Exercise 1: Coefficients from the definition: the n-th derivative at a, over n factorial

If $f$ has derivatives of every order at $a$ , its Taylor series centred at $a$ is $\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$ , and its Maclaurin series is the case $a = 0$ . The method from the definition: differentiate until a pattern appears, prove the pattern, evaluate AT $a$ , divide by $n!$ , and keep the powers of $(x - a)$ , never of $x$ .

This is the route of last resort, used when no series of the table can be adapted. It is still the one an exam asks for when it says from the definition.

a) Find the Taylor series of $e^x$ centred at $a = 3$ from the definition, then check it with $e^x = e^3 e^{x-3}$ .
b) Find the Taylor series of $\ln x$ centred at $a = 2$ : find $f^{(n)}(x)$ for $n \ge 1$ , prove the pattern, and write the general term.
c) Find the radius and the interval of convergence of the series of b), and check the series against the table entry for $\ln(1 + u)$ .
d) Find the first four nonzero terms of the Taylor series of $\cos x$ centred at $a = \frac{\pi}{3}$ .
e) Find the Taylor series of $p(x) = x^3 - 2x$ centred at $a = 1$ . Why is it finite, and why does it equal $p(x)$ for every $x$ ?

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Answers

a) $e^x = \sum_{n=0}^{\infty}\frac{e^3}{n!}(x-3)^n$ , for every $x$
b) $\ln x = \ln 2 + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\,2^n}(x-2)^n$
c) $R = 2$ , interval $(0, 4]$
d) $\frac{1}{2} - \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right) - \frac{1}{4}\left(x - \frac{\pi}{3}\right)^2 + \frac{\sqrt{3}}{12}\left(x - \frac{\pi}{3}\right)^3$
e) $-1 + (x-1) + 3(x-1)^2 + (x-1)^3$ , which expands back to $x^3 - 2x$

a) Every derivative of $e^x$ is $e^x$ , so $f^{(n)}(3) = e^3$ for every $n$ and $c_n = \frac{e^3}{n!}$ . The series is $\sum_{n=0}^{\infty}\frac{e^3}{n!}(x-3)^n = e^3\left(1 + (x-3) + \frac{(x-3)^2}{2} + \dots\right)$ . Ratio test: $\left|\frac{c_{n+1}(x-3)^{n+1}}{c_n(x-3)^n}\right| = \frac{|x-3|}{n+1} \to 0 < 1$ for every $x$ , so $R = \infty$ . Check: $e^x = e^3 \cdot e^{x-3}$ , and putting $u = x - 3$ in the table series $e^u = \sum\frac{u^n}{n!}$ gives exactly the same coefficients. Two routes, one series: this is the uniqueness theorem at work, since a function that equals a power series in $(x - a)$ near $a$ can only have $\frac{f^{(n)}(a)}{n!}$ as coefficients. The trap is to write $\sum\frac{e^3}{n!}x^n$ : the centre is lost, and at $x = 3$ that series would give $e^3 \cdot e^3$ instead of $e^3$ .

b) $f(x) = \ln x$ , $f'(x) = x^{-1}$ , $f''(x) = -x^{-2}$ , $f'''(x) = 2x^{-3}$ , $f^{(4)}(x) = -6x^{-4}$ . Pattern: $f^{(n)}(x) = (-1)^{n-1}(n-1)!\,x^{-n}$ for $n \ge 1$ . Proof by induction: true for $n = 1$ ; if true for $n$ , then $f^{(n+1)}(x) = (-1)^{n-1}(n-1)! \cdot (-n)x^{-n-1} = (-1)^n n!\,x^{-(n+1)}$ , which is the formula for $n + 1$ . At $a = 2$ : $f^{(n)}(2) = \frac{(-1)^{n-1}(n-1)!}{2^n}$ , so $c_n = \frac{(-1)^{n-1}(n-1)!}{2^n\,n!} = \frac{(-1)^{n-1}}{n\,2^n}$ , since $n! = n \cdot (n-1)!$ . The term $n = 0$ is $f(2) = \ln 2$ and is written apart, because the pattern only holds from $n = 1$ (at $n = 0$ it would contain $(-1)!$ ). So $\ln x = \ln 2 + \frac{x-2}{2} - \frac{(x-2)^2}{8} + \frac{(x-2)^3}{24} - \dots$ Check of $c_2$ : $f''(2) = -\frac{1}{4}$ , divided by $2! = 2$ , gives $-\frac{1}{8}$ .

c) Ratio test on $a_n = \frac{(-1)^{n-1}(x-2)^n}{n\,2^n}$ : $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-2|}{2} \cdot \frac{n}{n+1} \to \frac{|x-2|}{2}$ . Convergence for $|x - 2| < 2$ , divergence for $|x - 2| > 2$ : $R = 2$ . Each endpoint gets its own test. At $x = 4$ : $\frac{(x-2)^n}{2^n} = 1$ and the series is $\sum\frac{(-1)^{n-1}}{n}$ , which converges by the alternating series test. At $x = 0$ : $\frac{(-2)^n}{2^n} = (-1)^n$ , the terms are $\frac{(-1)^{n-1}(-1)^n}{n} = -\frac{1}{n}$ , minus the harmonic series, which diverges. Interval $(0, 4]$ . Table check: $\ln x = \ln\left(2\left(1 + \frac{x-2}{2}\right)\right) = \ln 2 + \ln(1 + u)$ with $u = \frac{x-2}{2}$ , and $\ln(1+u) = \sum_{n\ge1}\frac{(-1)^{n-1}u^n}{n}$ for $-1 < u \le 1$ , that is $0 < x \le 4$ : same coefficients, same interval, and the table GUARANTEES that the series equals $\ln x$ there. The radius is the distance from the centre $2$ to $0$ , where $\ln$ stops existing; the series cannot pass $x = 4$ either, although $\ln$ is smooth at $5$ , because a power series converges on an interval symmetric about its centre.

d) The derivatives of $\cos$ cycle: $\cos x$ , $-\sin x$ , $-\cos x$ , $\sin x$ , then $\cos x$ again. At $\frac{\pi}{3}$ : $\frac{1}{2}$ , $-\frac{\sqrt{3}}{2}$ , $-\frac{1}{2}$ , $\frac{\sqrt{3}}{2}$ . Dividing by $0!, 1!, 2!, 3!$ : $c_0 = \frac{1}{2}$ , $c_1 = -\frac{\sqrt{3}}{2}$ , $c_2 = -\frac{1}{4}$ , $c_3 = \frac{\sqrt{3}}{12}$ . None is zero, so the first four nonzero terms are the terms of degree $0$ to $3$ : $\frac{1}{2} - \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right) - \frac{1}{4}\left(x - \frac{\pi}{3}\right)^2 + \frac{\sqrt{3}}{12}\left(x - \frac{\pi}{3}\right)^3$ . The trap is to copy the Maclaurin pattern of $\cos$ , which has only even powers because $\sin 0 = 0$ : around $\frac{\pi}{3}$ the odd derivatives do not vanish, and the copy that writes $\frac{1}{2} - \frac{1}{4}\left(x - \frac{\pi}{3}\right)^2 + \dots$ loses half the terms. Check with the addition formula, $u = x - \frac{\pi}{3}$ : $\cos x = \frac{1}{2}\cos u - \frac{\sqrt{3}}{2}\sin u = \frac{1}{2}\left(1 - \frac{u^2}{2}\right) - \frac{\sqrt{3}}{2}\left(u - \frac{u^3}{6}\right) + \dots$ , the same four terms.

e) $p(1) = -1$ ; $p'(x) = 3x^2 - 2$ , $p'(1) = 1$ ; $p''(x) = 6x$ , $p''(1) = 6$ , so $c_2 = \frac{6}{2!} = 3$ ; $p'''(x) = 6$ , $c_3 = \frac{6}{3!} = 1$ ; and $p^{(n)} = 0$ for $n \ge 4$ . The series is $-1 + (x-1) + 3(x-1)^2 + (x-1)^3$ : every coefficient from $n = 4$ on is $0$ , so the sum is finite and there is nothing to converge. Expanding: $(x-1)^3 + 3(x-1)^2 + (x-1) - 1 = x^3 - 3x^2 + 3x - 1 + 3x^2 - 6x + 3 + x - 1 - 1 = x^3 - 2x$ . The Taylor series of a polynomial is the same polynomial, rewritten in powers of $(x - a)$ , and it is valid for every $x$ . The coefficient formula does not know the difference between a finite and an infinite series.

Exercise 2: The table, and substitution into it: the interval moves with the variable

No derivative is computed in this exercise. Every series comes from the table of the reminder, by substituting an expression for the variable and then multiplying by a power of $x$ . What the substitution changes is the INTERVAL: a table series valid for $u$ in some interval becomes valid for the $x$ that put $u$ in that interval, endpoints included or excluded exactly as in the table.

Recall $e^u = \sum_{n\ge0}\frac{u^n}{n!}$ and $\cos u = \sum_{n\ge0}\frac{(-1)^n u^{2n}}{(2n)!}$ for every $u$ , $\ln(1+u) = \sum_{n\ge1}\frac{(-1)^{n-1}u^n}{n}$ for $-1 < u \le 1$ , $\arctan u = \sum_{n\ge0}\frac{(-1)^n u^{2n+1}}{2n+1}$ for $-1 \le u \le 1$ .

a) Find the Maclaurin series of $x e^{-x^2}$ , its general term and its radius of convergence.
b) Find the Maclaurin series of $\ln(1 + 2x)$ and its interval of convergence, endpoints included.
c) Find the Maclaurin series of $x^2\arctan(x^3)$ and its interval of convergence.
d) Find a power series equal to $\cos\sqrt{x}$ for $x \ge 0$ . Why is it a series in whole powers of $x$ , although $\sqrt{x}$ is not?
e) Find the exact sums of $\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n}}{9^n(2n)!}$ and of $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}$ .

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Answers

a) $\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{n!}$ , $R = \infty$
b) $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^n x^n}{n}$ on $\left(-\frac{1}{2}, \frac{1}{2}\right]$
c) $\sum_{n=0}^{\infty}\frac{(-1)^n x^{6n+5}}{2n+1}$ on $[-1, 1]$
d) $\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{(2n)!}$ : $\cos$ has only even powers, and $(\sqrt{x})^{2n} = x^n$
e) $\cos\frac{\pi}{3} = \frac{1}{2}$ and $\sqrt{3}\arctan\frac{1}{\sqrt{3}} = \frac{\pi\sqrt{3}}{6}$

a) $e^u = \sum\frac{u^n}{n!}$ with $u = -x^2$ : $u^n = (-x^2)^n = (-1)^n x^{2n}$ , so $e^{-x^2} = \sum_{n\ge0}\frac{(-1)^n x^{2n}}{n!}$ . Multiplying by $x$ : $x e^{-x^2} = \sum_{n\ge0}\frac{(-1)^n x^{2n+1}}{n!} = x - x^3 + \frac{x^5}{2} - \frac{x^7}{6} + \dots$ The table series holds for every $u$ , and $u = -x^2$ is defined for every $x$ , so $R = \infty$ . The traps: writing $-x^{2n}$ , which loses the alternation, or $(2n)!$ in the denominator, which confuses the result with a cosine. The substitution replaces $u$ by the WHOLE expression $-x^2$ , in parentheses, and leaves the $n!$ alone.

b) With $u = 2x$ : $\ln(1 + 2x) = \sum_{n\ge1}\frac{(-1)^{n-1}(2x)^n}{n} = \sum_{n\ge1}\frac{(-1)^{n-1}2^n x^n}{n} = 2x - 2x^2 + \frac{8x^3}{3} - 4x^4 + \dots$ The table requires $-1 < u \le 1$ , that is $-1 < 2x \le 1$ , so $-\frac{1}{2} < x \le \frac{1}{2}$ . The endpoints are inherited: at $x = \frac{1}{2}$ , $u = 1$ , the series is the alternating harmonic series, which converges by the alternating series test; at $x = -\frac{1}{2}$ , $u = -1$ , it is $-\sum\frac{1}{n}$ and diverges. Interval $\left(-\frac{1}{2}, \frac{1}{2}\right]$ . The trap is to copy the interval $(-1, 1]$ of the table: at $x = 1$ the terms $\frac{2^n}{n}$ do not even tend to $0$ . The factor $2^n$ in the coefficients is what shrinks the radius, as the ratio test confirms: $\frac{2^{n+1}|x|^{n+1}/(n+1)}{2^n|x|^n/n} \to 2|x|$ .

c) With $u = x^3$ : $u^{2n+1} = (x^3)^{2n+1} = x^{6n+3}$ , so $\arctan(x^3) = \sum_{n\ge0}\frac{(-1)^n x^{6n+3}}{2n+1}$ and $x^2\arctan(x^3) = \sum_{n\ge0}\frac{(-1)^n x^{6n+5}}{2n+1} = x^5 - \frac{x^{11}}{3} + \frac{x^{17}}{5} - \dots$ Exponents MULTIPLY under a power: $(x^3)^{2n+1}$ is $x^{6n+3}$ , not $x^{2n+4}$ . Quick check on the first term: $\arctan(x^3)$ starts like $x^3$ , and $x^2 \cdot x^3 = x^5$ . Interval: $|x^3| \le 1$ exactly when $|x| \le 1$ , so $[-1, 1]$ , both endpoints included because the arctangent series converges at $u = \pm1$ (alternating series with $\frac{1}{2n+1}$ decreasing to $0$ ).

d) With $u = \sqrt{x}$ , $x \ge 0$ : $u^{2n} = (\sqrt{x})^{2n} = x^n$ , so $\cos\sqrt{x} = \sum_{n\ge0}\frac{(-1)^n x^n}{(2n)!} = 1 - \frac{x}{2} + \frac{x^2}{24} - \frac{x^3}{720} + \dots$ , with $R = \infty$ . The square root disappears because the cosine series contains only EVEN powers of $u$ , and an even power of $\sqrt{x}$ is a whole power of $x$ . With $\sin\sqrt{x} = \sqrt{x} - \frac{x\sqrt{x}}{6} + \dots$ it would not happen: the odd powers keep a $\sqrt{x}$ . One more caution: the series converges for $x < 0$ too, but $\cos\sqrt{x}$ has no meaning there. The equality is claimed only where the substitution was made, here $x \ge 0$ .

e) First sum: $\frac{\pi^{2n}}{9^n} = \frac{\pi^{2n}}{3^{2n}} = \left(\frac{\pi}{3}\right)^{2n}$ , so the sum is $\sum\frac{(-1)^n}{(2n)!}\left(\frac{\pi}{3}\right)^{2n} = \cos\frac{\pi}{3} = \frac{1}{2}$ . The gesture: identify the table series by the DENOMINATOR, here $(2n)!$ with alternating signs and even powers, which is the cosine; then gather everything else into one power of one number. Second sum: the denominator $2n + 1$ with alternating signs points to the arctangent, which needs $u^{2n+1}$ . With $u = \frac{1}{\sqrt{3}}$ , $u^{2n+1} = \frac{1}{3^n\sqrt{3}}$ , so $\sum\frac{(-1)^n}{(2n+1)3^n} = \sqrt{3}\sum\frac{(-1)^n u^{2n+1}}{2n+1} = \sqrt{3}\arctan\frac{1}{\sqrt{3}} = \sqrt{3} \cdot \frac{\pi}{6} = \frac{\pi\sqrt{3}}{6}$ , legitimate because $u = \frac{1}{\sqrt{3}}$ lies in $[-1, 1]$ . Plausibility: $\frac{3.14 \times 1.73}{6} \approx 0.91$ , and the first terms give $1 - \frac{1}{9} + \frac{1}{45} \approx 0.91$ .

Exercise 3: The binomial series: 1 + x to a power that is not a whole number

For every real $k$ and $|x| < 1$ : $(1 + x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n$ , where $\binom{k}{0} = 1$ and $\binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}$ , a product of exactly $n$ factors on top. When $k$ is a positive integer this is the binomial theorem; otherwise the series never stops.

The series is for $(1 + x)^k$ : the constant inside the parentheses must be $1$ before anything is expanded.

a) Compute $\binom{1/2}{3}$ . Then show that $\binom{-2}{n} = (-1)^n(n+1)$ and deduce the Maclaurin series of $\frac{1}{(1+x)^2}$ .
b) Write the Maclaurin series of $\sqrt{1 + x}$ up to the term in $x^3$ , and check it by squaring.
c) Find the first three terms of the Maclaurin series of $\sqrt[3]{8 + x}$ and its radius of convergence.
d) Find the Maclaurin series of $\frac{1}{\sqrt{1 - x^2}}$ , general term included, and deduce the Maclaurin series of $\arcsin x$ up to $x^7$ .
e) Show that the binomial series stops for $k = 4$ and is then valid for every $x$ , while for $k = \frac{1}{2}$ its radius is exactly $1$ .

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Answers

a) $\binom{1/2}{3} = \frac{1}{16}$ ; $\frac{1}{(1+x)^2} = \sum_{n=0}^{\infty}(-1)^n(n+1)x^n$ for $|x| < 1$
b) $1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \dots$ ; its square is $1 + x + 0x^2 + 0x^3 + \dots$
c) $2 + \frac{x}{12} - \frac{x^2}{288} + \dots$ , $R = 8$
d) $\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n}$ ; $\arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \dots$
e) $\binom{4}{n} = 0$ for $n \ge 5$ ; for $k = \frac{1}{2}$ the ratio $\frac{|k-n|}{n+1}|x| \to |x|$ , so $R = 1$ .

a) $\binom{1/2}{3} = \frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)\left(\frac{1}{2} - 2\right)}{3!} = \frac{\frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right)}{6} = \frac{3/8}{6} = \frac{1}{16}$ . Three factors on top, $k$ , $k - 1$ , $k - 2$ , as many as $n = 3$ ; the usual slip is a fourth factor $k - 3$ . Next, $\binom{-2}{n} = \frac{(-2)(-3)\cdots(-2-n+1)}{n!} = \frac{(-1)^n \cdot 2 \cdot 3\cdots(n+1)}{n!} = (-1)^n\frac{(n+1)!}{n!} = (-1)^n(n+1)$ . So $\frac{1}{(1+x)^2} = (1+x)^{-2} = \sum_{n\ge0}(-1)^n(n+1)x^n = 1 - 2x + 3x^2 - 4x^3 + \dots$ for $|x| < 1$ . It is the series obtained by differentiating $\frac{1}{1+x} = \sum(-x)^n$ and changing the sign: two routes, one series.

b) $\binom{1/2}{0} = 1$ , $\binom{1/2}{1} = \frac{1}{2}$ , $\binom{1/2}{2} = \frac{\frac{1}{2} \cdot \left(-\frac{1}{2}\right)}{2} = -\frac{1}{8}$ , $\binom{1/2}{3} = \frac{1}{16}$ . So $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \dots$ for $|x| < 1$ . Check by squaring, collecting the coefficients of each degree: $x^0$ : $1$ ; $x^1$ : $2 \cdot \frac{1}{2} = 1$ ; $x^2$ : $2\left(-\frac{1}{8}\right) + \left(\frac{1}{2}\right)^2 = 0$ ; $x^3$ : $2 \cdot \frac{1}{16} + 2 \cdot \frac{1}{2} \cdot \left(-\frac{1}{8}\right) = \frac{1}{8} - \frac{1}{8} = 0$ . The square is $1 + x$ up to degree $3$ , as it must be. The alternating signs from $x^2$ on come from the factors $k - 1, k - 2, \dots$ , all negative when $k = \frac{1}{2}$ .

c) Factor first: $(8 + x)^{1/3} = 8^{1/3}\left(1 + \frac{x}{8}\right)^{1/3} = 2(1 + u)^{1/3}$ with $u = \frac{x}{8}$ . With $\binom{1/3}{1} = \frac{1}{3}$ and $\binom{1/3}{2} = \frac{\frac{1}{3} \cdot \left(-\frac{2}{3}\right)}{2} = -\frac{1}{9}$ : $2\left(1 + \frac{1}{3} \cdot \frac{x}{8} - \frac{1}{9} \cdot \frac{x^2}{64} + \dots\right) = 2 + \frac{x}{12} - \frac{x^2}{288} + \dots$ Valid for $|u| < 1$ , that is $|x| < 8$ : $R = 8$ . The trap is to apply the binomial coefficients to $(8 + x)^{1/3}$ directly and write $2 + \frac{x}{3} + \dots$ The first-derivative test catches it at once: $f'(x) = \frac{1}{3}(8 + x)^{-2/3}$ , so $f'(0) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$ , which must be $c_1$ .

d) $\frac{1}{\sqrt{1 - x^2}} = (1 + u)^{-1/2}$ with $u = -x^2$ . Then $\binom{-1/2}{n}(-1)^n = \frac{\frac{1}{2} \cdot \frac{3}{2}\cdots\frac{2n-1}{2}}{n!} = \frac{1 \cdot 3 \cdots (2n-1)}{2^n\,n!}$ , and multiplying top and bottom by $2 \cdot 4 \cdots (2n) = 2^n n!$ gives $\frac{(2n)!}{4^n(n!)^2}$ . So $\frac{1}{\sqrt{1 - x^2}} = \sum_{n\ge0}\frac{(2n)!}{4^n(n!)^2}x^{2n} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \dots$ for $|x| < 1$ . Every coefficient is positive: the sign $(-1)^n$ of the binomial coefficient and the sign $(-1)^n$ of $u^n$ cancel. Since $\arcsin x = \int_0^x\frac{dt}{\sqrt{1 - t^2}}$ (with $\arcsin 0 = 0$ ), integrating term by term inside the radius: $\arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \dots$ , general term $\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$ . Check: $\arcsin(\sin x) = x$ , and $\left(x - \frac{x^3}{6}\right) + \frac{x^3}{6} + \dots = x$ up to degree $3$ .

e) For $k = 4$ , $\binom{4}{n}$ contains the factor $4 - 4 = 0$ as soon as $n \ge 5$ , so every coefficient from $n = 5$ on vanishes and $(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$ : a finite sum, valid for every $x$ . For $k = \frac{1}{2}$ no factor $k - j$ is ever $0$ , and $\frac{\binom{k}{n+1}}{\binom{k}{n}} = \frac{k - n}{n + 1}$ , so the ratio test gives $\left|\frac{k - n}{n + 1}\right||x| \to |x|$ : convergence for $|x| < 1$ , divergence for $|x| > 1$ , $R = 1$ exactly. The same computation gives $R = 1$ for every $k$ that is not a nonnegative integer. The radius is the distance from $0$ to $-1$ , where $(1 + x)^k$ stops being differentiable (or defined), even though nothing happens to $\sqrt{1 + x}$ at $x = 1$ .

Exercise 4: Products, quotients and compositions: build the series, never differentiate five times

Series multiply, divide and compose like polynomials, with two rules. In a PRODUCT, the coefficient of $x^n$ collects every pair of degrees that add up to $n$ : $\left(\sum a_i x^i\right)\left(\sum b_j x^j\right) = \sum_n\left(\sum_{i=0}^{n}a_i b_{n-i}\right)x^n$ . In a COMPOSITION $f(g(x))$ , the inner series must tend to $0$ , so that its powers start at higher and higher degrees and only finitely many of them reach the degree asked for.

Asked up to $x^4$ , expand every factor up to $x^4$ , no further, and drop every product of degree above $4$ .

a) Find the Maclaurin series of $e^x\sin x$ up to $x^5$ .
b) Find the Maclaurin series of $\tan x$ up to $x^5$ by dividing the series of $\sin x$ by that of $\cos x$ .
c) Find the Maclaurin series of $e^{\sin x}$ up to $x^4$ .
d) Find the Maclaurin series of $\ln(\cos x)$ up to $x^4$ .
e) Check d) by integrating the series of $-\tan x$ found in b), then give the fourth derivative of $\ln(\cos x)$ at $0$ .

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Answers

a) $x + x^2 + \frac{x^3}{3} - \frac{x^5}{30}$ (the $x^4$ coefficient is $0$ )
b) $x + \frac{x^3}{3} + \frac{2x^5}{15}$
c) $1 + x + \frac{x^2}{2} - \frac{x^4}{8}$ (the $x^3$ coefficient is $0$ )
d) $-\frac{x^2}{2} - \frac{x^4}{12}$
e) $-\int_0^x\left(t + \frac{t^3}{3}\right)dt = -\frac{x^2}{2} - \frac{x^4}{12}$ ; the fourth derivative is $-2$

a) $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \dots$ and $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ Collecting: $x^1$ : $1 \cdot 1 = 1$ . $x^2$ : $x \cdot x$ , coefficient $1$ . $x^3$ : $1 \cdot \left(-\frac{1}{6}\right) + \frac{1}{2} \cdot 1 = \frac{1}{3}$ . $x^4$ : $x \cdot \left(-\frac{x^3}{6}\right) + \frac{x^3}{6} \cdot x$ , coefficient $-\frac{1}{6} + \frac{1}{6} = 0$ . $x^5$ : $\frac{1}{120} + \frac{1}{2}\left(-\frac{1}{6}\right) + \frac{1}{24} = \frac{1 - 10 + 5}{120} = -\frac{1}{30}$ . So $e^x\sin x = x + x^2 + \frac{x^3}{3} - \frac{x^5}{30} + \dots$ The zero coefficient of $x^4$ is not an accident to be doubted: it says that the fourth derivative of $e^x\sin x$ at $0$ is $0$ , which differentiating four times by the product rule confirms, after a page.

b) $\tan x$ is odd, so write $\tan x = a_1x + a_3x^3 + a_5x^5 + \dots$ and require $\left(a_1x + a_3x^3 + a_5x^5\right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) = x - \frac{x^3}{6} + \frac{x^5}{120}$ . Degree $1$ : $a_1 = 1$ . Degree $3$ : $a_3 - \frac{a_1}{2} = -\frac{1}{6}$ , so $a_3 = \frac{1}{3}$ . Degree $5$ : $a_5 - \frac{a_3}{2} + \frac{a_1}{24} = \frac{1}{120}$ , so $a_5 = \frac{1}{120} + \frac{1}{6} - \frac{1}{24} = \frac{1 + 20 - 5}{120} = \frac{2}{15}$ . So $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ Writing the unknown series and matching coefficients IS the long division, and it is less error-prone. The trap is to divide coefficient by coefficient, $\frac{-1/6}{-1/2} = \frac{1}{3}$ for $x^3$ by luck, then $\frac{1/120}{1/24} = \frac{1}{5}$ for $x^5$ , which is false: a quotient of series is not the quotient of the coefficients.

c) The inner function $u = \sin x = x - \frac{x^3}{6} + \dots$ tends to $0$ , so $e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + \frac{u^4}{24} + \dots$ may be used, with $u^2 = x^2 - \frac{x^4}{3} + \dots$ , $u^3 = x^3 + \dots$ (next term of degree $5$ ), $u^4 = x^4 + \dots$ , and $u^5$ of degree $5$ , dropped. Sum: $1 + \left(x - \frac{x^3}{6}\right) + \frac{1}{2}\left(x^2 - \frac{x^4}{3}\right) + \frac{x^3}{6} + \frac{x^4}{24} = 1 + x + \frac{x^2}{2} + 0 \cdot x^3 - \frac{x^4}{8}$ . The trap is to replace $\sin x$ by $x$ too early: that gives $e^x$ , with a wrong $\frac{x^3}{6}$ and a wrong $\frac{x^4}{24}$ . Each power of $u$ must be expanded to degree $4$ before the terms are added. Direct check of the $0$ : $f' = \cos x\,e^{\sin x}$ , $f'' = \left(\cos^2 x - \sin x\right)e^{\sin x}$ , $f''' = \left(-2\sin x\cos x - \cos x + \cos^3 x - \sin x\cos x\right)e^{\sin x}$ , and at $0$ this is $(-1 + 1)e^0 = 0$ .

d) $\ln(1 + u)$ needs $u \to 0$ , and $\cos x \to 1$ : so write $\cos x = 1 + u$ with $u = -\frac{x^2}{2} + \frac{x^4}{24} - \dots$ , which does tend to $0$ . Then $\ln(\cos x) = u - \frac{u^2}{2} + \frac{u^3}{3} - \dots$ , with $u^2 = \frac{x^4}{4} + \dots$ and $u^3$ of degree $6$ , dropped. So $\ln(\cos x) = -\frac{x^2}{2} + \frac{x^4}{24} - \frac{x^4}{8} + \dots = -\frac{x^2}{2} - \frac{x^4}{12} + \dots$ The trap is to put $u = \cos x$ itself into $\ln(1 + u)$ : $\ln(1 + \cos x)$ is a different function, and even $\ln(\cos x)$ read as $\ln(1 + (\cos x - 1))$ only works because $\cos x - 1$ is small. No constant term: $\ln(\cos 0) = \ln 1 = 0$ , a first check. The function is even, and so is the series.

e) $\frac{d}{dx}\ln(\cos x) = \frac{-\sin x}{\cos x} = -\tan x$ , so $\ln(\cos x) = \ln(\cos 0) - \int_0^x\tan t\,dt = -\int_0^x\left(t + \frac{t^3}{3} + \dots\right)dt = -\frac{x^2}{2} - \frac{x^4}{12} - \dots$ , term-by-term integration being valid inside the radius. Same result as d) by an independent route. The coefficient of $x^4$ is $-\frac{1}{12}$ , so the fourth derivative at $0$ is $4! \cdot \left(-\frac{1}{12}\right) = -2$ . By hand: $f' = -\tan x$ , $f'' = -\sec^2 x$ , $f''' = -2\sec^2 x\tan x$ , $f^{(4)} = -4\sec^2 x\tan^2 x - 2\sec^4 x$ , which is $-2$ at $0$ . Four differentiations against one line of series.

Exercise 5: Reading the series backwards: the n-th derivative is n factorial times the coefficient

The coefficient formula works in both directions. As soon as a series of $f$ centred at $a$ is known, by the table or by the constructions of Exercise 4, every derivative at the centre is read from it: $f^{(n)}(a) = n!\,c_n$ . This is how an exam asks for $f^{(9)}(0)$ without anyone differentiating nine times.

The uniqueness theorem is what makes it legitimate: whatever route produced the series, if it equals $f$ near $a$ , its coefficients ARE $\frac{f^{(n)}(a)}{n!}$ .

a) For $f(x) = e^{x^3}$ , find $f^{(9)}(0)$ and $f^{(10)}(0)$ .
b) For $f(x) = x^2e^{-x}$ , find $f^{(21)}(0)$ .
c) Let $g(x) = \frac{\sin x}{x}$ for $x \ne 0$ and $g(0) = 1$ . Find $g'(0)$ , $g''(0)$ and $g^{(4)}(0)$ .
d) A function $h$ has the Taylor series $\sum_{n=0}^{\infty}\frac{n+1}{3^n}(x-2)^n$ at $a = 2$ . Find $h(2)$ , $h'(2)$ and $h'''(2)$ .
e) A function $k$ has the Taylor series $4 - 3(x-1)^2 + 5(x-1)^3 + \dots$ at $a = 1$ . What does it say about $k$ at $x = 1$ ?

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Answers

a) $f^{(9)}(0) = \frac{9!}{3!} = 60480$ ; $f^{(10)}(0) = 0$
b) $f^{(21)}(0) = -\frac{21!}{19!} = -420$
c) $g'(0) = 0$ , $g''(0) = -\frac{1}{3}$ , $g^{(4)}(0) = \frac{1}{5}$
d) $h(2) = 1$ , $h'(2) = \frac{2}{3}$ , $h'''(2) = \frac{8}{9}$
e) $k(1) = 4$ , $k'(1) = 0$ , $k''(1) = -6 < 0$ : a local maximum of value $4$

a) $e^u = \sum\frac{u^n}{n!}$ with $u = x^3$ : $e^{x^3} = \sum_{n\ge0}\frac{x^{3n}}{n!}$ . The power $x^9$ comes from $3n = 9$ , $n = 3$ : $c_9 = \frac{1}{3!} = \frac{1}{6}$ . So $f^{(9)}(0) = 9! \cdot \frac{1}{6} = \frac{362880}{6} = 60480$ . The power $x^{10}$ would need $3n = 10$ , which has no whole solution: $c_{10} = 0$ and $f^{(10)}(0) = 0$ . Two traps, side by side: answering $\frac{1}{6}$ , the coefficient, instead of the derivative, and dividing by $9!$ because the power is $9$ , when the factorial of the table belongs to $n = 3$ . Differentiating instead: $f' = 3x^2e^{x^3}$ , and the ninth derivative has dozens of terms.

b) $e^{-x} = \sum_{m\ge0}\frac{(-1)^m x^m}{m!}$ , so $x^2e^{-x} = \sum_{m\ge0}\frac{(-1)^m x^{m+2}}{m!}$ . The power $x^{21}$ comes from $m = 19$ : $c_{21} = \frac{(-1)^{19}}{19!} = -\frac{1}{19!}$ . So $f^{(21)}(0) = 21! \cdot \left(-\frac{1}{19!}\right) = -21 \cdot 20 = -420$ . The sign comes from $(-1)^{19}$ , odd. The factorials cancel down to a small integer, which is why exams choose such pairs: an answer still containing $21!$ signals a slip.

c) For $x \ne 0$ : $\frac{\sin x}{x} = \frac{1}{x}\left(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots\right) = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \dots$ , and at $x = 0$ the series gives $1 = g(0)$ . So the series equals $g$ everywhere and, by uniqueness, it is the Maclaurin series of $g$ . Then $g'(0) = 1! \cdot 0 = 0$ (no term in $x$ ; $g$ is even), $g''(0) = 2! \cdot \left(-\frac{1}{6}\right) = -\frac{1}{3}$ and $g^{(4)}(0) = 4! \cdot \frac{1}{120} = \frac{1}{5}$ . By the quotient rule this would be hopeless: the formula $\frac{\sin x}{x}$ is not defined at $0$ , so every derivative at $0$ would need a limit of a difference quotient. The series settles in one line what the quotient rule cannot even start.

d) Here $c_n = \frac{n+1}{3^n}$ . At $x = 2$ only the term $n = 0$ survives: $h(2) = c_0 = 1$ . Then $h'(2) = 1!\,c_1 = \frac{2}{3}$ and $h'''(2) = 3!\,c_3 = 6 \cdot \frac{4}{27} = \frac{8}{9}$ . The traps: forgetting the $3!$ and answering $\frac{4}{27}$ , or substituting $x = 2$ into the coefficients, which are numbers: the variable lives only in $(x - 2)^n$ . Independent check: with $u = \frac{x-2}{3}$ the series is $\sum(n+1)u^n = \frac{1}{(1-u)^2}$ (part a) of Exercise 3 with $-u$ ), so $h(x) = \frac{9}{(5-x)^2}$ . Then $h(2) = 1$ , $h'(x) = \frac{18}{(5-x)^3}$ gives $\frac{2}{3}$ , and $h'''(x) = \frac{216}{(5-x)^5}$ gives $\frac{216}{243} = \frac{8}{9}$ .

e) $c_0 = 4$ , $c_1 = 0$ , $c_2 = -3$ , $c_3 = 5$ . So $k(1) = 4$ , $k'(1) = 1!\,c_1 = 0$ : $x = 1$ is a critical point. $k''(1) = 2!\,c_2 = -6 < 0$ : by the second derivative test, $k$ has a local MAXIMUM at $1$ , of value $4$ . The cubic term does not change this: near $x = 1$ , $-3(x-1)^2$ dominates $5(x-1)^3$ . The usual slip is to read $k''(1) = -3$ off the coefficient; the sign, and so the conclusion, survive the slip here, but a question asking for $k''(1)$ does not. For the record, $k'''(1) = 3! \cdot 5 = 30$ .

Exercise 6: Limits by series: expand until something survives

At a $\frac{0}{0}$ form, series do in one line what L'Hôpital's rule does in three rounds: replace each function by its series, simplify, and read the limit on the lowest power that SURVIVES. The only decision is how far to expand: far enough for the numerator and the denominator to each have a nonzero term, and every piece to the same degree.

The figure shows $y = \frac{\sin x - x}{x^3}$ on $[-8, 8]$ . The formula is undefined at $0$ , yet the curve plainly heads for a value somewhere near $-0.17$ .

a) $\lim_{x\to0}\frac{\sin x - x}{x^3}$
b) $\lim_{x\to0}\frac{\tan x - \sin x}{x^3}$
c) $\lim_{x\to0}\frac{\ln(1+x) - x + \frac{x^2}{2}}{x - \sin x}$
d) Find the constants $a$ and $b$ for which $\lim_{x\to0}\frac{\sin 2x + ax + bx^3}{x^5}$ is finite, and give the limit.
e) $\lim_{x\to0}\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right)$

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Answers

a) $-\frac{1}{6}$
b) $\frac{1}{2}$
c) $2$
d) $a = -2$ , $b = \frac{4}{3}$ , limit $\frac{4}{15}$
e) $\frac{1}{3}$

a) $\sin x - x = \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots\right) - x = -\frac{x^3}{6} + \frac{x^5}{120} - \dots$ , so $\frac{\sin x - x}{x^3} = -\frac{1}{6} + \frac{x^2}{120} - \dots \to -\frac{1}{6}$ , the value the figure suggests. The trap is to stop too early: with $\sin x \approx x$ the numerator becomes $0$ and the copy concludes that the limit is $0$ . The approximation killed exactly the term that decides. Rule: keep expanding until something survives the cancellation. By L'Hôpital's rule the same limit takes three rounds, each one a new $\frac{0}{0}$ .

b) From Exercise 4, $\tan x = x + \frac{x^3}{3} + \dots$ , and $\sin x = x - \frac{x^3}{6} + \dots$ The difference is $\left(\frac{1}{3} + \frac{1}{6}\right)x^3 + \dots = \frac{x^3}{2} + \dots$ (the next term is of degree $5$ ), so the limit is $\frac{1}{2}$ . Both functions start with $x$ , and a one-term expansion of each gives $\frac{0}{x^3}$ again: here too the $x^3$ terms are the ones that decide, and each must be exact.

c) Numerator: $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ , so $\ln(1+x) - x + \frac{x^2}{2} = \frac{x^3}{3} - \frac{x^4}{4} + \dots$ Denominator: $x - \sin x = \frac{x^3}{6} - \frac{x^5}{120} + \dots$ Both start at $x^3$ : $\frac{\frac{x^3}{3} + \dots}{\frac{x^3}{6} + \dots} = \frac{\frac{1}{3} - \frac{x}{4} + \dots}{\frac{1}{6} + \dots} \to \frac{1/3}{1/6} = 2$ . When the lowest surviving powers on top and bottom are the same, the limit is the ratio of their coefficients; when the top starts higher, the limit is $0$ ; when the bottom starts higher, the quotient is unbounded.

d) $\sin 2x = 2x - \frac{(2x)^3}{6} + \frac{(2x)^5}{120} - \dots = 2x - \frac{4x^3}{3} + \frac{4x^5}{15} - \dots$ So the numerator is $(2 + a)x + \left(b - \frac{4}{3}\right)x^3 + \frac{4}{15}x^5 + \dots$ and the quotient is $\frac{2 + a}{x^4} + \frac{b - 4/3}{x^2} + \frac{4}{15} + \dots$ It has a finite limit only if the first two coefficients vanish: $a = -2$ and $b = \frac{4}{3}$ , and then the limit is $\frac{4}{15}$ . The traps are in the powers of $2$ : $(2x)^3 = 8x^3$ , not $2x^3$ , and $\frac{32}{120} = \frac{4}{15}$ . By L'Hôpital's rule this question would take five rounds, with $a$ and $b$ still unknown.

e) This is $\infty - \infty$ , and series cannot be used on it as it stands: first make ONE fraction. $\frac{1}{\sin^2 x} - \frac{1}{x^2} = \frac{x^2 - \sin^2 x}{x^2\sin^2 x}$ . Then $\sin^2 x = \left(x - \frac{x^3}{6} + \dots\right)^2 = x^2 - 2 \cdot x \cdot \frac{x^3}{6} + \dots = x^2 - \frac{x^4}{3} + \dots$ , so the numerator is $\frac{x^4}{3} + \dots$ and the denominator $x^2\left(x^2 - \dots\right) = x^4 + \dots$ The limit is $\frac{1}{3}$ . The trap is the cross term: squaring $x - \frac{x^3}{6}$ gives $-\frac{x^4}{3}$ , twice the product, and the copy that writes $x^2 - \frac{x^4}{6}$ finds $\frac{1}{6}$ .

Exercise 7: Integrals with no elementary antiderivative: the series is the answer

Neither $e^{-x^2}$ nor $\frac{\sin x}{x}$ has an antiderivative that can be written with the usual functions. Their integrals exist all the same, and a power series gives them EXACTLY: expand the integrand, integrate term by term on an interval inside the interval of convergence, and the result is a series valid on the same interval.

The figure shows $y = e^{-x^2}$ , with the area from $0$ to $1$ shaded. Nothing in this exercise asks for a decimal: how many terms give a required accuracy is the question of the next chapter.

a) Find the Maclaurin series of $F(x) = \int_0^x e^{-t^2}\,dt$ , with its general term and radius of convergence.
b) Deduce $\int_0^1 e^{-x^2}\,dx$ as the sum of a numerical series and write its first four terms as fractions. Show by an elementary inequality that the value lies between $\frac{2}{3}$ and $1$ .
c) Let $\text{Si}(x) = \int_0^x\frac{\sin t}{t}\,dt$ . Explain why the integrand causes no trouble at $t = 0$ , then give the Maclaurin series of Si.
d) Find the series of $\int_0^x\frac{\arctan t}{t}\,dt$ with its interval of validity, then express $\int_0^{1/2}\frac{\arctan t}{t}\,dt$ as a numerical series.
e) A student writes $\int_0^\infty e^{-t^2}\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_0^\infty t^{2n}\,dt$ . Explain why this is meaningless, although the series of $e^{-t^2}$ converges for every $t$ .

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Answers

a) $F(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{n!\,(2n+1)}$ , $R = \infty$
b) $\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\,(2n+1)} = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \dots$ , between $\frac{2}{3}$ and $1$
c) $\frac{\sin t}{t} \to 1$ , a removable point; $\text{Si}(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)\,(2n+1)!}$
d) $\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)^2}$ for $|x| \le 1$ ; $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2\,2^{2n+1}}$
e) Each $\int_0^\infty t^{2n}\,dt$ diverges: term by term integration holds on bounded intervals only.

a) $e^{-t^2} = \sum_{n\ge0}\frac{(-1)^n t^{2n}}{n!}$ for every $t$ . Integrating term by term on $[0, x]$ , legitimate on any bounded interval since the radius is infinite: $F(x) = \sum_{n\ge0}\frac{(-1)^n}{n!}\int_0^x t^{2n}\,dt = \sum_{n\ge0}\frac{(-1)^n x^{2n+1}}{n!\,(2n+1)} = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \dots$ Integration does not change the radius: $R = \infty$ . No constant appears because $F(0) = 0$ . Each denominator has TWO factors, $n!$ from the exponential and $2n + 1$ from integrating $t^{2n}$ ; dropping either one is the usual slip. This series is the function: up to the factor $\frac{2}{\sqrt{\pi}}$ it is the error function of statistics and heat diffusion, and there is no other formula to find.

b) $\int_0^1 e^{-x^2}\,dx = F(1) = \sum_{n\ge0}\frac{(-1)^n}{n!\,(2n+1)} = 1 - \frac{1}{3} + \frac{1}{2 \cdot 5} - \frac{1}{6 \cdot 7} + \dots = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \dots$ This IS the exact value, and an exam that asks for the integral as a series expects exactly this answer. Plausibility, without any error bound: $e^{-u} \ge 1 - u$ for every $u$ (the tangent line at $0$ lies under the convex exponential) and $e^{-x^2} \le 1$ , so $1 - x^2 \le e^{-x^2} \le 1$ on $[0, 1]$ and $\frac{2}{3} \le \int_0^1 e^{-x^2}\,dx \le 1$ . The first two terms of the series are exactly the lower bound, and the shaded region of the figure indeed fills most of the unit square.

c) As $t \to 0$ , $\frac{\sin t}{t} \to 1$ : the integrand has a removable point at $0$ , it extends continuously by the value $1$ , and $\text{Si}(x)$ is an ordinary integral, not an improper one. From Exercise 5 c), $\frac{\sin t}{t} = \sum_{n\ge0}\frac{(-1)^n t^{2n}}{(2n+1)!}$ for every $t$ . Integrating: $\text{Si}(x) = \sum_{n\ge0}\frac{(-1)^n x^{2n+1}}{(2n+1)\,(2n+1)!} = x - \frac{x^3}{18} + \frac{x^5}{600} - \dots$ , since $3 \cdot 3! = 18$ and $5 \cdot 5! = 600$ , with $R = \infty$ . Si is odd, as the integral of an even function from $0$ . The solution figure shows $\text{Si}(\pi)$ as the area of the first arch of $\frac{\sin t}{t}$ .

d) $\arctan t = \sum_{n\ge0}\frac{(-1)^n t^{2n+1}}{2n+1}$ for $|t| \le 1$ , so $\frac{\arctan t}{t} = \sum_{n\ge0}\frac{(-1)^n t^{2n}}{2n+1}$ for $0 < |t| \le 1$ , with value $1$ at $t = 0$ as before. Integrating: $\int_0^x\frac{\arctan t}{t}\,dt = \sum_{n\ge0}\frac{(-1)^n x^{2n+1}}{(2n+1)^2}$ , radius $1$ inherited from the arctangent, and the series also converges at $x = \pm1$ , absolutely, by comparison with $\sum\frac{1}{n^2}$ : interval $[-1, 1]$ . At $x = \frac{1}{2}$ , inside the radius: $\int_0^{1/2}\frac{\arctan t}{t}\,dt = \sum_{n\ge0}\frac{(-1)^n}{(2n+1)^2\,2^{2n+1}} = \frac{1}{2} - \frac{1}{72} + \frac{1}{800} - \dots$ The square in $(2n+1)^2$ collects one $2n + 1$ from the arctangent and one from the integration.

e) Term-by-term integration is justified on a BOUNDED interval inside the interval of convergence. On $[0, \infty)$ every term is an improper integral that diverges, $\int_0^\infty t^{2n}\,dt = \infty$ , so the right-hand side is a sum of infinities with alternating signs, not a number. The integral itself converges (by comparison with $e^{-t}$ for $t \ge 1$ ) and equals $\frac{\sqrt{\pi}}{2}$ , a value that comes from another method and is usually admitted in MATH 141. The underlying reason: every partial sum of the series is a polynomial, and a nonzero polynomial never has a finite integral on $[0, \infty)$ . The series describes $e^{-t^2}$ perfectly on each bounded interval and says nothing about the area of an infinite tail.

Exercise 8: Five statements to correct

Each statement below was written on a MATH 141 practice paper, and each one is false. Say what is wrong, give the correct statement, and settle it with a short computation or a counterexample.

For a), the figure shows the function $f(x) = e^{-1/x^2}$ for $x \ne 0$ , $f(0) = 0$ .

a) Every derivative of $f$ at $0$ is $0$ (admitted). So the Maclaurin series of $f$ is $0$ , it converges for every $x$ , and therefore $f(x) = 0$ for every $x$ .
b) Since $\sin(x^2) = x^2 - \frac{x^6}{6} + \dots$ , the sixth derivative of $\sin(x^2)$ at $0$ is $-\frac{1}{6}$ .
c) Putting $x = 2$ in $\sqrt{1+x} = \sum_{n=0}^{\infty}\binom{1/2}{n}x^n$ gives $\sqrt{3} = \sum_{n=0}^{\infty}\binom{1/2}{n}2^n = 1 + 1 - \frac{1}{2} + \frac{1}{2} - \dots$
d) The Taylor series of $\frac{1}{x}$ centred at $a = 1$ is $\sum_{n=0}^{\infty}(-1)^n x^n$ .
e) Two series are multiplied by multiplying the coefficients of equal powers, so $e^x\cos x = 1 + 0 \cdot x - \frac{1}{4}x^2 + \dots$

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Answers

a) False: $f(1) = \frac{1}{e} \ne 0$ . The series converges everywhere but equals $f$ only at $0$ .
b) False: $f^{(6)}(0) = 6! \cdot \left(-\frac{1}{6}\right) = -120$ .
c) False: $x = 2$ is outside $|x| < 1$ and the series diverges; $\sqrt{3} = 2\sum_{n=0}^{\infty}\binom{1/2}{n}\left(-\frac{1}{4}\right)^n$ .
d) False: powers of $x - 1$ , $\frac{1}{x} = \sum_{n=0}^{\infty}(-1)^n(x-1)^n$ on $(0, 2)$ .
e) False: collect every pair of degrees; $e^x\cos x = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + \dots$

a) FALSE. The Maclaurin series is indeed $\sum 0 \cdot x^n = 0$ , and it does converge for every $x$ , but to $0$ , which is not $f(x)$ : $f(1) = e^{-1} = \frac{1}{e} \ne 0$ , and $f(x) \to 1$ as $|x| \to \infty$ . The figure shows how this is possible: $f$ is so flat at $0$ that $f(0.3) = e^{-1/0.09} = e^{-11.1}$ is below $0.00002$ , and every derivative vanishes there, yet the function is not zero. Correct statement: a Taylor series can converge everywhere and still not converge to its function. The equalities of the table are THEOREMS, proved by showing that the remainder tends to $0$ , which is the subject of the next chapter; for a new function, either prove it the same way or build the series from the table.

b) FALSE: the coefficient is not the derivative. The coefficient of $x^6$ is $c_6 = -\frac{1}{6}$ , so $f^{(6)}(0) = 6!\,c_6 = 720 \cdot \left(-\frac{1}{6}\right) = -120$ . Correct statement: the sixth derivative of $\sin(x^2)$ at $0$ is $-120$ . By the same reading, every odd derivative at $0$ is $0$ , since $\sin(x^2)$ has only even powers.

c) FALSE: the binomial series equals $(1 + x)^k$ only for $|x| < 1$ . At $x = 2$ the ratio of consecutive terms is $\left|\frac{1/2 - n}{n + 1}\right| \cdot 2 \to 2 > 1$ , so the terms grow ( $1, 1, -\frac{1}{2}, \frac{1}{2}, -\frac{5}{8}, \dots$ already refuse to shrink) and the series diverges: it equals nothing. Correct route: rewrite the number so that the variable falls inside the interval. $\sqrt{3} = \sqrt{4 - 1} = 2\sqrt{1 - \frac{1}{4}}$ , and $x = -\frac{1}{4}$ is inside: $\sqrt{3} = 2\sum_{n\ge0}\binom{1/2}{n}\left(-\frac{1}{4}\right)^n = 2\left(1 - \frac{1}{8} - \frac{1}{128} - \dots\right)$ , whose first three terms already give $\frac{111}{64} \approx 1.73$ . The same gesture, factoring out the nearest perfect square, is what Exercise 3 c) did with $\sqrt[3]{8 + x}$ .

d) FALSE: a series centred at $1$ is written in powers of $(x - 1)$ . From the definition, $f^{(n)}(x) = (-1)^n n!\,x^{-n-1}$ , so $f^{(n)}(1) = (-1)^n n!$ and $c_n = (-1)^n$ . From the table, $\frac{1}{x} = \frac{1}{1 + (x-1)} = \sum_{n\ge0}(-1)^n(x-1)^n$ for $|x - 1| < 1$ , that is $0 < x < 2$ . The series of the statement is that of $\frac{1}{1 + x}$ centred at $0$ , a different function; $\frac{1}{x}$ itself has no Maclaurin series at all, being undefined at $0$ .

e) FALSE: the coefficient of $x^n$ in a product collects EVERY pair of degrees adding to $n$ . With $e^x$ : $1, 1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}$ and $\cos x$ : $1, 0, -\frac{1}{2}, 0, \frac{1}{24}$ : $x^1$ : $1 \cdot 0 + 1 \cdot 1 = 1$ ; $x^2$ : $1 \cdot \left(-\frac{1}{2}\right) + 1 \cdot 0 + \frac{1}{2} \cdot 1 = 0$ ; $x^3$ : $0 - \frac{1}{2} + 0 + \frac{1}{6} = -\frac{1}{3}$ ; $x^4$ : $\frac{1}{24} + 0 - \frac{1}{4} + 0 + \frac{1}{24} = -\frac{1}{6}$ . So $e^x\cos x = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + \dots$ The claimed $0 \cdot x$ fails the simplest test: the derivative of $e^x\cos x$ at $0$ is $e^0(\cos 0 - \sin 0) = 1$ , so $c_1 = 1$ .

Exercise 9: Goals in a hockey game: sums and products of exponential series

A model used for rare, independent events (goals in a game, typos on a page, decays in a radioactive sample) gives the probability that exactly $k$ events occur as $\frac{e^{-\lambda}\lambda^k}{k!}$ , for $k = 0, 1, 2, \dots$ , where $\lambda > 0$ is a fixed number. A hockey analyst uses it for the number $K$ of goals the home team scores in one game, with $\lambda = 3$ , so $p_k = \frac{e^{-3}3^k}{k!}$ ; the figure shows $p_0$ to $p_8$ in percent. The visiting team scores $L$ goals with $\lambda = 2$ , so $q_j = \frac{e^{-2}2^j}{j!}$ , and the two are independent: the probability that $K = i$ and $L = j$ is $p_iq_j$ (admitted).

Every question below is a series to recognize or to multiply, the gestures of this chapter. No calculator: exact answers, then an order of magnitude from $e \approx 2.72$ .

a) Show that $\sum_{k=0}^{\infty}p_k = 1$ .
b) Find the exact probability that the home team scores an even number of goals ( $0$ counts as even). Is it more or less than $\frac{1}{2}$ ?
c) The probability that the game has $n$ goals in total is $\sum_{i=0}^{n}p_iq_{n-i}$ . Recognize it as the coefficient of $x^n$ in the product of two exponential series, and show that it equals $\frac{e^{-5}5^n}{n!}$ .
d) Write the probability of a tie, $K = L$ , as a numerical series, give its first four terms, and estimate it from $e^5 \approx 148$ .
e) Compute $\sum_{k=0}^{\infty}\frac{p_k}{k+1}$ exactly.

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Answers

a) $e^{-3}\sum_{k\ge0}\frac{3^k}{k!} = e^{-3}e^{3} = 1$
b) $\frac{1 + e^{-6}}{2}$ , slightly more than $\frac{1}{2}$ (about $0.501$ )
c) coefficient of $x^n$ in $e^{3x}e^{2x} = e^{5x}$ : $\frac{e^{-5}5^n}{n!}$
d) $e^{-5}\sum_{k\ge0}\frac{6^k}{(k!)^2} = e^{-5}\left(1 + 6 + 9 + 6 + \dots\right)$ , about $\frac{1}{6}$
e) $\frac{1 - e^{-3}}{3}$ , about $0.32$

a) $e^{-3}$ does not depend on $k$ and comes out of the sum: $\sum_{k\ge0}p_k = e^{-3}\sum_{k\ge0}\frac{3^k}{k!}$ . The remaining series is the table series of $e^u$ at $u = 3$ , valid for every real $u$ : it equals $e^3$ . So $\sum p_k = e^{-3}e^3 = 1$ : the probabilities of the model add up to $1$ , whatever $\lambda$ . The chapter in one line: a series with $k!$ in the denominator and no alternating sign is an exponential.

b) The probability is $e^{-3}\sum_{k\text{ even}}\frac{3^k}{k!}$ . The even terms are isolated by adding the series of $e^3$ and $e^{-3}$ : $e^3 + e^{-3} = \sum_{k\ge0}\frac{\left(1 + (-1)^k\right)3^k}{k!}$ , where odd terms cancel and even terms double. So $\sum_{k\text{ even}}\frac{3^k}{k!} = \frac{e^3 + e^{-3}}{2}$ , and the probability is $e^{-3} \cdot \frac{e^3 + e^{-3}}{2} = \frac{1 + e^{-6}}{2}$ . It is slightly MORE than $\frac{1}{2}$ : $e^2 \approx 7.4$ , so $e^6 \approx 7.4^3 \approx 400$ and the probability is about $\frac{1}{2} + \frac{1}{800} \approx 0.501$ . The trap: even powers over even factorials look like a cosine, but $\cos 3$ has ALTERNATING signs; here every term is positive.

c) $\sum_{i=0}^{n}p_iq_{n-i} = e^{-5}\sum_{i=0}^{n}\frac{3^i}{i!} \cdot \frac{2^{n-i}}{(n-i)!}$ . Now $e^{3x} = \sum_i\frac{3^i}{i!}x^i$ and $e^{2x} = \sum_j\frac{2^j}{j!}x^j$ , and the coefficient of $x^n$ in their product collects every pair $i + j = n$ : it is exactly $\sum_{i=0}^{n}\frac{3^i}{i!} \cdot \frac{2^{n-i}}{(n-i)!}$ . But $e^{3x}e^{2x} = e^{5x} = \sum_n\frac{5^n}{n!}x^n$ , and by uniqueness of the coefficients the sum equals $\frac{5^n}{n!}$ . So the total number of goals follows the same model with $\lambda = 5$ : $\frac{e^{-5}5^n}{n!}$ . Check at $n = 1$ : $p_0q_1 + p_1q_0 = e^{-5}(2 + 3) = 5e^{-5}$ . The product rule of this chapter did the combinatorics: expanding $(3 + 2)^n$ by the binomial theorem gives the same identity.

d) A tie means $K = L = k$ for some $k$ , so the probability is $\sum_{k\ge0}p_kq_k = e^{-5}\sum_{k\ge0}\frac{6^k}{(k!)^2} = e^{-5}\left(1 + 6 + \frac{36}{4} + \frac{216}{36} + \dots\right) = e^{-5}\left(1 + 6 + 9 + 6 + \dots\right)$ . The series converges (ratio $\frac{6}{(k+1)^2} \to 0$ ), but it is NOT a table series: the square of the factorial has no counterpart in the table, and no elementary closed form exists. The series is the exact answer, the same situation as the integrals of Exercise 7. Order of magnitude: the next terms are $\frac{9}{4}$ and about $0.5$ , so the sum is roughly $25$ , and $\frac{25}{148} \approx \frac{1}{6}$ : about one game in six is tied at the end of regulation time in this model.

e) $\frac{p_k}{k+1} = \frac{e^{-3}3^k}{(k+1)\,k!} = \frac{e^{-3}3^k}{(k+1)!}$ . Multiply and divide by $3$ to complete the power: $\sum_{k\ge0}\frac{3^k}{(k+1)!} = \frac{1}{3}\sum_{k\ge0}\frac{3^{k+1}}{(k+1)!} = \frac{1}{3}\sum_{j\ge1}\frac{3^j}{j!} = \frac{e^3 - 1}{3}$ , the term $j = 0$ , equal to $1$ , being MISSING from the exponential series. So $\sum\frac{p_k}{k+1} = e^{-3}\frac{e^3 - 1}{3} = \frac{1 - e^{-3}}{3}$ . With $e^3 \approx 20$ , it is about $\frac{1 - 0.05}{3} \approx 0.32$ . Forgetting the missing term gives exactly $\frac{1}{3}$ : close, and wrong. The shift of index is only legitimate once the new starting value, here $j = 1$ , is written.

Exercise 10: A final exam problem: one function, every use of its series

Let $f(x) = \frac{e^x - 1}{x}$ for $x \ne 0$ and $f(0) = 1$ . The figure shows its graph: the formula has no value at $0$ , and the definition fills the hole. This problem chains the gestures of the chapter on a single function, as MATH 141 finals often do.

a) Find the Maclaurin series of $f$ and its radius of convergence. Why does it equal $f$ at $x = 0$ as well?
b) Find $f^{(n)}(0)$ for every $n$ . In particular, what is $f^{(10)}(0)$ ?
c) Differentiate the series term by term and evaluate at $x = 2$ to find the exact sum of $\sum_{n=1}^{\infty}\frac{n\,2^{n-1}}{(n+1)!}$ .
d) Express $\int_0^1 f(x)\,dx$ as a numerical series.
e) Compute $\lim_{x\to0}\frac{f(x) - 1}{x}$ and $\lim_{x\to0}\frac{f(x) - 1 - \frac{x}{2}}{x^2}$ . What does the first limit say about $f'(0)$ ?

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Answers

a) $f(x) = \sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}$ , $R = \infty$ ; at $0$ the series gives $1 = f(0)$
b) $f^{(n)}(0) = \frac{1}{n+1}$ ; $f^{(10)}(0) = \frac{1}{11}$
c) $f'(2) = \frac{e^2 + 1}{4}$
d) $\sum_{n=1}^{\infty}\frac{1}{n \cdot n!}$
e) $\frac{1}{2}$ , which is $f'(0)$ ; then $\frac{1}{6}$

a) $e^x - 1 = \sum_{n\ge1}\frac{x^n}{n!}$ : the term $n = 0$ , equal to $1$ , cancels. For $x \ne 0$ , dividing by $x$ : $f(x) = \sum_{n\ge1}\frac{x^{n-1}}{n!} = \sum_{m\ge0}\frac{x^m}{(m+1)!} = 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \dots$ At $x = 0$ the series gives $1$ , which is $f(0)$ by definition. So the series equals $f$ at EVERY $x$ , and by uniqueness it is the Maclaurin series of $f$ . Ratio test: $\frac{|x|^{m+1}/(m+2)!}{|x|^m/(m+1)!} = \frac{|x|}{m+2} \to 0$ , so $R = \infty$ . The value $f(0) = 1$ was not chosen at random: any other value would make $f$ discontinuous at $0$ , and no power series could equal it.

b) $c_n = \frac{1}{(n+1)!}$ , so $f^{(n)}(0) = n!\,c_n = \frac{n!}{(n+1)!} = \frac{1}{n+1}$ . In particular $f^{(10)}(0) = \frac{1}{11}$ . By the quotient rule this would mean ten differentiations of $\frac{e^x - 1}{x}$ , and then a limit at $0$ where the formula is undefined. Small cases as a check: $f(0) = 1 = \frac{1}{1}$ , and $f'(0) = \frac{1}{2}$ , which part e) finds again by a limit.

c) Term-by-term differentiation, valid inside the radius, here everywhere: $f'(x) = \sum_{n\ge1}\frac{n\,x^{n-1}}{(n+1)!}$ . Directly, for $x \ne 0$ : $f'(x) = \frac{x e^x - (e^x - 1)}{x^2} = \frac{xe^x - e^x + 1}{x^2}$ . At $x = 2$ the two expressions agree: $\sum_{n\ge1}\frac{n\,2^{n-1}}{(n+1)!} = f'(2) = \frac{2e^2 - e^2 + 1}{4} = \frac{e^2 + 1}{4}$ , about $\frac{7.4 + 1}{4} \approx 2.1$ . Plausibility: the first four terms, $\frac{1}{2} + \frac{4}{6} + \frac{12}{24} + \frac{32}{120}$ , already add up to about $1.93$ , and all terms are positive. The gesture: a series whose general term has an extra factor $n$ in front of a known coefficient is the DERIVATIVE of a known series, evaluated at a point.

d) $f$ is continuous on $[0, 1]$ thanks to $f(0) = 1$ , so the integral is an ordinary one; but $\frac{e^x - 1}{x}$ has no elementary antiderivative. Integrating the series term by term on $[0, 1]$ : $\int_0^1 f(x)\,dx = \sum_{m\ge0}\frac{1}{(m+1)!}\int_0^1 x^m\,dx = \sum_{m\ge0}\frac{1}{(m+1)!\,(m+1)} = \sum_{n\ge1}\frac{1}{n \cdot n!} = 1 + \frac{1}{4} + \frac{1}{18} + \frac{1}{96} + \dots$ That series is the exact answer. Plausibility: $f$ increases from $f(0) = 1$ to $f(1) = e - 1 \approx 1.72$ , so the integral lies between $1$ and $1.72$ , and the first terms, about $1.32$ , sit in that range.

e) $\frac{f(x) - 1}{x} = \frac{\frac{x}{2} + \frac{x^2}{6} + \dots}{x} = \frac{1}{2} + \frac{x}{6} + \dots \to \frac{1}{2}$ . Since $f(0) = 1$ , this quotient is the difference quotient $\frac{f(x) - f(0)}{x - 0}$ : its limit IS $f'(0)$ , so $f'(0) = \frac{1}{2}$ , as b) said with $n = 1$ . Next, $\frac{f(x) - 1 - \frac{x}{2}}{x^2} = \frac{1}{6} + \frac{x}{24} + \dots \to \frac{1}{6}$ , which is $c_2 = \frac{f''(0)}{2}$ with $f''(0) = \frac{1}{3}$ from b). By L'Hôpital's rule the second limit would require the second derivative of a quotient, twice, before the $\frac{0}{0}$ goes away.

Corrected exercises: Taylor and Maclaurin series (MATH 141)

Course recap

Part A: the basics (/50)

Exercise 1: Coefficients from the definition: the n-th derivative at a, over n factorial

Exercise 2: The table, and substitution into it: the interval moves with the variable

Exercise 3: The binomial series: 1 + x to a power that is not a whole number

Exercise 4: Products, quotients and compositions: build the series, never differentiate five times

Exercise 5: Reading the series backwards: the n-th derivative is n factorial times the coefficient

Part B: problems and reasoning (/50)

Exercise 6: Limits by series: expand until something survives

Exercise 7: Integrals with no elementary antiderivative: the series is the answer

Exercise 8: Five statements to correct

Exercise 9: Goals in a hockey game: sums and products of exponential series

Exercise 10: A final exam problem: one function, every use of its series

See also

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