- #1
- 1,752
- 143
This is from the teacher's notes
[tex]f(x) = \frac{x}{{9 + x^2 }} = \frac{x}{9} \cdot \frac{1}{{1 - \left( { - \frac{{x^2 }}{9}} \right)}} = \frac{x}{9}\sum\limits_{n = 1}^\infty {\left( { - \frac{{x^2 }}{9}} \right)^n } = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n x^{2n + 1} }}{{9^{n + 1} }}} [/tex]
I can see distributing the n inside the parenthesis, to -1, x^2, and 9. But what's the justification for chaning it to n+1 for x^2 and for 9?
The next step is
This converges for [tex]\left| { - \frac{{x^2 }}{9}} \right| < 1\,\,\,\,\,\,or\,\,\,\,x^2 < 9\,\,\,\,\,\,\,\,\,\,\, - 3 < x < 3[/tex]
(−3, 3)
So what was the point in doing the last step in my first tex, if she just resorted to the 2nd to last step to determine the interval of convergence?
Homework Statement
[tex]f(x) = \frac{x}{{9 + x^2 }} = \frac{x}{9} \cdot \frac{1}{{1 - \left( { - \frac{{x^2 }}{9}} \right)}} = \frac{x}{9}\sum\limits_{n = 1}^\infty {\left( { - \frac{{x^2 }}{9}} \right)^n } = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n x^{2n + 1} }}{{9^{n + 1} }}} [/tex]
I can see distributing the n inside the parenthesis, to -1, x^2, and 9. But what's the justification for chaning it to n+1 for x^2 and for 9?
The next step is
This converges for [tex]\left| { - \frac{{x^2 }}{9}} \right| < 1\,\,\,\,\,\,or\,\,\,\,x^2 < 9\,\,\,\,\,\,\,\,\,\,\, - 3 < x < 3[/tex]
(−3, 3)
So what was the point in doing the last step in my first tex, if she just resorted to the 2nd to last step to determine the interval of convergence?