyle \frac{f''(0)}{2!} = \frac32, as the table says.

Example 7

Day 2: a square root inside cosine

Branka wants a power series for \displaystyle f(x) = \cos\sqrt{x}, x \ge 0, the value f'(0) (from the right), and the sum \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}\,n}{(2n)!}.

Solution

Substitute \displaystyle u = \sqrt{x} into the cosine series. Since u^{2n} = x^n for x \ge 0,

\begin{aligned} \cos\sqrt{x} &= \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{(2n)!} \\ &= 1 - \frac{x}{2} + \frac{x^2}{24} - \frac{x^3}{720} + \cdots \end{aligned}

for every x \ge 0. The square roots are gone: this is an ordinary power series in x, and the ratio \displaystyle \frac{\lvert x \rvert}{(2n+1)(2n+2)} \to 0 shows it converges for every real x.

The closed form is awkward at 0: the chain rule gives \displaystyle f'(x) = -\frac{\sin\sqrt{x}}{2\sqrt{x}}, which is \displaystyle \frac00 at x = 0. The series has no such trouble. Differentiating term by term,

f'(x) = \sum_{n=1}^{\infty} \frac{(-1)^n\,n\,x^{n-1}}{(2n)!} = -\frac12 + \frac{x}{12} - \cdots

so \displaystyle f'(0) = -\frac12, the right-hand derivative of \displaystyle \cos\sqrt{x} at 0. In the same way \displaystyle f''(0) = 2! \cdot \frac{1}{24} = \frac{1}{12}.

At x = 1 the two forms of f' give \displaystyle -\frac{\sin 1}{2} = \sum_{n=1}^{\infty} \frac{(-1)^n\,n}{(2n)!}. Changing every sign,

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}\,n}{(2n)!} = \frac{\sin 1}{2} \approx 0.420735

The first four terms, \displaystyle \frac12, \displaystyle -\frac{2}{24}, \displaystyle \frac{3}{720} and \displaystyle -\frac{4}{40{,}320}, already add to 0.420734.

Boundaries

A Taylor series needs every derivative at the center. \displaystyle \sqrt{x} has no Maclaurin series: it is not differentiable at 0. Example 7 works because the substituted series is a power series in x, not in \displaystyle \sqrt{x}.
Convergence is not equality. A Taylor series can converge everywhere and still sum to something other than its function. The standard example is f(x) = e^{-1/x^2} with f(0) = 0: every derivative of f at 0 is 0 (stated here without proof), so its Maclaurin series is 0 for every x, while f(x) > 0 for every x \ne 0. Equality for the functions in the table is proved here for e^x and in The Taylor remainder and the Lagrange error bound for sine and cosine.
Each series holds only on its own set. \ln(1+x) and \arctan x need \lvert x \rvert \le 1 (with x = -1 excluded for the logarithm), and \displaystyle \frac{1}{1-x} needs \lvert x \rvert < 1. After a substitution u = g(x), the new series holds exactly where g(x) lies in the old set (Example 5).
A substituted series can outlive its formula. The series for \displaystyle \cos\sqrt{x} converges for x < 0 too, but there it sums to a different function (Exercise 23).
Products are guaranteed inside both open intervals. At an endpoint where one series converges only conditionally, the product needs a separate argument.
Uniqueness is at one center. A function has at most one power series centered at a; centered at a different point it has a different one (Example 3).

Where this goes next

The Taylor remainder and the Lagrange error bound proves that the series for \sin x and \cos x equal their functions at every x and bounds the error of each partial sum, which turns a Taylor polynomial into an estimate with a guarantee. The binomial series adds the last common series, for (1+x)^k with any real k. Using series to evaluate limits and integrals uses the series built here to evaluate limits and to compute definite integrals of functions such as \displaystyle \cos\sqrt{x} that have no antiderivative in closed form, to any accuracy wanted.

Exercises

Day 1: series from derivatives

For Exercises 1 to 5, find the Taylor series from the derivatives of the function at the center, and its interval of convergence.

2^x, centered at 0
e^x, centered at -2
\sin x, centered at \pi
\dfrac{e^x + e^{-x}}{2}, centered at 0
\ln x, centered at 1
Find \displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{3^n}{n!}.
Find \displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n \pi^{2n}}{36^n\,(2n)!}.

Day 1: reasoning

Show that every Taylor coefficient of \displaystyle \frac{1}{1-x} at 0 is 1, and explain why uniqueness alone already guaranteed this.
Use the ratio test to show that the series \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} converges for every real x.
Explain why the Maclaurin series of an even function has no odd powers.

Day 2: building series

For Exercises 11 to 15, find the Maclaurin series by building it from the table, and where it equals the function.

x^3 e^{2x}
\cos(x^3)
\ln(1 - 2x)
\arctan(x^2)
\dfrac{e^x - e^{-x}}{2}, by adding two series
Find the Maclaurin series of e^x \ln(1+x) through x^5.
Let f(x) = x\arctan(2x). Find f^{(7)}(0) and f^{(8)}(0).
Let f(x) = e^{x^3}. Find f^{(12)}(0).
Find the function \displaystyle \sum_{n=1}^{\infty} \frac{n\,x^n}{n!} in closed form, and then \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n\,n}{3^n\,n!}.
Find a power series for \displaystyle\int_0^x e^{-t^4}\,dt and where it is valid.
Find \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\,3^n}.
Find \displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\,3^n}.

Day 2: reasoning

Branka says that since \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{(2n)!} converges for every real x, it equals \displaystyle \cos\sqrt{x} for every real x. Explain what is wrong, and find the sum for x < 0.
A function g has g(0) = 1, satisfies g'(x) = 2x\,g(x), and equals a power series \sum c_n x^n on the whole line. Find every c_n and identify g.

Answers

\sum_{n=0}^{\infty} \dfrac{(\ln 2)^n x^n}{n!}, all real x. By Derivatives of exponential functions, f^{(k)}(x) = (\ln 2)^k 2^x, so f^{(k)}(0) = (\ln 2)^k; the ratio \displaystyle \frac{\lvert x \rvert \ln 2}{n+1} \to 0.
\sum_{n=0}^{\infty} \dfrac{e^{-2}(x+2)^n}{n!}, all real x. Every derivative at -2 is e^{-2}; the ratio \displaystyle \frac{\lvert x+2 \rvert}{n+1} \to 0.
\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}(x-\pi)^{2n+1}}{(2n+1)!}, all real x. The derivatives at \pi run 0, -1, 0, 1 and repeat, the negative of the pattern at 0.
\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n)!}, all real x. The kth derivative is \displaystyle \frac{e^x + (-1)^k e^{-x}}{2}, which is 1 at 0 for even k and 0 for odd k; the ratio \displaystyle \frac{x^2}{(2n+1)(2n+2)} \to 0.
\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}(x-1)^n}{n} on (0, 2]. \displaystyle f^{(k)}(x) = \frac{(-1)^{k-1}(k-1)!}{x^k}, so \displaystyle c_k = \frac{(-1)^{k-1}}{k} and the radius is 1. At x = 2 the series is the alternating harmonic series, convergent; at x = 0 every term is \displaystyle -\frac1n, divergent.
e^3, the series of e^x at x = 3.
\displaystyle \frac{\sqrt3}{2}. The terms are \displaystyle \frac{(-1)^n (\pi/6)^{2n}}{(2n)!}, the cosine series at \displaystyle \frac{\pi}{6}.
\displaystyle f^{(k)}(x) = \frac{k!}{(1-x)^{k+1}} by induction, so f^{(k)}(0) = k! and c_k = 1. The geometric series \sum x^n already equals \displaystyle \frac{1}{1-x} on (-1, 1), and a function has only one power series at 0, so it had to be the Taylor series.
For x \ne 0, the ratio of the sizes of consecutive terms is \displaystyle \frac{x^{2n+2}}{(2n+2)!} \cdot \frac{(2n)!}{x^{2n}}, which simplifies to \displaystyle \frac{x^2}{(2n+1)(2n+2)} and tends to 0 < 1, so the series converges absolutely; at x = 0 it is 1.
If f(x) = \sum c_n x^n near 0 and f(-x) = f(x), then \sum c_n x^n = \sum (-1)^n c_n x^n near 0, and uniqueness gives c_n = (-1)^n c_n, so c_n = 0 for odd n.
\sum_{n=0}^{\infty} \dfrac{2^n x^{n+3}}{n!}, all real x. Substitute u = 2x into e^u, then multiply by x^3.
\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{6n}}{(2n)!}, all real x. Substitute u = x^3 into \cos u.
-\sum_{n=1}^{\infty} \dfrac{2^n x^n}{n} on \displaystyle \left[-\frac12, \frac12\right). Substitute u = -2x into \ln(1+u); -1 < -2x \le 1 means \displaystyle -\frac12 \le x < \frac12.
\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{4n+2}}{2n+1} on [-1, 1]. Substitute u = x^2; -1 \le x^2 \le 1 means \lvert x \rvert \le 1.
\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{(2n+1)!}, all real x. \displaystyle e^x - e^{-x} = \sum \frac{(1 - (-1)^n)x^n}{n!}: even terms cancel, odd terms double.
\displaystyle x + \frac12 x^2 + \frac13 x^3 + 0\cdot x^4 + \frac{3}{40}x^5, for \lvert x \rvert < 1. Multiply \displaystyle e^x = \sum \frac{x^j}{j!} by \displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots (terms through \displaystyle \frac{x^5}{5}). The x^4 coefficient is \displaystyle \frac16 - \frac14 + \frac13 - \frac14, which is 0, and the x^5 coefficient is \displaystyle \frac1{24} - \frac1{12} + \frac16 - \frac14 + \frac15, which is \displaystyle \frac{3}{40}.
f^{(7)}(0) = 0 and f^{(8)}(0) = -737{,}280. \displaystyle x\arctan(2x) = \sum \frac{(-1)^n 2^{2n+1} x^{2n+2}}{2n+1} has only even powers; for x^8, n = 3 and \displaystyle c_8 = -\frac{128}{7}, so \displaystyle f^{(8)}(0) = 8! \cdot \left(-\frac{128}{7}\right) = -5760 \cdot 128.
19{,}958{,}400. \displaystyle e^{x^3} = \sum \frac{x^{3n}}{n!}, so \displaystyle c_{12} = \frac{1}{4!} and \displaystyle f^{(12)}(0) = \frac{12!}{4!}.
x e^x, and the sum is \displaystyle -\frac13 e^{-1/3}. \displaystyle \frac{n}{n!} = \frac{1}{(n-1)!}, so the series is \displaystyle x\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = x e^x; then take \displaystyle x = -\frac13.
\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{4n+1}}{n!\,(4n+1)}, all real x. Substitute u = -t^4 into e^u and integrate term by term from 0.
\displaystyle \ln\frac43, the series of \ln(1+x) at \displaystyle x = \frac13.
\displaystyle \frac{\pi\sqrt3}{6}. At \displaystyle x = \frac{1}{\sqrt3}, \displaystyle x^{2n+1} = \frac{1}{\sqrt3\,3^n}, so the arctangent series at that point is \displaystyle \frac{1}{\sqrt3} times the series asked for. Since \displaystyle \arctan\frac{1}{\sqrt3} = \frac{\pi}{6}, the sum is \displaystyle \sqrt3 \cdot \frac{\pi}{6}.
The substitution \displaystyle u = \sqrt{x} is only available for x \ge 0; convergence of the series says nothing about which function it sums to. For x = -s^2 with s > 0, the terms are \displaystyle \frac{(-1)^n (-1)^n s^{2n}}{(2n)!} = \frac{s^{2n}}{(2n)!}, so by Exercise 4 the sum is \displaystyle \frac{e^{\sqrt{-x}} + e^{-\sqrt{-x}}}{2}, which is greater than 1 and so is not the cosine of any real number.
\displaystyle c_{2k} = \frac{1}{k!}, odd c_n = 0, and g(x) = e^{x^2}. Term by term, \sum n c_n x^{n-1} = \sum 2 c_n x^{n+1}. Matching constants gives c_1 = 0; matching x^m for m \ge 1 gives (m+1)c_{m+1} = 2c_{m-1}. With c_0 = g(0) = 1, the odd coefficients are 0 and \displaystyle c_{2k} = \frac{c_{2k-2}}{k}, so \displaystyle c_{2k} = \frac{1}{k!} and \displaystyle g(x) = \sum \frac{(x^2)^k}{k!} = e^{x^2}.