Coordinated Calculus

1. \(\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k\) for \(-1 \lt x \lt 1\text{.}\)

2. \(f(x) = g(-x^2) = \sum_{k=0}^{\infty} (-1)^k x^{2k}\text{.}\)

3. \(-1 \lt x \lt 1\text{.}\)

1. Recall that

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

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

2. Substituting \(-x^2\) for \(x\) in the power series expansion for \(g(x)\) gives

   \begin{align*} f(x) \amp = g(-x^2)\\ \amp = \frac{1}{1-(-x^2)}\\ \amp = \sum_{k=0}^{\infty} \left(-x^2\right)^k\\ \amp = \sum_{k=0}^{\infty} (-1)^k x^{2k}\text{.} \end{align*}

3. This power series expansion for \(f(x)\) will be valid as long as \(-1 \lt (-x)^2 \lt 1\) or for \(-1 \lt x \lt 1\text{.}\)

We'll begin with the Taylor series for \(\sin(x)\) about \(x=0\text{,}\) and build the desired series from there.

Next, substitute \(t^3\) in for \(x\text{:}\)

Now multiply the entire series by \(t\text{:}\)

And finally, subtract t^4:

Now that we've got the final series, we can write this with \(\Sigma\)-notation, as well:

with interval of convergence \((-1/2, 1/2) \text{.}\)

We write \(f(t)=\frac{1}{\sqrt{1+2t}} = (1+2t)^{-\frac{1}{2}}.\) Note that the Taylor series about \(0 \) for the function \((1+x)^{-\frac{1}{2}} \) is given by

This series coverges for \(x \) in the interval \((-1, 1) \text{.}\) Substituting \(2t \) for \(x \) yields

which simplifies to

For the series to coverge, we need \(2t \) to fall in the interval \((-1, 1) \text{,}\) which means the interval of convergence is \((-1/2, 1/2) \text{.}\)

1. \(f'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots + (-1)^k \frac{x^{k-1}}{(k-1)!} + \cdots \text{.}\)

2. \(\frac{d}{dx} \cos(x) = -\sin(x) \text{.}\)

3. \(f''(x) = - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots \text{.}\)

1. Note that

   \begin{equation*} f'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots + (-1)^k \frac{x^{k-1}}{(k-1)!} + \cdots\text{.} \end{equation*}

2. We recognize \(f(x)\) as \(\cos(x)\) and \(f'(x)\) as \(-\sin(x)\text{.}\) This gives the known differentiation formula

   \begin{equation*} \frac{d}{dx} \cos(x) = -\sin(x)\text{.} \end{equation*}

3. We see that

   \begin{equation*} f''(x) = - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\text{,} \end{equation*}

   which is the series expansion for \(-\cos(x)\) as expected.

1. \(\int \sin(x^2) \, dx = \sum_{k=0}^{\infty} (-1)^k\frac{x^{4k+3}}{(2k+1)!(4k+3)} + C \text{.}\)

2. \(\int_0^1 \sin(x^2) \, dx = \sum_{k=0}^{\infty} (-1)^k\frac{1}{(2k+1)!(4k+3)} \text{.}\) Use \(n = 1\) to generate the desired estimate.

1. Substituting \(x^2\) for \(x\) in the Taylor series expansion for \(\sin(x) \) gives us

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

   as the Taylor series for \(\sin(x^2)\) centered at \(x=0\text{.}\) We can integrate a Taylor series term by term on its interval of convergence, so

   \begin{align*} \int \sin(x^2) \, dx &= \int \sum_{k=0}^{\infty} (-1)^k\frac{x^{2(2k+1)}}{(2k+1)!} \, dx\\ &= \sum_{k=0}^{\infty} \int (-1)^k\frac{x^{2(2k+1)}}{(2k+1)!} \, dx\\ &= \sum_{k=0}^{\infty} \int (-1)^k\frac{x^{4k+2}}{(2k+1)!} \, dx\\ &= \sum_{k=0}^{\infty} (-1)^k\frac{x^{4k+3}}{(2k+1)!(4k+3)} + C\text{.} \end{align*}

2. To evaluate \(\int_0^1 \sin(x^2) \ dx\text{,}\) we evaluate the antiderivative \(\sum_{k=0}^{\infty} (-1)^k\frac{x^{4k+3}}{(2k+1)!(4k+2)}\) at \(1\) and \(0\) and subtract. So

   \begin{align*} \int_0^1 \sin(x^2) \, dx &= \left. \sum_{k=0}^{\infty} (-1)^k\frac{x^{4k+3}}{(2k+1)!(4k+3)}\right|_{0}^{1}\\ &= \sum_{k=0}^{\infty} (-1)^k\frac{1}{(2k+1)!(4k+3)}\text{.} \end{align*}

   The resulting series is an alternating series, so we can find a value of \(n\) that makes the \(n\)th partial sum of the series approximate the sum of the series to \(3\) decimal places by determining a value of \(n\) such that

   \begin{equation*} a_{n+1} = \frac{1}{(2(n+1)+1)!(4(n+1)+3)} \lt 0.001\text{.} \end{equation*}

   This is not an inequality that we can solve algebraically, so we use trial and error. Doing so shows that \(n = 1\) will work.

Use the Taylor series expansion for \(\frac{1}{1+x}\) centered at \(x=0\text{.}\)

\(\ln(1+x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{k+1}}{k+1} \text{.}\) for \(-1 \lt x \lt 1\text{.}\)

Using the Taylor series

for \(-1 \lt x \lt 1\) we have that

Integrating the left and right sides of this last equation gives us

for \(-1 \lt x \lt 1\text{.}\) Since \(\ln(1) = 0\) we have that

and so \(C = 0\text{.}\) We conclude that

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

1. While we could differentiate \(\arctan(x)\) repeatedly and look for patterns in the derivative values at \(x = 0\) in an attempt to find the Maclaurin series for \(\arctan(x)\) from the definition, it turns out to be far easier to use a known series in an insightful way. By thinking of the right-hand side as a geometric series, we can see that

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

   for \(-1 \lt x \lt 1\text{.}\) Recall that

   \begin{equation*} \frac{d}{dx} \left[ \arctan(x) \right] = \frac{1}{1+x^2}\text{,} \end{equation*}

   and therefore

   \begin{equation*} \int \frac{1}{1+x^2} \ dx = \arctan(x) + C\text{.} \end{equation*}

   It follows that we can integrate the series for \(\frac{1}{1+x^2}\) term by term to obtain the power series expansion for \(\arctan(x)\text{.}\) Doing so, we find that

   \begin{align*} \arctan(x) \amp = \int \left( \sum_{k=0}^{\infty} (-1)^kx^{2k} \right) \ dx\\ \amp = \sum_{k=0}^{\infty} \left( \int (-1)^k x^{2k} \ dx \right)\\ \amp = \left( \sum_{k=0}^{\infty} (-1)^k\frac{x^{2k+1}}{2k+1} \right) + C\text{.} \end{align*}

   The Power Series Differentiation and Integration Theorem tells us that this equality is valid for at least \(-1 \lt x \lt 1\text{.}\)

   To find the value of the constant \(C\text{,}\) we can use the fact that \(\arctan(0) = 0\text{.}\) So

   \begin{equation*} 0 = \arctan(0) = \left( \sum_{k=0}^{\infty} (-1)^k\frac{0^{2k+1}}{2k+1} \right) + C = C\text{,} \end{equation*}

   and we must have \(C = 0\text{.}\) Therefore,

   \begin{equation} \arctan(x) = \sum_{k=0}^{\infty} (-1)^k\frac{x^{2k+1}}{2k+1}\label{E-arctanseries}\tag{7.26} \end{equation}

   for at least \(-1 \lt x \lt 1\text{.}\)

   It is a straightforward exercise to check that the power series

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

   converges both when \(x = -1\) and when \(x = 1\text{;}\) in each case, we have an alternating series with terms \(\frac{1}{2k+1}\) that decrease to \(0\text{,}\) and thus the interval of convergence for the series expansion for \(\arctan(x)\) in Equation(7.26) is \(-1 \le x \le 1\text{.}\)

2. To make the computations a little easier, let's write out the first five terms of the Taylor series explicitly:

   \begin{equation*} \arctan(x)= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9} - \cdots \end{equation*}

   With \(x=1\text{,}\) we have

   \begin{equation*} \arctan(1)=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots \end{equation*}

   So, if we use just the first term to approximate, we would have

   \begin{equation*} \pi = 4\arctan(1) \approx 4(1)=4, \end{equation*}

   giving the table entry for \(n=1\text{.}\) For \(n=2\text{,}\) we use the first two terms:

   \begin{equation*} \pi = 4 \arctan(1) \approx 4\left(1-\frac{1}{3}\right)=4\left(\frac{2}{3}\right) \approx 2.6667. \end{equation*}

   Continuing, for \(n=3\text{:}\)

   \begin{equation*} \pi \approx 4\left(1-\frac{1}{3}+\frac{1}{5}\right) = 4 \left(\frac{13}{15}\right)\approx 3.4667. \end{equation*}

   The completed table is as follows.

   \(n\)	Approximate Value
   1	4
   2	2.6667
   3	3.4667
   4	2.8952
   5	3.3397

   Since we know that \(\pi\) is about 3.14159, we haven't gotten extremely accurate yet, but we are getting there, and continuing to use more terms would yield more accuracy.

1. \(1-\cos(y) \leq \ln(1+y^2)\leq \sin(y^2) \)

2. About 0.84145.

1. We'll need to build all three Taylor series, and then compare them. There's a brief outline of each construction below.

   For \(\sin(y^2)\text{:}\)

   \begin{align*} \sin(x)\amp=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} + \frac{x^9}{9!} - \cdots \\ \sin(y^2) \amp=y^2 - \frac{y^6}{3!} +\frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \frac{x^{18}}{9!} \end{align*}

   For \(1-\cos(y)\text{:}\)

   \begin{align*} \cos(y)\amp=1-\frac{y^2}{2!}+\frac{y^4}{4!}-\frac{y^6}{6!} +\frac{y^8}{8!} - \cdots \\ 1-\cos(y) \amp= \frac{y^2}{2!} - \frac{y^4}{4!} + \frac{y^6}{6!} - \frac{y^8}{8!} + \cdots \end{align*}

   For \(\ln(1+y^2)\)

   \begin{align*} \ln(1+t) \amp= t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4} + \frac{t^5}{5} - \cdots \\ \ln(1+y^2)\amp= y^2 - \frac{y^4}{2}+\frac{y^6}{3}-\frac{y^8}{4} + \frac{y^{10}}{5} - \cdots \end{align*}

   Now, let's use, say, the degree six taylor polynomials as an estimate, and compare the three series. For values of \(y\) near 0, we have

   \begin{align*} \sin(y^2) \amp\approx y^2 - \frac{y^6}{3!} \\ 1-\cos(y) \amp\approx \frac{y^2}{2!} - \frac{y^4}{4!} + \frac{y^6}{6!}\\ \ln(1+y^2) \amp\approx y^2 - \frac{y^4}{2} + \frac{y^6}{3} \end{align*}

   We can compare term-by-term and decide which is smaller, or we can plug in a value near 0 to all three approximations; say, \(y=0.5\)

   \begin{align*} \sin((0.5)^2) \amp \approx (0.5)^2 - \frac{(0.5)^6}{3!} \approx 0.2474 \\ 1-\cos(0.5) \amp\approx \frac{(0.5)^2}{2!} - \frac{(0.5)^4}{4!} + \frac{(0.5)^6}{6!} \approx 0.1224\\ \ln(1+(0.5)^2) \amp\approx (0.5)^2 - \frac{(0.5)^4}{2} + \frac{(0.5)^6}{3} \approx 0.2234 \end{align*}

   Thus, the order is \(1-\cos(y) \leq \ln(1+y^2)\leq \sin(y^2) \)

2. We'll plug in \(y=1\) to the Taylor series for \(sin(y^2)\text{:}\)

   \begin{equation*} \sin(1) \approx 1^2 - \frac{1^6}{3!} + \frac{1^{10}}{5!} - \frac{1^{14}}{7!} \approx 0.84145 \end{equation*}

The function is \(f(x)=\cos(x)\) and the sum of the series is \(\cos(2)\text{.}\)

This series looks a lot like

where \(x=2\) has been plugged in. We can recognize that this is the Taylor series for \(\cos(x)\) centered at 0, and since it converges for all real numbers \(x\text{,}\) the sum of the series must be \(\cos(2)\text{.}\)

1. \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1 \)

2. \(\lim_{x \to 0} \frac{1-\cos(x^{2})}{x^{4}} = \frac{1}{2}\)

3. \(\lim_{x \to 0} \frac{\ln(x^{6}+1)}{5x^{6}} = \frac{1}{5}\)

1. The Taylor series for \(\sin(x) \) about \(0 \) is \(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots \text{.}\) Therefore,

   \begin{align*} \lim_{x \to 0}\frac{\sin(x)}{x}=\amp \lim_{x \to 0}\frac{x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots}{x} \\ =\amp \lim_{x \to 0}1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\cdots \\ =\amp 1 \end{align*}

2. The Taylor series for \(\cos(x) \) about \(0 \) is \(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots \text{.}\) By composing with \(x^{2}\) and subtracting the resulting series from \(1 \text{,}\) we obtain

   \begin{equation*} 1-\cos(x^{2}) = 1-(1-\frac{x^{4}}{2!}+\frac{x^{8}}{4!}-\cdots)= \frac{x^{4}}{2!}-\frac{x^{8}}{4!}+\cdots \end{equation*}
   \begin{align*} \lim_{x \to 0}\frac{1-\cos(x^{2})}{x^{4}}=\amp \lim_{x \to 0}\frac{\frac{x^{4}}{2!}-\frac{x^{8}}{4!}+\cdots}{x^{4}} \\ =\amp \lim_{x \to 0} \frac{1}{2}-\frac{x^{4}}{4!}+\cdots \\ =\amp \frac{1}{2} \end{align*}

3. The Taylor series for \(\ln(x+1) \) about \(0 \) is \(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots \text{.}\) By composing with \(x^{6} \text{,}\) we reach

   \begin{align*} \lim_{x \to 0}\frac{\ln(x^{6}+1)}{5x^{6}} = \amp \lim_{x \to 0}\frac{x^{6}-\frac{x^{12}}{2}+\frac{x^{18}}{3}-\cdots}{5x^{6}} \\ =\amp\lim_{x \to 0}\frac{1}{5}-\frac{x^{6}}{10}+\frac{x^{12}}{15}-\cdots \\ =\amp \frac{1}{5} \end{align*}