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MarcLeclair Group Title

How can you rewrite the function as a power series? f(x) = x/ (1 + 2x^2) I know I have to take out the x so x ( 1 / (1 + 2x^2) however I'm stuck afterwards

  • one year ago
  • one year ago

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  1. Jhannybean Group Title
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    @terenzreignz

    • one year ago
  2. satellite73 Group Title
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    \[\frac{1}{1-x}=1+x+x^2+x^3+...\] \[\frac{1}{1+x}=1-x+x^2-x^3+...\] \[\frac{1}{1+2x^2}=1-2x^2+(2x^2)^2-(2x^2)^3+...\]

    • one year ago
  3. MarcLeclair Group Title
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    So would I write the sum as \[\sum_{n=1}^{\infty} (-1)^{n} (x)^{2n+1}\]

    • one year ago
  4. satellite73 Group Title
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    you need a 2 inside the parentheses

    • one year ago
  5. MarcLeclair Group Title
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    Oh yeah my bad. But how can I find the radius of convergence if I can't use the ratio test afterwards? I mean there's no division going on. Only thing I can think of is the root test but its negative, so the root test won't work perfectly.

    • one year ago
  6. MarcLeclair Group Title
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    I thought power series was just sum x^n

    • one year ago
  7. MarcLeclair Group Title
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    well the representation

    • one year ago
  8. MarcLeclair Group Title
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    Well I took the x and added it to the 2x^(2n+1) because the x was "outside" of the sum. So i just added it and thats what I did. I don't feel like writting the equation. I don't know if you follow what I'm saying.It was my guess really.

    • one year ago
  9. satellite73 Group Title
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    converges if \(|2x|<1\) i.e if \(|x|<\frac{1}{2}\)

    • one year ago
  10. MarcLeclair Group Title
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    Ah So i just take it right away. Makes sense. Thanks a lot^^

    • one year ago
  11. satellite73 Group Title
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    also you should start your summation at \(n=0\) i think

    • one year ago
  12. satellite73 Group Title
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    to be more precise i guess if converges if \(|2x^2|<1\) but same answer

    • one year ago
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