Preview Activity 8.2.1.
Let \(f(x) = \sin(x)\) and let \(T_3(x) = c_0 + c_1 x + c_2 x^2 + c_3x^3\text{.}\) Our goal is to find the values of \(c_0, \ldots, c_3\) that make the sine function and its derivative values agree with those of the cubic polynomial \(T_3\) at \(a = 0\) and to study the resulting degree \(3\) approximation of the sine function.
(a)
As in previous work, the derivatives of \(T_3(x)\) and their respective values at \(a = 0\) are those shown below.
Compute the first three derivatives of \(f(x) = \sin(x)\) and evaluate them at \(a = 0\) accordingly, recording your results in the blanks provided below.
\begin{align*}
f(x) \amp= \sin(x) \amp T_3(x) \amp= c_0 + c_1 x + c_2 x^2 + c_3 x^3 \\
f'(x) \amp= \fillinmath{XXXXX} \amp T_3'(x) \amp= c_1 + 2c_2 x + 3c_3 x^2 \\
f''(x) \amp= \fillinmath{XXXXX} \amp T_3''(x) \amp= 2c_2 + 6c_3x \\
f'''(x) \amp= \fillinmath{XXXXX} \amp T_3'''(x) \amp= 6c_3 \\
\amp \\
f(0) \amp= \fillinmath{XXX} \amp T_3(0) \amp= c_0 \\
f'(0) \amp= \fillinmath{XXX} \amp T_3'(0) \amp= c_1 \\
f''(0) \amp= \fillinmath{XXX} \amp T_3''(0) \amp= 2c_2 \\
f'''(0) \amp= \fillinmath{XXX} \amp T_3'''(0) \amp= 6c_3
\end{align*}
(b)
Now, set \(T_3(0) = f(0)\text{,}\) \(T_3'(0) = f'(0)\text{,}\) \(T_3''(0) = f''(0)\text{,}\) and \(T_3'''(0) = f'''(0)\text{.}\) What are the resulting values of \(c_0, \ldots, c_3\text{?}\) What is the resulting formula for \(T_3(x)\text{?}\)
(c)
Recall that the tangent line approximation \(T_1\) to \(f(x) = \sin(x)\) at \(a = 0\) is \(T_1(x) = x\text{.}\) Use appropriate computing technology to plot the cubic approximation \(T_3(x)\) you found in (b) along with \(f(x)\) and \(T_1(x)\) in the same window shown in Figure 8.2.2.
(d)
What do you observe about the approximation of \(f(x)\) by \(T_3(x)\) compared to the approximation of \(f(x)\) by \(T_1(x)\text{?}\) For example, how do \(f(1) - T_1(1)\) and \(f(1) - T_3(1)\) compare?
(e)
In our work so far, we’ve seen that raising the degree of polynomial improves the approximation of the original function \(f(x)\text{.}\) If we wanted to find a degree \(9\) approximation to \(f(x) = \sin(x)\text{,}\) we’d need to find a polynomial whose value and first \(9\) derivatives match those of the sine function at \(a = 0\text{.}\)
Fill in the blanks below to determine the next six derivatives of \(f(x) = \sin(x)\text{,}\) up to \(f^{(9)}(x)\text{.}\)
\begin{align*}
f(x) \amp= \sin(x) \amp f^{(5)}(x) \amp= \fillinmath{XXXXX} \\
f'(x) \amp= \cos(x) \amp f^{(6)}(x) \amp= \fillinmath{XXXXX} \\
f''(x) \amp= -\sin(x) \amp f^{(7)}(x) \amp= \fillinmath{XXXXX} \\
f'''(x) \amp= -\cos(x) \amp f^{(8)}(x) \amp= \fillinmath{XXXXX} \\
f^{(4)}(x) \amp= \fillinmath{XXXXX} \amp f^{(9)}(x) \amp= \fillinmath{XXXXX}
\end{align*}
What pattern do you notice in the derivatives of \(f(x)=\sin(x)\text{?}\)
