Calculus 2
- Thread starter Marie1309
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You need to give us something to start with.
https://www.freemathhelp.com/forum/threads/read-before-posting.41538/
You can't have no idea at all.
https://www.freemathhelp.com/forum/threads/read-before-posting.41538/
You can't have no idea at all.
HallsofIvy
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The denominator, \(\displaystyle (4x^2+1)^3\), is 0 when \(\displaystyle 4x^2+ 1=0\) so when \(\displaystyle x= \pm\frac{1}{2}i\). What does that tell you about the radius of convergence?
Also, you should know that the sum of the geometric series, \(\displaystyle \sum_{n= 0}^\infty r^n\), is \(\displaystyle \frac{1}{1- r}\). Thinking of this as the cube of \(\displaystyle \frac{1}{1- (-4x^2)}\) that woud be the sum of the geometric series \(\displaystyle \sum_{n= 0}^\infty (-4x^2)^n\). The first three terms of that are \(\displaystyle 1- 4x^2+ 16x^4\). Cube that.
Also, you should know that the sum of the geometric series, \(\displaystyle \sum_{n= 0}^\infty r^n\), is \(\displaystyle \frac{1}{1- r}\). Thinking of this as the cube of \(\displaystyle \frac{1}{1- (-4x^2)}\) that woud be the sum of the geometric series \(\displaystyle \sum_{n= 0}^\infty (-4x^2)^n\). The first three terms of that are \(\displaystyle 1- 4x^2+ 16x^4\). Cube that.
