Corrected exercises: Taylor polynomials and error bounds (MATH 141)

The derivatives of $\sin x$ cycle: $\sin x$, $\cos x$, $-\sin x$, $-\cos x$, $\sin x$, $\dots$, so at $0$ they take the values $0, 1, 0, -1, 0, 1, 0, \dots$ Hence $T_1(x) = 0 + 1 \cdot x = x$, $T_3(x) = x + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 = x - \frac{x^3}{6}$, and $T_5(x) = x - \frac{x^3}{6} + \frac{1}{5!}x^5 = x - \frac{x^3}{6} + \frac{x^5}{120}$. Since $f^{(4)}(0) = \sin 0 = 0$, the term of degree $4$ is $0$ and $T_4 = T_3$; likewise $f^{(6)}(0) = -\sin 0 = 0$ gives $T_6 = T_5$. This is not a detail: $T_3$ is ALSO $T_4$, and Exercises 2, 4 and 6 show that using it as $T_4$ gives an error bound twenty to fifty times better.

Reading the figure: the orange line $T_1 = x$ leaves the sine curve visibly around $\lvert x\rvert \approx 1$; the green cubic $T_3$ holds until about $\lvert x\rvert \approx 2$; the red $T_5$ until about $\lvert x\rvert \approx 3$. For small $x > 0$, $T_1$ is ABOVE $\sin x$, $T_3$ BELOW, $T_5$ ABOVE again: each new term overcorrects the previous one, because the terms of the sine series alternate in sign. At $x = 1$: $T_1(1) = 1$, $T_3(1) = \frac{5}{6} \approx 0.833$, $T_5(1) = \frac{101}{120} \approx 0.842$, and $\sin 1$ lies between the last two. Exercise 4 turns this alternation into a bound.

$f(x) = \ln x$, $f'(x) = \frac{1}{x}$, $f''(x) = -\frac{1}{x^2}$, $f'''(x) = \frac{2}{x^3}$; at $1$: $0$, $1$, $-1$, $2$. So $T_3(x) = 0 + (x - 1) - \frac{1}{2!}(x - 1)^2 + \frac{2}{3!}(x - 1)^3 = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3}$. Two slips cost marks here: expanding in powers of $x$ instead of $x - 1$ (the polynomial is centred at $1$, keep the powers of $x - 1$), and forgetting the $k!$, which turns the last coefficient into $2$. With $u = x - 1$ one recognizes the first terms of $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$, a free check.

$f(x) = x^{1/2}$, $f'(x) = \frac{1}{2}x^{-1/2}$, $f''(x) = -\frac{1}{4}x^{-3/2}$. At $4$: $f(4) = 2$, $f'(4) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$, $f''(4) = -\frac{1}{4} \cdot \frac{1}{8} = -\frac{1}{32}$. So $T_2(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32 \cdot 2}(x - 4)^2 = 2 + \frac{x - 4}{4} - \frac{(x - 4)^2}{64}$. Check: $T_2(4) = 2 = f(4)$; $T_2'(x) = \frac{1}{4} - \frac{x - 4}{32}$ gives $T_2'(4) = \frac{1}{4} = f'(4)$; $T_2''(x) = -\frac{1}{32} = f''(4)$. The classic error is the coefficient $-\frac{1}{32}$ instead of $-\frac{1}{64}$: then $T_2'' = -\frac{1}{16}$ and the third condition fails, which is exactly how to catch it.

A Taylor polynomial knows $f$ only through $f(a), f'(a), \dots, f^{(n)}(a)$: it copies the shape of $f$ at $0$ and extrapolates. Far from $0$ the polynomial is dominated by its highest power and runs off to $\pm\infty$, while $\sin x$ stays between $-1$ and $1$, so no $T_n$ can follow the sine on the whole real line. Raising the degree widens the zone where the fit is good (about $1$, $2$, $3$ for $n = 1, 3, 5$), and near $0$ the gap shrinks like the first neglected power: $x^3$, then $x^5$, then $x^7$. Quantifying that gap, with a bound that is GUARANTEED rather than read on a picture, is the job of Taylor's inequality.

Here $n = 2$, $a = 0$, $d = 0.5$ and $f'''(x) = e^x$. On $[-0.5, 0.5]$ the exponential is increasing, so its largest value is at the RIGHT end: $M = e^{0.5} = \sqrt{e}$. No calculator needed to bound it: $e < 4$, so $\sqrt{e} < 2$. Then $\lvert R_2(x)\rvert \le \frac{2}{3!}(0.5)^3 = \frac{2}{6} \cdot \frac{1}{8} = \frac{1}{24} \approx 0.042$ for all $\lvert x\rvert \le 0.5$. A sharper $M$ is allowed, $1.65^2 = 2.7225 > e$ gives $\sqrt{e} < 1.65$ and the bound $\frac{1.65}{48} \approx 0.034$, but any number ABOVE the true maximum is a valid $M$. The one thing that is not allowed is a number below it.

$f(x) = \ln(1 + x)$, $f'(x) = \frac{1}{1 + x}$, $f''(x) = -\frac{1}{(1 + x)^2}$, $f'''(x) = \frac{2}{(1 + x)^3}$. On $[-0.5, 0.5]$, $1 + x$ ranges over $[0.5, 1.5]$ and $\frac{2}{(1 + x)^3}$ is DECREASING, so its maximum is at the LEFT end: $M = \frac{2}{(0.5)^3} = 16$, the red point of the figure. Taylor's inequality gives $\lvert R_2(x)\rvert \le \frac{16}{6}(0.5)^3 = \frac{1}{3}$. Weak, but guaranteed. Now the tempting shortcut $M = f'''(0) = 2$ would claim $\lvert R_2\rvert \le \frac{2}{6} \cdot \frac{1}{8} = \frac{1}{24} \approx 0.042$. Test it at $x = -0.5$: $f(-0.5) = \ln\frac{1}{2} = -\ln 2 \approx -0.693$ and $T_2(-0.5) = -0.5 - 0.125 = -0.625$, so the true error is $\ln 2 - \frac{5}{8} \approx 0.068$, larger than $0.042$. A bound that the function itself violates is not a bound; on a final it is worth zero, whatever the arithmetic.

On $[0, 0.5]$ the same decreasing function $\frac{2}{(1 + x)^3}$ has its maximum at $x = 0$, the left end of THIS interval, so $M = 2$ is now correct and $\lvert R_2(x)\rvert \le \frac{2}{6}(0.5)^3 = \frac{1}{24}$ for $0 \le x \le 0.5$. At $x = 0.5$ the true error is $\ln 1.5 - 0.375 \approx 0.031$, below $\frac{1}{24}$ as it must be. The function, the polynomial and the formula are those of b); only the interval changed, and with it the worst point. This is why every $M$ on a copy is written with its interval: $\lvert f'''(x)\rvert \le 2$ for $0 \le x \le 0.5$.

With $n = 2$: $f'''(x) = \sin x$, and on $[-0.2, 0.2]$, $\lvert\sin x\rvert \le \lvert x\rvert \le 0.2$, so $M = 0.2$ and $\lvert R_2\rvert \le \frac{0.2}{6}(0.2)^3 = \frac{0.0016}{6} = \frac{1}{3750} \approx 0.00027$. (With the lazy $M = 1$: $\frac{0.008}{6} \approx 0.0013$, also correct.) With $n = 3$: since $\cos$ has no $x^3$ term, $T_3 = T_2$, and the remainder is ALSO $R_3$, controlled by $f^{(4)}(x) = \cos x$, $\lvert\cos x\rvert \le 1$: $\lvert R_3\rvert \le \frac{1}{24}(0.2)^4 = \frac{0.0016}{24} = \frac{1}{15000} \approx 0.000067$. Both bounds are true; the second is four times better than the first and twenty times better than the lazy one. The true error at $0.2$ is about $0.0000666$: the $n = 3$ bound is practically exact.

The worst point of $\lvert f^{(n+1)}\rvert$ is found like any maximum on a closed interval, and in practice $\lvert f^{(n+1)}\rvert$ is monotone on it, so the maximum sits at an END. Exponentials $e^{x}$: increasing, the RIGHT end (a). Powers of $\frac{1}{1 + x}$, which is where $\ln(1 + x)$ and $(1 + x)^k$ lead: decreasing, the LEFT end (b, c). Sine and cosine: $M = 1$ always works, and a smaller $M$ is available near $0$ (d). Two rules follow. Never take $M$ at the centre unless the centre IS the worst point, as in c). And when $T_n = T_{n+1}$, use $n + 1$ in the inequality: the bound gains a full power of $\lvert x - a\rvert$.

$e^x$ is its own derivative, so every coefficient is $\frac{1}{k!}$ and $T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$. At $x = 0.2 = \frac{1}{5}$: $T_3(0.2) = 1 + \frac{1}{5} + \frac{1}{50} + \frac{1}{750}$. Over the common denominator $750$: $\frac{750 + 150 + 15 + 1}{750} = \frac{916}{750} = \frac{458}{375}$, that is $1.221333\ldots$ Working in fractions keeps the arithmetic exact; the decimal is written only at the end.

$n = 3$, so we need $M \ge \lvert f^{(4)}(x)\rvert = e^x$ for $0 \le x \le 0.2$. The exponential is increasing: the worst point is $x = 0.2$, and $M = e^{0.2}$, the very number we are approximating. It is bounded by something simpler. Crude and valid: $e^{0.2} < e < 3$, which gives $\lvert R_3\rvert \le \frac{3(0.2)^4}{24} = \frac{3 \cdot 0.0016}{24} = 0.0002$. Better: $1.25^2 = 1.5625$, $1.25^4 \approx 2.44$, $1.25^5 \approx 3.05 > e$, so $e^{0.2} < 1.25$ and $\lvert R_3\rvert \le \frac{1.25 \cdot 0.0016}{24} = \frac{0.002}{24} = \frac{1}{12000} \approx 0.000083$. What is NOT acceptable is $M = e^0 = 1$, the value at the centre, which sits at the wrong end of the interval.

The remainder has a definite sign here. All the terms of the series of $e^{0.2}$ are positive, so the part left out, $\frac{(0.2)^4}{24} + \frac{(0.2)^5}{120} + \cdots$, is positive: $T_3(0.2)$ is an UNDERESTIMATE. (Equivalently, the Lagrange form of the remainder is $\frac{e^z}{24}(0.2)^4 > 0$ for some $z$ between $0$ and $0.2$.) So $\frac{458}{375} < e^{0.2} < \frac{458}{375} + \frac{1}{12000}$, i.e. $1.22133 < e^{0.2} < 1.22142$. The interval is the answer a marker wants: it is guaranteed, and its width says how good the approximation is.

$f(x) = (1 + x)^{1/3}$, $f'(x) = \frac{1}{3}(1 + x)^{-2/3}$, $f''(x) = -\frac{2}{9}(1 + x)^{-5/3}$; at $0$: $1$, $\frac{1}{3}$, $-\frac{2}{9}$. So $T_2(x) = 1 + \frac{x}{3} - \frac{2}{9} \cdot \frac{x^2}{2} = 1 + \frac{x}{3} - \frac{x^2}{9}$, the first terms of the binomial series with $k = \frac{1}{3}$. At $x = 0.1$: $1 + \frac{1}{30} - \frac{1}{900} = \frac{900 + 30 - 1}{900} = \frac{929}{900} \approx 1.032222$.

$f'''(x) = \frac{10}{27}(1 + x)^{-8/3}$, positive and DECREASING on $[0, 0.1]$: the worst point is the left end, $M = f'''(0) = \frac{10}{27}$ (here the centre IS the worst point, and that has to be said, not assumed). Then $\lvert R_2(0.1)\rvert \le \frac{10}{27} \cdot \frac{(0.1)^3}{3!} = \frac{10}{27 \cdot 6000} = \frac{1}{16200} \approx 0.000062$. Sign: $R_2(0.1) = \frac{f'''(z)}{6}(0.1)^3$ with $f''' > 0$, so the error is positive and $T_2$ is an UNDERESTIMATE: $\frac{929}{900} < \sqrt[3]{1.1} < \frac{929}{900} + \frac{1}{16200}$. The figure agrees: for $x > 0$ the green parabola $T_2$ runs below the curve, while the tangent line $T_1$ runs above it. If $M$ is evaluated at $x = 0.1$ instead, one gets $\frac{1}{16200}(1.1)^{-8/3}$, about $0.000048$; the true error is about $0.000058$ (a machine confirms it afterwards), so that number is NOT a bound. The correct bound is barely above the true error, which is why the worst point matters.

$\sin 0.1 = 0.1 - \frac{(0.1)^3}{3!} + \frac{(0.1)^5}{5!} - \cdots$, and $T_3(0.1) = 0.1 - \frac{0.001}{6} = \frac{600 - 1}{6000} = \frac{599}{6000} \approx 0.0998333$. Hypotheses, one by one: the series alternates; its terms $b_k = \frac{(0.1)^{2k+1}}{(2k + 1)!}$ are positive and decreasing, since $\frac{b_{k+1}}{b_k} = \frac{(0.1)^2}{(2k + 2)(2k + 3)} < 1$; and $b_k \to 0$. So $\lvert \sin 0.1 - T_3(0.1)\rvert \le b_2 = \frac{(0.1)^5}{120} = \frac{10^{-5}}{120} = \frac{1}{12\,000\,000} \approx 8.3 \times 10^{-8}$. The first omitted term is POSITIVE, so the error $\sin 0.1 - T_3(0.1)$ is positive: $T_3(0.1)$ is too SMALL, and $\frac{599}{6000} < \sin 0.1 < \frac{599}{6000} + \frac{1}{12\,000\,000}$. Seven correct decimals from two terms.

Taylor's inequality with $n = 3$: $\lvert f^{(4)}(x)\rvert = \lvert\sin x\rvert \le 1$, so $\lvert R_3(0.1)\rvert \le \frac{(0.1)^4}{24} = \frac{1}{240\,000} \approx 4.2 \times 10^{-6}$, true but fifty times weaker. With $n = 4$, legitimate because $T_4 = T_3$ for the sine (Exercise 1): $\lvert f^{(5)}\rvert = \lvert\cos x\rvert \le 1$ and $\lvert R_4(0.1)\rvert \le \frac{(0.1)^5}{120}$, exactly the alternating bound. The alternating estimate does automatically what Taylor's inequality does only if you remember that the even terms are zero, and it adds the SIGN, which Taylor's inequality never gives.

$\cos 0.5 = 1 - \frac{(0.5)^2}{2} + \frac{(0.5)^4}{24} - \cdots$ alternates, with terms $\frac{(0.5)^{2k}}{(2k)!}$ decreasing (ratio $\frac{0.25}{(2k + 1)(2k + 2)} < 1$) to $0$. $T_2(0.5) = 1 - \frac{1}{8} = \frac{7}{8}$. First omitted term: $\frac{(0.5)^4}{24} = \frac{1}{16 \cdot 24} = \frac{1}{384} \approx 0.0026$, positive. Hence $\frac{7}{8} < \cos 0.5 < \frac{7}{8} + \frac{1}{384}$, that is $0.875 < \cos 0.5 < 0.8777$. The true value is about $0.87758$: the error, $0.00258$, is almost exactly the first omitted term, because the terms after it are tiny. That is why this estimate is often SHARPER in practice: it uses the next term itself, while Taylor's inequality uses the worst value of a derivative over a whole interval.

$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ for $\lvert x\rvert \le 1$. At $x = \frac{1}{2}$ the terms $b_k = \frac{(1/2)^{2k+1}}{2k + 1}$ are positive, decreasing (numerator divided by $4$, denominator larger) and tend to $0$. Two terms: $\frac{1}{2} - \frac{1}{24} = \frac{11}{24} \approx 0.4583$. First omitted term: $\frac{(1/2)^5}{5} = \frac{1}{160} = 0.00625$, positive: $\frac{11}{24} < \arctan\frac{1}{2} < \frac{11}{24} + \frac{1}{160}$. The figure shows the mechanism: $S_1 = \frac{1}{2}$ above the dashed value, $S_2 = \frac{11}{24}$ below, and each partial sum lands on the other side, closer, so the value is always trapped between two consecutive sums. Taylor's inequality would need a bound on the fifth derivative of $\arctan x$ on $[0, \frac{1}{2}]$, a rational function whose numerator is a polynomial of degree four: no one computes it on an exam, and the alternating estimate makes it unnecessary.

$e^{-0.2} = 1 - 0.2 + \frac{(0.2)^2}{2} - \frac{(0.2)^3}{6} + \cdots$ alternates, terms $\frac{(0.2)^k}{k!}$ decreasing to $0$: YES. $e^{0.2}$: every term is positive, the series does not alternate: NO, which is why Exercise 3 needed Taylor's inequality. $\ln 1.2 = \ln(1 + 0.2) = 0.2 - \frac{(0.2)^2}{2} + \frac{(0.2)^3}{3} - \cdots$ alternates, terms $\frac{(0.2)^k}{k}$ decreasing to $0$: YES. $\ln 0.8 = \ln(1 - 0.2) = -0.2 - \frac{(0.2)^2}{2} - \frac{(0.2)^3}{3} - \cdots$: every term is negative: NO. The check is on the series AT THE POINT, not on the function: the same $\ln(1 + x)$ alternates at $x = 0.2$ and not at $x = -0.2$.