anonymous
  • anonymous
Find the radius of convergence \[ y= C_0\sum_{0}^{\infty}x^{2n} + C_1\sum_{0}^{\infty}x^{2n+1} \] I keep getting stuck on this part of a problem, can someone work it out for me and show all the steps. Thanks!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
the constants are unimportant. the radius of convergence of \(\sum_{n=0}^{\infty}x^n\) is \(|x|<1\) as it is a geometric series i don't think it makes any difference that you split off the even and odd exponents, unless i am missing something
anonymous
  • anonymous
Well the book is saying that the answer is 1. I keep getting 0 (1/infty) when I do it.
abb0t
  • abb0t
can you show your work?

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anonymous
  • anonymous
\[ \lim_{n \rightarrow \infty} | \frac{C_n}{C_{n+1}} | \] \[ \lim_{n \rightarrow \infty} | \frac{x^{2n} + x^{2n+1}}{x^{2n+2} + x^{2n+3}} | \] \[ \lim_{n \rightarrow \infty} | \frac{x^{2n}(1 + x)}{x^{2n}( x^2+ x^{3})} | \] \[ \lim_{n \rightarrow \infty} | \frac{1 + x}{x^2+ x^{3}} | = 0\]
anonymous
  • anonymous
BTW the book shows that this should be \[ \frac{c_0 + c_1x}{1-x^2}\] Which I don't understand how they got that.
anonymous
  • anonymous
It's not too bad: \[c_0 \sum_{n=0}^{\infty}x^{2n}+c_1 \sum_{n=0}^{\infty}x^{2n+1}=(c_0 + c_1 x)\sum_{n=0}^{\infty}x^{2n}\] If |x| < 1 then: \[\sum_{m=0}^{\infty}x^m=\frac{1}{1-x} \textrm{with }\sum_{m=0}^{\infty} \left( x^2\right)^m=\frac{1}{1-x^2} \textrm{for } |x|<1\] Giving: \[\frac{c_0+c_1 x}{1-x^2}\]
anonymous
  • anonymous
Oh wow, thank you!

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