Calculus Early Transcendentals: Integral & Multi-Variable Calculus for Social Sciences

Find the radius and interval of convergence for each series. In part c), do not attempt to determine whether the endpoints are in the interval of convergence.

1. \(\ds\sum_{n=0}^\infty n x^n\)

   Answer
   \(R=1\text{,}\) \(I=(-1,1)\)
   Solution
   Applying the Ratio Test to the power series gives

   \begin{equation*} \begin{split} L \amp= \lim_{n\to\infty} \left\vert \frac{(n+1)x^{n+1}}{nx^n}\right\vert\\ \amp= \lim_{n\to\infty} \left\vert x \frac{n+1}{n}\right\vert\\ \amp= |x| \lim_{n\to\infty} \left\vert 1+ \frac{1}{n}\right\vert\\ \amp= |x|\end{split} \end{equation*}

   Hence, The radius of convergence is \(R=1\text{.}\) We now investigate what happens when \(x=-1\text{:}\) the series

   \begin{equation*} \sum_{n=0}^{\infty} (-1)^n n \end{equation*}

   is a divergent series, and similarly for

   \begin{equation*} \sum_{n=0}^{\infty} n. \end{equation*}

   Therefore, the endpoints \(x=\pm 1\) are not included in the interval of convergence. The interval of convergence is thus \(I=(-1,1)$\text{.}\)

2. \(\ds\sum_{n=0}^\infty {x^n\over n!}\)

   Answer
   \(R=\infty\text{,}\) \(I=(-\infty,\infty)\)
   Solution

   Applying the Ratio Test to the power series gives

   \begin{equation*} \begin{split} L \amp = \lim_{n\to\infty} \left\vert \frac{x^{n+1}}{(n+1)!} \frac{n!}{x^n}\right\vert \\ \amp = \lim_{n\to\infty} \left\vert \frac{x^{n+1}}{x^n} \frac{1}{n+1}\right\vert \\ \amp = \lim_{n\to\infty} \left\vert \frac{x}{n+1} \right\vert \\ \amp = |x| \lim_{n\to\infty} \frac{1}{n+1}= 0. \end{split} \end{equation*}

   And so the series

   \begin{equation*} \sum_{n=0}^\infty \frac{x^n}{n!} \end{equation*}

   converges for all \(x\text{,}\) i.e., the radius of convergence is \(R=\infty\) and the interval of convergence is \((-\infty,\infty)\text{.}\)

3. \(\ds\sum_{n=1}^\infty {n!\over n^n}(x-2)^n\)

   Answer
   \(R=e\text{,}\) \(I=(2-e,2+e)\)
   Solution
   We apply the Ratio Test:

   \begin{equation*} \begin{split} L \amp= \lim_{n\to\infty} \left\vert \frac{(n+1)!n^n(x-2)^{n+1}}{n!n^n(x-2)^n}\right\vert\\ \amp= |x-2| \lim_{n\to\infty} \left\vert \frac{(n+1)n^n}{(n+1)^{n+1}}\right\vert\\ \amp= |x-2| \lim_{n\to\infty} \left\vert \frac{n^n}{(n+1)^n}\right\vert\\ \amp= |x-2|\lim_{n\to\infty} \left(\frac{1}{1+1/n}\right)^{1/n} = |x-2| \frac{1}{e}. \end{split} \end{equation*}

   Therefore, the radius of convergence is \(R= e\text{.}\) It turns out the endpoints are not included in the interval of convergence: \((2-e,2+e)\text{.}\)

4. \(\ds\sum_{n=1}^\infty {(n!)^2\over n^n}(x-2)^n\)

   Answer
   \(R=0\text{,}\) converges only when \(x=2\)
   Solution
   We apply the Ratio Test:

   \begin{equation*} \begin{split} L \amp= |x-2| \lim_{n\to\infty} \left(\frac{(n+1)!^2}{(n+1)^{n+1}} \frac{n^n}{n!^2}\right)\\ \amp= |x-2|\lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} \frac{(n+1)!^2}{n!^2}\right)\\ \amp= |x-2|\lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} (n+1)^2\right)\\ \amp= |x-2|\lim_{n\to\infty} \left(\frac{n^n}{(n+1)^(n-1)}\right) \end{split} \end{equation*}

   Now take the logarithm of this limit:

   \begin{equation*} \log\left( \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{(n-1)}}\right)\right) = \lim_{n\to\infty} n\log n - (n-1)\log(n+1) = \infty. \end{equation*}

   Therefore, the series converges only if \(x=2\text{.}\) Therefore, the radius of convergence is \(R=0\) and the interval of convergence is \(I=2\text{.}\)

5. \(\ds\sum_{n=1}^\infty {(x+5)^n\over n(n+1)}\)

   Answer
   \(R=1\text{,}\) \(I=[-6,-4]\)
   Solution

   We use the Ratio Test to determine the radius of convergence:

   \begin{equation*} \begin{split} L \amp = \lim_{n\to\infty} \left\vert\frac{(x+5)^{n+1}}{(n+1)(n+2)} \frac{n(n+1)}{(x+5)^n} \right\vert\\ \amp = \lim_{n\to\infty} \left\vert \frac{(x+5) n}{n+2} \right\vert \\ \amp = |x+5| \lim_{n\to\infty} \frac{n}{n+2} \\ \amp = |x+5| \lim_{n\to\infty} \frac{1}{1+2/n} = |x+5| \end{split} \end{equation*}

   So the radius of convergence is \(R=1\text{,}\) and the power series diverges for all \(x \lt -6\) and \(x > -4\text{.}\) We now determine the convergence at the endpoints. For \(x=-6\text{,}\) the series becomes

   \begin{equation*} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n(n+1)} \end{equation*}

   which converges by the Alternating Series Test. When \(x=-4\) the series becomes

   \begin{equation*} \sum_{n=0}^{\infty} \frac{1}{n(n+1)}\text{.} \end{equation*}

   This series converges since

   \begin{equation*} \frac{1}{n(n+1)} \lt \frac{1}{n^2}\text{.} \end{equation*}

   Therefore, the interval of convergence is \([-6,-4]\text{.}\)

6. \(\ds\sum_{n=1}^\infty {n!\over n^n}x^n\)

   Answer
   \(R=e\text{,}\) \(I=(-e,e) \)
   Solution
   We apply the Ratio Test:

   \begin{equation*} \begin{split} L \amp= |x| \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} \frac{(n+1)!}{n!}\right)\\ \amp= |x| \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} (n+1)\right)\\ \amp= |x| \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^n}\right)\\ \amp= |x| \lim_{n\to\infty} \frac{1}{(1+\frac{1}{n})^n} = |x| \frac{1}{e}. \end{split} \end{equation*}

   Threfore, the radius of convergence is \(R=e \text{.}\) We now test whether or not the endpoints are included in the interval of convergence. Since neither

   \begin{equation*} \sum_{n=1}^{\infty} (x-e)^n \frac{n!}{n} \end{equation*}

   nor

   \begin{equation*} \sum_{n=1}^{\infty} (x+e)^n \frac{n!}{n} \end{equation*}

   converge, the interval of convergence is \(I=(-e,e) \text{.}\)

Find a series representation for

1. \(\ln 2\)

   Answer
   the alternating harmonic series
   Solution
   We know that we can write

   \begin{equation*} \ln|1-x| = \sum_{n=0}^{\infty} \frac{-1}{n+1} x^{n+1}, \end{equation*}

   for \(x \in [-1,1)\text{.}\) Therefore,

   \begin{equation*} \ln(2) = \ln|1-(-1)| = \sum_{n=0}^{\infty} \frac{-1}{n+1} (-1)^{n+1} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}, \end{equation*}

   which is the alternating harmonic series.

2. \(\ln (9/2)\)

   Answer
   \(\ds\sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{7}{9}\right)^{n}\)
   Solution
   We know that we can write

   \begin{equation*} \ln|1-x| = \sum_{n=0}^{\infty} \frac{-1}{n+1} x^{n+1}, \end{equation*}

   for \(x \in [-1,1)\text{.}\) Therefore, let \(\ln(9/2) = -\ln(2/9)\text{.}\) Then

   \begin{equation*} \ln(2/9) = -\ln|1-(7/9)| = -\sum_{n=0}^{\infty} \frac{-1}{n+1} (7/9)^{n+1} = \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{7}{9}\right)^{n}. \end{equation*}

Find a power series representation for the following functions.

1. \(\ds 1/(1-x)^2\)

   Answer
   \(\ds\sum_{n=0}^\infty (n+1)x^n\)
   Solution

   We know that

   \begin{equation*} \frac{1}{(1-x)^2} = \diff{}{x} \frac{1}{1-x}\text{,} \end{equation*}

   and that

   \begin{equation*} \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \end{equation*}

   for \(|x| \lt 1\text{.}\) Combining this information, we find the following power series representation:

   \begin{equation*} \frac{1}{(1-x)^2} = \diff{}{x} \sum_{n=0}^{\infty} x^n = \sum_{n=-1}^{\infty} nx^{n-1} = \sum_{n=0}^{\infty} (n+1)x^n\text{,} \end{equation*}

   for \(|x| \lt 1\text{.}\)

2. \(\ds 2/(1-x)^3\)

   Answer
   \(\ds\sum_{n=0}^\infty (n+1)(n+2)x^n\)
   Solution
   We have

   \begin{equation*} \frac{2}{(1-x)^3} = \diff{}{x} \frac{1}{(1-x)^2}. \end{equation*}

   Therefore,

   \begin{equation*} \frac{2}{(1-x)^3} = \diff{}{x} \sum_{n=0}^{\infty} (n+1) x^n = \sum_{n=-1}^{\infty} n(n+1)x^{n-1} = \sum_{n=0}^{\infty} (n+1)(n+2) x^n, \end{equation*}

   for \(|x| \lt 1\text{.}\)

3. \(\ds 1/(1-x)^3\)

4. \(\ds\int\ln(1-x)\,dx\)

   Answer
   \(\ds C+\sum_{n=0}^\infty {-1\over (n+1)(n+2)}x^{n+2}\)
   Solution
   We know that

   \begin{equation*} \ln|1-x| = \sum_{n=1}^{\infty} \frac{-1}{n} x^n. \end{equation*}

   Therefore,

   \begin{equation*} \int \ln|1-x| \,dx = \int \sum_{n=1}^{\infty} \frac{-1}{n+1} x^{n+1}\,dx = C + \sum_{n=0}^{\infty} \frac{-1}{(n+1)(n+2)}x^{n+2}, \end{equation*}

   for \(|x| \lt 1\text{.}\)

For each function, find the Maclaurin series and the radius of convergence.

1. \(f(x)=\cos x\)

   Answer
   \(\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}\text{,}\) \(R=\infty\text{.}\)
   Solution
   We compute:

   \begin{equation*} \begin{split} f(0) \amp= \cos (0) = 1\\ f'(0) \amp= -\sin(0) = 0\\ f''(0) \amp= -\cos(0) = -1 \\ f'''(0) \amp= \sin(0) = 0\\ f^{(iv)}(0) \amp= \cos(0) = 1 \vdots \end{split} \end{equation*}

   Therefore, our Maclaurin series expansion is

   \begin{equation*} \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}. \end{equation*}

   We now compute the radius of convegence using the Ratio Test:

   \begin{equation*} L = |x|^2 \lim_{n\to\infty} \left\vert \frac{1}{(2n+1)(2n+2)}\right\vert = 0. \end{equation*}

   Hence, \(R=\infty.\)

2. \(f(x)=\ds e^x\)

   Answer
   \(\ds\sum_{n=0}^{\infty} \frac{x^n}{n!}\text{,}\) \(R=\infty.\)
   Solution
   We compute:

   \begin{equation*} \begin{split} f(0) \amp= e^{(0)} = 1\\ f'(0) \amp= e^{(0)} = 1\\ f''(0) \amp= e^{(0)} = 1 \\ \vdots \end{split} \end{equation*}

   Therefore, our Maclaurin series expansion is

   \begin{equation*} e^{x} = 1 + x+ \frac{x^2}{2!} + \frac{x^43}{3!} + \dots = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \end{equation*}

   We now compute the radius of convegence using the Ratio Test:

   \begin{equation*} L = |x| \lim_{n\to\infty} \left\vert \frac{1}{n+1}\right\vert = 0. \end{equation*}

   Hence, \(R=\infty.\)

3. \(f(x)=\dfrac{1}{1+x}\)

   Answer
   \begin{equation*} \sum_{n=0}^{\infty} (-1)^nx^n \text{ for } |x| \lt 1 \end{equation*}
   Solution

   The simplest method for determining the Maclaurin series for \(f(x) = \dfrac{1}{1+x}\) is to use a known series expansion. For example, we know that

   \begin{equation*} \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \text{ for } |x| \lt 1\text{,} \end{equation*}

   and so if we make the substitution \(x=-x\text{,}\) we find

   \begin{equation*} \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^nx^n \text{ for } |x| \lt 1\text{.} \end{equation*}

   We could similarly use the fact that

   \begin{equation*} \frac{1}{1+x} = \diff{}{x} \ln(1+x) = \diff{}{x}\left(\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}\right)\text{,} \end{equation*}

   or we could develop the desired Maclaurin series from scratch: The Maclaurin series expansion of a function \(f\) is given by

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

   and so we calculate

   \begin{equation*} \begin{split} f(x) \amp = \frac{1}{1+x} \\ f'(x) \amp = -\frac{1}{(1+x)^2} \\ f''(x) \amp = \frac{2}{(1+x)^3} \\ f'''(x) \amp = -\frac{6}{(1+x)^4} \\ \amp \vdots \\ f^{(n)}(x) \amp = \frac{(-1)^{n-1}n!}{(1+x)^{n+1}} \end{split} \end{equation*}

   Thus,

   \begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n n!}{(1+0)^{n+1} n!}x^n =\sum_{n=0}^{\infty} (-1)^n x^n\text{.} \end{equation*}

   We compute the radius of convergence using the Ratio Test:

   \begin{equation*} \begin{split} L \amp = \lim_{n\to\infty} \left\vert \frac{(-1)^{n+1}x^{n+1}}{(-1)^n x^n} \right\vert \\ \amp = \lim_{n\to\infty} \left\vert -x \right\vert = x, \end{split} \end{equation*}

   and so \(R=1\) as we found above.

4. \(f(x)=\sin(2x)\)

   Answer
   \(\ds \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1} x^{2n+1}}{(2n+1)!}\text{,}\) \(R=\infty.\)
   Solution
   We use the fact that

   \begin{equation*} \sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \end{equation*}

   with \(R=\infty\text{.}\) Now let \(x=2y\text{.}\) Then,

   \begin{equation*} \sin(2y) = \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1} y^{2n+1}}{(2n+1)!}, \end{equation*}

   with \(R=\infty\text{.}\)

For each function, find the Taylor series centred at \(a\) and the radius of convergence.

1. \(f(x) = \dfrac{1}{x}\text{,}\) \(a=5\)

   Answer
   \(\ds\sum_{n=0}^\infty (-1)^n{(x-5)^n\over 5^{n+1}}\text{,}\) \(R=5\)
   Solution

   Use the formula for the Taylor Series expansion of \(f\) about \(a=5\text{:}\)

   \begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(5)}{n!}(x-5)^n \end{equation*}

   Therefore, we compute

   \begin{equation*} \begin{array}{ll} f(x) = \dfrac{1}{x} \amp f(5) = \dfrac{1}{5} \\[1.5ex] f'(x) = -\dfrac{1}{x^2} \amp f'(5) = -\dfrac{1}{25}\\[1.5ex] f''(x) = \dfrac{2}{x^3} \amp f''(5) = \dfrac{2}{125} \\[1.5ex] f'''(x) = -\dfrac{6}{x^4} \amp f'''(5) = -\dfrac{6}{625} \end{array} \end{equation*}

   and so

   \begin{equation*} f^{(n)}(5) = \dfrac{(-1)^{n}n!}{5^{n+1}} \text{ for } n \geq 0\text{.} \end{equation*}

   Thus, the Taylor Series expansion for \(f(x)=\dfrac{1}{x}\) about \(a=5\) is

   \begin{equation*} \frac{1}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{5^{n+1}}(x-5)^n\text{.} \end{equation*}

   To determine the radius of convergence, we use the Ratio Test:

   \begin{equation*} \begin{split} L \amp = \lim_{n\to\infty} \left\vert \frac{(-1)^{n+1}(x-5)^{n+1}}{5^{n+2}} \frac{5^{n+1}}{(-1)^n (x-5)^n} \right\vert \\ \amp = \lim_{n\to\infty} \left\vert \frac{-(x-5)}{5} \right\vert \\ \amp = \frac{1}{5} (x-5) \end{split} \end{equation*}

   Therefore, \(R=5\text{.}\)

2. \(f(x)=\ln x\text{,}\) \(a=1\)

   Answer
   \(\ds\sum_{n=1}^\infty (-1)^{n-1}{(x-1)^n\over n}\text{,}\) \(R=1\)
   Solution

   Use the formula for the Taylor Series expansion of \(f\) about \(a=1\text{,}\)

   \begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n \end{equation*}

   Notice that \(f(1) = \ln(1) = 0\text{,}\) and so we have

   \begin{equation*} f(x) = 0 + \sum_{n=1}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n\text{.} \end{equation*}

   We compute the first few derivatives evaluated at \(1\) to discern a pattern:

   \begin{equation*} \begin{array}{ll} f'(x) = \dfrac{1}{x} \amp f'(1) = 1\\[1.5ex] f''(x) = -\dfrac{1}{x^2} \amp f''(1) = -1 \\[1.5ex] f'''(x) = \dfrac{2}{x^3} \amp f'''(1) = 2 \end{array} \end{equation*}

   and so

   \begin{equation*} f^{(n)}(1) = (-1)^{n-1}(n-1)! \text{ for } n \geq 1\text{.} \end{equation*}

   Therefore, about \(a=1\) we have

   \begin{equation*} \ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (x-1)^n \end{equation*}

   with radius of convergence \(R=1\text{.}\)

3. \(f(x)=\ln x\text{,}\) \(a=2\)

   Answer
   \(\ds\ln(2)+\sum_{n=1}^\infty (-1)^{n-1}{(x-2)^n\over n 2^n}\text{,}\) \(R=2\)
   Solution
   Use the formula for the Taylor Series expansion of \(f\) about \(a=2\text{,}\)

   \begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n. \end{equation*}

   We compute:

   \begin{equation*} \begin{array}{ll} f'(x) = \dfrac{1}{x} \amp f'(2) = \frac{1}{2}\\[1.5ex] f''(x) = -\dfrac{1}{x^2} \amp f''(2) = -\frac{1}{4} \\[1.5ex] f'''(x) = \dfrac{2}{x^3} \amp f'''(2) = \frac{1}{4}\\[1.5ex] f^{(iv)}(x) = \frac{-6}{x^4} \amp f^{(iv)}(2) = \frac{-3}{8} \end{array} \end{equation*}

   Therefore, we have

   \begin{equation*} f(x) = \ln(2) + \frac{x-2}{2} - \frac{1}{8} (x-2)^2 + \frac{1}{24} (x-2)^3 - \frac{1}{64} (x-2)^4 + \dots \end{equation*}

   Hence,

   \begin{equation*} \ln(x) = \ln(2) + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(x-2)^n}{n2^n} \end{equation*}

   with radius of convergence \(R=2\text{.}\)

4. \(f(x)=\ds \dfrac{1}{x^2}\text{,}\) \(a=1\)

   Answer
   \(\ds\sum_{n=0}^\infty (-1)^n(n+1)(x-1)^n\text{,}\) \(R=1\)
   Solution
   Use the formula for the Taylor Series expansion of \(f\) about \(a=1\text{,}\)

   \begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n. \end{equation*}

   We compute:

   \begin{equation*} \begin{array}{ll} f'(x) = \dfrac{-2}{x^3} \amp f'(1) = -2\\[1.5ex] f''(x) = -\dfrac{6}{x^4} \amp f''(1) = 6 \\[1.5ex] f'''(x) = \dfrac{-24}{x^5} \amp f'''(1) = -24 \end{array} \end{equation*}

   Therefore, we have

   \begin{equation*} f(x) = 1-2(x-1)+3(x-1)^2-4(x-1)^3 + \dots \end{equation*}

   Hence,

   \begin{equation*} \frac{1}{x^2} = \sum_{n=0}^{\infty} (-1)^{n}(n-1) (x-1)^n \end{equation*}

   with radius of convergence \(R=1\text{.}\)

Find the first four terms of the Maclaurin series for the following functions:

1. \(f(x) = \tan x\)

   Answer
   \begin{equation*} \tan(x) = x + \frac{x^3}{3} +\frac{2 x^5}{15} + \frac{17 x^7}{215} + \dots \end{equation*}
   Solution
   We differentiate until we get gour nonzero terms:

   \begin{equation*} \begin{array}{ll} f(x) = \tan(x) \amp f(0) = 0 \\[1.5ex] f'(x) = \sec^2(x) \amp f'(0) = 1\\[1.5ex] f''(x) = 2\tan(x)\sec^2(x) \amp f''(0) =0\\[1.5ex] f'''(x) = 2\sec^2(x) \bigl(2\tan^2(x) + \sec^2(x)\bigr) \amp f'''(0) = 2\\[1.5ex] f^{(iv)}(x) = 8\tan(x)\sec^2(x)\bigl(\tan^2(x) + 2\sec^2(x)\bigr) \amp f^{(iv)}(0) = 0 \\[1.5ex] f^{(v)}(x) = 8\sec^2(x)\bigl(2\tan^4(x) + 2\sec^4(x) + 11\tan^2(x)\sec^2(x)\bigr) \amp f^{(v)}(0) = 16\\[1.5ex] f^{(vi)}(x) = \dots \amp f^{(vi)}(0) = 0 \\[1.5ex] f^{(vii)}(x) =\dots \amp f^{(vii)}(0) = 272 \end{array} \end{equation*}

   Hence, the first four terms in the Maclaurin series are

   \begin{equation*} \begin{split} \tan(x)\amp = \frac{x}{1!} + 2\frac{x^3}{3!} + 16 \frac{x^5}{5!} + 272 \frac{x^7}{7!} + \dots\\ \amp = x + \frac{x^3}{3} +\frac{2 x^5}{15} + \frac{17 x^7}{215} + \dots \end{split} \end{equation*}

2. \(\ds f(x)=x\cos (x^2)\)

   Answer
   \begin{equation*} x\cos(x^2) = x - \frac{x^5}{2} + \frac{x^9}{24} - \frac{x^13}{720} + \dots \end{equation*}
   Solution
   We differentiate until we get gour nonzero terms:

   \begin{equation*} \begin{array}{ll} f(x) = x\cos(x^2) \amp f(0) = 0\\[1.5ex] f'(x) = \cos(x^2) - 2 x^2 \sin(x^2) \amp f'(0) = 1\\[1.5ex] f''(x) = -6 x \sin(x^2) - 4 x^3 \cos(x^2) \amp f''(0) = 0 \\[1.5ex] f'''(x) = 2 (4 x^4 - 3) \sin(x^2) - 24 x^2 \cos(x^2)\amp f'''(0) = 0 \\[1.5ex] f^{(iv)}(x) = 4 (4 x^4 - 15) x \cos(x^2) + 80 x^3 \sin(x^2) \amp f^{(iv)}(0) = 0 \\[1.5ex] f^{(v)}(x) = 8 (45 - 4 x^4) x^2 \sin(x^2) + 60 (4 x^4 - 1) \cos(x^2) \amp f^{(v)}(0) = -60 \\[1.5ex] \end{array} \end{equation*}

   Eventually, we find that

   \begin{equation*} x\cos(x^2) = x - \frac{x^5}{2} + \frac{x^9}{24} - \frac{x^13}{720} + \dots \end{equation*}

3. \(\ds f(x) = xe^{-x}\)

   Answer
   \begin{equation*} \begin{split} xe^{-x} \amp= x - x^2 + \frac{x^3}{2} - \frac{x^4}{6} + \dots \end{split} \end{equation*}
   Solution
   We differentiate until we get gour nonzero terms:

   \begin{equation*} \begin{array}{ll} f(x) = xe^{-x} \amp f(0) = 0 \\[1.5ex] f'(x) = e^{-x} (1 - x) \amp f'(0) = 1 \\[1.5ex] f''(x) = e^{-x} (x - 2) \amp f''(0) = -2\\[1.5ex] f'''(x) = -e^{-x} (x - 3)\amp f'''(0) = 3\\[1.5ex] f^{(iv)}(x) = e^{-x} (x - 4) \amp f^{(iv)}(0) = -4 \end{array} \end{equation*}

   Therefore, we have

   \begin{equation*} \begin{split} xe^{-x} \amp= \frac{x}{1!} - \frac{2x^2}{2!} + \frac{3x^3}{3!} - \frac{4x^4}{4!} + \dots\\ \amp= x - x^2 + \frac{x^3}{2} - \frac{x^4}{6} + \dots \end{split} \end{equation*}

Find the 5th degree Taylor polynomial for \(f(x)=\sin x\) around \(a=0\text{.}\)

1. Use this Taylor polynomial to approximate \(\sin (0.1)\text{.}\) Answer
   \(T_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}\text{,}\) so \(\sin (0.1)\approx T_5(0.1)\approx 0.10016675\)
2. Use a calculator to find \(\sin (0.1)\text{.}\) How does this compare to our approximation in part (a)? Answer
   \(T_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}\text{,}\) so \(\sin (0.1)=0.0998334\ldots\) using a calculator. Our approximation is accurate to \(0.10016675-0.0998334\ldots =0.000\bar{3}\text{.}\)

Find the 3rd degree Taylor polynomial for \(f(x)=\frac{1}{1-x}-1\) around \(a=0\text{.}\) Explain why this approximation would not be useful for calculating \(f(5)\text{.}\) Answer

Consider \(f(x)=\ln x\) around \(a=1\text{.}\)

1. Find a general formula for \(f^{(n)}(x)\) for \(n\geq 1\text{.}\)

   Answer
   \(f^{(n)}(x)=\frac{(-1)^{(n-1)}(n-1)!}{x^n}\)
   Solution

   We compute the first few derivatives to discern a pattern:

   \begin{equation*} \begin{split} f'(x) = \dfrac{1}{x}\\ f''(x) = -\dfrac{1}{x^2} \\ f'''(x) = \dfrac{2}{x^3} \end{split} \end{equation*}

   and so

   \begin{equation*} f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^{n}} \text{ for } n \geq 1\text{.} \end{equation*}

2. Find a general formula for the Taylor Polynomial, \(T_n(x)\text{.}\)

   Answer
   \(T_n(x)=\ln (1)+\displaystyle\sum_{i=1}^{n} \frac{\big(\frac{(-1)^{(i-1)}(i-1)!}{1^n}\big)}{i!}(x-1)^i=\displaystyle\sum_{i=1}^{n} \bigg(\frac{(-1)^{(i-1)}(i-1)!}{i!}\bigg)(x-1)^i\) since \(\ln (1)=0\) and \(1^n=1\text{.}\)
   Solution

   The general Taylor polynomial of \(f(x) = \ln x\) centred at \(a=1\) is

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

   From part (a), we know that

   \begin{equation*} f^{(n)}(1) = (-1)^{n-1}(n-1)! \end{equation*}

   and so

   \begin{equation*} T_n(x) = \sum_{i=1}^n \frac{f^{(i)}(1)}{i!}(x-1)^i = \frac{(-1)^{i-1}(i-1)!}{i!}(x-1)^i\text{.} \end{equation*}

Approximate \(\ln(1.3)\) to accuracy of at least 0.0001. Answer

Find a polynomial approximation for each of the functions on the given interval within the stated error.

1. \(f(x) = \cos x\text{,}\) \([0,\pi]\text{,}\) \(\ds \pm 10^{-3}\)

   Solution

   We wish to approximate \(f(x) = \cos x\) on the interval \(I=[0,\pi]\text{.}\) By Taylor's Theorem,

   \begin{equation*} \cos x = \sum_{n=0}^N \frac{f^{n}(\pi/2)}{n!}(x-\pi/2)^n + \frac{f^{N+1}(z)}{(N+1)!}(x-\pi/2)^{N+1}\text{,} \end{equation*}

   where we took \(a=\pi/2\text{.}\) We need to determine the number of terms \(N\) required. By Taylor's Theorem, the remainder term is of the form

   \begin{equation*} R_n = \frac{f^{(N+1)}(z)}{(N+1)!} (x-\pi/2)^{N+1} \end{equation*}

   for some \(z\) between \(x\) and \(\pi/2\text{.}\) Therefore, we require

   \begin{equation*} |R_n| = \left\vert \frac{f^{(N+1)}(z)}{(N+1)!} \left(x-\frac{\pi}{2}\right)^{N+1}\right\vert \leq 0.01 \end{equation*}

   We know that \(|f^{(N+1)}(z)| \leq 1\) for all \(z\text{,}\) and that for all \(x \in I\) we have that \(|(x-\pi/2)^{N+1}| \leq (\pi/2)^{N+1}\text{.}\) Thus, we have

   \begin{equation*} |R_n| \leq \left\vert \frac{(\pi/2)^{N+1}}{(N+1)!} \right\vert \leq 0.01\text{.} \end{equation*}

   We find that \(R_6 \approx 0.005 \lt 0.01\text{,}\) and so

   \begin{equation*} \begin{split} \cos x \amp= \sum_{n=0}^{6} \frac{f^{(n)}(\pi/2)}{n!} \left(x-\frac{\pi}{2}\right)^n \pm 0.01\\ \amp= -\left(x-\frac{\pi}{2}\right)+\frac{\left(x-\frac{\pi}{2}\right)^3}{3!}-\frac{\left(x-\frac{\pi}{2}\right)^5}{5!} \pm 0.01. \end{split} \end{equation*}

   We verify the result with the graph below.

   

2. \(f(x) = \ln x\text{,}\) \([1/2,3/2]\text{,}\) \(\ds 10^{-3}\) Answer

   We need 1000 terms.
   Solution
   We wish to approximate \(f(x) = \ln x\) on the interval \(I=[1/2,3/2]\text{.}\) By Taylor's Theorem,

   \begin{equation*} \ln x = \sum_{n=0}^N \frac{f^{n}(1)}{n!}(x-1)^n + \frac{f^{N+1}(z)}{(N+1)!}(x-1)^{N+1}, \end{equation*}

   where we took \(a=1\) (the midpoint of \(I\)). We know that the derivatives of \(\ln(x)\) evaluated at 1 take the form

   \begin{equation*} \left\vert f^{(N)}(x) \right\vert = \left\vert \frac{(N-1)!}{x^N}\right\vert \implies \left\vert f^{(N+1)}(z) \right\vert \leq 2^{N+1} N! \end{equation*}

   for all \(z \in I\text{.}\)Therefore,

   \begin{equation*} \begin{split} |R_N| \amp \leq \left\vert \frac{2^{N+1} N!}{(N+1)!} (x-1)^{N+1} \right\vert\\ \amp = \frac{2^{N+1}}{N+1} \left\vert(x-1)^{N+1}\right\vert \\ \amp \leq \frac{2^{N+1}}{N+1} \left(\frac{1}{2^{N+1}}\right)\\ \amp = \frac{1}{N+1} \end{split} \end{equation*}

   for all \(x \in I \text{.}\) Hence,

   \begin{equation*} |R_N| \leq \frac{1}{N+1}. \end{equation*}

   Thus, if we need an accuracy of at least \(10^{-3}\text{,}\) this means we need at least 1000 terms in the Taylor Series approximation.

3. \(f(x) = \ln x\text{,}\) \([1,3/2]\text{,}\) \(\ds 10^{-3}\) Answer

   We need at least 8 terms.
   Solution
   We wish to approximate \(f(x) = \ln x\) on the interval \(I=[1,3/2]\text{.}\) By Taylor's theorem,

   \begin{equation*} \ln x = \sum_{n=0}^N \frac{f^{n}(1)}{n!}(x-1)^n + \frac{f^{N+1}(z)}{(N+1)!}(x-1)^{N+1}, \end{equation*}

   where we took \(a=1\text{.}\) We know that the derivatives of \(\ln(x)\) evaluated at 1 take the form

   \begin{equation*} \left\vert f^{(N)}(x) \right\vert = \left\vert \frac{(N-1)!}{x^N}\right\vert \implies \left\vert f^{(N+1)}(z) \right\vert \leq N! \end{equation*}

   for all \(z \in I\text{.}\) Therefore,

   \begin{equation*} \begin{split} |R_N| \amp \leq \left\vert \frac{N!}{(N+1)!} (x-1)^{N+1}\right\vert\\ \amp= \frac{1}{N+1} \left\vert(x-1)^{N+1}\right\vert \\ \amp \leq \frac{2^{N+1}}{N+1} \left(\frac{1}{2^{N+1}}\right)\\ \amp = \frac{1}{(N+1) 2^{N+1}} \end{split} \end{equation*}

   for all \(x \in I\text{.}\) Thus, if we need an accuracy of at least \(10^{-3}\text{,}\) this means we need at least 8 terms in the Taylor Series approximation.

Show that each function is equal to its Taylor series for all \(x\) by showing that \(\lim\limits_{n\to\infty}R_n(x) = 0\text{.}\)

1. \(f(x) = \cos x\) Solution

   We compute

   \begin{equation*} |R_N| = \left\vert \frac{f^{(N+1)}(z)}{(N+1)!} (x-a)^N\right\vert. \end{equation*}

   Now, since

   \begin{equation*} \left\vert f^{(N+1)}(z) \right\vert \leq 1 \end{equation*}

   for all \(z\text{,}\) we must have

   \begin{equation*} 0 \leq |R_N| \leq \left\vert \frac{(x-a)^N}{(N+1)!}\right\vert. \end{equation*}

   Now, since

   \begin{equation*} \lim_{N\to\infty} \left\vert\frac{(x-a)^N}{(N+1)!}\right\vert = 0, \end{equation*}

   by the Squeeze Theorem, we must have that

   \begin{equation*} \lim_{N\to\infty} |R_N| = 0. \end{equation*}

2. \(f(x) = e^x\) Solution

   We compute

   \begin{equation*} |R_N| = \left\vert \frac{f^{(N+1)}(z)}{(N+1)!} (x-a)^N\right\vert = \left\vert \frac{e^z}{(N+1)!} (x-a)^N\right\vert . \end{equation*}

   Therefore,

   \begin{equation*} \lim_{N\to\infty} |R_N| = e^z \lim_{N\to\infty} \frac{(x-a)^N}{(N+1)!} = 0. \end{equation*}