Week 11 Activity

Section 11.4 Week 11 Activity

Subsection 11.4.1 Power Series

Activity 11.4.1.

Determine the radius of convergence and domain of definition for this power series.

\begin{equation*} \sum_{n=0}^\infty \frac{n}{n^3 + 1} x^n \end{equation*}

Solution.

I’ll use the limit test for this radius of convergence, since none of the coefficients are zero. I need to simplify the nested fraction, expand the cubed binomial, and evaluate the limit asymptotically.

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\begin{align*} R \amp = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| \\ \amp = \lim_{n \rightarrow \infty} \left| \frac{\frac{n}{n^3+1}}{\frac{(n+1)}{(n+1)^3+1}} \right| \\ \amp = \lim_{n \rightarrow \infty} \frac{n((n+1)^3+1}{(n+1)(n^3+1)} = \lim_{n \rightarrow \infty} \frac{n^4 + 3n^3 + 2n + 2}{n^4 + n^3 + n + 1} = 1 \end{align*}

The radius of convergence is \(R = 1\text{.}\) Since the centre point is \(\alpha = 0\text{,}\) the domain of definition is \((-1,1)\text{.}\)

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Activity 11.4.2.

Determine the radius of convergence and domain of definition for this power series.

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\begin{equation*} \sum_{n=0}^\infty \frac{1}{n!} (x-4)^n \end{equation*}

Solution.

I’ll use the limit test for this radius of convergence, since none of the coefficients are zero.

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{\frac{1}{n!}}{\frac{1}{(n+1)!}} \right| = \lim_{n \rightarrow \infty} \frac{(n+1)!}{n!} = \lim_{n \rightarrow \infty} n = \infty \end{equation*}

The radius of convergence is \(R = \infty\text{.}\) The domain of definition is all real numbers.

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Activity 11.4.3.

Determine the radius of convergence and domain of definition for this power series.

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\begin{equation*} \sum_{n=1}^\infty \frac{(-1)^n}{n} (x-3)^n \end{equation*}

Solution.

I’ll use the limit test for this radius of convergence, since none of the coefficients are zero.

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{\frac{(-1)^n}{n}}{\frac{(-1)^{n+1}}{n+1}} \right| = \lim_{n \rightarrow \infty} \frac{n+1}{n} = 1 \end{equation*}

The radius of convergence is \(R = 1\text{.}\) Since the centre point is \(\alpha = 3\text{,}\) the domain of definition is \((2,4)\text{.}\)

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Activity 11.4.4.

Determine the radius of convergence and domain of definition for this power series.

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

Solution.

I’ll use the limit test for this radius of convergence, since none of the coefficients are zero. I’ll use L’Hôpital’s rule for this limit, which conveniently gets rid of the logarithms. (The limit has type \(\frac{\infty}{\infty}\text{,}\) so L’Hôpital’s rule is justified.

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{\ln n}{\ln (n+1)} \right| = \lim_{n \rightarrow \infty} \frac{\frac{1}{n}}{\frac{1}{n+1}} = \lim_{n \rightarrow \infty} \frac{n+1}{n} = 1 \end{equation*}

The radius of convergence is \(R = 1\text{.}\) Since the centre point is \(\alpha = 0\text{,}\) the domain of definition is \((-1,1)\text{.}\)

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Activity 11.4.5.

Determine the radius of convergence and domain of definition for this power series.

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

Solution.

This series has only even terms; all the odd terms are zero. Since there are even terms, a direct use of either the limit calculation for the radius of convergence or the ratio test leads to division by zero problems. However, I can adjust a bit. I can think of this as a power series in \(x^2\text{.}\) If I do that, then all the terms are non-zero and I can use the limit. The result will give a radius that applies to \(x^2\text{.}\) Here is the limit.

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{\frac{1}{n!}}{\frac{1}{(n_1)!}} \right| = \lim_{n \rightarrow \infty} \frac{(n+1)!}{n!} = \lim_{n \rightarrow \infty} n+1 = \infty \end{equation*}

The radius of convergence is \(R = \infty\text{.}\) Therefore, the series is defined everywhere. Technically, since I made this a series in \(x^2\text{,}\) the limit shows that \(x^2\) can be any real number. That immediatly implies that the original input \(x\) can be any real number as well.

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Activity 11.4.6.

Determine the radius of convergence and domain of definition for this power series.

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

Solution.

I’ll use the limit test for this radius of convergence, since none of the coefficients are zero. This limit is all about factorial algebra. I’ll simplify the nested fraction and then group similar factorial terms together. Then I’ll cancel the factorials and look at the remaining polynomial terms.

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\begin{align*} R \amp = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \frac{\frac{n!(n+2)!}{(2n+3)!}}{\frac{(n+1)!(n+3)!}{(2n+5)!}}\\ \amp = \lim_{n \rightarrow \infty} \frac{n!}{(n+1)!} \frac{(n+2)!}{(n+3)!} \frac{(2n+5)!}{(2n+3)!} \\ \amp = \lim_{n \rightarrow \infty} \frac{(2n+4)(2n+5)}{(n+1)(n+3)} = \lim{n \rightarrow \infty} \frac{4n^2 + 18n + 20}{n^2 + 4n + 3} = 4 \end{align*}

After simplifying the factorial, the limit is a quadratic in both the denominator and numerator. Since these have the same asymptotic order, the limit is the ratio of the leading coefficients. The radius of convergence is \(R = 4\text{.}\) Since the centre point is \(\alpha = -2\text{,}\) the domain of definition is \((-6,2)\text{.}\)

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Subsection 11.4.2

Activity 11.4.7.

Calculate the Taylor series for \(f(x) = \cosh x\) with centre point \(\alpha = 0\text{.}\) Determine its radius of convergence.

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Solution.

First I need the derivatives of the function. I’ll take the first few derivatives.

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\begin{align*} f(x) \amp = \cosh x \\ \frac{d}{dx} f(x) \amp = \sinh x \\ \frac{d^2}{dx^2} f(x) \amp = \cosh x\\ \frac{d^3}{dx^3} f(x) \amp = \sinh x \\ \frac{d^4}{dx^4} f(x) \amp = \cosh x\\ \frac{d^5}{dx^5} f(x) \amp = \sinh x \\ \frac{d^6}{dx^6} f(x) \amp = \cosh x \end{align*}

Then I’ll evaluate at the centre point.

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\begin{align*} f(0) \amp = 1\\ \frac{d}{dx} f(x) \Bigg|_{x=0} \amp = 0\\ \frac{d^2}{dx^2} f(x) \Bigg|_{x=0} \amp = 1\\ \frac{d^3}{dx^3} f(x) \Bigg|_{x=0} \amp = 0\\ \frac{d^4}{dx^4} f(x) \Bigg|_{x=0} \amp = 1\\ \frac{d^5}{dx^5} f(x) \Bigg|_{x=0} \amp = 0\\ \frac{d^6}{dx^6} f(x) \Bigg|_{x=0} \amp = 1 \end{align*}

Then I need to find a pattern. Here, the pattern is relatively obvious: a sequences of alternating zeros and ones. Only the even terms are non-zero, so I’ll only write the even terms in the final Taylor series. To write the even terms, I use \((2n)\) as \(n \in \NN\text{.}\)

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\begin{equation*} \frac{d^{2n}}{dx^{2n}} f(x) \Bigg|_{x=0} = 1 \end{equation*}

After describing a pattern for the derivatives at the centre point, I put that pattern into the standard form of the Taylor series.

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

I take the limit of the ratio of the coefficients to determine the radius of convergence. (In this case, since there are only even terms, the resulting radius will be a bound on \(x^2\) instead of \(x\text{.}\))

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{\frac{1}{2n!}}{\frac{1}{(2n+2)!}} \right| = \lim_{n \rightarrow \infty} \frac{(2n+2)!}{2n!} = \lim_{n \rightarrow \infty} (2n+1)(2n+2) = \infty \end{equation*}

The radius of convergence for \(x^2\) is \(R = \infty\text{.}\) Since \(x^2\) can be any real number, \(x\) can also be any real number, to the radius of convergence for the series is, indeed, infinity.

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Activity 11.4.8.

Calculate the Taylor series for \(f(x) = \frac{1}{x}\) with centre point \(\alpha = 1\text{.}\) Determine its radius of convergence.

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Solution.

First I need the derivatives of the function. I’ll take the first few derivatives.

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\begin{align*} f(x) \amp = \frac{1}{x} \\ \frac{d}{dx} f(x) \amp = \frac{-1}{x^2} \\ \frac{d^2}{dx^2} f(x) \amp = \frac{2}{x^3} \\ \frac{d^3}{dx^3} f(x) \amp = \frac{-(2)(3)}{x^4}\\ \frac{d^4}{dx^4} f(x) \amp = \frac{(2)(3)(4)}{x^5} \\ \frac{d^5}{dx^5} f(x) \amp = \frac{-(2)(3)(4)(5)}{x^6}\\ \frac{d^6}{dx^6} f(x) \amp = \frac{(2)(3)(4)(5)(6)}{x^7} \end{align*}

Then I’ll evaluate at the centre point.

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\begin{align*} f(1) \amp = 1\\ \frac{d}{dx} f(x) \Bigg|_{x=1} \amp = -1\\ \frac{d^2}{dx^2} f(x) \Bigg|_{x=1} \amp = 2\\ \frac{d^3}{dx^3} f(x) \Bigg|_{x=1} \amp = -(2)(3)\\ \frac{d^4}{dx^4} f(x) \Bigg|_{x=1} \amp = (2)(3)(4)\\ \frac{d^5}{dx^5} f(x) \Bigg|_{x=1} \amp = -(2)(3)(4)(5)\\ \frac{d^6}{dx^6} f(x) \Bigg|_{x=1} \amp = (2)(3)(4)(5)(6) \end{align*}

Then I need to find a pattern. The sign is alternating, so I need a power of \((-1)\text{.}\) Since the even terms are positive, \((-1)^n\) will work. After the sign, it looks like I’m building a factorial. The degree of the derivative matches the degree of the factorial.

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\begin{equation*} \frac{d^n}{dx^n} f(x) \Bigg|_{x=1} = (-1)^n n! \end{equation*}

After describing a pattern for the derivatives at the centre point, I put that pattern into the standard form of the Taylor series.

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

I take the limit of the ratio of the coefficients to determine the radius of convergence.

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{(-1)^n}{(-1)^{n+1}} \right| = \lim_{n \rightarrow \infty} 1 = 1 \end{equation*}

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

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Activity 11.4.9.

Calculate the Taylor series for \(f(x) = x^7 + 8x^6 - 19x^5 + 12x^4 + 3x^3 - 19x^2 + 4x + 41 \) with centre point \(\alpha = 0\text{.}\) Determine its radius of convergence.

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Solution.

This is a bit of a trick question. A polynomial is already in the form of a Taylor series centred at \(\alpha = 0\text{.}\) Since series representations are unique, there is no other way to write this polynomial as a series. There is nothing to do here. If you did do the steps, you would recover exactly the same expression and you would find that all coefficients for \(x^8\) and above would be zero. The radius of convergence is \(\infty\text{,}\) since a polynomial is defined everywhere.

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Activity 11.4.10.

Calculate the Taylor series for \(f(x) = \ln (x+3)\) with centre point \(\alpha = 0\text{.}\) Determine its radius of convergence.

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Solution.

First I need the derivatives of the function. I’ll take the first few derivatives.

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\begin{align*} f(x) \amp = \ln (x+3)\\ \frac{d}{dx} f(x) \amp = \frac{1}{x+3}\\ \frac{d^2}{dx^2} f(x) \amp = \frac{-1}{(x+3)^2}\\ \frac{d^3}{dx^3} f(x) \amp = \frac{2}{(x+3)^3} \\ \frac{d^4}{dx^4} f(x) \amp = \frac{-(2)(3)}{(x+3)^4}\\ \frac{d^5}{dx^5} f(x) \amp = \frac{(2)(3)(4)}{(x+3)^5}\\ \frac{d^6}{dx^6} f(x) \amp = \frac{-(2)(3)(4)(5)}{(x+3)^6} \end{align*}

Then I’ll evaluate at the centre point.

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\begin{align*} f(0) \amp = \ln 3 \\ \frac{d}{dx} f(x) \Bigg|_{x=0} \amp = \frac{1}{3} \\ \frac{d^2}{dx^2} f(x) \Bigg|_{x=0} \amp = \frac{-1}{3^2} \\ \frac{d^3}{dx^3} f(x) \Bigg|_{x=0} \amp = \frac{2}{3^3} \\ \frac{d^4}{dx^4} f(x) \Bigg|_{x=0} \amp = \frac{-(2)(3)}{3^4} \\ \frac{d^5}{dx^5} f(x) \Bigg|_{x=0} \amp = \frac{(2)(3)(4)}{3^5} \\ \frac{d^6}{dx^6} f(x) \Bigg|_{x=0} \amp = \frac{-(2)(3)(4)(5)}{3^6} \end{align*}

Then I need to find a pattern. The signs are alternating and the odd terms are positive, so I need \((-1)^{n+1}\text{.}\) I am buliding a factorial in the numerator, but the degree of the factorial is one less than the degree of the derivatives, so I’ll need \((n-1)!\text{.}\) Finally, there is a power of \(3\) in the denominator and the exponent is the same as the degree of the derivative. Also, the first term doesn’t fit the pattern, so I’ll have to remember to keep that term seperate.

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\begin{equation*} \frac{d^n}{dx^n} f(x) \Bigg|_{x=0} = \frac{(-1)^{n+1}(n-1)!}{3^n} \end{equation*}

After describing a pattern for the derivatives at the centre point, I put that pattern into the standard form of the Taylor series. I need to remember that the first term doesn’t fit the pattern and must be written apart from the sigma notation.

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\begin{equation*} f(x) = \ln 3 + \sum_{n=1}^\infty \frac{(-1)^{n+1}(n-1)!}{3^n n!} x^n = \ln 3 + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n3^n} x^n \end{equation*}

I take the limit of the ratio of the coefficients to determine the radius of convergence.

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\begin{equation*} R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \rightarrow \infty} \left| \frac{\frac{(-1)^{n+1}}{n3^n}}{\frac{(-1)^{n+2}}{(n+1)3^{n+1}}} \right| = \lim_{n \rightarrow \infty} \frac{(n+1)3^{n+1}}{n3^n} = \lim_{n \rightarrow \infty} 3 \frac{n+1}{n} = 3 \end{equation*}

The radius of convergence is \(R = 3\text{.}\)

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Subsection 11.4.3 Conceptual Review Questions

What is a radius of convergence?

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How do series represent functions?

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Why can series define all sorts of new non-elementary functions?

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