Science:Math Exam Resources/Courses/MATH105/April 2016/Question 04 (b)

MATH105 April 2016

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Question 04 (b)
Let $f(x)=\sum \limits _{k=1}^{\infty }{\frac {x^{k}}{3^{k}(k)}}$, $g(x)=f^{\prime }(x)={\frac {df}{dx}}$, and $R$ be the radius of convergence of the power series $f(x)$. Find $R$ and $g(1)$.

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To find the radius of convergence, consider the ratio test.

Hint 2

If a function $f$ has of the power series form $f(x)=\sum _{n=1}^{\infty }c_{n}(x-a)^{n}$ and a point $x_{0}$ is in the interval of convergence, then the derivative of $f$ at $x_{0}$ is $f'(x_{0})={\frac {d}{dx}}\left(\sum _{n=1}^{\infty }c_{n}(x-a)^{n}\right){\bigg |}_{x=x_{0}}=\sum _{n=1}^{\infty }{\bigg (}{\frac {d}{dx}}c_{n}(x-a)^{n}{\bigg |}_{x=x_{0}}{\bigg )}=\sum _{n=1}^{\infty }nc_{n}(x_{0}-a)^{n-1}.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can rewrite the function $f(x)$ in the power series form as follows: $f(x)=\sum _{k=1}^{\infty }a_{k}$, where $a_{k}={\frac {x^{k}}{k3^{k}}}.$ In order to find the radius of convergence for a power series, we first apply the ratio test. So, $L=\lim \limits _{k\to \infty }{\Big |}{\frac {a_{k+1}}{a_{k}}}{\Big |}=\lim \limits _{k\to \infty }{\Big |}{\frac {\frac {x^{k+1}}{(k+1)3^{k+1}}}{\frac {x^{k}}{k3^{k}}}}{\Big |}=\lim \limits _{k\to \infty }{\Big |}{\frac {k}{k+1}}\cdot {\frac {3^{k}}{3^{k+1}}}\cdot x{\Big |}={\frac {1}{3}}|x|\lim \limits _{k\to \infty }{\Big |}{\frac {k}{k+1}}{\Big |}={\frac {1}{3}}|x|\cdot 1={\frac {1}{3}}|x|.$ Then, the power series converges when ${\frac {1}{3}}|x|<1$, (i.e., $|x|<3$), while it diverges when ${\frac {1}{3}}|x|>1$, (i.e., $|x|>3$). Therefore, the radius of convergence is ${\color {blue}R=3}.$ Now, we consider the second part. From the first part, we know that the interval of convergence of $f$ is $(-3,3)$, and $1$ is in this interval, also $f$ is differentiable at $1$ and its derivative is the derivative of the power series $f(x)$ term by term; $g(x)=f'(x)={\frac {df}{dx}}={\frac {d}{dx}}\sum _{k=1}^{\infty }{\frac {1}{k3^{k}}}x^{k}=\sum _{k=1}^{\infty }{\frac {d}{dx}}{\frac {1}{k3^{k}}}x^{k}=\sum _{k=1}^{\infty }{\frac {1}{3^{k}}}x^{k-1}.$ We now compute $g(1):$ ${\color {blue}g(1)}=\sum _{k=1}^{\infty }{\frac {1}{3^{k}}}(1)^{k-1}=\sum _{k=1}^{\infty }{\frac {1}{3^{k}}}={\frac {1}{3}}+{\frac {1}{3^{2}}}+{\frac {1}{3^{3}}}+{\frac {1}{3^{4}}}+...={\frac {\frac {1}{3}}{1-{\frac {1}{3}}}}={\frac {\frac {1}{3}}{\frac {2}{3}}}{\color {blue}={\frac {1}{2}}}.$ Note that the last series is a geometric series with the first term ${\frac {1}{3}}$ and the common ratio ${\frac {1}{3}}.$ To sum up, the answers are $\color {blue}{R=3,g(1)={\frac {1}{2}}}$.