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Active Calculus

Section 8.2 Taylor Polynomials

In Activity 8.1.3, we investigated a degree 3 polynomial approximation T3(x) to f(x)=ex near a=0. To do so, we chose the conditions T3(0)=f(0), T3(0)=f(0), T3(0)=f(0), and T3(0)=f(0). Starting with
 1 
We used the notation k0,,k3 for the constants in our earlier work; here we will use c0,,c3.
T3(x)=c0+c1x+c2x2+c3x3 and f(x)=ex and finding the needed derivatives and evaluating them at a=0 led us to the results in Table 8.2.1.
Table 8.2.1. Formulas and values for f(x)=ex and T3(x)=c0+c1x+c2x2+c3x3 and their derivatives.
f(x)= ex T3(x)= c0+c1x+c2x2+c3x3
f(x)= ex T3(x)= c1+2c2x+3c3x2
f(x)= ex T3(x)= 2c2+6c3x
f(x)= ex T3(x)= 6c3
f(0)= 1 T3(0)= c0
f(0)= 1 T3(0)= c1
f(0)= 1 T3(0)= 2c2
f(0)= 1 T3(0)= 6c3
Equating the function and derivative values of f and T at a=0, we found that
c0=1,c1=1,2c2=1,6c3=1
and therefore
c0=1,c1=1,c2=12,c3=16
so that
T3(x)=1+x+12x2+16x3.
Moreover, plotting f and T3 near a=0, we see in the Figure 8.2.2 that the interval of accuracy for a tolerance of 0.1 is about 1.2x1.2.
Figure 8.2.2. The function f(x)=ex and its degree 3 Taylor approximation T3(x)=1+x+12x2+16x3 near the point (0,f(0)).
More precisely, we might compute that |f(1.16)T3(1.16)|0.0608 and |f(1.16)T3(1.16)|0.0970, which helps us see the degree 3 approximation is quite accurate on an interval wider than |x|<1.
In this section, we will learn how we can more systematically find degree n approximations for functions such as f(x)=ex that have at least n derivatives, as well as how to center the approximation at a value other than a=0.

Preview Activity 8.2.1.

PTX:ERROR: WeBWorK problem webwork-PA-8a-2 with seed 357 is either empty or failed to compile Use -a to halt with full PG and returned content

Subsection 8.2.1 Taylor polynomials

In our work so far in Chapter 8, we have found several different approximations of two important functions: ex and sin(x). In Section 8.1, we saw that near a=0
  • ex1+x (the degree 1 approximation);
  • ex1+x+12x2 (the degree 2 approximation);
  • ex1+x+12x2+16x3 (the degree 3 approximation); and
  • ex1+x+12x2+16x3+124x4 (the degree 4 approximation).
Moreover, we found that as the degree of the approximation increased, the polynomial approximation got more accurate by being closer to ex at each fixed value of x as well as on a wider interval.
In Preview Activity 8.2.1, we saw further that
  • sin(x)x (the degree 1 approximation); and
  • sin(x)x16x3 (the degree 3 approximation).
We can also notice that the linear approximation is embedded within the quadratic approximation, and the quadratic approximation within the cubic one, and so on. These observations naturally lead us to consider approximations of arbitrary degree n in order to generate more accurate approximations for any function with a sufficient number of derivatives. We thus define the Taylor polynomial of degree n centered at a=0.

Definition 8.2.3.

Let n be a natural number and let f be a function with at least n derivatives at a=0. The degree n Taylor polynomial of f centered at a=0 is the function
Tn(x)=c0+c1x+c2x2++cnxn
that satisfies Tn(0)=f(0), Tn(0)=f(0), Tn(0)=f(0), , Tn(n)(0)=f(n)(0).
By definition, Tn is the polynomial whose function and first n derivative values at a=0 match the function value and all n derivative values of f at a=0.
The n+1 equations
Tn(0)=f(0),Tn(0)=f(0),Tn(0)=f(0),,Tn(n)(0)=f(n)(0)
enable us to determine the coefficients c0,c1,,cn in terms of the values of the various derivatives of f. First, we take n derivatives of Tn(x), and assemble those in Table 8.2.4. As we do so, we choose not to combine products of numbers that arise in order to see certain patterns in the coefficients.
Table 8.2.4. The first n derivatives of a degree n Taylor polynomial.
Tn(x)= c0+c1x+c2x2+c3x3+c4x4++cnxn
Tn(x)= c1+2c2x+3c3x2+4c4x3++ncnxn1
Tn(x)= 2c2+32c3x+43c4x2++n(n1)cnxn2
Tn(x)= 321c3+432c4x++n(n1)(n2)cnxn3
Tn(4)(x)= 4321c4++n(n1)(n2)(n3)cnxn4
Tn(n)(x)= n(n1)(n2)(n3)321cn
Next, we (1) evaluate each of the derivatives at a=0, and then (2) set the result equal to the corresponding derivative value of f evaluated at a=0, which ultimately enables us to determine the coefficients c0,c1,,cn. These two steps are summarized in Table 8.2.5.
Table 8.2.5. Using the defining properties of a degree n Taylor polynomial to find equations involving the coefficients ck.
Tn(k)(0) what follows from Tn(k)(0)=f(k)(0)
Tn(0)=c0 c0=f(0)
Tn(0)=c1 c1=f(0)
Tn(0)=2c2 2c2=f(0)
T(0)=321c3 321c3=f(0)
T(4)(0)=4321c4 4321c4=f(4)(0)
T(n)(0)=n(n1)321cn n(n1)321cn=f(n)(0)
We see a natural pattern that results from taking the kth derivative of a degree k term. For example, the repeated derivatives of x4 are 4x3, 43x2, 432x, and finally 4321. By the time we get to the fourth derivative of x4, only a constant remains, and that constant is the factorial
 2 
For any positive whole number n, its factorial, n! is the product of all of the positive whole numbers less than or equal to n: n!=n(n1)(n2)321.
d4dx4[x4]=4321=4!
From the rightmost column of Table 8.2.5, we now see how the values of c0,c1,,cn are determined by the values of the various derivatives evaluated at a=0, each scaled by a corresponding factorial. In particular, solving each equation in the rightmost column of Table 8.2.5 for ck, we see that
c0=f(0),c1=f(0),c2=f(0)2!,c3=f(0)3!,c4=f(4)(0)4!,,cn=f(n)(0)n!
This enables us to find the Taylor polynomial of degree n for any function f by finding the values of f(0),f(0),,f(n)(0) and using these numbers to determine c0,c1,,cn. We summarize our recent work as follows.

Finding the degree n Taylor polynomial of f centered at a=0.

If f is a function with at least n derivatives at a=0, then the degree n Taylor polynomial of f centered at a=0, Tn(x), is
Tn(x)=c0+c1x+c2x2++cnxn
where each coefficient ck is given by
ck=f(k)(0)k!.
It is important to remember that the Taylor polynomial then provides us with an approximation of f near a=0. In particular,
f(x)Tn(x)=f(0)+f(0)x+f(0)2!x2++f(n)(0)n!xn.

Example 8.2.6. The degree 5 Taylor polynomial of f(x)=ln(1+x).

Find the degree 5 Taylor polynomial centered at a=0 for the function f(x)=ln(1+x).
Solution.
To find the degree 5 Taylor polynomial, we need to compute f(0), f(0), f(0), , f(5)(0), so we first find the first through fifth derivatives of f in the left column of Table 8.2.7, and then evaluate those derivatives at a=0 in the right column.
Table 8.2.7. Finding the derivatives of f(x)=ln(1+x) at a=0.
f(x)=ln(1+x) f(0)=ln(1)=0
f(x)=11+x=(1+x)1 f(0)=(1)1=1
f(x)=(1)(1+x)2 f(0)=(1)(1)2=1
f(x)=(2)(1)(1+x)3 f(0)=(2)(1)(1)3=(2)(1)
f(4)(x)=(3)(2)(1)(1+x)4 f(4)(0)=(3)(2)(1)
f(5)(x)=(4)(3)(2)(1)(1+x)5 f(5)(0)=(4)(3)(2)(1)
When finding the coefficients of a Taylor polynomial, it is often helpful to not combine products such as (3)(2)(1) and (4)(3)(2)(1) into a single number, in order to better observe patterns; indeed, by not combining the constants that arise in higher derivatives of f(x)=ln(1+x), we see patterns of alternating signs and factorials that arise. From the right column of Table 8.2.7 and the fact that ck=f(k)(0)k!, we see that
  • c0=f(0)=0
  • c1=f(0)=1
  • c2=12!f(0)=12!=12
  • c3=13!f(0)=(2)(1)3!=13
  • c4=14!f(4)(0)=(3)(2)(1)4!=14
  • c5=15!f(5)(0)=(4)(3)(2)(1)5!=15
so that the degree 5 Taylor approximation of f(x)=ln(1+x) at a=0 is
T5(x)=1x12x2+13x314x4+15x5.
Thus, we have found the approximation
ln(1+x)1x12x2+13x314x4+15x5,
and plotting T5(x) along with f(x) and T1(x) in Figure 8.2.8, we see how much better the degree 5 approximation is than the tangent line approximation.
Figure 8.2.8. The function f(x)=ln(1+x) and its degree 5 Taylor approximation T5(x)=x12x2+13x314x4+15x5 near the point (0,f(0)), along with T1(x)=x.
From our work in Example 8.2.6, we can also see a pattern that arises and thus we expect the general degree n Taylor polynomial centered at a=0 for f(x)=ln(1+x) to be
Tn(x)=x12x2+13x3+(1)n+11nxn.
For many familiar functions, a pattern emerges in their derivatives that enables us to find the general form of the degree n Taylor polynomial.

Activity 8.2.2.

Let f(x)=cos(x). Through the questions that follow, we seek to find the degree n Taylor polynomial for f(x) centered at a=0.
  1. Determine the first 8 derivatives of f(x)=cos(x) and evaluate each at a=0. Summarize your work by filling in all the blanks in Table 8.2.9.
    Table 8.2.9. Finding the derivatives of f(x)=cos(x) at a=0.
    f(x)= cos(x) f(0)= cos(0)=1
    f(x)= f(0)=
    f(x)= f(0)=
    f(x)= f(0)=
    f(4)(x)= f(4)(0)=
    f(5)(x)= f(5)(0)=
    f(6)(x)= f(6)(0)=
    f(7)(x)= f(7)(0)=
    f(8)(x)= f(8)(0)=
  2. Use your work in (a) along with the fact that the coefficients of the Taylor polynomial are determined by ck=f(k)(0)k! to find T8(x).
  3. Based on the pattern you observe in Table 8.2.9, what do you expect to be the formula for T10(x)?
  4. In Figure 8.2.10, we see f(x)=cos(x) and T2(x) plotted on the same axes. Add T4(x) and T6(x) to the figure by plotting those two functions on along with f(x) and T2(x). What do you notice?
    Figure 8.2.10. The function f(x)=cos(x) and its degree 2 Taylor approximation T2(x)=112x2 near the point (0,f(0)).
  5. Build a spreadsheet similar to the one in Table 8.1.12 and Table 8.1.13 from Activity 8.1.4, but do so using Δx=0.2, a start value of x=2, and the functions f(x)=cos(x), T2(x), T4(x), and T6(x). The first six columns of your spreadsheet should begin as shown in Table 8.2.11,
    Table 8.2.11. Comparing f(x)=cos(x) and its degree 2, 4, and 6 approximations near a=0.
    Δx x f(x) T2(x) T4(x) T6(x)
    0.2 2.0 0.41615 1.00000 0.33333 0.42222
    0.2 1.8 0.22720 0.62000 0.18260 0.22984
    and the last three columns of your spreadsheet should begin as follows:
    Table 8.2.12. The absolute error between f(x)=cos(x) and its degree 2, 4, and 6 approximations.
    |f(x)T2(x)| |f(x)T4(x)| |f(x)T6(x)|
    0.58385 0.08281 0.00608
    0.39280 0.04460 0.00263
  6. For about what interval of x-values is it true that |f(x)T2(x)|<0.1? How does the interval of x-values change if we instead consider where |f(x)T4(x)|<0.1? |f(x)T6(x)|<0.1?
In Activity 8.2.2, we found that the degree 6 Taylor approximation of the cosine function at a=0 is
T6(x)=112!x2+14!x416!x6,
cos(x)112!x2+14!x416!x6.
We can do similar work to extend what we found in Preview Activity 8.2.1 and observe that
sin(x)T7(x)=x13!x3+15!x517!x7,
using the degree 7 Taylor approximation of the sine function at a=0. Furthermore, as we saw in Section 8.1, the degree 4 Taylor approximation of ex provides us with the estimate
exT4(x)=1+x+12!x2+13!x3+14!x4.
Thus, we now have higher degree Taylor approximations for ex, sin(x), and cos(x) that exhibit interesting patterns in their coefficients that we can use to easily find higher degree approximations that are even more accurate. Indeed, these approximations are what computational devices use to find numerical estimates for quantities such as e2, sin(1), and cos(1.2). For example,
sin(1)T7(1)=113!13+15!1517!17=424150400.84146825
which is a remarkably accurate estimate of sin(1)=0.84147098 given that the estimate only involves the sum of four rational numbers.

Subsection 8.2.2 Taylor polynomial approximations centered at an arbitrary value a

In all of our work so far in Chapter 8, we have focused on approximating functions such as f(x)=ex, f(x)=sin(x), f(x)=ln(1+x), and f(x)=cos(x) near a=0. But we could instead be interested in the behavior of some function f near a=5, or be interested in a function f that wasn’t even defined at a=0. Thus, we next consider how we can generalize our earlier work to Taylor polynomial approximations centered at any value a.
From our early studies in Section 1.8, we know that at any input value x=a where a function f has a first derivative, f has a tangent line approximation
L(x)=f(a)+f(a)(xa)
that satisfies f(x)L(x) for x values near a. Provided that f has a second derivative at x=a, we can build a quadratic approximation near a for f, similar to the one we found at a=0 for f(x)=ex in Activity 8.1.3. In addition, as long as f has a third derivative at x=a, we can even find a cubic approximation (just as we did at a=0 in Activity 8.1.4), and so on.
In developing such approximations centered at any value x=a, our guiding principle is the same as with our work at a=0: we’ll require that at the input value a, the original function’s output and its derivatives’ outputs match the corresponding approximation’s output and derivatives’ output.
Building on the form of the tangent line approximation, which we now denote T1(x),
T1(x)=f(a)+f(a)(xa),
it is natural for us to consider a quadratic approximation of form
T2(x)=c0+c1(xa)+c2(xa)2,
a cubic approximation of form
T3(x)=c0+c1(xa)+c2(xa)2+c3(xa)3,
and so on. Indeed, we define the degree n Taylor Polynomial centered at a more generally as follows.

Definition 8.2.21.

Let n be a natural number and let f be a function with at least n derivatives at a. The degree n Taylor polynomial of f centered at a is the function
Tn(x)=c0+c1(xa)+c2(xa)2++cn(xa)n
that satisfies Tn(a)=f(a), Tn(a)=f(a), Tn(a)=f(a), , Tn(n)(a)=f(n)(a).
Similar to the situation when a=0, it follows that we can find the coefficients ck of the Taylor polynomial in terms of the various derivatives of f evaluated at a.

Finding the degree n Taylor polynomial of f centered at a.

If f is a function with at least n derivatives at a, then the degree n Taylor polynomial of f centered at a, Tn(x), is
Tn(x)=c0+c1(xa)+c2(xa)2++cn(xa)n
where each coefficient ck is given by
ck=f(k)(a)k!.
As with approximations centered at a=0, the Taylor polynomial now provides us with an approximation of f near a. In particular,
f(x)Tn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2++f(n)(a)n!(xa)n.

Activity 8.2.3.

Let f(x)=ln(x), and recall that f is only defined for x>0. As such, we can’t consider the tangent line (or any other) approximation at a=0. Instead, we choose to work with an approximation to f(x)=ln(x) centered at a=1 and will find the degree 4 Taylor polynomial approximation
T4(x)=c0+c1(x1)+c2(x1)2+c3(x1)3+c4(x1)4.
  1. Determine f(x), f(x), f(x), and f(4)(x), and then compute f(1), f(1), f(1), and f(4)(1). Enter your results in in Table 8.2.22.
    Table 8.2.22. Finding the derivatives of f(x)=ln(x) at a=1.
    f(x)= ln(x) f(1)= 0
    f(x)= f(1)=
    f(x)= f(1)=
    f(x)= f(1)=
    f(4)(x)= f(4)(1)=
  2. Use your work in (a) along with the fact that the coefficients of the Taylor polynomial are determined by ck=f(k)(a)k! to determine
    T4(x)=c0+c1(x1)+c2(x1)2+c3(x1)3+c4(x1)4.
  3. In Figure 8.2.23, we see f(x)=ln(x) and its tangent line, T1(x)=x1 plotted on the same axes. Add T4(x) to the figure. What do you notice?
    Figure 8.2.23. The function f(x)=ln(x) and its degree 1 Taylor approximation T1(x)=x1 near the point (1,f(1)).
  4. Compute |f(x)T4(x)| for several different x values (you might find it helpful to use a slider in Desmos in the variable k to experiment with |f(k)T4(k)|); for approximately what values of x is it true that |f(x)T4(x)|<0.1?
  5. Use the pattern you observe in Table 8.2.22 to conjecture formulas for T5(x) and T6(x).
    For about what interval of x-values is it true that |f(x)T5(x)|<0.1? What about |f(x)T6(x)|<0.1? How is this different from what we observed with f(x)=cos(x) in Activity 8.2.2?

Activity 8.2.4.

This activity builds on Activity 8.2.3, and only changes one key thing: the location where the approximation is centered. Again, we let f(x)=ln(x), and recall that f is only defined for x>0. Here, we choose to work with an approximation centered at a=2, and find the degree 4 Taylor polynomial approximation
T4(x)=c0+c1(x2)+c2(x2)2+c3(x2)3+c4(x2)4.
  1. Recall f(x), f(x), f(x), and f(4)(x) from Table 8.2.22 in Activity 8.2.3, and then compute f(2), f(2), f(2), and f(4)(2). Enter your results in in Table 8.2.28.
    Table 8.2.28. Finding the derivatives of f(x)=ln(x) at a=2.
    f(x)= ln(x) f(2)= ln(2)
    f(x)= f(2)=
    f(x)= f(2)=
    f(x)= f(2)=
    f(4)(x)= f(4)(2)=
  2. Use your work in (a) along with the fact that the coefficients of the Taylor polynomial are determined by ck=f(k)(a)k! to determine T4(x)=c0+c1(x2)+c2(x2)2+c3(x2)3+c4(x2)4.
  3. In Figure 8.2.29, we see f(x)=ln(x) and its tangent line, T1(x)=ln(2)+12(x2) plotted on the same axes. Add T4(x) to the figure. What do you notice?
    Figure 8.2.29. The function f(x)=ln(x) and its degree 1 Taylor approximation T1(x)=ln(2)+12(x2) near the point (2,f(2)).
  4. Compute |f(x)T4(x)| for several different x values (you might find it helpful to use a slider in Desmos); for approximately what values of x is it true that |f(x)T4(x)|<0.1?
  5. Use the pattern you observe in Table 8.2.28 to conjecture formulas for T5(x) and T6(x).
    For about what interval of x-values is it true that |f(x)T5(x)|<0.1? What about |f(x)T6(x)|<0.1? How is this different from what we observed with the Taylor approximations centered at a=1 in Activity 8.2.3? How is it similar?

Subsection 8.2.3 Summary

  • Provided that a function f(x) has n derivatives at a selected input value x=a, we can find a degree n polynomial Tn(x) that approximates f(x) near a by requiring that Tn(a)=f(a), Tn(a)=f(a), Tn(a)=f(a), , Tn(n)(a)=f(n)(a).
  • When a=0, the degree n polynomial approximation, Tn(x), to a function f(x), centered at a=0, is a polynomial of the form
    Tn(x)=c0+c1x+c2x2++cnxn
    and it follows that the coefficients ck are determined by the values of the various derivatives of f(x) evaluated at 0 according to the formula
    ck=f(k)(0)k!.
    Moreover, for x near a=0,
    f(x)Tn(x)=f(0)+f(0)x+f(0)2!x2++f(n)(0)n!xn.
  • Just as we can consider any function f that has n derivatives at a=0 and find approximations centered there, we can also consider any input value a at which those n derivatives exist, and find a polynomial approximation that satisfies Tn(a)=f(a), Tn(a)=f(a), Tn(a)=f(a), , Tn(n)(a)=f(n)(a).
    At such a value a, the degree n Taylor polynomial of f centered at a has form
    Tn(x)=c0+c1(xa)+c2(xa)2++cn(xa)n
    and it follows that the coefficients ck are determined by the values of the various derivatives of f(x) evaluated at a according to the formula
    ck=f(k)(a)k!.
    Moreover, for x near a,
    f(x)Tn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2++f(n)(a)n!(xa)n.

Exercises 8.2.4 Exercises

1.

Find the Taylor polynomials of degree n approximating cos(3x) for x near 0:
For n=2, T2(x)=
For n=4, T4(x)=
For n=6, T6(x)=

2.

Calculate the Taylor polynomials T2(x) and T3(x) centered at x=π2 for f(x)=cos(x).

3.

Calculate the Taylor polynomials T2(x) and T3(x) centered at x=9 for f(x)=ln(x+1).
T2(x)=
T3(x)=T2(x)+

4.

Compute T2(x) at x=1 for y=ex and use a calculator to compute the error |exT2(x)| at x=0.3.
T2(x)=
|exT2(x)| =

5.

Find the second-degree Taylor polynomial for f(x)=2x24x+6 about x=0.
T2(x)=
What do you notice about your polynomial?

6.

Suppose g is a function which has continuous derivatives, and that g(2)=5,g(2)=4, g(2)=1, g(2)=4.
(a) What is the Taylor polynomial of degree 2 for g near 2?
T2(x)=
(b) What is the Taylor polynomial of degree 3 for g near 2?
T3(x)=
(c) Use the two polynomials that you found in parts (a) and (b) to approximate g(1.9).
With T2, g(1.9)
With T3, g(1.9)

7.

Suppose we know the following information about a function f:
f(0)=2,f(0)=3,f(0)=1,f(0)=0,f(4)(0)=3.
  1. Determine T4(x), the degree 4 Taylor polynomial of f that is centered at a=0.
  2. Use T4(x) to estimate f(0.5).
  3. State each of the Taylor polynomials T3(x), T2(x), and T1(x).

8.

In Exercise 8.2.4.5, we found that the degree 2 Taylor polynomial centered at a=0 of a quadratic function is the quadratic function itself. In this exercise, we explore how changing the center of the approximation offers additional insight into the function.
Let f(x)=12x22x+5, and let a=2 be the center at which we will find a degree 2 Taylor polynomial approximation of f.
  1. By finding f(x), f(x), f(2), f(2), and f(2), determine T2(x), the degree 2 Taylor polynomial approximation of f that is centered at a=2.
  2. Plot both f(x) and T2(x) on the same axes. What do you observe?
  3. What does the algebraic form of T2(x) tell you about the original function f(x)?
  4. What is the tangent line approximation to f(x) at a=2? What is special about the function’s behavior at this input value?

9.

Recall that we found in Preview 8.2.1 and subsequent work that sin(x)x13!x3+15!x517!x7, which is the degree 7 Taylor approximation centered at 0. And in Activity 8.2.2, we found that the degree 6 Taylor approximation centered at a=0 for cos(x) is cos(x)112!x2+14!x416!x6.
In this exercise, we investigate Taylor polynomial approximations of f(x)=sin(x) centered at a=π2.
  1. By finding the appropriate derivatives of f(x)=sin(x) and evaluating them at a=π2, determine the degree 6 Taylor polynomial approximation of sin(x) centered at a=π2.
  2. How is your result in (a) similar to the degree 6 Taylor polynomial of cos(x) that is centered at a=0?
  3. Recall the trigonometric identity that states sin(x)=cos(xπ2). How does this identity help explain what you found in (a) and (b)?

10.

In Example 8.2.6, we found that
f(x)=ln(1+x)T5(x)=1x12x2+13x314x4+15x5,
where T5 is the degree 5 Taylor approximation of ln(1+x) centered at a=0.
In Activity 8.2.3, we found that
g(x)=ln(x)P5(x)=1(x1)12(x1)2+13(x1)314(x1)4+15(x1)5,
where P5 is the degree 5 Taylor approximation of ln(x) centered at a=1. (Here we are using “T5” and “P5” to distinguish between these two degree 5 polynomial approximations of the two different functions f(x)=ln(1+x) and g(x)=ln(x), centered at two different values.)
  1. Note that f(0.5)=ln(1.5). Use T5(x) appropriately to find an estimate of ln(1.5).
  2. Observe that g(1.5)=ln(1.5). Use P5(x) to estimate ln(1.5).
  3. Are the estimates of ln(1.5) generated by T5(x) and P5(x) the same or different? Why do you think this happens?
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