b7-power series: problem 6\n(2 points)\nuse eq. (1) from the text to expand the function into a power series…

b7-power series: problem 6\n(2 points)\nuse eq. (1) from the text to expand the function into a power series with center ( c = 0 ) and determine the set of ( x ) for which the expansion is valid.\n( f(x)=\frac{1}{2+x^{6}} )\n( \frac{1}{2+x^{6}}=sum_{n = 0}^{infty} )\nthe interval of convergence is:\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 20 attempts remaining.\npage generated october 15, 2025, 4:23:45 pm edt\nwebwork © 1996 - 2025 | theme: math4 | ww_version 2.20 | pg_version 2.20\nthe webwork project

b7-power series: problem 6\n(2 points)\nuse eq. (1) from the text to expand the function into a power series with center ( c = 0 ) and determine the set of ( x ) for which the expansion is valid.\n( f(x)=\frac{1}{2+x^{6}} )\n( \frac{1}{2+x^{6}}=sum_{n = 0}^{infty} )\nthe interval of convergence is:\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 20 attempts remaining.\npage generated october 15, 2025, 4:23:45 pm edt\nwebwork © 1996 - 2025 | theme: math4 | ww_version 2.20 | pg_version 2.20\nthe webwork project

Answer

Explanation:

Step1: Rewrite the function

We know the formula for the geometric series (\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}), for (|t|\lt1). Rewrite (f(x)=\frac{1}{2+x^{6}}) as (f(x)=\frac{1}{2(1+\frac{x^{6}}{2})}=\frac{1}{2}\cdot\frac{1}{1-(-\frac{x^{6}}{2})}).

Step2: Apply the geometric series formula

Let (t =-\frac{x^{6}}{2}). Then (\frac{1}{2}\cdot\frac{1}{1-(-\frac{x^{6}}{2})}=\frac{1}{2}\sum_{n = 0}^{\infty}\left(-\frac{x^{6}}{2}\right)^{n}). Using the property ((ab)^{n}=a^{n}b^{n}), we get (\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{2^{n + 1}}x^{6n}).

Step3: Find the interval of convergence

For the geometric series (\sum_{n = 0}^{\infty}t^{n}), the condition is (|t|\lt1). Here (t=-\frac{x^{6}}{2}), so (\left|-\frac{x^{6}}{2}\right|\lt1). Simplify (\left|\frac{x^{6}}{2}\right|\lt1), which gives (|x^{6}|\lt2), then (|x|\lt2^{\frac{1}{6}}).

Answer:

The power - series expansion is (\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2^{n + 1}}x^{6n}) and the interval of convergence is ((-2^{\frac{1}{6}},2^{\frac{1}{6}}))