EXAMPLE IX.A.1: Suppose  ${\displaystyle f\left(x\right)=x^2+x^3 +x^{100}}$ and we wish to find ${\displaystyle f\left(0.5\right)}$.

Of course the exact answer is ${\displaystyle .25+.125+{\left(.5\right)}^{100}}$, but if we only need an estimate of the answer it is easy to settle for .375. The error we make in using this estimate is relatively small [namely ${\displaystyle {\left(.5\right)}^{100}}$]. In this example we should note two things.

• First, the computation for the estimate is theoretically easy because the function involved is a polynomial; and
• second, the estimate was found by evaluating a related polynomial, ${\displaystyle g\left(x\right)= x^2+x^3}$ .

Taylor's Theory-Objective and Key Ideas: The main concerns of Taylor's theory for estimating function values are

1. to find estimating polynomials for a given function and
2. to measure the error in using these polynomials to estimate the desired values for the given function.

Taylor's theory generalizes these two ideas that use the derivative to estimate.

To focus our discussion more specifially, in this section we will consider only  ${\displaystyle f\left(x\right)=e^x}$and begin by trying to estimate ${\displaystyle f\left(1\right)=e}$ illu strating the key ideas of Taylor's theory.
At ${\displaystyle x=0}$, we find that ${\displaystyle f\left(0\right)=1}$, ${\displaystyle f\text{'}\left(0\right)=1}$, and since ${\displaystyle f\text{'}\left(x\right)=f\left(x\right)}$ we have that ${\displaystyle f^{\left(n\right)}\left(0\right)=1}$ for all natural numbers ${\displaystyle n}$. We now consider our first result, typical of the Taylor's theory approach to estimation using polynomials and derivative information.

PROPOSITION IX.A.1: If  ${\displaystyle P_n\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{2\cdot 3}+\frac{x^4}{4!}+...+\frac{x^n}{n!}}$ then
${\displaystyle P_n\left(x\right)}$ is a polynomial of degree n so that ${\displaystyle P_n\left(0\right)=P_n\text{'}\left(0\right)=P_n\text{'}\text{'}\left(0\right)=...=P_n^{\left(n\right)}\left(0\right)=1}$ and ${\displaystyle P_n\left(b\right)}$ is approximately equal to ${\displaystyle e^b}$.
In fact, if we let ${\displaystyle R_n=e^b-P_n\left(b\right)}$, then for some ${\displaystyle c}$ between ${\displaystyle 0}$ and ${\displaystyle b}$, ${\displaystyle R_n=e^c\frac{b^{n+1}}{(n+1)!}}$ .

GeoGebra: Table, Graph, and Mapping Diagram of ${\displaystyle P_n\left(x\right)}$ and ${\displaystyle R_n\left(x\right)}$
 

Estimating e: Before we proceed to justify this result, we'll apply this result using ${\displaystyle n=5}$ to estimate the value of ${\displaystyle e[=e^1]}$.
First, ${\displaystyle P_5\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{2\cdot 3}+\frac{x^4}{4!}+\frac{x^5}{5!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}}$ , so according to the proposition, ${\displaystyle e}$ is approximately equal to ${\displaystyle P_5\left(1\right)=1+1+\frac{1^2}{2}+\frac{1^3}{6}+\frac{1^4}{24}+\frac{1^5}{120}\approx 2.716667}$ and ${\displaystyle R_5=e-P_5\left(1\right)}$ where ${\displaystyle R_5=e^c\frac{1^6}{6!}=e^c\frac{1}{720}}$ for some ${\displaystyle c}$ between ${\displaystyle 0}$ and ${\displaystyle 1}$. Thus, ${\displaystyle e}$ is approximately ${\displaystyle 2.716667}$ and the error in using this estimate is ${\displaystyle R_5}$. Since ${\displaystyle c}$ is between ${\displaystyle 0}$ and ${\displaystyle 1}$, ${\displaystyle e^c}$ is between ${\displaystyle 1}$ and ${\displaystyle e}$, so we can deduce that ${\displaystyle R_5}$, the difference between e and the estimate, ${\displaystyle P_5\left(1\right)}$, is no greater than ${\displaystyle \frac{e}{720}}$. Using estimates of ${\displaystyle e}$ from our earlier work in Chapter VI we know that so our error ${\displaystyle R_5\left(1\right)}$ is no larger than ${\displaystyle \frac{3}{720}=\frac{1}{240}\approx 0.004167}$.This compares roughly well with the error from the GeoGebra applet that shows an error after setting the slider to  find ${\displaystyle R_5\left(1\right)=0.00162}$.

A more accurate estimate can be obtained by using a larger value for ${\displaystyle n}$. Try using ${\displaystyle n=6}$ and ${\displaystyle 7}$ to see the improvement. [This can be done progressively using the GeoGebra  applet above or download a spreadsheet which you can examine now or later.]

Proof of Proposition: We'll begin our proof by finding ${\displaystyle P_n\text{'}\left(x\right)}$.

${\displaystyle P_n\text{'}\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{2\cdot 3}+\frac{x^4}{4!}+...+\frac{n{x}^{n-1}}{n!}}$
${\displaystyle =1+x+\frac{x^2}{2}+\frac{x^3}{2\cdot 3}+\frac{x^4}{4!}+...+\frac{x^{n-1}}{(n-1)!}=P_{n-1}\left(x\right)}$ 

From this it follows easily that  ${\displaystyle P_n\left(0\right)=P_n\text{'}\left(0\right)=P_n\text{'}\text{'}\left(0\right)=...=P_n^{\left(n\right)}\left(0\right)=1}$.

Now suppose that ${\displaystyle b}$ is not  ${\displaystyle 0}$ and let ${\displaystyle R_n=e^b-P_n\left(b\right)}$.
We need only justify the formula for evaluating ${\displaystyle R_n}$. For convenience we'll write ${\displaystyle R=R_n}$ and let

Furthermore ${\displaystyle g\text{'}\left(t\right)=e^t{P}_n(b-t)-e^t{P}_n\text{'}(b-t)-R\frac{(n+1){(b-t)}^n}{b^{n+1}}=e^t[P_n(b-t)-P_n\text{'}(b-t)]-R\frac{(n+1){(b-t)}^n}{b^{n+1}}}$

 But ${\displaystyle P_n\left(x\right)-P_n\text{'}\left(x\right)=\frac{x^n}{n!}}$, so${\displaystyle g\text{'}\left(t\right)=e^t\frac{{(b-t)}^n}{n!}-R\frac{(n+1){(b-t)}^n}{b^{n+1}}}$.

Now we apply the Mean Value (or Rolle's) Theorem to the function g, we can say that there is a number ${\displaystyle c}$ between ${\displaystyle 0}$ and ${\displaystyle b}$ where ${\displaystyle g\text{'}\left(c\right)=0}$. Thus 
${\displaystyle 0=e^c\frac{{(b-c)}^n}{n!}-R\frac{(n+1){(b-c)}^n}{b^{n+1}}}$and  ${\displaystyle e^c\frac{{(b-c)}^n}{n!}=R\frac{(n+1){(b-c)}^n}{b^{n+1}}}$. Solving this last equation for R gives
${\displaystyle R_n=R=e^c\frac{b^{n+1}}{(n+1)!}}$.

Note: Since the exponential function has a positive derivative for all ${\displaystyle x}$, the function is increasing for all ${\displaystyle x}$.
(i) If then and .
(iia) If ${\displaystyle 0>c>b}$ then ${\displaystyle 1>e^c>e^b>2^b}$ and if ${\displaystyle n}$ is odd then ${\displaystyle \frac{b^{n+1}}{(n+1)!}}$ is positive and thus ${\displaystyle \frac{b^{n+1}}{(n+1)!}>R_n=R=e^c\frac{b^{n+1}}{(n+1)!}>2^b\frac{b^{n+1}}{(n+1)!}>0}$.
(iib) If ${\displaystyle 0>c>b}$ then ${\displaystyle 1>e^c>e^b>2^b}$ and if ${\displaystyle n}$ is even then ${\displaystyle \frac{b^{n+1}}{(n+1)!}}$ is negative and thus .
Putting this information about the quality of ${\displaystyle R_n}$ together we can see the following:
For all ${\displaystyle x>0}$ and for any ${\displaystyle n}$, ${\displaystyle e^x>P_n\left(x\right)}$,
while for the polynomials differ in relation to ${\displaystyle e^x}$:
When ${\displaystyle n}$ is odd then ${\displaystyle e^x>P_n\left(x\right)}$;
when ${\displaystyle n}$ is even then ${\displaystyle P_n\left(x\right)>e^x}$.

Estimating the value of e is not the only use for the previous proposition. Before turning to the more general Taylor's theory in the next section, here are two more examples of its application in estimating definite integrals related to e^x.

EXAMPLE IX.A.2: Estimate ${\displaystyle {\int }_0^1{e}^{x}dx=e-1}$. Use this to estimate ${\displaystyle e}$.

Solution: Using Proposition IX.1 with ${\displaystyle n=4}$ for each x between ${\displaystyle 0}$ and ${\displaystyle 1}$ we have
where ${\displaystyle R_4\left(x\right)=e^c\frac{x^5}{120}}$
 for some ${\displaystyle c}$ with .

Thus. [Remember .] and so for all ${\displaystyle x}$ between ${\displaystyle 0}$ and ${\displaystyle 1}$. Now we use the monotone property of the definite integral [Ch. V . ** ] to obtain the estimate:
, so

Therefore ${\displaystyle e}$ is approximately ${\displaystyle 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}\approx 2.716667}$ as in the first estimate we saw in the note after the statement of the theorem, and the error in using this estimate is less than ${\displaystyle \frac{1}{240}}$ as we also saw previously.

The next example follows a similar analysis but applied to a more difficult yet important integral for probability and statistics. EXAMPLE IX.A.3: Estimate ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$. Solution: For each ${\displaystyle t}$ between ${\displaystyle 0}$ and ${\displaystyle 1}$, ${\displaystyle -t^2}$ is between ${\displaystyle -1}$ and ${\displaystyle 0}$. Let ${\displaystyle x=-t^2}$. By Proposition IX.1 with ${\displaystyle n=4}$ for each ${\displaystyle x}$ between ${\displaystyle -1}$ and ${\displaystyle 0}$ we have where ${\displaystyle R_4\left(x\right)=e^c\frac{x^5}{120}}$ for some ${\displaystyle c}$ with ${\displaystyle 0>c>x}$. Since , , and since we have that  ${\displaystyle 0>R_4\left(x\right)>\frac{x^5}{120}}$.
GeoGebra Estimates for  ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$. Check off box to show estimate.

Substituting ${\displaystyle -t^2}$ for ${\displaystyle x}$ ,we see that

Now since ${\displaystyle 0>R_4(-t^2)>\frac{{(-t^2)}^5}{120}=\frac{-t^{10}}{120}}$ we have ${\displaystyle {\int }_0^1{P}_4(-t^2)dt>{\int }_0^1{e}^{-t^2}dt>{\int }_0^1{P}_4(-t^2)-\frac{t^{10}}{120}dt}$.

By evaluating these integrals we obtain

Therefore, ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$ is approximately equal to ${\displaystyle 1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}\approx 0.747486...}$;

and this is an overestimate by no more than ${\displaystyle \frac{1}{1320}}$.

Comment: In both of these examples we have been able to estimate a definite integral involving the exponential function by using a Taylor polynomial of degree 4. It should be apparent that by using a higher degree for the estimating polynomial, the error term will become smaller and we will obtain a more precise estimate. The systematic pattern in these polynomials should allow you to find more precise estimates for the last example without much difficulty.

Go on to Chapter IX.B

1. Use the Taylor polyonmial for ${\displaystyle e^x}$ of degree ${\displaystyle 4}$ to estimate the following:

(a) ${\displaystyle e^2}$ (b) ${\displaystyle e^3}$ (c) ${\displaystyle e^{.5}}$ (d) ${\displaystyle e^{-1}}$ (e) ${\displaystyle e^{3.14}}$. [Use GeoGebra or  Spreadsheet helper.]
2. Estimate e using the Taylor polynomial of degree n where n is (a) 6 (b) 7 (c) 8 (d) 10.

In each of these estimates discuss the size of the error term R_n.  [Use GeoGebra or  Spreadsheet helper.]
3. What value of n should be used so that the Taylor polynomial of degree n will give an estimate of e that is within .000001 of the exact value of e? Explain your result using Proposition XI.A.1
   
4. Use the Taylor polynomial for e^x of degree 5 to estimate ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$. Discuss the error in this approximation.
5. Use the trapezoidal rule and Simpson's rule with n = 6 to estimate ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$ .
6. Use the Taylor polynomial for ${\displaystyle e^x}$ of degree 6 to estimate ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$. Discuss the error in this approximation.
7. On the same graph sketch the graph of ${\displaystyle e^x}$ along with those for the Taylor polynomials for ${\displaystyle e^x}$ of degree 1,2,3,4 and 5. [You can do this by use the trace feature on the graph of ${\displaystyle P_n}$ in the GeoGebra applet.] Discuss the graphical interpretation of ${\displaystyle R_n}$. [Here is Java sketch solution for n = 1 to 5]
8. What value of n should be used so that the Taylor polynomial of degree n will give an estimate of  ${\displaystyle {\int }_0^1{e}^{-t^2}dt}$ that is within ${\displaystyle 0.000001}$ of the exact value? Explain your result.
9. Use the Taylor polynomial for ${\displaystyle e^x}$ of degree 4 to estimate the area of the region in the plane contained by the lines ${\displaystyle X=0}$, ${\displaystyle X=1}$, the X- axis and the graph of ${\displaystyle y=1+x{e}^{-x}}$ . [Hint: First find a polynomial to estimate ${\displaystyle e^{-x}}$ .] Discuss the error in this approximation.
10. Another Polynomial Estimate: Consider the functions ${\displaystyle \frac{1}{1-x}}$ and ${\displaystyle \frac{1}{1+x}}$.
1. Show that ${\displaystyle 1=(1-x)(1+x+x^2+x^3+x^4+x^5)+R}$ where ${\displaystyle R=x^6}$ .
2. Show that ${\displaystyle 1=(1+x)(1-x+x^2-x^3+x^4-x^5)+R}$ where ${\displaystyle R=x^6}$ .
3. Show that when where ${\displaystyle R_1=\frac{x^6}{1-x}}$
4. Show that when where ${\displaystyle R_2=\frac{x^6}{1+x}}$.
5. Use the definite integral and the previous equations to estimate ln(.9) = ln (1 - .1) and ln(1.1) = ln(1 + .1). [ This approach to estimating the natural logarithm was used by Isaac Newton to give very accurate estimates of ln(2), ln(3), etc.]. Discuss the error in your estimate based on the integrals of R₁ and R₂.
6. Generalize this for higher degree polynomials and estimating ln(.8), ln(1.2), ln(.99) and ln(1.01).