Exercises — Taylor's remainder theorem — error estimation

The error bound (from the parent note): $|R_n(x)|\le \frac{M}{(n+1)!}|x-a{|}^{n+1},M={\max }_{t\text{ between }a,x}|f^{(n+1)}(t)|.$ Read it as "Next derivative, Next factorial, Next power" — everything is $n+1$, one step beyond the polynomial degree $n$.

• $R_n(x)=f(x)-P_n(x)$ is the exact error.
• $P_n$ is the degree-$n$ Taylor polynomial.
• The bound comes from the Lagrange form $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a{)}^{n+1}$, where $c$ is unknown, so we replace $f^{(n+1)}(c)$ by its worst-case size $M$.

WHAT the symbols mean. $n$ is the degree of the polynomial you kept. The bound always uses one step beyond $n$.

• Derivative needed: $f^{(n+1)}=f^{(5)}$ — the fifth derivative.
• Factorial: $(n+1)!=5!=120$.
• Power: $|x-a{|}^5$.

WHY $n+1$ and not $n$. The polynomial $P_4$ already matches $f$ up to the 4th derivative; the first thing it gets wrong is the 5th derivative term, so that is what controls the leftover error.

Derivatives at $0$: $f(0)=\cos 0=1$, $f^{\prime }(0)=-\sin 0=0$, $f^{\prime \prime }(0)=-\cos 0=-1$. $P_2(x)=1+0\cdot x-\frac{1}{2}{x}^2=1-\frac{x^2}{2}.$ $R_2(x)=\cos x-\left(1-\frac{x^2}{2}\right)$ — the exact gap between the true cosine and this parabola.

$f(x)=e^x$, $a=0$, $n=2$, $x=0.2$.

• $f^{(3)}(t)=e^t$. On $[0,0.2]$ it is increasing, biggest at $t=0.2$: $e^{0.2}\approx 1.221<1.5$. Take $M=1.5$.
• Why an over-estimate? We want a safe upper bound; picking $M$ a little large only makes the guaranteed error a little large, never wrong. $|R_2(0.2)|\le \frac{1.5}{3!}(0.2{)}^3=\frac{1.5}{6}(0.008)=\mathrm{0.002.}$ So $P_2(0.2)=1.22$ is correct to within $0.002$. (True $e^{0.2}=1.22140\dots$, actual error $\approx 0.00140$ — inside the bound. ✓)

$f(x)=\sin x$, $a=0$, $n=3$.

• $f^{(4)}(t)=\sin t$, and $|\sin t|\le 1$ always, so $M=1$ (no interval work needed — the cleanest possible bound). $|R_3(0.5)|\le \frac{1}{4!}(0.5{)}^4=\frac{0.0625}{24}\approx 2.60\times 10^{-3}.$ $P_3(0.5)=0.5-0.020833=0.479167$; true $\sin 0.5=0.479426$, actual error $\approx 2.6\times 10^{-4}$. ✓

Derivatives: $f^{\prime }(x)=(1+x{)}^{-1}$, $f^{\prime \prime }(x)=-(1+x{)}^{-2}$, $f^{\prime \prime \prime }(x)=2(1+x{)}^{-3}$.

• We need $M=\max |f^{\prime \prime \prime }(t)|$ on $[0,0.3]$. Since $2(1+t{)}^{-3}$ decreases as $t$ grows, its max is at $t=0$: $M=2(1{)}^{-3}=2$.
• Why the left endpoint? $(1+t{)}^{-3}$ shrinks when $t$ rises, so the worst (largest) case is the smallest $t$. $|R_2(0.3)|\le \frac{2}{3!}(0.3{)}^3=\frac{2}{6}(0.027)=\mathrm{0.009.}$

$n=1$, so we need $f^{(2)}(t)=-\sin t$, giving $|f^{(2)}(t)|=|\sin t|$ on $[0,\pi ]$.

• WHERE is the max? $|\sin t|$ is zero at both endpoints ($\sin 0=0$, $\sin \pi =0$) and rises to $1$ at the interior point $t=\pi /2$. So the maximum is strictly inside — $M=1$ at $t=\pi /2$, not at an endpoint. Look at the figure: the peak (red dot) sits in the middle, not at the edges.

  

• Why endpoints would fail here: plugging $t=0$ or $t=\pi$ gives $|f^{(2)}|=0$, which would falsely claim zero error — clearly wrong, since $\sin \pi =0$ but $P_1(\pi )=\pi \ne 0$. $|R_1(\pi )|\le \frac{1}{2!}|\pi -0{|}^2=\frac{{\pi }^2}{2}\approx \mathrm{4.93.}$ A large but honest bound: the linear approximation is genuinely terrible that far out, and only the interior maximum captures it.

$f(x)=e^x$, $a=0$, $x=0.5$. On $[0,0.5]$, $f^{(n+1)}(t)=e^t\le e^{0.5}<2$, so $M=2$. $|R_n(0.5)|\le \frac{2}{(n+1)!}(0.5{)}^{n+1}.$ Demand $<10^{-5}$. Test values:

• $n=4$: $\frac{2(0.5{)}^5}{5!}=\frac{2(0.03125)}{120}=5.2\times 10^{-4}$ — too big.
• $n=5$: $\frac{2(0.5{)}^6}{6!}=\frac{2(0.015625)}{720}=4.34\times 10^{-5}$ — still too big.
• $n=6$: $\frac{2(0.5{)}^7}{7!}=\frac{2(0.0078125)}{5040}=3.10\times 10^{-6}<10^{-5}$. ✓ Answer: $n=6$ (terms up to $x^6/6!$). Note how the small factor $(0.5{)}^{n+1}$ helps you — far fewer terms than the $x=1$ case in the parent note.

$f^{(5)}(t)=\sin t$ (up to sign), so $M=1$. $|R_4(x)|\le \frac{1}{5!}|x{|}^5=\frac{|x{|}^5}{120}\le 10^{-3}.$ Solve for $|x|$. Multiply both sides by $120$: $|x{|}^5\le 0.12$. Now take the fifth root of both sides — why is that legal? The function $u\mapsto u^{1/5}$ is increasing for $u\ge 0$, so it preserves the inequality direction; and $|x|\ge 0$, so its fifth root is well defined. Hence $|x|\le 0.12^{1/5}.$ Compute: $0.12^{1/5}=e^{(\ln 0.12)/5}=e^{-0.4241}\approx 0.6544$. Answer: $|x|\lesssim 0.654$ guarantees $10^{-3}$ accuracy. (Because the bound uses $|x|$, this range is symmetric: it works for $x=+0.6$ and $x=-0.6$ — see L4.3.)

Here $x=0.5$ and $a=0.5$, so $x-a=0$. Every Taylor polynomial gives $P_n(0.5)=f(0.5)$ exactly. $|R_1(0.5)|\le \frac{M}{2!}|0.5-0.5{|}^2=\frac{M}{2}\cdot 0=0.$ Zero error — because $|x-a{|}^{n+1}=0$. This is the degenerate/limiting case: the closer your centre is to the target, the smaller $|x-a|$ and the more brutally the power $|x-a{|}^{n+1}$ crushes the error. It is the whole reason Taylor approximations are "local."

$a=1$, $x=1.5$, so $x-a=0.5$, $n=3$. Derivatives of $\ln x$: $f^{\prime }=x^{-1}$, $f^{\prime \prime }=-x^{-2}$, $f^{\prime \prime \prime }=2x^{-3}$, $f^{(4)}=-6x^{-4}$.

• $M={\max }_{[1,1.5]}|-6x^{-4}|=6x^{-4}$. This decreases in $x$, so max at $x=1$: $M=6$.
• Why $x=1$? $x^{-4}$ shrinks as $x$ grows, so the largest value is at the left endpoint. $|R_3(1.5)|\le \frac{6}{4!}(0.5{)}^4=\frac{6}{24}(0.0625)=\mathrm{0.015625.}$ (For reference $P_3(x)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}$, giving $P_3(1.5)=0.5-0.125+0.041\overline{6}=0.41\overline{6}$; true $\ln 1.5=0.405465$, actual error $\approx 0.0112<0.0156$. ✓)

$f(x)=e^x$, $a=0$, $n=2$, target $x=-0.4$. The interval "between $a$ and $x$" is now $[-0.4,0]$ — it runs to the left of the centre.

• WHAT $M$ is here. $f^{(3)}(t)=e^t$ on $[-0.4,0]$ is increasing, so its max is at the right end $t=0$: $e^0=1$. Take $M=1$.
• Why the right end this time? $e^t$ grows with $t$, and $t=0$ is the largest $t$ in $[-0.4,0]$ — the mirror image of the L2.1 reasoning.
• WHY the sign disappears. The bound uses $|x-a{|}^{n+1}=|-0.4-0{|}^3=|-0.4{|}^3=(0.4{)}^3$. The absolute value erases the sign — a target $0.4$ units to the left gives the exact same power as $0.4$ units to the right. That is why we never worry about which side of the centre we are on. $|R_2(-0.4)|\le \frac{1}{3!}(0.4{)}^3=\frac{0.064}{6}\approx 1.067\times 10^{-2}.$ Check: $P_2(-0.4)=1-0.4+0.08=0.68$; true $e^{-0.4}=0.670320$, actual error $\approx 0.00968<0.01067$. ✓

From the parent note: on $[0,1]$, $f^{(n+1)}(t)=e^t\le e<3$, so $M=3$ and $|R_n(1)|\le \frac{3}{(n+1)!}<10^{-6}⟹(n+1)!>3\times 10^6.$ Since $9!=362880<3\times 10^6$ but $10!=3628800>3\times 10^6$, we need $n+1=10$, i.e. $n=9$ (terms up to $x^9/9!$).

The wrong alternative, stated plainly. The exact Lagrange form is $R_9(1)=\frac{e^c}{10!}\cdot 1^{10}$ for some unknown $c\in (0,1)$. A tempting shortcut is to guess a specific value of $c$ — say $c=0$, giving $e^0=1$ and error $\approx \frac{1}{10!}$ — and call that "the error." Why it is wrong: the theorem only promises that some $c$ exists; it never tells you which one. If the true $c$ is closer to $1$, the real factor $e^c$ is up to $e\approx 2.718$, nearly $3\times$ your guess — so a guessed $c$ can under-report the error and break your guarantee. Why the bound wins: replacing $e^c$ by its safe maximum $M=3$ over the whole interval gives a value that is valid no matter where $c$ hides. You trade a slightly larger (but certified) number for a promise that can never be violated.

Every derivative of $\sin x$ is $\pm \sin$ or $\pm \cos$, so $|f^{(n+1)}(t)|\le 1$ for all $t$ and all $n$. Hence $M=1$ works uniformly: $|R_n(x)|\le \frac{1}{(n+1)!}|x{|}^{n+1}.$ WHY this $\to 0$. Fix any real $x$. The quantity $\frac{|x{|}^{n+1}}{(n+1)!}$ is the general term of the convergent series $\sum \frac{|x{|}^k}{k!}=e^{|x|}$; terms of a convergent series must tend to $0$. Therefore $|R_n(x)|\to 0$ for every $x$. Conclusion: the factorial always beats the power, no matter how big $|x|$ is — the series converges everywhere (infinite radius of convergence). $■$

Lagrange (from our formula), $n=3$: $|R_3(1)|\le \frac{M}{4!}|1{|}^4=\frac{1}{24}\approx 0.0417$ (using $M=1$). Alternating Series bound: the Maclaurin series $\sin 1=1-\frac{1}{3!}+\frac{1}{5!}-\cdots$ is alternating with decreasing terms, so the error after the $x^3$ term is at most the first omitted term $\frac{1}{5!}=\frac{1}{120}\approx 0.00833$. Which is tighter: the alternating bound $0.00833$ is about 5× smaller than $0.0417$. WHY. The Lagrange bound uses the worst-case derivative $M=1$ blindly; the alternating theorem exploits the cancellation between $+$ and $-$ terms, which is extra information. When a series alternates, prefer that theorem; when it does not, Lagrange is your general tool.

Exact Lagrange form: $R_n(x)=\frac{e^c}{(n+1)!}{x}^{n+1}$ for some $c$ between $0$ and $x$. As $x\to 0$, that $c\to 0$ too, so $e^c\to 1$. Thus $R_n(x)=\frac{e^c}{(n+1)!}{x}^{n+1}=\underset{\text{tends to a constant}}{\underbrace{\frac{1+o(1)}{(n+1)!}}}{x}^{n+1}.$ The dominant behaviour is $x^{n+1}$ times a bounded factor, which is exactly the meaning of $R_n(x)=O(x^{n+1})$: the error shrinks like the next power of $x$ as $x\to 0$. The factor $\frac{1}{(n+1)!}$ is the leading constant hidden inside the Big-O.