Corrected exercises: power series and Taylor series, 201-SN3-RE (201-NYB)

Ratio Test: $\left|\frac{x^{n+1}}{n+1}\cdot\frac{n}{x^{n}}\right|=|x|\frac{n}{n+1}\to|x|$: convergence for $|x|<1$, so $R=1$. Endpoints: at $x=1$, the harmonic series $\sum\frac{1}{n}$ DIVERGES; at $x=-1$, the alternating harmonic series CONVERGES (Alternating Series Test). Interval: $[-1, 1)$: closed on one side, open on the other, and it took both tests to know it.

$\sum n!x^{n}$: ratio $(n+1)|x|\to\infty$ for every $x\ne 0$: $R=0$, the series lives only at the point $x=0$: a stillborn infinite polynomial. $\sum\frac{x^{n}}{n!}$: ratio $\frac{|x|}{n+1}\to 0<1$ for EVERY $x$: $R=\infty$, convergence everywhere: it is the series of the exponential. The two extremes bracket every possible case: a radius can be zero, finite or infinite, depending on whether the coefficients blow up or collapse.

Ratio: $\frac{|x-2|}{3}\cdot\frac{n}{n+1}\to\frac{|x-2|}{3}<1$, that is $|x-2|<3$: $R=3$, centred at 2, so $(-1, 5)$ for now. Endpoints: at $x=5$: $\sum\frac{3^{n}}{n3^{n}}=\sum\frac{1}{n}$ diverges; at $x=-1$: $\sum\frac{(-3)^{n}}{n3^{n}}=\sum\frac{(-1)^{n}}{n}$ converges. Interval: $[-1, 5)$. It is the series of a), shifted and stretched: same behaviour at the endpoints, recentred.

Inside, $|x-a|<R$ means that the limiting ratio is strictly less than 1: the Ratio Test concludes ABSOLUTE convergence, the most robust kind. At the endpoints, the ratio is exactly 1: the Ratio Test falls silent (its blind spot), and each endpoint becomes an ordinary NUMERICAL series that must be judged with the tools of the previous chapter: the p-test, comparison, the Alternating Series Test. The whole interval except two points is settled in one line; the two remaining points hold all the subtlety.

$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$ for $|x|<1$: the geometric series, mother of all power series, the one in the figure. Substitution $x\to-x^{2}$: $\frac{1}{1+x^{2}}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}=1-x^{2}+x^{4}-\cdots$, valid for $|x|<1$: no derivative to compute, the substitution inherits everything.

Term-by-term differentiation (allowed inside the radius): $\frac{1}{(1-x)^{2}}=\sum_{n=1}^{\infty}nx^{n-1}$, so $\frac{x}{(1-x)^{2}}=\sum_{n=1}^{\infty}nx^{n}$. At $x=\frac{1}{2}$: $\sum\frac{n}{2^{n}}=\frac{1/2}{(1/2)^{2}}=2$: the sum that the Ratio Test declared convergent without giving its value falls here in one line. Power series are the machine that COMPUTES the sums that the tests merely announce.

$\ln(1+x)=\int_{0}^{x}\frac{dt}{1+t}$ and $\frac{1}{1+t}=\sum(-1)^{n}t^{n}$: integrating term by term, $\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n}}{n}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots$. At $x=1$ (endpoint included, by the Alternating Series Test): $1-\frac{1}{2}+\frac{1}{3}-\cdots=\ln 2$: the value taken for granted in the chapter on alternating series receives its proof here.

$\arctan x=\int_{0}^{x}\frac{dt}{1+t^{2}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots$. At $x=1$: $\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$: Leibniz's formula, $\pi$ written with the reciprocals of the odd numbers. Maximum beauty, minimum efficiency: the error after $n$ terms is of order $\frac{1}{2n+1}$, and it takes HUNDREDS of terms for three decimals (the first five give 0.835 for a true 0.785). Real computations of $\pi$ use arctangents of small arguments, where the same series races along exponentially fast.

All the derivatives of $e^{x}$ equal $e^{x}$, so $f^{(n)}(0)=1$ for every $n$: $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots$, valid everywhere ($R=\infty$, exercise 1b). Term-by-term differentiation: $\frac{d}{dx}\frac{x^{n}}{n!}=\frac{x^{n-1}}{(n-1)!}$: each term becomes its predecessor, the series slides onto itself: it is its own derivative, the property that DEFINES the exponential. At $x=1$, it gives $e=\sum\frac{1}{n!}$.

The derivatives of $\sin$ go round like a carousel: $\sin,\cos,-\sin,-\cos,\ldots$; at 0: $0, 1,0,-1,\ldots$: only the ODD orders survive, with alternating signs: $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots$. For $\cos$: $1, 0,-1, 0,\ldots$: EVEN orders: $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots$. The structure of the exponents IS the parity of the functions: $\sin(-x)=-\sin x$ requires odd powers, $\cos(-x)=\cos x$ even ones: the series wears the symmetry on its face.

At $a=1$: $f=\ln x$, $f'=x^{-1}$, $f''=-x^{-2}$, $f'''=2x^{-3}$, hence $f^{(n)}(1)=(-1)^{n+1}(n-1)!$ for $n\ge 1$. Coefficients: $\frac{f^{(n)}(1)}{n!}=\frac{(-1)^{n+1}}{n}$: $\ln x=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-1)^{n}}{n}$. It is EXACTLY the series of $\ln(1+u)$ from exercise 2c with $u=x-1$: the two routes (derivatives computed, or geometric series integrated) lead to the same expansion: the Taylor series of a function is unique.

$\frac{d}{dx}\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots\right)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots$: the cosine series, term for term. No accident: the Taylor series of a function is UNIQUE, and term-by-term differentiation is legitimate inside the radius; since $(\sin)'=\cos$, their series must differentiate into each other. This web of consistencies (derivatives, parities, special values) is the best detector of computational errors in the chapter: a wrong Maclaurin series always gives itself away somewhere.

Write $e^{x}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}+\cdots$ and $\sin x=x-\frac{x^{3}}{6}+\cdots$, then multiply, collecting by degree and forgetting no cross term. Degree 1: $1\times x=x$. Degree 2: $x\times x=x^{2}$. Degree 3: $\frac{x^{2}}{2}\times x=\frac{x^{3}}{2}$ and $1\times\left(-\frac{x^{3}}{6}\right)=-\frac{x^{3}}{6}$, total $\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$. Degree 4: $\frac{x^{3}}{6}\times x=\frac{x^{4}}{6}$ and $x\times\left(-\frac{x^{3}}{6}\right)=-\frac{x^{4}}{6}$, total $\frac{1}{6}-\frac{1}{6}=0$. So $e^{x}\sin x=x+x^{2}+\frac{x^{3}}{3}+0\cdot x^{4}+\cdots$. The coefficient of $x^{4}$ is EXACTLY zero, and that is no computational accident: it means that the fourth derivative of $e^{x}\sin x$ vanishes at 0, which can be checked with the Leibniz rule for the derivative of a product. Numerical check at $x=0.2$: the product equals $0.2426553$ and the approximation $0.2426667$, the gap coming from the $x^{5}$ term.

Composition. Set $u=\sin x=x-\frac{x^{3}}{6}+\cdots$ and substitute into $e^{u}=1+u+\frac{u^{2}}{2}+\frac{u^{3}}{6}+\frac{u^{4}}{24}+\cdots$. First you need the powers of $u$ up to degree 4: $u=x-\frac{x^{3}}{6}$; $u^{2}=x^{2}-\frac{x^{4}}{3}$ (the cross term $2\cdot x\cdot\left(-\frac{x^{3}}{6}\right)$ gives the $x^{4}$ term); $u^{3}=x^{3}$ and $u^{4}=x^{4}$, the corrections being of degree higher than 4. Substituting: $e^{\sin x}=1+\left(x-\frac{x^{3}}{6}\right)+\frac{1}{2}\left(x^{2}-\frac{x^{4}}{3}\right)+\frac{x^{3}}{6}+\frac{x^{4}}{24}$. Collect: degree 3, $-\frac{1}{6}+\frac{1}{6}=0$; degree 4, $-\frac{1}{6}+\frac{1}{24}=-\frac{1}{8}$. Hence $e^{\sin x}=1+x+\frac{x^{2}}{2}+0\cdot x^{3}-\frac{x^{4}}{8}+\cdots$. Numerical check at $x=0.2$: the function equals $1.2197786$ and the approximation $1.2198000$. The direct computation would have required FOUR successive derivatives of $e^{\sin x}$, each one producing, through the chain rule and the product rule, an expression longer than the last; the fourth runs over several lines. Composing series does the same job in two lines, and that is the whole point of the method. One precaution is essential: composition is only legitimate if $u\to 0$ as $x\to 0$, which is indeed the case for $\sin x$; composing with a quantity that does not tend to 0 would make no sense.

Write $\tan x=\frac{\sin x}{\cos x}$ and look for $\tan x=a_{1}x+a_{3}x^{3}+a_{5}x^{5}+\cdots$, writing only odd powers since the tangent is an odd function. The division is carried out by writing $\sin x=\tan x\cdot\cos x$, that is $x-\frac{x^{3}}{6}+\frac{x^{5}}{120}=\left(a_{1}x+a_{3}x^{3}+a_{5}x^{5}\right)\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\right)$, then by matching coefficients degree by degree. Degree 1: $a_{1}=1$. Degree 3: $a_{3}-\frac{a_{1}}{2}=-\frac{1}{6}$, so $a_{3}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$. Degree 5: $a_{5}-\frac{a_{3}}{2}+\frac{a_{1}}{24}=\frac{1}{120}$, so $a_{5}=\frac{1}{120}+\frac{1}{6}-\frac{1}{24}=\frac{1+20-5}{120}=\frac{16}{120}=\frac{2}{15}$. Hence $\tan x=x+\frac{x^{3}}{3}+\frac{2x^{5}}{15}+\cdots$. Numerical check at $x=0.1$: the series gives $0.10033467$ and the calculator $0.10033467$, identical to eight decimal places. Notice that the coefficients of the tangent follow no simple formula, unlike those of the sine and cosine: they involve the Bernoulli numbers, which is why the tangent series is never asked from memory but always by division.

With $x=0.1$: $\sqrt{1.1}\approx 1+\frac{0.1}{2}-\frac{0.01}{8}+\frac{0.001}{16}=1+0.05-0.00125+0.0000625=1.0488125$. The series is alternating with decreasing terms, so the error is bounded by the first neglected term, which is $\frac{5x^{4}}{128}=\frac{5\times 10^{-4}}{128}\approx 3.91\times 10^{-6}$. Exact value: $\sqrt{1.1}=1.0488088$, so the actual error is $3.65\times 10^{-6}$, indeed below the bound and very close to it, the signature of a well-behaved alternating series. Four terms are therefore enough for five decimals. This binomial series is worth knowing: it extends Newton's binomial theorem to non-integer exponents, the sum then becoming infinite instead of stopping, and its radius of convergence is 1. It is what justifies the approximation $\sqrt{1+x}\approx 1+\frac{x}{2}$, everywhere in physics, for instance in relativistic expansions at low speed.

Taylor's theorem with the Lagrange remainder: if $f$ is $(n+1)$ times differentiable on an interval containing $a$ and $x$, then $f(x)=T_{n}(x)+R_{n}(x)$, where $T_{n}$ is the Taylor polynomial of degree $n$ at $a$, and there exists a point $c$ STRICTLY BETWEEN $a$ and $x$ such that $R_{n}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$. Since this point $c$ is unknown, we bound: if $\left|f^{(n+1)}(t)\right|\le M$ for every $t$ in the interval $[a,x]$ (or $[x,a]$ if $x<a$), then $\left|R_{n}(x)\right|\le\frac{M\left|x-a\right|^{n+1}}{(n+1)!}$. This is the crucial point of the write-up: $M$ must bound the derivative of order $n+1$ on the WHOLE interval between the centre and the point of evaluation, and not only at the centre. Taking $M=\left|f^{(n+1)}(a)\right|$ is the classic error, and it makes the bound false as soon as the derivative increases. Note finally the structure of the remainder: it is exactly the next term of the expansion, with the derivative evaluated at an intermediate point instead of the centre.

$T_{3}(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$, so $T_{3}(0.5)=1+0.5+0.125+0.0208333=1.6458333$. Lagrange remainder with $n=3$: the fourth derivative of $e^{x}$ is $e^{x}$, which is increasing, so on $[0, 0.5]$ it is bounded by $e^{0.5}\approx 1.6487$, and we can conveniently take $M=2$, a simple and safe bound. Then $\left|R_{3}\right|\le\frac{2\times(0.5)^{4}}{4!}=\frac{2\times 0.0625}{24}\approx 0.0052$. Actual error: $e^{0.5}=1.6487213$, so the gap is $0.0028879$. The bound is respected, and it overestimates the error by a factor of about 1.8. That is normal and healthy: an error bound is meant to be GUARANTEED, not tight. Here the overestimate comes from two combined sources: the choice $M=2$ instead of $1.649$, and the fact that the unknown point $c$ actually lies closer to 0 than to 0.5.

We want $e=e^{1}$, so $a=0$ and $x=1$. The derivative of order $n+1$ of $e^{x}$ is $e^{x}$, bounded on $[0, 1]$ by $e^{1}=e\approx 2.718$; so we take $M=3$, a simple and safe bound, chosen precisely to avoid using the value of $e$ that we are trying to compute, which would be circular. The condition becomes $\frac{3}{(n+1)!}\le 10^{-6}$, that is $(n+1)!\ge 3\times 10^{6}$. Now $9!=362{,}880$, not enough, and $10!=3{,}628{,}800\ge 3\times 10^{6}$ ✓. So we need $n+1=10$, that is $n=9$: the polynomial of degree 9, i.e. TEN terms from $1$ to $\frac{x^{9}}{9!}$. Check: $\frac{3}{10!}\approx 8.27\times 10^{-7}<10^{-6}$, whereas $\frac{3}{9!}\approx 8.27\times 10^{-6}$ would not be enough. Ten terms for six decimals of $e$: factorial convergence is formidably efficient, compared with the hundreds of terms that Leibniz's formula needed for $\pi$ in exercise 2.

Alternating series bound: the sine series is alternating with decreasing terms at $x=0.5$, so the error after $T_{5}$ is bounded by the first neglected term, $\frac{0.5^{7}}{7!}=\frac{0.0078125}{5040}\approx 1.55\times 10^{-6}$. Lagrange bound with $n=5$: all the derivatives of the sine are $\pm\sin$ or $\pm\cos$, so $M=1$ always works, and $\left|R_{5}\right|\le\frac{1\times 0.5^{6}}{6!}=\frac{0.015625}{720}\approx 2.17\times 10^{-5}$. The alternating bound is therefore FOURTEEN times better, and the actual error, $1.54\times 10^{-6}$, comes very close to it. The clever choice: since the sine expansion contains NO term of degree 6, its coefficient being zero, the polynomial $T_{6}$ is identical to $T_{5}$. We are therefore perfectly entitled to apply Lagrange with $n=6$ while having computed only $T_{5}$, which gives $\left|R_{6}\right|\le\frac{1\times 0.5^{7}}{7!}\approx 1.55\times 10^{-6}$: exactly the alternating bound. The two methods coincide as soon as parity is exploited. The general lesson holds for the exam: when both bounds are available, take the smaller one, and check whether the expansion skips degrees, because each missing degree gives a free notch of precision in the Lagrange bound.

$T_{n}$ is the unique polynomial of degree at most $n$ that shares with $f$, at the point of contact, its VALUE and its first $n$ DERIVATIVES: same position, same slope, same concavity, and so on. $T_{1}(x)=f(a)+f'(a)(x-a)$ is exactly the TANGENT LINE: the linear approximation you already know. The Taylor chapter is simply the logical continuation: after the tangent line, the osculating parabola, then better and better.

$T_{1}=x$ (the tangent line at 0) sticks on roughly $[-0.5, 0.5]$, then heads straight on while the sine curves away. $T_{3}$ follows faithfully up to about $|x|\approx 1.5$ and already reproduces the first hump, before plunging. $T_{5}$ hugs the curve beyond $|x|\approx 2.5$, humps included. Each odd degree added pushes back the point where the polynomial breaks away: locally, the gain is a factor of $\frac{x^{2}}{(n+2)(n+3)}$ per step: spectacular near 0, more and more expensive far from 0.

$T_{5}(0.5)=0.5-\frac{0.125}{6}+\frac{0.03125}{120}=0.479427$. The sine series is alternating with decreasing terms: the error is bounded by the first neglected term, $\frac{0.5^{7}}{5040}\approx 1.6\times 10^{-6}$. Five decimals guaranteed ($\sin 0.5=0.4794255\ldots$: the actual error, $1.5\times 10^{-6}$, comes very close to the bound). Three terms, six digits: this is precisely the algorithm of a calculator, angle reduction included.

The error in $\sin\theta\approx\theta$ is dominated by $\frac{\theta^{3}}{6}$: at $\theta=0.1745$, absolute error $\approx 8.9\times 10^{-4}$, that is a RELATIVE error $\frac{\theta^{2}}{6}\approx 0.5$%. The pendulum equation $\ddot{\theta}=-\frac{g}{L}\sin\theta$ becomes linear, $\ddot{\theta}=-\frac{g}{L}\theta$, with a period independent of the amplitude, off by barely 0.2% at 10° (the error on the period is about $\frac{\theta^{2}}{16}$). All of physics' “small angles”, “small oscillations” and “low amplitudes” is a first term of a Taylor series, knowingly accepted: this chapter is the mathematical guarantee behind those shortcuts.

$e^{u}=\sum\frac{u^{n}}{n!}$ with $u=-x^{2}$: $e^{-x^{2}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}=1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{6}+\cdots$ (valid everywhere). Term-by-term integration: $\int_{0}^{0.5}e^{-x^{2}}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}(0.5)^{2n+1}}{n!(2n+1)}=0.5-\frac{0.5^{3}}{3}+\frac{0.5^{5}}{10}-\frac{0.5^{7}}{42}+\cdots$.

Four terms: $0.5-0.0416667+0.0031250-0.0001860=0.461272$. Alternating series: error bounded by the fifth term, $\frac{0.5^{9}}{216}\approx 9\times 10^{-6}$: four reliable decimals (true value $0.461281$). This integral is THE Gaussian: the normal distribution in statistics, heat diffusion, measurement errors: the most integrated function in science has no elementary antiderivative, and everyone integrates it anyway, with series and tables.

$\frac{\sin x}{x}=1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+\cdots$ (divide the sine series by $x$: the singularity at 0 disappears, the function equals 1 there). $\int_{0}^{1}\frac{\sin x}{x}dx=1-\frac{1}{3\cdot 3!}+\frac{1}{5\cdot 5!}-\frac{1}{7\cdot 7!}+\cdots=1-0.055556+0.001667-0.000028=0.946083$, error bounded by $\frac{1}{9\cdot 9!}\approx 3\times 10^{-7}$: six decimals in four terms. This function (the sinc function) governs diffraction and signal processing.

The series CONVERGES to the exact value: giving it, with an explicit error bound that can be made as small as you like, is giving the number itself, exactly as “$\sqrt{2}$” names a number that no finite decimal writes out, but which can be approached at will. $\pi$, $e$, $\ln 2$ have no other status: limits of series, truncated at the desired precision. A “closed form” is a cultural comfort, not a mathematical requirement: the chapter on series simply widens the list of acceptable exact answers.

$\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots$, so $\sin x-x=-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots$ and $\frac{\sin x-x}{x^{3}}=-\frac{1}{6}+\frac{x^{2}}{120}-\cdots\to-\frac{1}{6}$. With L'Hôpital: THREE successive applications (the $\frac{0}{0}$ forms follow one another), with three differentiations not to be botched. The series gives more than a limit: the whole expansion, and therefore also the SPEED of approach.

$1-\cos x=\frac{x^{2}}{2}-\frac{x^{4}}{24}+\cdots$: the limit is $\frac{1}{2}$. $e^{x}-1-x=\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots$: the limit is $\frac{1}{2}$ too. The moral: subtracting from a function the beginning of its expansion reveals the next term: these limits merely READ a Taylor coefficient, $\frac{f''(0)}{2}$ in both cases.

“Equivalent” means: same first term of the expansion ($\frac{\sin x}{x}\to 1$), and therefore interchangeable in products and quotients. The DANGER: subtractions. In $\frac{\sin x-x}{x^{3}}$, replacing $\sin x$ by $x$ would give 0: WRONG, because the subtraction cancels precisely the first terms, and the limit lives in the term of order 3, which the equivalence threw away. Safety rule: in a difference, expand one order further than what cancels; the bare equivalence is reserved for factors.

The actual steps: 1) conversion to radians: $37°=0.6458$ rad; 2) argument REDUCTION: bring the angle close to 0 using the symmetries of the circle (periodicity, $\sin(\pi-x)$, etc.): here it is already there; 3) evaluation of a Taylor-type POLYNOMIAL (optimized, degree 7 to 9 or so) in a few multiplications; 4) precision guaranteed by the remainder bound, pushed below the last digit displayed. No table, no magic circle: the “sin” key runs a polynomial from this chapter. Taylor series are not an end-of-semester curiosity: they are the numerical engine of every “special” function in every processor.