微积分I(标准路径) · free preview

§8.1 Extending local linearization (continued)

在 $a = 0$ 附近的切线对函数 $f(x) = e^x$(在 $a = 0$ 附近)的近似效果如何?

1. As the degree of the approximation increases

As the degree of the approximation increases

2. Summary

Summary

3. In Preview Activity 8.1

In Preview Activity 8.1, we built a spreadsheet that computed the differences between f(x)f(x) and T1(x)T_1(x) for xx-values between −1-1 and 11, spaced 0.10.1 units apart. Your spreadsheet started like the one shown in the table in Preview Activity 8.1.

Next, we build an updated version of this spreadsheet that computes similar differences between ff and the three higher degree approximations we have found. In particular, we now want to have columns for Δx\Delta x, xx, f(x)f(x), T1(x)T_1(x), T2(x)T_2(x), T3(x)T_3(x), and T4(x)T_4(x), plus the absolute differences ∣f(x)−T1(x)∣|f(x) - T_1(x)|, ∣f(x)−T2(x)∣|f(x) - T_2(x)|, ∣f(x)−T3(x)∣|f(x) - T_3(x)|, and ∣f(x)−T4(x)∣|f(x) - T_4(x)|. Hint: when building your entries, note that you can think of T2(x)T_2(x) as T2(x)=T1(x)+12x2T_2(x) = T_1(x) + \frac{1}{2}x^2, and similarly view T3(x)T_3(x) as “T2(x)T_2(x) plus one more term”.

Include at least 55 digits of accuracy beyond the decimal. The first seven columns of your spreadsheet might start like this:

Δx\Delta x | xx | f(x)f(x) | T1(x)T_1(x) | T2(x)T_2(x) | T3(x)T_3(x) | T4(x)T_4(x)

0.10.1 | −1.0-1.0 | 0.367870.36787 | 0.000000.00000 | 0.500000.50000 | 0.333330.33333 | 0.375000.37500

0.10.1 | −0.9-0.9 | 0.406570.40657 | 0.100000.10000 | 0.505000.50500 | 0.383500.38350 | 0.410830.41083

The next four columns of your spreadsheet should begin as follows:

∣f(x)−T1(x)∣|f(x)-T_1(x)| | ∣f(x)−T2(x)∣|f(x)-T_2(x)| | ∣f(x)−T3(x)∣|f(x)-T_3(x)| | ∣f(x)−T4(x)∣|f(x)-T_4(x)|

0.367870.36787 | 0.132120.13212 | 0.034540.03454 | 0.007120.00712

0.306570.30657 | 0.098430.09843 | 0.023070.02307 | 0.004260.00426

4. We call the value of…

We call the value of ∣f(x)−T2(x)∣|f(x) - T_2(x)| the absolute error of the quadratic approximation of ff at the value xx. What is the absolute error of the quadratic approximation at x=−1x = -1? at x=1x = 1?

5. What is the absolute…

What is the absolute error of the cubic (degree 33) approximation, T3(x)T_3(x), at x=−1x = -1? at x=1x = 1?

6. What is the absolute…

What is the absolute error of the quartic (degree 44) approximation, T4(x)T_4(x), at x=−1x = -1? at x=1x = 1?

7. Study your spreadshe…

Study your spreadsheet for trends that you notice as the value of xx changes or the degree nn of the approximation changes. What are your observations?

8. Investigate the erro…

Investigate the errors in the various approximations for a wider interval of xx-values. For example, you might consider starting at x=−2x = -2 with Δx=0.2\Delta x = 0.2. What do you notice?

9. Throughout our work …

Throughout our work in Section 8.1, we have focused on approximating the function f(x)=exf(x) = e^x. In this exercise, we change the function of interest to f(x)=13x3+14x2−2x−1f(x) = \frac{1}{3}x^3 + \frac{1}{4}x^2 - 2x - 1, and consider the linear and quadratic approximations to ff near a=0a = 0.

  • Determine f′(x)f'(x) and f′′(x)f''(x) and enter their formulas below.
f(x)=13x3+14x2−2x−1f′(x)=f′′(x)=\begin{aligned} f(x) &= \frac{1}{3}x^3 + \frac{1}{4}x^2 - 2x - 1 \\ f'(x) &= \\ f''(x) &= \end{aligned}
  • Next, compute f(0)f(0), f′(0)f'(0), and f′′(0)f''(0) and enter those values below.
f(0)=f′(0)=f′′(0)=\begin{aligned} f(0) &= \\ f'(0) &= \\ f''(0) &= \end{aligned}
  • Use your work so far to determine the formula for T1(x)T_1(x), the tangent line approximation to f(x)f(x) at a=0a =0 (which satisfies T1(0)=f(0)T_1(0) = f(0) and T1′(0)=f′(0)T_1'(0) = f'(0)). - Let T2(x)T_2(x) be the quadratic approximation to f(x)f(x) near a=0a = 0 that satisfies T2(0)=f(0)T_2(0) = f(0), T2′(0)=f′(0)T_2'(0) = f'(0), and T2′′(0)=f′′(0)T_2''(0) = f''(0). You might start by letting T2(x)=c0+c1x+c2x2T_2(x) = c_0 + c_1 x + c_2 x^2, and creating an updated table like the one shown below.
f(x)=x3/3+x2/4−2x−1f′(x)=f′′(x)=f(0)=f′(0)=f′′(0)=\begin{aligned} f(x) &= x^3/3 + x^2/4 - 2x - 1 \\ f'(x) &= \\ f''(x) &= \\ & \\ f(0) &= \\ f'(0) &= \\ f''(0) &= \end{aligned}
T2(x)=c0+c1x+c2x2T2′(x)=c1+2c2xT2′′(x)=T2(0)=T2′(0)=T2′′(0)=\begin{aligned} T_2(x) &= c_0 + c_1 x + c_2 x^2 \\ T_2'(x) &= c_1 + 2c_2 x \\ T_2''(x) &= \\ & \\ T_2(0) &= \\ T_2'(0) &= \\ T_2''(0) &= \end{aligned}

Use your work in in the table above to find the formula for T2(x)T_2(x) that satisfies T2(0)=f(0)T_2(0) = f(0), T2′(0)=f′(0)T_2'(0) = f'(0), and T2′′(0)=f′′(0)T_2''(0) = f''(0). - Plot f(x)f(x), T1(x)T_1(x), and T2(x)T_2(x) on the same axes, centered at a=0a = 0. What do you notice?

10. In this exercise

In this exercise, we extend our work in Exercise 8.1.1. We continue to consider the function f(x)=13x3+14x2−2x−1f(x) = \frac{1}{3}x^3 + \frac{1}{4}x^2 - 2x - 1, but now build the cubic (degree 33) approximation to ff near a=0a = 0.

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