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mf1021 Group Title

find the radius of convergence of the taylor series around x=0 for ln(1/(1+2x))

  • 2 years ago
  • 2 years ago

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  1. amistre64 Group Title
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    i spose we would nee to generate the power series for it first

    • 2 years ago
  2. amistre64 Group Title
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    \[[ln\frac{1}{1+2x}]'=\frac{1/(1+2x)'}{1/(1+2x)}\] \[[ln\frac{1}{1+2x}]'=\frac{-2/(1+2x)^2}{1/(1+2x)}\] \[[ln\frac{1}{1+2x}]'=\frac{-2/(1+2x)}{1/1}\to\ -\frac{2}{1+2x}\]

    • 2 years ago
  3. amistre64 Group Title
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    maybe?

    • 2 years ago
  4. amistre64 Group Title
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    lol -ln(1+2x) = -2/(1+2x)

    • 2 years ago
  5. amistre64 Group Title
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    -2+4x-8x^2+16x^3-32x^4+64x^5 ... ----------------- 1+2x ) -2 (-2-4x) 4x (4x+8x^2) -8x^2 (-8x^2-16x^3)

    • 2 years ago
  6. afranklin12 Group Title
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    2 i think

    • 2 years ago
  7. experimentX Group Title
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    somewhere i had seen something like this integration dln(1+2x)/dx and expand it as power series.

    • 2 years ago
  8. amistre64 Group Title
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    soo, a relevant power series would be:\[ln(\frac{1}{1+2x})=\int\sum (-1)^{n+1}\ 2^{2n}x^{n}dx\]

    • 2 years ago
  9. amistre64 Group Title
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    \[ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(-1)^{n+1}}{n+1}x^{n+1}\]

    • 2 years ago
  10. amistre64 Group Title
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    forgot the 2^n

    • 2 years ago
  11. amistre64 Group Title
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    http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5E%28n%2B1%29x%5E%28n%2B1%29%292%5En%2F%28n%2B1%29+from+0+to+inf soo close

    • 2 years ago
  12. amistre64 Group Title
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    2^(n+1)

    • 2 years ago
  13. amistre64 Group Title
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    \[ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(-1)^{n+1}2^{n+1}}{n+1}x^{n+1}\] \[lim\frac{(-1)^{n+1}2^{n+1}x^{n+1}}{n+1}\frac{n}{(-1)^{n}2^{n}x^{n}}\] \[lim\frac{-2x}{n+1}\frac{n}{1}\] \[|x|lim\frac{-2n}{n+1} = -2\]

    • 2 years ago
  14. amistre64 Group Title
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    prolly shoulda taken the - with the x :)

    • 2 years ago
  15. amistre64 Group Title
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    all non - ns should be vacated \[|-2x|\ lim\frac{n}{n+1}=2x\] \[2x<1;\ x<\frac{1}{2}\]

    • 2 years ago
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