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Amy Ferguson Moncure

Calculus BC ยท Unit 10

Error Bounds

An alternating tail is at most the next term. A Taylor remainder is at most M times the next power over the next factorial.

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Key Ideas Easy Medium Hard

\(\left|R_n(x)\right|\le \dfrac{M\,|x-a|^{n+1}}{(n+1)!}\)

What to Remember

Alternating error

If the alternating series test applies, the error after some terms is at most the first omitted term. The true sum lies on the side of that next term.

How many terms

Ask for the first omitted term to be smaller than the allowed error. Check whether the inequality is strict.

Lagrange remainder

If \(|f^{(n+1)}(z)|\le M\) between \(a\) and \(x\), then

\(\left|R_n(x)\right|\le \dfrac{M\,|x-a|^{n+1}}{(n+1)!}\).

The bound is not the error

An alternating bound or a Lagrange bound is a guarantee. The exact error can be smaller.

Level 1

Easy

The next term of an alternating harmonic series.

Example
For \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}\), compute \(s_4\) and give an error bound.

\(s_4=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\)

\(s_4=\dfrac{12}{12}-\dfrac{6}{12}+\dfrac{4}{12}-\dfrac{3}{12}\)

\(s_4=\dfrac{7}{12}\)

The first omitted term is \(\dfrac{1}{5}\).

The error is at most \(\dfrac{1}{5}\).

Answer: \(s_4=\dfrac{7}{12}\), error at most \(\dfrac{1}{5}\)

1
How many terms of \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}\) guarantee an error smaller than \(0.01\)?
Show Answer

If the last term kept has index \(n\), the next term is \(1/(n+1)\).

\(\dfrac{1}{n+1}<0.01\)

\(n+1>100\)

The smallest integer is \(n+1=101\).

\(n\ge 100\)

Answer: \(n\ge 100\)

2
Approximate \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2}\) by its first three terms, and bound the error.
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\(s_3=1-\dfrac{1}{4}+\dfrac{1}{9}\)

\(s_3=\dfrac{36}{36}-\dfrac{9}{36}+\dfrac{4}{36}\)

\(s_3=\dfrac{31}{36}\)

The next term has absolute value \(\dfrac{1}{16}\).

The error is at most \(\dfrac{1}{16}\).

Answer: \(\dfrac{31}{36}\), error at most \(\dfrac{1}{16}\)

3
For the alternating harmonic series, \(s_2=\dfrac{1}{2}\). What error bound does the next term give?
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The next term is \(\dfrac{1}{3}\).

The error is at most \(\dfrac{1}{3}\).

Answer: \(\dfrac{1}{3}\)

4
The partial sum \(s_4=\dfrac{7}{12}\) of the alternating harmonic series stops just before a positive term. Is \(s_4\) too high or too low?
Show Answer

The next term is \(+1/5\).

Adding a positive term increases the sum.

\(s_4\) is smaller than the true sum.

Answer: too low

5
An alternating series that meets the test is approximated by stopping just before a term of absolute value \(1/8\). Which bound is justified?
A. \(1/8\)
B. \(1/16\)
C. \(1/64\)
D. no bound is available
Show Answer

The error is at most the first omitted term.

That term has absolute value \(1/8\).

Answer: A. \(1/8\)

Level 2

Medium

A Lagrange bound with a simple M, or a geometric tail.

Example
The Maclaurin polynomial of degree 3 for \(\sin x\) is used at \(x=0.5\). Bound the remainder. Use \(|f^{(4)}(z)|\le 1\).

\(\left|R_3(0.5)\right|\le \dfrac{M\,|0.5|^{4}}{4!}\)

\(M=1\) and \(4!=24\)

\(\left|R_3(0.5)\right|\le \dfrac{(1/2)^4}{24}\)

\(=\dfrac{1/16}{24}\)

\(=\dfrac{1}{384}\)

Answer: \(\dfrac{1}{384}\)

1
Bound the degree-2 Maclaurin remainder for \(e^x\) at \(x=1\). On \([0,1]\), use \(e^z\le 3\).
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\(\left|R_2(1)\right|\le \dfrac{3\cdot 1^{3}}{3!}\)

\(3!=6\)

\(\left|R_2(1)\right|\le \dfrac{3}{6}\)

\(\left|R_2(1)\right|\le \dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

2
Bound the degree-2 Maclaurin remainder for \(\cos x\) at \(x=0.2\). Use \(M=1\).
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\(\left|R_2(0.2)\right|\le \dfrac{(0.2)^3}{3!}\)

\((0.2)^3=\dfrac{1}{125}\)

\(\left|R_2(0.2)\right|\le \dfrac{1/125}{6}\)

\(=\dfrac{1}{750}\)

Answer: \(\dfrac{1}{750}\)

3
How many terms of \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^3}\) guarantee an error smaller than \(0.001\)?
Show Answer

The next term after index \(n\) is \(1/(n+1)^3\).

\(\dfrac{1}{(n+1)^3}<0.001\)

\((n+1)^3>1000\)

\(10^3=1000\), so \(n+1=10\) is not small enough.

\(n+1\ge 11\), hence \(n\ge 10\).

Answer: \(n\ge 10\)

4
The geometric series \(\sum_{n=0}^{\infty}\left(\dfrac{1}{3}\right)^n\) is approximated by the terms \(n=0,1,2\). Find the exact remainder.
Show Answer

The first omitted term is \((1/3)^3=\dfrac{1}{27}\).

\(r=\dfrac{1}{3}\)

\(R=\dfrac{\frac{1}{27}}{1-\frac{1}{3}}\)

\(R=\dfrac{\frac{1}{27}}{\frac{2}{3}}\)

\(R=\dfrac{1}{18}\)

Answer: \(\dfrac{1}{18}\)

5
Why is the alternating-series error bound not available for \(\sum 1/n^2\)?
Show Answer

Every term is positive.

The alternating error bound needs alternating signs and a decreasing absolute value.

A different remainder, such as an integral remainder, would be the right tool.

Answer: the series is not alternating

Level 3

Hard

A tighter factorial, or a remainder centered away from zero.

Example
Bound the degree-3 Maclaurin remainder for \(e^x\) at \(x=0.2\). On \([0,0.2]\), use \(M=2\).

\(\left|R_3(0.2)\right|\le \dfrac{2\cdot(0.2)^4}{4!}\)

\((0.2)^4=\dfrac{1}{625}\) and \(4!=24\)

\(\left|R_3(0.2)\right|\le \dfrac{2}{625\cdot 24}\)

\(=\dfrac{2}{15000}\)

\(=\dfrac{1}{7500}\)

Answer: \(\dfrac{1}{7500}\)

1
Bound the degree-5 Maclaurin remainder for \(\sin x\) at \(x=1/2\). Use \(M=1\).
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\(\left|R_5(1/2)\right|\le \dfrac{(1/2)^6}{6!}\)

\(6!=720\) and \((1/2)^6=\dfrac{1}{64}\)

\(\left|R_5\right|\le \dfrac{1}{64\cdot 720}\)

\(=\dfrac{1}{46080}\)

Answer: \(\dfrac{1}{46080}\)

2
The series \(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{n!}\) is approximated by the terms through \(n=3\). Bound the error.
Show Answer

The first omitted term is the \(n=4\) term.

\(\left|\dfrac{(-1)^4}{4!}\right|=\dfrac{1}{24}\)

The absolute values \(1/n!\) decrease, and the signs alternate.

The error is at most \(\dfrac{1}{24}\).

Answer: \(\dfrac{1}{24}\)

3
Bound the degree-2 Taylor remainder for \(\ln x\) centered at 1, evaluated at \(x=1.2\). On \([1,1.2]\), \(|f'''(z)|=2/z^3\le 2\).
Show Answer

\(\left|R_2(1.2)\right|\le \dfrac{2\cdot(0.2)^3}{3!}\)

\((0.2)^3=\dfrac{1}{125}\)

\(\left|R_2\right|\le \dfrac{2}{125\cdot 6}\)

\(=\dfrac{2}{750}\)

\(=\dfrac{1}{375}\)

Answer: \(\dfrac{1}{375}\)

4
The degree-1 approximation \(1+x\) is used for \(e^x\) at \(x=1/2\). Bound the remainder with \(e^z\le 2\) on \([0,1/2]\).
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\(\left|R_1(1/2)\right|\le \dfrac{2\cdot(1/2)^2}{2!}\)

\(=\dfrac{2\cdot(1/4)}{2}\)

\(=\dfrac{1}{4}\)

Answer: \(\dfrac{1}{4}\)

5
A Lagrange bound uses \(M=1\), \(|x-a|=1/2\), and \(n=3\). Write the bound and simplify.
Show Answer

\(\left|R_3\right|\le \dfrac{1\cdot(1/2)^4}{4!}\)

\(=\dfrac{1}{16\cdot 24}\)

\(=\dfrac{1}{384}\)

Answer: \(\dfrac{1}{384}\)

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