Course Notes for Calculus III

I use the limit form for calculating radius of convergence.

The radius of convergence is 1.

I use the limit form for calculating radius of convergence.

The radius of convergence is \(\infty\text{,}\) meaning that the power series converges for all real numbers.

I use the limit form for calculating radius of convergence.

The radius of convergence is 1.

I use the limit form for calculating radius of convergence.

The numerator and denominator have the same asymptotic order and the leading coefficients are both 1, so the limit is 1. The radius of convergence is 1.

I use the limit form for calculating radius of convergence.

The radius of convergence is 1.

I use the limit form for calculating radius of convergence.

The asymptotic order of the numerator is greater than the denominator, so the limit is \(\infty\text{.}\) The radius of convergence is \(\infty\text{,}\) which means that the power series converges for all real numbers.

I use the limit form for calculating radius of convergence.

The numerator has a higher asymptotic order, so the limit is \(\infty\text{.}\) The radius of convergence is \(\infty\text{,}\) meaning that the power series converges for all real numbers.

I use the limit form for calculating radius of convergence.

The numerator has higher asymptotic order, so the limit is \(\infty\text{.}\) The radius of convergence is \(\infty\text{.}\)

I use the limit form for calculating radius of convergence.

Both denominator and numerator have the same asymptotic order and the leading coefficients are both 1, so the limit is 1. (Note that the cosine term, since it doesn't grow at all, has no effect on the asymptotic order.) The radius of convergence is 1.

I follow the steps in the process. First, I find the first few derivatives of the function.

Then I evaluate at the centre point.

I try to set a pattern for these derivatives evaluated at the centre point. There is an alterning sign, which can be represented by \((-1)^{n+1}\text{.}\) I need the exponent \((n+1)\) since the terms are even when the index is odd. Then the numerator is building up a factorial, stoping at one place less than the index: I can write this as \((n-1)!\text{.}\) Finally, the denomniator is constructing a power of 3 with the exponent equal to the index.

Finally, I put this into the form of a Taylor series. Here is the general form.

I put in the centre points and the pattern of the derivatives to determine the Taylor series. However, the first term doens't fit the general pattern for this series, so I have to write the first term outside the \(\Sigma\) notation.

I follow the steps in the process. First, I find the first few derivatives of the function.

Then I evaluate at the centre point.

I try to set a pattern for these derivatives evaluated at the centre point. There is an alternating patterns of zeros and positive/negative ones. Since all the odd terms are zeros, I can ignore the indices \((2n+1)\) and just use the indices \((2n)\text{.}\) Then when \(n\) in the \((2n)\) is even, there are positive values, and when it is odd, there are negative values. Therefore, I can match indices \((2n)\) with a derivative value of \((-1)^n\)

Finally, I put this into the form of a Taylor series. Here is the general form.

I put in the centre points and the pattern of our derivatives to determine the Taylor series. Since the odd indicies are zero, I simply won't include them in the sum.

This is a trick question. I could do all the steps above, and they would certainly work. I wouldn't get a pattern, but I would explicitly get the first five derivatives and all higher derivatives would be zero. I could put those derivatives into the form and calcluate the result. However, what I would get from that process would be exactly what I started with.

This makes an important point. A polynomial is already a Taylor series (at least centred at zero). Since Taylor series are unique

1. Inside the radius \(4\pi\text{,}\) what is the error of the 35th order approximation?

   The radius \(d = 4\pi\) and the order \(n=36\) are already given. To use the Lagrange error bound, I need to calculate \(M\text{.}\) Thankfully, the trig functions are easy to work with here: due to the cycle of derivatives, I can take \(M=1\) for any order. I'll move on by writing the Lagrange error bound.

   \begin{equation*} |R_{35}(x)| \leq \frac{M}{36!} |x-2\pi|^{36} \end{equation*}

   This still has \(|x-2\pi|\text{,}\) giving the error for specific points in the interval. The distance I can set the variable from the centre point is the radius, so the largest possible value of \(|x-2\pi|\) is \(d=4 \pi\text{.}\) I'll make that replacement to get an error bound that applies to the whole interval.

   \begin{equation*} |R_{35}(x)| \leq \frac{1}{36!} (4\pi)^{36} \doteq 0.01002 \end{equation*}

   The error bound is just slightly larger than \(\frac{1}{100}\text{.}\)

2. What radius is required for a 10th order approximation with error bounded by \(10^{-6}\text{?}\)

   As with the first part, I can set \(M=1\text{.}\) The order is given (\(n=10\)), but \(d\) is unknown. Let me start by writing the Lagrange error bound.

   \begin{equation*} |R_{10}(x)| \leq \frac{M}{11!} |x-2\pi|^{11} \end{equation*}

   The maximum value of \(|x-2\pi|\) is the distance \(d\) out from the centre point, so I can replace \(|x - 2\pi\) with \(d\) to get an error bound which applies to the whole interval.

   \begin{equation*} |R_{10}(x)| \leq \frac{1}{11!} d^{11} \end{equation*}

   I want the error to be less that \(10^{-6}\text{,}\) so I need to solve this inequality.

   \begin{align*} \frac{1}{11!} d^{11} \amp \leq \frac{1}{1000000} \\ d^{11} \amp \leq \frac{11!}{1000000} \\ d \amp \leq \sqrt[11]{\frac{11!}{1000000}} \doteq 1.398 \end{align*}

   I conclude that a radius of approximately \(1.398\) is possible while the 10th order approximation has error less than \(10^{-6}\text{.}\)

3. Inside the radius \(2\pi\text{,}\) what order is needed for an error less than \(10^{-1}\text{.}\)

   As with the first two parts, I can set \(M=1\text{.}\) I don't know the order \(n\text{,}\) but I do know \(d=2\pi\) and that the error should be less than \(10^{-1}\text{.}\) I'll start by writing the Lagrange error bound.

   \begin{equation*} |R_{n}(x)| \leq \frac{M}{(n+1)!} |x-2\pi|^{n+1} \end{equation*}

   The maximum value of \(|x-2\pi|\) is the radius, which is \(2\pi\text{.}\) I can make that replacement to get an error bound for the whole interval.

   \begin{equation*} |R_{n}(x)| \leq \frac{1}{(n+1)!} (2\pi)^{n+1} \end{equation*}

   I want this error to be less that \(\frac{1}{10}\text{,}\) so I need to solve this inequality.

   \begin{equation*} \frac{(2\pi)^{n+1}}{(n+1)!} \leq \frac{1}{10} \end{equation*}

   Solving this directly is very difficult due to the factorial term: I don't have algebraic techniques than can deal with the factorial along with the \(n\) in the exponent. However, since I know \(n\) must be a positive integer, I can guess until I find an order that works. I'll put approximate values in the order column in this table.

   \begin{align*} \amp \text{Order} \amp \amp \text{Error Bound} \\ \amp n=4 \amp \amp |R_5(x)| \leq 85.5 \\ \amp n=9 \amp \amp |R_{10}(x)| \leq 26.4 \\ \amp n=14 \amp \amp |R_{15}(x)| \leq 0.718 \\ \amp n=15 \amp \amp |R_{16}(x)| \leq 0.282 \\ \amp n=16 \amp \amp |R_{17}(x)| \leq 0.104 \\ \amp n=17 \amp \amp |R_{18}(x)| \leq 0.0363 \end{align*}

   The 17th order is the first order which has error less than \(\frac{1}{10}\text{.}\)

1. Inside the radius \(\frac{1}{4} \text{,}\) what is the error of the 10th order approximation?

   The radius \(d = \frac{1}{4}\) and the order \(n=10\) are given. To use the Lagrange error bound, I also need the bound of the eleventh derivative. There is a nice pattern of derivatives for \(\ln x\text{,}\) which I used for calculating Taylor series in previous activities. The eleventh derivative is

   \begin{equation*} \frac{d^{11}}{dx^{11}} \ln x = \frac{10!}{x^{11}} \end{equation*}

   I am looking at the interval \(\left( \frac{3}{4}, \frac{5}{4} \right)\text{.}\) This derivative is a decreasing function on this interval, so it has its maximum at the lowest point. I evaluate the eleventh derivative at \(x = \frac{3}{4}\) to get \(M\text{.}\)

   \begin{equation*} M = \frac{d^{11}}{dx^{11}} \ln x \Big|_{x=\frac{3}{4}} = \frac{10!}{\left( \frac{3}{4} \right)^{11}} = \frac{(4^{11})(10!)}{3^{11}} \end{equation*}

   Then I put \(M\text{,}\) \(d\) and \(n\) into the Lagrange error bound.

   \begin{equation*} |R_{10}(x)| \leq \frac{M}{11!} |x-1|^{11} = \frac{(4^{11})(10!)}{(3^{11})(11!)} |x-1|^{11} = \frac{4^{11}}{(3^{11})(11)} |x-1|^{11} \end{equation*}

   Finally, the largest value of \(|x-1|\) on the interval in equation is the radius \(\frac{1}{4}\text{.}\) I can replace \(|x-1|\) with this radius for a bound that applies to the whole interval.

   \begin{equation*} |R_{10}(x)| \leq \frac{4^{11}}{(3^{11})(11)} \frac{1}{4^{11}} = \frac{1}{(3^{11}){11}} = \frac{1}{1948617} \doteq 0.000000513 \end{equation*}

   The error of the 10th order approximation is less than \(0.0000005\text{.}\)

2. What radius is required for a 8th order approximation with error bounded by \(10^{-3}\text{?}\)

   To use Lagrange error bound, I first need \(M\text{.}\) Using the same logic as the first part, the 9th derivative is

   \begin{equation*} \frac{d^{9}}{dx^{9}} \ln x = \frac{8!}{x^{9}} \end{equation*}

   This is a decreasing function, so \(M\) is maximized at the smallest point on the interval. Since I'm trying to find an unknown radius \(d\text{,}\) the smallest point on the interval is unknown and must simply be written. \(1-d\text{.}\)

   \begin{equation*} M = \frac{d^{9}}{dx^{9}} \ln x \Big|_{x=(1-d)} = \frac{8!}{(1-d)^9} \end{equation*}

   I can put this in the Lagrange error bound with \(n=8\text{.}\)

   \begin{equation*} |R_{8}(x)| \leq \frac{M}{9!} |x-1|^{9} = \frac{1}{9 (1-d)^9} |x-1|^9 \end{equation*}

   The maximum value of \(|x-1|\) will be at the futherest point in the interval. This distance is simply the radius \(d\) that I am looking for.

   \begin{equation*} |R_{8}(x)| \leq \frac{1}{9 (1-d)^9} d^9 \end{equation*}

   So I need a radius \(d\) such that \(\frac{d^9}{9(1-d)^9} \leq \frac{1}{1000}\text{.}\) I can solve this inequality. (I will assume \(d \in (0,1)\text{,}\) so that multiplication by \((1-d)\) preserves the inequality.)

   \begin{align*} \frac{d^9}{9(1-d)^9} \amp \leq \frac{1}{1000} \\ \frac{1000}{9} \amp \leq \left(\frac{1-d}{d} \right)^9 = \left( \frac{1}{d} - 1 \right)^9 \\ \sqrt[9]{\frac{1000}{9}} \amp \leq \frac{1}{d} - 1\\ \sqrt[9]{\frac{1000}{9}} + 1 \amp \leq \frac{1}{d} \\ \frac{1}{\sqrt[9]{\frac{1000}{9}} + 1} \amp \geq d \end{align*}

   If I approximate the left side, I get \(d \lt 0.372\text{.}\) This is the size of the interval that fits the desire order and error bound.

3. Inside the radius \(\frac{1}{2}\text{,}\) what order is needed for an error less than \(10^{-2}\text{.}\)

   I know \(d\) and the desired error, but \(n\) is unknown. I have to use the general expression for the \(n\)th derivative.

   \begin{equation*} \frac{d^n}{dx^n} \ln x = \frac{(-1)^{n-1}(n-1)!}{x^n} \end{equation*}

   \(M\) is the bound on the \((n+1)\)st derivative on the interval. The derivative is always increasing, so the derivative is largest at \(x= \frac{1}{2}\text{.}\)

   \begin{equation*} M = \left| \frac{d^{n+1}}{dx^{n+1}} \ln x \Big|_{x=\frac{1}{2}} \right| = \frac{n!}{(\frac{1}{2})^{n+1}} = 2^{n+1} n! \end{equation*}

   I put this in the error bound. The maximum value of \(|x-1|\) on the interval is \(\frac{1}{2}\text{.}\)

   \begin{equation*} |R_n(x)| \leq \frac{M}{(n+1)!} |x-1|^{n+1} \leq \frac{2^{n+1} n!}{(n+1)!} \left( \frac{1}{2} \right)^{n+1} = \frac{1}{n+1} \end{equation*}

   I need to choose \(n\) such that \(\frac{1}{n+1} \leq \frac{1}{100}\text{,}\) so \(n+1 \geq 100\text{,}\) which is \(n \geq 99\text{.}\) I need at least the 99th order approximation.