6. a function f has derivatives of all orders...
6. a function f has derivatives of all orders for all real numbers x. a portion of the graph of f is shown above, along with the line tangent to the graph of f at x = 0. selected derivatives of f at x = 0 are given in the table above. (a) write the third - degree taylor polynomial for f about x = 0. p3(x)=3 - 4x+(3x^2/2!)-(23x^3/3!) (b) write the first three nonzero terms of the maclaurin series for e^x. write the second - degree taylor polynomial for e^xf(x) about x = 0. 1 + x+(x^2/2!) (c) let h be the function defined by h(x)=∫0^x f(t)dt. use the taylor polynomial found in part (a) to find an approximation for h(1). (d) it is known that the maclaurin series for h converges to h(x) for all real numbers x. it is also known that the individual terms of the series for h(1) alternate in sign and decrease in absolute value to 0. use the alternating series error bound to show that the approximation found in part (c) differs from h(1) by at most 0.45.
Answer
# Answer:
(a) $P_3(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}$. From the graph, the slope of the tangent - line at $x = 0$ gives $f^{\prime}(0)=-4$, $f^{\prime\prime}(0) = 3$, $f^{(3)}(0)=-\frac{23}{2}$. And assume $f(0) = 3$. So $P_3(x)=3-4x+\frac{3}{2}x^{2}-\frac{23}{12}x^{3}$.
(b) The Maclaurin series for $e^{x}=1 + x+\frac{x^{2}}{2!}+\cdots$. The second - degree Taylor polynomial for $e^{x}f(x)$:
\[
\begin{align*}
e^{x}f(x)&\approx(1 + x+\frac{x^{2}}{2})(3-4x+\frac{3}{2}x^{2})\\
&=3-4x+\frac{3}{2}x^{2}+3x-4x^{2}+\frac{3}{2}x^{3}+\frac{3}{2}x^{2}-2x^{3}+\frac{3}{4}x^{4}\\
&=3 - x+\frac{3 - 8+3}{2}x^{2}+(\frac{3}{2}-2)x^{3}+\frac{3}{4}x^{4}\\
&=3 - x - x^{2}+\frac{-1}{2}x^{3}+\frac{3}{4}x^{4}\approx3 - x - x^{2}
\end{align*}
\]
(c) If $h(x)=\int_{0}^{x}f(t)dt$, and $P_3(t)=3-4t+\frac{3}{2}t^{2}-\frac{23}{12}t^{3}$, then $h(1)\approx\int_{0}^{1}(3-4t+\frac{3}{2}t^{2}-\frac{23}{12}t^{3})dt$.
\[
\begin{align*}
\int_{0}^{1}(3-4t+\frac{3}{2}t^{2}-\frac{23}{12}t^{3})dt&=\left[3t-4\times\frac{t^{2}}{2}+\frac{3}{2}\times\frac{t^{3}}{3}-\frac{23}{12}\times\frac{t^{4}}{4}\right]_0^1\\
&=3 - 2+\frac{1}{2}-\frac{23}{48}\\
&=1+\frac{1}{2}-\frac{23}{48}\\
&=\frac{48 + 24-23}{48}\\
&=\frac{49}{48}\approx1.02
\end{align*}
\]
(d) The Maclaurin series for $h(x)$ is an alternating series. The error bound for an alternating series $\sum_{n = 0}^{\infty}(-1)^{n}a_{n}$ (where $a_{n}\gt0$, $a_{n+1}\leq a_{n}$, and $\lim_{n\rightarrow\infty}a_{n}=0$) is $|E_N|\leq a_{N + 1}$. The Taylor polynomial $P_3(x)$ used in part (c) is the sum of the first 4 terms of the Maclaurin series of $h(x)$ (since we integrated the Taylor polynomial of $f(x)$). The next - term of the Maclaurin series of $h(x)$ (after integrating the fourth - degree term of the Taylor polynomial of $f(x)$) comes from integrating the fourth - degree term of the Taylor polynomial of $f(x)$. The Taylor polynomial of $f(x)$ used in part (a) is $P_3(x)$, and if we consider the next non - included term of the Taylor series of $f(x)$ (the fourth - derivative term), when we integrate to get $h(x)$, the error $E$ in approximating $h(1)$ using the Taylor polynomial from part (c) is bounded by $\left|\int_{0}^{1}\frac{f^{(4)}(0)}{4!}t^{4}dt\right|$. Given $f^{(4)}(0)=54$, then $\left|\int_{0}^{1}\frac{54}{4!}t^{4}dt\right|=\left|\frac{54}{24}\times\frac{t^{5}}{5}\right]_0^1=\frac{54}{120}=0.45$.
# Explanation:
## Step1: Recall Taylor polynomial formula
$P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}$, for $a = 0$ (Maclaurin), $P_3(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}$. Get values from graph and table.
## Step2: Multiply Maclaurin series of $e^{x}$ and Taylor polynomial of $f(x)$
Multiply $(1 + x+\frac{x^{2}}{2})$ and $(3-4x+\frac{3}{2}x^{2})$ and keep terms up to degree 2.
## Step3: Integrate Taylor polynomial to approximate $h(1)$
Use $h(1)\approx\int_{0}^{1}P_3(t)dt$, integrate term - by - term: $\int(3-4t+\frac{3}{2}t^{2}-\frac{23}{12}t^{3})dt=3t - 2t^{2}+\frac{1}{2}t^{3}-\frac{23}{48}t^{4}+C$.
## Step4: Apply alternating series error bound
For an alternating series $\sum(-1)^{n}a_{n}$, error $|E_N|\leq a_{N + 1}$. Integrate the next non - included term of the Taylor series of $f(x)$ to find the error bound for $h(1)$.