tutorial exercise\nfind a power series representation for the function. determine the interval of…

tutorial exercise\nfind a power series representation for the function. determine the interval of convergence. (g)\nf(x) = 1 / (6 + x)\nstep 1\nwe need to express f(x) = 1 / (6 + x) in the form 1 / (1 - r) and then use the following equation.\n1 / (1 - r) = ∑(n = 0 to ∞)(r^n)\nstep 2\nwe can re - write f(x) = 1 / (6 + x) as f(x) = 1/6 · 1 / (1 - ())

tutorial exercise\nfind a power series representation for the function. determine the interval of convergence. (g)\nf(x) = 1 / (6 + x)\nstep 1\nwe need to express f(x) = 1 / (6 + x) in the form 1 / (1 - r) and then use the following equation.\n1 / (1 - r) = ∑(n = 0 to ∞)(r^n)\nstep 2\nwe can re - write f(x) = 1 / (6 + x) as f(x) = 1/6 · 1 / (1 - ())

Answer

Explanation:

Step1: Rewrite the function in the form of $\frac{1}{1 - r}$

We start with $f(x)=\frac{1}{6 + x}$. Factor out 6 from the denominator: $f(x)=\frac{1}{6(1+\frac{x}{6})}=\frac{1}{6}\cdot\frac{1}{1-(-\frac{x}{6})}$. Here, $r =-\frac{x}{6}$.

Step2: Use the geometric - series formula

The geometric - series formula is $\frac{1}{1 - r}=\sum_{n = 0}^{\infty}r^{n}$, for $|r|\lt1$. Substituting $r =-\frac{x}{6}$ into the formula, we get $f(x)=\frac{1}{6}\sum_{n = 0}^{\infty}(-\frac{x}{6})^{n}=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{6^{n + 1}}x^{n}$.

Step3: Find the interval of convergence

We know that the geometric series $\sum_{n = 0}^{\infty}r^{n}$ converges when $|r|\lt1$. Since $r=-\frac{x}{6}$, we set up the inequality $\left|-\frac{x}{6}\right|\lt1$. This simplifies to $|x|\lt6$. To check the endpoints: When $x = 6$, the series becomes $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n+1}}\cdot6^{n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6}$, which is a divergent alternating series (by the divergence test, $\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\frac{(-1)^{n}}{6}$ does not exist). When $x=-6$, the series becomes $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n + 1}}\cdot(-6)^{n}=\sum_{n = 0}^{\infty}\frac{1}{6}$, which is a divergent series (it is a constant - term series with non - zero constant terms).

Answer:

The power - series representation of $f(x)=\frac{1}{6 + x}$ is $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n+1}}x^{n}$, and the interval of convergence is $(-6,6)$.