example video example express 1/(1 + x^4) as the sum of a power series and find the interval of convergence…
example video example express 1/(1 + x^4) as the sum of a power series and find the interval of convergence. solution we have the following equation. 1/(1 - x)=1 + x + x^2 + x^3+...=∑(n = 0 to ∞)x^n |x| < 1 replacing x by -x^4 in the equation above, we have the following. 1/(1 + x^4)=1/(1 - (-x^4))=∑(n = 0 to ∞)( )^n =∑(n = 0 to ∞)(-1)^nx^4n = 1 - x^4+ - x^12 + x^16 -... because this is a geometric series, it converges when |-x^4| <, that is, x^4 < or |x| <. therefore the interval of convergence is the open interval. (of course, have determined the radius of convergence by applying the ratio test, but that much work is unnecessary here.) need help? read it submit answer
Answer
Explanation:
Step1: Substitute $-x^4$ into geometric - series formula
We know that $\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}$ for $|t|<1$. Substituting $t=-x^{4}$, we get $\frac{1}{1-(-x^{4})}=\sum_{n = 0}^{\infty}(-x^{4})^{n}=\sum_{n = 0}^{\infty}(- 1)^{n}x^{4n}$.
Step2: Expand the series
$\sum_{n = 0}^{\infty}(-1)^{n}x^{4n}=1 - x^{4}+x^{8}-x^{12}+x^{16}-\cdots$.
Step3: Find the convergence condition
For a geometric series $\sum_{n = 0}^{\infty}a\cdot r^{n}$ (here $a = 1$ and $r=-x^{4}$), it converges when $|r|<1$. So we need $|-x^{4}|<1$, which simplifies to $x^{4}<1$. Taking the fourth - root of both sides, we get $|x|<1$.
Answer:
The power series is $\sum_{n = 0}^{\infty}(-1)^{n}x^{4n}$ and the interval of convergence is $(-1,1)$.