9.6 Error in Taylor polynomials

$R(10)=f(10)-F(10)=-3-5$

(a) False (b) True (c) True (d) True

From the given information,

So, (a) is false (since 8 is not less than or equal to 7), while (b), (c), and (d) are true.

Remark: $R(x)$ is the error in our approximation. As mentioned in the text, we almost never know $R$ exactly, but we can give a bound. We don't need the tightest bound–just a reasonable one that is easy to calculate. If we were dealing with real functions and approximations, we might not know that $|R(10)|=8$, but if we knew it was at most 9, that would be a pretty decent approximation.

Often in this section, we will make simplifying assumptions to get a bound that is easy to calculate. But, don't go overboard! It is a true statement to say that our absolute error is at most 100, but this statement would probably not be very helpful as a bound.

Equation 9.6.5 tells us

for some $c$ strictly between 0 and 2.

Equation 9.6.5 gives us the bound $|f(2)-T_3(2)|<6$. A calculator tells us actually $|f(2)-T_3(2)|\approx 1.056$.

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=3$, $a=0$, $x=2$, and $f^{(4)}(c)=e^c$, so

Since $c$ is strictly between 0 and 2, $e^c < e^2$:

but this isn't a number we really know. Indeed: $e^2$ is the very number we're trying to approximate. So, we use the estimation $e < 3$:

We conclude that the error $|f(2)-T_3(2)|$ is less than 6.

Now we'll get a more exact idea of the error using a calculator. (Calculators will also only give approximations of numbers like $e$, but they are generally very good approximations.)

So, our actual answer was only off by about 1.

Remark: $1<6$, so this does not in any way contradict our bound $|f(2)-T_3(2)|<6$.

You are approximating a third-degree polynomial with a fifth-degree Taylor polynomial. You should be able to tell how good your approximation will be without a long calculation.

$|f(37)-T(37)|=0$

Whenever you approximate a polynomial with a Taylor polynomial of greater or equal degree, your Taylor polynomial is exactly the same as the function you are approximating. So, the error is zero.

Draw a picture–it should be clear how the two approximations behave.

You do, you clever goose!

The constant approximation gives

while the linear approximation gives

Since $-1 \leq \sin(33)\leq 1$, the constant approximation is better. (But both are a little silly.)

In this case, Equation 9.6.5 tells us that

for some $c$ strictly between 11 and 11.5.

$|f(11.5)-T_5(11.5)|<\dfrac{9}{7\cdot 2^6}<0.02$

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=5$, $a=11$, $x=11.5$, and $f^{(6)}(c)=\dfrac{6!(2c-5)}{c+3}$.

for some $c$ in $(11,11.5)$. We don't know exactly which $c$ this is true for, but since we know that $c$ lies in $(11,11.5)$, we can provide bounds.

With this bound, we see

Our error is less than 0.02.

In this case, Equation 9.6.5 tells us that $\left|f(0.1)-T_2(0.1)\right| = \left|\dfrac{f^{(3)}(c)}{3!}(0.1-0)^{3}\right|$ for some $c$ strictly between 0 and 0.1.

$\left|f(0.1)-T_2(0.1)\right| < \dfrac{1}{1125}$

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=2$, $a=0$, and $x=0.1$, so

for some $c$ in $(0,0.1)$.

We will find $f'''(x)$, and use it to give an upper bound for

when $c$ is in $(0,0.1)$.

When $0 < c < \dfrac{1}{10}$, also $0 < c < \dfrac{\pi}{6}$, so:

With these bounds in mind for secant and tangent, we return to the expression we found for our error.

The error is less than $\dfrac{1}{1125}$.

In our case, Equation 9.6.5 tells us

for some $c$ between $-\dfrac{1}{4}$ and 0.

$\left|f\left(-\dfrac{1}{4}\right)-T_5\left(-\dfrac{1}{4}\right)\right| < \dfrac{1}{6\cdot 4^6}<0.00004$

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=5$, $a=0$, and $x=-\dfrac{1}{4}$, so

for some $c$ in $\left(-\frac{1}{4},0\right)$. We'll need to know the sixth derivative of $f(x)$.

Plugging in $\left|f^{(6)}(c)\right|=\dfrac{5!}{(1-c)^{6}}$:

for some $c$ in $\left(-\frac{1}{4},0\right)$.

We're interested in an upper bound for the error: we want to know the worst case scenario, so we can say that the error is no worse than that. We need to know what the biggest possible value of $\dfrac{1}{6\cdot 4^6 \cdot (1-c)^6}$ is, given $-\dfrac{1}{4}<c<0$. That means we want to know the biggest possible value of $\dfrac{1}{(1-c)^6}$. This corresponds to the smallest possible value of $(1-c)^6$, which in turn corresponds to the smallest absolute value of $1-c$.

Finally, we conclude

In this case, Equation 9.6.5 tells us that $\left|f(30)-T_3(30)\right| = \left|\dfrac{f^{(4)}(c)}{4!}(30-32)^{4}\right|$ for some $c$ strictly between 30 and 32.

Your answer may vary. One reasonable answer is
$\left|f(30)-T_3(30)\right|<\dfrac{14}{5^7\cdot9\cdot15}<0.000002$. Another reasonable answer is
$\left|f(30)-T_3(30)\right|<\dfrac{14}{5^7\cdot9}<0.00002$.

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=3$, $a=30$, and $x=32$, so

for some $c$ in $(30,32)$.

We will now find $f^{(4)}(x)$. Then we can give an upper bound on $|f(30)-T_3(30)| =\dfrac{2}{3}\left|f^{(4)}(c)\right|$ when $c \in (30,32)$.

Using this,

Since $30 < c< 32$,

This isn't a number we know. We're trying to find the error in our estimation of $\sqrt[5]{30}$, but $\sqrt[5]{30}$ shows up in our error. From here, we have to be a little creative to get a bound that actually makes sense to us. There are different ways to go about it. You could simply use $30^{\tfrac{4}{5}}>1$. We will be a little more careful, and use the following estimation:

We conclude $\left|f(30)-T_3(30)\right|<0.000002$.

In our case, Equation 9.6.5 tells us
$\left|f\left(0.01\right)-T_1\left(0.01\right)\right| = \left|\dfrac{f^{(2)}(c)}{2!}\left(0.01-\frac{1}{\pi}\right)^2\right|$ for some $c$ between 0.01 and $\dfrac{1}{\pi}$.

Equation 9.6.5 gives the bound $|f(0.01)-T_n(0.01)| \leq 100^2\left(\frac{100}{\pi}-1\right)^2$.

A more reasonable bound on the error is that it is less than 5.

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=1$, $a=\dfrac{1}{\pi}$, and $x=0.01$, so

for some $c$ in $\left(\frac{1}{100},\frac{1}{\pi}\right)$.

Let's find $f''(x)$.

Now we can plug in a better expression for $f''(c)$:

for some $c$ in $\left(\frac{1}{100},\frac{1}{\pi}\right)$.

What we want to do now is find an upper bound on this expression containing $c$,
$\dfrac{1}{2}\left(\dfrac{100-\pi}{100\pi}\right)^2\cdot \dfrac{\left|2c\cos\left(\frac{1}{c}\right)-\sin\left(\frac{1}{c}\right)\right|}{c^4}$ .

Equation 9.6.5 gives the bound $|f(0.01)-T_1(0.01)| \leq \frac{3\cdot 100^2}{2}\left(\frac{100}{\pi}-1\right)^2$.

The bound above works out to approximately fourteen million. One way to understand why the bound is so high is that $\sin\left(\frac{1}{x}\right)$ moves about crazily when $x$ is near zero–it moves up and down incredibly fast, so a straight line isn't going to approximate it very well at all.

That being said, because $\sin\left(\frac{1}{x}\right)$ is still “sine of something," we know $-1 \leq f\left(0.01\right) \leq 1$. To get a better bound on the error, let's find $T_1(x)$.

Now that we know $T_1(0.01)$, and we know $-1 \leq f(0.01) \leq 1$, we can give the bound

A more reasonable bound on the error is that it is less than 5.

Still more reasonably, we would not use $T_1(x)$ to evaluate $\sin(100)$ approximately. We would write $\sin(100) = \sin(100-32\pi)$ and approximate the right hand side, which is roughly $\sin(-\pi/6)$.

Using Equation 9.6.5, $\left|f\left(\dfrac{1}{2}\right)-T_2\left(\dfrac{1}{2}\right)\right| = \left|\dfrac{f^{(3)}(c)}{3!}\left(\dfrac{1}{2}-0\right)^{3}\right|$ for some $c$ in $\left(0,\dfrac{1}{2}\right)$.

Using Equation 9.6.5,

The actual error is

which is about 0.02.

Equation 9.6.5 tells us that, when $T_n(x)$ is the $n$th degree Taylor polynomial for a function $f(x)$ about $x=a$, then

for some $c$ strictly between $x$ and $a$. In our case, $n=2$, $a=0$, and $x=\dfrac{1}{2}$, so

for some $c$ in $\left(0,\frac{1}{2}\right)$.
The next task that suggests itself is finding $f^{(3)}(x)$.

Since $\left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right| =\dfrac{\left|f^{(3)}(c)\right|}{3!\cdot2^3}$ for some $c$ in $\left(0,\frac{1}{2}\right)$,

for some $c$ in $\left(0,\frac{1}{2}\right)$.

We want to know what is the worst case scenario-what's the biggest this expression can be. So, now we find an upper bound on $\dfrac{1+2c^2}{48\left(\sqrt{1-c^2}\right)^5}$ when $0 \leq c \leq \dfrac{1}{2}$. Remember that our bound doesn't have to be exact, but it should be relatively easy to calculate.