\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Infinite Series - Remainder and Error Bounds for Power and Taylor Series

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Once you have found the infinite Taylor Series, you will often be asked to approximate the original function with a finite number of terms from the Taylor Series. Doing so introduces error since the finite Taylor Series does not exactly represent the original function. To handle this error we write the function like this.

\(\displaystyle{ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + . . . + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) }\)

where \(R_n(x)\) is the remainder. Notice we are cutting off the series after the n-th derivative and \(R_n(x)\) represents the rest of the series.

Lagrange's formula for this remainder term is \[ R_n(x) = \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!} \] This looks very similar to the equation for the Taylor series terms . . . and it is, except for one important item. Notice that in the numerator, we evaluate the \(n+1\) derivative at \(z\) instead of \(a\). So, what is the value of \(z\)? \(z\) takes on a value between \(a\) and \(x\), but, and here's the key, we don't know exactly what that value is. So this remainder can never be calculated exactly. However, since we know that \(z\) is between \(a\) and \(x\), we can determine an upper bound on the remainder and be confident that the remainder will never exceed this upper bound. So how do we do that?

Upper Bound on the Remainder (Error)

We usually consider the absolute value of the remainder term \(R_n\) and call it the upper bound on the error, also called Taylor's Inequality. \[ \abs{R_n(x)} \leq \frac{\abs{f^{(n+1)}(z)(x-a)^{n+1}} }{(n+1)!} \]

How To Choose z

To find an upper bound on this error, we choose the value of \(z\) using these rules.
1. If the \(n+1\) derivative contains a sine or cosine term, we replace the sine or cosine term with one, since the maximum value of sine or cosine is one. This seems somewhat arbitrary but most calculus books do this even though this could give a much larger upper bound than could be calculated using the next rule. [ As usual, check with your instructor to see what they expect. ]
2. If we do not have a sine or cosine term, we calculate \(\abs{f^{(n+1)}(z)}\) and then choose the value of \(z\) between \(a\) and the \(x\)-value that we are estimating that makes this term a maximum. Sometimes, we need to find the critical points and find the one that is a maximum. Since we have a closed interval, either \([a,x]\) or \([x,a]\), we also have to consider the end points. Many times, the maximum will occur at one of the end points, but not always.

Okay, so what is the point of calculating the error bound? The point is that once we have calculated an upper bound on the error, we know that at all points in the interval of convergence, the truncated Taylor series will always be within \(\abs{R_n(x)}\) of the original function \(f(x)\). This \(\abs{R_n(x)}\) is a mathematical 'nearness' number that we can use to determine the number of terms we need to have for a Taylor series.

Here is a great video clip explaining the remainder and error bound on a Taylor series.

Dr Chris Tisdell - What is a Taylor polynomial? [17min-25secs]

video by Dr Chris Tisdell

Practice

For the questions related to finding an upper bound on the error, there are many (in fact, infinite) correct answers. However, for these problems, use the techniques above for choosing \(z\), unless otherwise instructed. Give all answers in exact form, if possible.

Basic

Find the first order Taylor polynomial for \( f(x) = \sqrt{1+x^2} \) about \(x=1\) and write an expression for the remainder.

Problem Statement

Find the first order Taylor polynomial for \( f(x) = \sqrt{1+x^2} \) about \(x=1\) and write an expression for the remainder.

Solution

383 video

video by PatrickJMT

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Find the first order Taylor polynomial for \( f(x) = \sqrt{1+x^2} \) about \(x=1\) and write an expression for the remainder.

Problem Statement

Find the first order Taylor polynomial for \( f(x) = \sqrt{1+x^2} \) about \(x=1\) and write an expression for the remainder.

Solution

384 video

video by PatrickJMT

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Intermediate

Show that \(\displaystyle{ \cos(x) = \sum_{n=0}^{\infty}{ (-1)^n\frac{x^{2n}}{(2n)!} } }\) holds for all \(x\).

Problem Statement

Show that \(\displaystyle{ \cos(x) = \sum_{n=0}^{\infty}{ (-1)^n\frac{x^{2n}}{(2n)!} } }\) holds for all \(x\).

Solution

355 video

video by PatrickJMT

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For \(\displaystyle{ f(x) = x^{2/3} }\) and \(a=1\); a) Find the third degree Taylor polynomial.; b) Use Taylors Inequality to estimate the accuracy of the approximation for \( 0.8 \leq x \leq 1.2 \).

Problem Statement

For \(\displaystyle{ f(x) = x^{2/3} }\) and \(a=1\); a) Find the third degree Taylor polynomial.; b) Use Taylors Inequality to estimate the accuracy of the approximation for \( 0.8 \leq x \leq 1.2 \).

Solution

356 video

video by PatrickJMT

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Use the 2nd order Maclaurin polynomial of \( e^x \) to estimate \( e^{0.3} \) and find an upper bound on the error.

Problem Statement

Use the 2nd order Maclaurin polynomial of \( e^x \) to estimate \( e^{0.3} \) and find an upper bound on the error.

Solution

385 video

video by PatrickJMT

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Determine an upper bound on the error for a 4th degree Maclaurin polynomial of \( f(x) = \cos(x) \) at \( \cos(0.1) \).

Problem Statement

Determine an upper bound on the error for a 4th degree Maclaurin polynomial of \( f(x) = \cos(x) \) at \( \cos(0.1) \).

Solution

386 video

video by MIP4U

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Determine the error in estimating \( e^{0.5} \) when using the 3rd degree Maclaurin polynomial.

Problem Statement

Determine the error in estimating \( e^{0.5} \) when using the 3rd degree Maclaurin polynomial.

Solution

387 video

video by MIP4U

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Estimate the remainder of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n^4+1}} } }\) using the first 10 terms.

Problem Statement

Estimate the remainder of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n^4+1}} } }\) using the first 10 terms.

Solution

1481 video

video by Krista King Math

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Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

For the questions related to finding an upper bound on the error, there are many (in fact, infinite) correct answers. However, for these problems, use the techniques above for choosing \(z\), unless otherwise instructed. Give all answers in exact form, if possible.

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