Taylor Series - Error Bounds

TheLagrange error boundof aTaylor polynomialgives theworst-case scenariofor the difference between the estimated value of the function as provided by the Taylor polynomial and the actual value of the function. This error bound $\big(R_n(x)\big)$is the maximum value of the $(n+1)^\text{th}$term of the Taylor expansion, where $M$is an upper bound of the $(n+1)^\text{th}$derivative for $a$

$R_n(x)=\frac{M}{(n+1)!}(x-a)^{n+1}.$

The $n^\text{th}$degreeTaylor polynomialat $x=a$is

$P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n.$

Since the Taylor approximation becomes more accurate as more terms are included, the $P_{n+1}(x)$polynomial must bemore accuratethan $P_n(x):$

$\begin{aligned} P_{n+1}(x) &= f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}\\ &= P_n(x) + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. \end{aligned}$

Since the difference between $P_n(x)$and $P_{n+1}(x)$is just that last term, the error of $P_n(x)$can be no larger than that term. In other words, the error $R_n$is

$R_n(x) = \max\left( \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} \right).$

Since $a$and $n$are constant in this formula, terms depending only on those constants and $x$are unaffected by the $\max$operator and can be pulled outside:

$R_n(x) =\frac{\max\big( f^{(n+1)}(a)\big)}{(n+1)!} (x-a)^{n+1}.$

The largest value obtainable by $f^{n+1}$could not possibly exceed the maximum value of that derivative between $a$and $x.$Call the $x$value that provides that maximum value $z$and the error becomes

$R_n(x)=\frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}.$

Let $M$be an upper bound on the $(n+1)^\text{th}$derivative of $f(x)$for the interval between $a$and $x$such that

$\big| f^{(n+1)}(z) \big| \leq M$

for all $z\in [a, x].$

The upper bound of the $(n+1)^\text{th}$derivative on the interval $[a, x]$will usually occur at $z=a$or $z=x.$If given a defined interval on which to find the error, test the endpoints of the interval.

What is the upper bound of the third derivative of $y = \sin(x)$on the interval $[0, 2\pi]?$

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The third derivative of $y=\sin(x)$is $y^{(3)} = -\cos(x),$which oscillates between -1 and 1. So $M=1$and

$\big| f^{(n+1)}(z) \big| \leq 1.\ _\square$

In order to compute the error bound, follow these steps:

• Step 1:Compute the $(n+1)^\text{th}$derivative of $f(x).$
• Step 2:Find the upper bound on $f^{(n+1)}(z)$for $z\in [a, x].$
• Step 3:Compute $R_n(x).$

Find the error bound of theMaclaurin polynomial $P_3\big(\frac{\pi}{2}\big)$for $f(x) = \sin(x).$

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The Maclaurin series is just a Taylor series centered at $a=0.$Follow the prescribed steps.

• Step 1:Compute the $(n+1)^\text{th}$derivative of $f(x):$
  Since $P_3$is being investigated, $n = 3,$so write down the $4^\text{th}$derivative of $f(x) = \sin(x):$ $f^{(4)}(x) = \sin(x).$

• Step 2:Find the upper bound on $f^{(n+1)}(z)$for $z\in [a, x]:$
  The Maclaurin series is centered on $a = 0$and $P_n(x)=P_3\big(\frac{\pi}{2}\big)$implies $x = \frac{\pi}{2}:$ $\begin{aligned} f^{(4)}(0) &= \sin(0)=0\\ f^{(4)}\left(\frac{\pi}{2}\right) &= \sin\left(\frac{\pi}{2}\right)=1. \end{aligned}$So $M=1.$

• Step 3:Compute $R_n(x):$
  $\{对齐}开始R_n (x) & = \压裂{M} {(n + 1) !} (x) ^ {n + 1}\\ R_3\left(\frac{\pi}{2}\right)&=\frac{1}{(3+1)!}\left(\frac{\pi}{2}-0\right)^{3+1}\\ &= \frac{\pi^4}{384}.\ _\square \end{aligned}$