Coordinated Calculus

We know that \(f'(x) = e^x\) and so \(f''(x) = e^x\) and \(f'''(x) = e^x\text{.}\) Thus,

So the third order Taylor polynomial of \(f(x) = e^x\) centered at \(x=0\) is

In general, for the exponential function \(f\) we have \(f^{(k)}(x) = e^x\) for every positive integer \(k\text{.}\) Thus, the \(k\)th term in the \(n\)th order Taylor polynomial for \(f(x)\) centered at \(x=0\) is

Therefore, the \(n\)th order Taylor polynomial for \(f(x) = e^x\) centered at \(x=0\) is

1. 1. \(f^{(k)}(0) = k! \text{.}\)

   2. \begin{equation*} P_n(x) = \sum_{k=0}^n x^k\text{.} \end{equation*}

2. 1. \(f^{k}(0) = 0\) if \(k\) is odd, and \(f^{2k}(0) = (-1)^k\text{.}\)

   2. \(P_n(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + (-1)^{n/2}\frac{x^n}{n!}\) if \(n\) is even and \(P_n(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + (-1)^{(n-1)/2}\frac{x^(n-1)}{(n-1)!}\) if \(n\) is odd.

3. 1. \(f^{k}(0) = 0\) if \(k\) is even and \(f^{2k+1}(0) = (-1)^k \text{.}\)

   2. \(P_n(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots + (-1)^{(n-1)/2}\frac{x^n}{n!}\) if \(n\) is odd and \(P_n(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots + (-1)^{n/2+1}\frac{x^{n-1}}{(n-1)!}\) if \(n\) is even.

1. 1. The first four derivatives of \(f(x)\) at \(x=0\) are

      \begin{align*} f(x) \amp = \frac{1}{1-x} \amp f(0) \amp = 1\\ f'(x) \amp = \frac{1}{(1-x)^2} \amp f'(0) \amp = 1\\ f''(x) \amp = \frac{2}{(1-x)^3} \amp f''(0) \amp = 2\\ f^{(3)}(x) \amp = \frac{3!}{(1-x)^4} \amp f^{(3)}(0) \amp = 3!\\ f^{(4)}(x) \amp = \frac{4!}{(1-x)^5} \amp f^{(4)}(0) \amp = 4!\text{.} \end{align*}

      It appears that the pattern is

      \begin{equation*} f^{(k)}(0) = k!\text{.} \end{equation*}

   2. The \(n\)th order Taylor polynomial for \(f\) at \(x=0\) is

      \begin{equation*} \sum_{k=0}^n \frac{f^{(k)}}{k!} x^k = \sum_{k=0}^n \frac{k!}{k!} x^k = \sum_{k=0}^n x^k\text{.} \end{equation*}

      This makes sense since \(f(x)\) is the sum of the geometric series with ratio \(x\text{,}\) so the \(n\)th order Taylor polynomial should just be the \(n\)th partial sum of this geometric series.

2. 1. The first four derivatives of \(f(x)\) at \(x=0\) are

      \begin{align*} f(x) \amp = \cos(x) \amp f(0) \amp = 1\\ f'(x) \amp = -\sin(x) \amp f'(0) \amp = 0\\ f''(x) \amp = -\cos(x) \amp f''(0) \amp = -1\\ f^{(3)}(x) \amp = \sin(x) \amp f^{(3)}(0) \amp = 0\\ f^{(4)}(x) \amp = \cos(x) \amp f^{(4)}(0) \amp = 1\text{.} \end{align*}

      It appears that the odd derivatives of \(f(x)\) are all plus or minus \(\sin(x)\) and so have values of 0 at \(x=0\) and the even derivatives are \(\pm \cos(x)\) and have alternating values of 1 and \(-1\) at \(x-0\text{.}\) Since the even numbers can be represented in the form \(2k\) where \(k\) is an integer we have \(f^{k}(0) = 0\) if \(k\) is odd and \(f^{2k}(0) = (-1)^k\text{.}\)

   2. Based on the previous part of this problem the \(n\)th order Taylor polynomial for \(\cos(x)\) is

      \begin{equation*} 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + (-1)^{n/2}\frac{x^n}{n!} \end{equation*}

      if \(n\) is even and

      \begin{equation*} 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + (-1)^{(n-1)/2}\frac{x^{n-1}}{(n-1)!} \end{equation*}

      if \(n\) is odd.

3. 1. The first four derivatives of \(f(x)\) at \(x=0\) are

      \begin{align*} f(x) \amp = \sin(x) \amp f(0) \amp = 0\\ f'(x) \amp = \cos(x) \amp f'(0) \amp = 1\\ f''(x) \amp = -\sin(x) \amp f''(0) \amp = 0\\ f^{(3)}(x) \amp = -\cos(x) \amp f^{(3)}(0) \amp = -1\\ f^{(4)}(x) \amp = \sin(x) \amp f^{(4)}(0) \amp = 0\text{.} \end{align*}

      It appears that the even derivatives of \(f(x)\) are all plus or minus \(\sin(x)\) and so have values of 0 at \(x=0\) and the odd derivatives are \(\pm \cos(x)\) and have alternating values of 1 and \(-1\) at \(x=0\text{.}\) Since the odd numbers can be represented in the form \(2k+1\) where \(k\) is an integer we have \(f^{k}(0) = 0 \) if \(k\) is even and \(f^{2k+1}(0) = (-1)^k\text{.}\)

   2. Based on the previous part of this problem the \(n\)th order Taylor polynomial for \(\sin(x)\) is

      \begin{equation*} x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots + (-1)^{(n-1)/2}\frac{x^n}{n!} \end{equation*}

      if \(n\) is odd and

      \begin{equation*} x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots + (-1)^{n/2+1}\frac{x^{n-1}}{(n-1)!} \end{equation*}

      if \(n\) is even.

1. \(P(x) = 1 + x + x^2 + x^3 + \cdots + x^n + \cdots\)

2. \(P(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \cdots + (-1)^{n}\frac{1}{(2n)!}x^{2n} + \cdots \text{.}\)

3. \(P(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots + (-1)^{n}\frac{1}{(2n+1)!}x^{2n+1} + \cdots \text{.}\)

4. \(P_n(x) = 1 + x + x^2 + x^3 + \cdots + x^n\)

1. For \(f(x) = \frac{1}{1-x}\text{,}\) its Taylor series is

   \begin{equation*} P(x) = 1 + x + x^2 + x^3 + \cdots + x^n + \cdots \end{equation*}

2. For \(f(x) = \cos(x)\text{,}\) its Taylor series is

   \begin{equation*} P(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \cdots + (-1)^{n}\frac{1}{(2n)!}x^{2n} + \cdots\text{.} \end{equation*}

3. For \(f(x) = \sin(x)\text{,}\) its Taylor series is

   \begin{equation*} P(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots + (-1)^{n}\frac{1}{(2n+1)!}x^{2n+1} + \cdots\text{.} \end{equation*}

4. For \(f(x) = \frac{1}{1-x}\text{,}\)

   \begin{equation*} P_n(x) = 1 + x + x^2 + x^3 + \cdots + x^n \end{equation*}

1. \begin{align*} P_1(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) = 1\\ P_2(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2\\ P_3(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3\\ &= 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = P_2(x)\\ P_4(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ P(x) &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4 - \frac{1}{6!}\left(x - \frac{\pi}{2} \right)^6 + \cdots \end{align*}
2. \begin{equation*} P_4(x) = 0 + 1(x-1) - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{6}{4!}(x-1)^4\text{.} \end{equation*}
   \begin{equation*} P(x) = 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 - \cdots \end{equation*}

1. For \(f(x) = \sin(x)\text{,}\) \(f'(x) = \cos(x)\text{,}\) \(f''(x) = -\sin(x)\text{,}\) \(f'''(x) = -\cos(x)\text{,}\) and \(f^{(4)}(x) = \sin(x)\text{.}\) Thus, \(f\left(\frac{\pi}{2} \right) = 1\text{,}\) \(f'\left(\frac{\pi}{2} \right) = 0\text{,}\) \(f''\left(\frac{\pi}{2} \right) = -1\text{,}\) \(f'''\left(\frac{\pi}{2} \right) = 0\text{,}\) and \(f^{(4)}\left(\frac{\pi}{2} \right) = 1\text{.}\) It follows that the first four Taylor polynomials of \(f\) are

   \begin{align*} P_1(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) = 1\\ P_2(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2\\ P_3(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3\\ &= 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = P_2(x)\\ P_4(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4 \end{align*}

   From the pattern, the Taylor series for \(f(x)\) centered at \(x = \frac{\pi}{2}\) is

   \begin{equation*} P(x) = 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4 - \frac{1}{6!}\left(x - \frac{\pi}{2} \right)^6 + \cdots \end{equation*}

   which is expected because of the repeating patterns in the derivatives of the sine function evaluated at \(\frac{\pi}{2}\text{.}\)

2. For \(f(x) = \ln(x)\text{,}\) \(f'(x) = x^{-1}\text{,}\) \(f''(x) = -x^{-2}\text{,}\) \(f'''(x) = 2x^{-3}\text{,}\) and \(f^{(4)}(x) = -6x^{-4}\text{.}\) It follows that \(f(1) = 0\text{,}\) \(f'(1) = 1\text{,}\) \(f''(1) = -1\text{,}\) \(f'''(1) = 2\text{,}\) and \(f^{(4)}(1) = -6\text{.}\) Thus, the fourth Taylor polynomial (in which we can see the polynomials of lower degree) is

   \begin{equation*} P_4(x) = 0 + 1(x-1) - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{6}{4!}(x-1)^4\text{.} \end{equation*}

   Simplifying the coefficients and seeing the pattern, it follows that the Taylor series for \(f(x)\) centered at \(x = 1\) is

   \begin{equation*} P(x) = 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 - \cdots \end{equation*}

1. It appears that as we increase the order of the Taylor polynomials, they fit the graph of \(f\) better and better over larger intervals.

2. It appears that as we increase the order of the Taylor polynomials, they fit the graph of \(f\) better and better over larger intervals.

3. The Taylor polynomials converge to \(\frac{1}{1-x}\) only on the interval \((-1,1)\text{.}\)

1. The graphs of the 10th (magenta), 20th (blue), and 30th (green) Taylor polynomials centered at \(0\) for \(e^x\) are shown below along with the graph of \(f(x)\) in red:

   

   It appears that as we increase the order of the Taylor polynomials, they fit the graph of \(f\) better and better over larger intervals. So it looks like the Taylor polynomials converge to \(e^x\) for every value of \(x\text{.}\)

2. The graphs of the 10th (magenta), 20th (blue), and 30th (green) Taylor polynomials centered at \(0\) for \(\cos(x)\) are shown below along with the graph of \(f(x)\) in red:

   

   It appears that as we increase the order of the Taylor polynomials, they fit the graph of \(f\) better and better over larger intervals. So it looks like the Taylor polynomials converge to \(\cos(x)\) for every value of \(x\text{.}\) Based on the \(n\)th order Taylor polynomials we found earlier for \(\cos(x)\text{,}\) the Taylor series for \(f(x)\) centered at \(0\) is

   \begin{equation*} \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}\text{.} \end{equation*}

3. The graphs of the 10th (magenta), 20th (blue), and 30th (green) Taylor polynomials centered at \(0\) for \(\frac{1}{1-x}\) are shown below along with the graph of \(f(x)\) in red:

   

   It appears that as we increase the order of the Taylor polynomials, they only fit the graph of \(f\) better and better over the interval \((-1,1)\) and appear to diverge outside that interval. So it looks like the Taylor polynomials converge to \(\frac{1}{1-x}\) only on the interval \((-1,1)\text{.}\)

   Based on the \(n\)th order Taylor polynomials we found earlier for \(\frac{1}{1-x}\text{,}\) the Taylor series for \(f(x)\) centered at \(0\) is

   \begin{equation*} \sum_{k=0}^{\infty} x^k\text{.} \end{equation*}

1. \begin{align*} f(0) \amp = a_0\\ f'(0) \amp = a_1\\ f''(0) \amp = 2!a_2\\ f^{(3)}(0) \amp = 3!a_3\\ f^{(4)}(0) \amp = 4!a_4 \end{align*}
2. \begin{align*} f'(x) \amp = \sum_{k=1}^{\infty} ka_kx^{k-1}\\ f''(x) \amp = \sum_{k=2}^{\infty} k(k-1)a_kx^{k-2}\\ f^{(3)}(x) \amp = \sum_{k=3}^{\infty} k(k-1)(k-2)a_kx^{k-3}\\ \vdots \amp \ \qquad \vdots\\ f^{(n)}(x) \amp = \sum_{k=n}^{\infty} k(k-1)(k-2) \cdots (k-n+1) a_kx^{k-n}\\ \vdots \amp \ \qquad \vdots \end{align*}

   So

   \begin{align*} f(0) \amp = a_0\\ f'(0) \amp = a_1\\ f''(0) \amp = 2!a_2\\ f^{(3)}(0) \amp = 3!a_3\\ \vdots \amp \ \qquad \vdots\\ f^{(k)}(0) \amp = k!a_k\\ \vdots \amp \ \qquad \vdots \end{align*}

   and

   \begin{equation*} a_k = \frac{f^{(k)}(0)}{k!} \end{equation*}

   for each \(k \geq 0\text{.}\) But these are just the coefficients of the Taylor series expansion of \(f\text{,}\) which leads us to the following observation.

1. Observe that

   \begin{align*} f'(x) \amp = \sum_{k=1}^{\infty} ka_kx^{k-1}\\ f''(x) \amp = \sum_{k=2}^{\infty} k(k-1)a_kx^{k-2}\\ f^{(3)}(x) \amp = \sum_{k=3}^{\infty} k(k-1)(k-2)a_kx^{k-3}\\ f^{(4)}(x) \amp = \sum_{k=4}^{\infty} k(k-1)(k-2)(k-3)a_kx^{k-4} \end{align*}

   and therefore

   \begin{align*} f(0) \amp = a_0\\ f'(0) \amp = a_1\\ f''(0) \amp = 2!a_2\\ f^{(3)}(0) \amp = 3!a_3\\ f^{(4)}(0) \amp = 4!a_4\text{.} \end{align*}

2. Since

   \begin{equation*} f^{(n)}(x) = \sum_{k=n}^{\infty} k(k-1)(k-2) \cdots (k-n+1) a_kx^{k-n} \end{equation*}

   every term of this series vanishes at \(x = 0\) except the first. Thus it follows \(f^{(n)}(0) = n(n-1)(n-2) \cdots (1) a_n\text{,}\) so \(f^{(n)}(0) = n! a_n\text{.}\)

3. Since \(a_k = \frac{f^{(k)}(0)}{k!}\) for each \(k \geq 0\text{,}\) we see that these are just the coefficients of the Taylor series expansion of \(f\text{,}\) and thus we get the unsurprising result that the coefficients of a power series are identical to the Taylor series of the power series.

Recall that the Ratio Test applies only to series of nonnegative terms. In this example, the variable \(x\) may have negative values. But we are interested in absolute convergence, so we apply the Ratio Test to the series

Now, observe that

for any value of \(x\text{.}\) So the Taylor series (7.23) converges absolutely for every value of \(x\text{,}\) and thus converges for every value of \(x\text{.}\)

1. \((-\infty, \infty)\text{.}\)

2. \((-\infty, \infty)\text{.}\)

3. The interval \((-1,1)\text{.}\)

1. Using the Ratio Test with the \(k\)th term \(\frac{|x|^{2k}}{(2k)!}\) we get

   \begin{align*} \lim_{k \to \infty} \frac{ \frac{|x|^{2(k+1)}}{(2(k+1))!} }{ \frac{|x|^{2k}}{(2k)!} } \amp = \lim_{k \to \infty} \frac{|x|^{2(k+1)}(2k)!}{|x|^{2k}(2(k+1))!}\\ \amp = \lim_{k \to \infty} \frac{|x|^{2}}{(2k+2)(2k+1)}\\ \amp = 0\text{.} \end{align*}

   So the interval of convergence of the Taylor series for \(f(x) = \cos(x)\) centered at \(x=0\) is \((-\infty, \infty)\text{.}\)

2. Using the Ratio Test with the \(k\)th term \(\frac{|x|^{2k+1}}{(2k+1)!}\) we get

   \begin{align*} \lim_{k \to \infty} \frac{ \frac{|x|^{2(k+1)+1}}{(2(k+1)+1)!} }{ \frac{|x|^{2k+1}}{(2k+1)!} } \amp = \lim_{k \to \infty} \frac{|x|^{2(k+1)+1}(2k+1)!}{|x|^{2k+1}(2(k+1)+1)!}\\ \amp = \lim_{k \to \infty} \frac{|x|^{2}}{(2k+3)(2k+2)}\\ \amp = 0\text{.} \end{align*}

   So the interval of convergence of the Taylor series for \(f(x) = \sin(x)\) centered at \(x=0\) is \((-\infty, \infty)\text{.}\)

3. Using the Ratio Test with the \(k\)th term \(|x|^{k}\) we get

   \begin{equation*} \lim_{k \to \infty} \frac{ |x|^{k+1} }{ |x|^{k} } = \lim_{k \to \infty} |x| = |x|\text{,} \end{equation*}

   So the series \(\sum_{k=0}^{\infty} x^k\) converges absolutely when \(|x| \lt 1\) or for \(-1 \lt x \lt 1\) and diverges when \(|x| \gt 1\text{.}\) Since the Ratio Test doesn't tell us what happens when \(x=1\text{,}\) we need to check the endpoints separately.

   • When \(x=1\) we have the series \(\sum_{k=0}^{\infty} 1\) which diverges since \(\lim_{k \to \infty} 1 \neq 0\text{.}\)

   • When \(x=-1\) we have the series \(\sum_{k=0}^{\infty} (-1)^k\) which diverges since \(\lim_{k \to \infty} (-1)^k\) does not exist.

   Therefore, the interval of convergence of the Taylor series for \(f(x) = \frac{1}{1-x}\) centered at \(x=0\) is \((-1,1)\text{.}\)

To answer this question we use \(f(x) = \sin(x)\text{,}\) \(c = 2\text{,}\) \(a=0\text{,}\) and \(n = 10\) in the Lagrange error bound formula. We also need to find an appropriate value for \(M\text{.}\) Note that the derivatives of \(f(x) = \sin(x)\) are all equal to \(\pm \sin(x)\) or \(\pm \cos(x)\text{.}\) Thus,

for any \(n\) and \(x\text{.}\) Therefore, we can choose \(M\) to be \(1\text{.}\) Then

So \(P_{10}(2)\) approximates \(\sin(2)\) to within at most \(0.00005130671797\text{.}\) A computer algebra system tells us that

with an actual difference of about \(0.0000500159\text{.}\)

\(n \ge 27\text{.}\)

In this example, if we can find a value of \(n\) so that

then we will have

Again we use \(f(x) = \sin(x)\text{,}\) \(c = 2\text{,}\) \(a=0\text{,}\) and \(M = 1\) from the previous example. So we need to find \(n\) to make

There is no good way to solve equations involving factorials, so we simply use trial and error, evaluating \(\frac{2^{n+1}}{(n+1)!}\) at different values of \(n\) until we get one we need.

So we need to use an \(n\) of at least 27 to ensure accuracy to 20 decimal places.

A computer algebra system gives

and we can see that these agree to 20 places.

Recall from the previous example that since \(f(x) = \sin(x)\text{,}\) we know

for any \(n\) and \(x\text{.}\) This allows us to choose \(M = 1\) in the Lagrange error bound formula. Thus,

for every \(x\text{.}\)

We showed in earlier work that the Taylor series \(\sum_{k=0}^{\infty} \frac{x^k}{k!}\) converges for every value of \(x\text{.}\) Because the terms of any convergent series must approach zero, it follows that

for every value of \(x\text{.}\) Thus, taking the limit as \(n \to \infty\) in the inequality(7.25), it follows that

As a result, we can now write

for every real number \(x\text{.}\)

1. Compare Example7.64.

2. 1. Use the fact that that \(|f^{(n)}(x)| \le e^c\) on the interval \([0,c]\) for any fixed positive value of \(c\text{.}\)

   2. Repeat the argument in (a) but replace \(e^c\) with \(1\text{,}\) and everything else holds in the same way.

   3. Combine the results of (a) and (b)

3. \(n = 28\text{.}\)

1. Compare Example7.64.

2. 1. Let \(x \ge 0\text{.}\) Since \(f(x) = e^x\text{,}\) \(f^{(n)}(x) = e^x\) for every natural number \(n\text{.}\) Since \(e^x\) is an increasing function, we know that \(|f^{(n)}(x)| \le e^c\) on the interval \([0,c]\) for any fixed positive value of \(c\text{.}\) Thus, by the Lagrange error formula, we can say that

      \begin{equation*} |P_n(x) - e^x| \le e^c \frac{x^{n+1}}{(n+1)!}\text{.} \end{equation*}

      Since the series \(\sum \frac{x^{n}}{n!}\) converges for every \(x\text{,}\) \(\frac{x^{n}}{n!} \to 0\) as \(x \to \infty\text{,}\) and thus \(\frac{x^{n+1}}{(n+1)!} \to 0\) as \(n \to \infty\) for every \(x\) in \([0,c]\text{.}\) Further, since \(e^c\) is a constant independent of \(n\text{,}\) \(e^c \frac{x^{n+1}}{(n+1)!} \to 0\) as well. Thus,

      \begin{equation*} \lim_{n \to \infty} |P_n(x) - e^x| = 0\text{,} \end{equation*}

      as desired.

   2. When \(x \lt 0\text{,}\) we know \(e^x \lt 1\text{.}\) Thus, we can repeat our argument in (a) but replace \(e^c\) with \(1\text{,}\) and everything else holds in the same way.

   3. Because we have shown that the Taylor series for \(e^x\) converges to \(e^x\) for both every nonnegative \(x\)-value and for every negative \(x\)-value, it follows that we have convergence for every value of \(x\text{.}\)

3. Since \(e^x\) is increasing on \([0,5]\) we know that \(e^x \lt e^5\) on \([0,5]\text{.}\) Now \(e^5 \lt 243\text{,}\) so

   \begin{equation*} \left|P_n(5) - e^5\right| \leq 243\frac{|5|^{n+1}}{(n+1)!}\text{.} \end{equation*}

   We want a value of \(n\)that makes this error term less than \(10^{-8}\text{.}\) Testing various values of \(n\) gives

   \begin{equation*} 243\frac{|5|^{28+1}}{(28+1)!} \approx 5.119146745 \times 10^{-9} \end{equation*}

   so we can choose \(n = 28\text{.}\) A computer algebra system shows that \(P_{28}(5) \approx 148.413159102551\) while \(e^5 \approx 148.413159102577\) and we can see that these two approximations agree to 8 decimal places.