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Loser66

  • one year ago

Find the radii of convergence of 1) \(\sum_{n=1}^\infty \dfrac{(-3i)^n}{n^3}z^n\) 2) \(\sum_{n=1}^\infty (\dfrac{2in +1}{3n -2i})^n z^n \) Please, help

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  1. Loser66
    • one year ago
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    @oldrin.bataku

  2. freckles
    • one year ago
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    I haven't done radii of convergence with complex numbers before but I think it could be similar do you have the answers?

  3. freckles
    • one year ago
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    \[|z|=|z i | \text{ since if } z=a+bi \text{ then } |z|=\sqrt{a^2+b^2} \text{ and } iz=ia-b \\ \text{ so } |iz|=\sqrt{(-b)^2+(a)^2}=\sqrt{b^2+a^2}=\sqrt{a^2+b^2} \\ \text{ so for the first one I think I can say } \\ |\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}|=|\lim_{n \rightarrow \infty} \frac{(-3i)^{n+1} z^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(-3i)^nz^n}| \\ =|\lim_{n \rightarrow \infty} (-3i)z (\frac{n}{n+1})^3| \\ =|(-3i)z (\lim_{n \rightarrow \infty} \frac{n}{n+1})^3| \text{ and we want this less than 1}\] you know where we can say |-3iz|=|3iz|=3|iz|=3|z| I believe

  4. freckles
    • one year ago
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    I think if I were you I would try the root test on the second one

  5. freckles
    • one year ago
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    that seem to work the key is knowing |iz|=|z|

  6. Loser66
    • one year ago
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    I am sorry, my computer is lagging.

  7. Loser66
    • one year ago
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    I don't have the answers for them.

  8. Loser66
    • one year ago
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    In class, we use either \(R = \dfrac{1}{limsup \sqrt[n]{|a_n|^n}}\) Or \(R =\lim|\dfrac{a_n}{a_{n+1}}|\) We don't put z inside the lim

  9. Loser66
    • one year ago
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    And we didn't have any example of the radius involve to multiple of i like this. We have |z -i| <1, that is z's are inside the circle center i with radius 1

  10. Loser66
    • one year ago
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    As you stated above , the lim =1, hence we just consider \(|\dfrac{-1}{3iz}| <1 \)

  11. freckles
    • one year ago
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    for the first one I get |z|<1/3 which means the radius is 1/3

  12. Loser66
    • one year ago
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    That throws me off also. We have R =\( lim |\dfrac{a_n}{a_{n+1}}|\)

  13. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=sum%28%28-3i%29%5En%2Fn%5E3*z%5En%2Cn%3D1..infty%29 I assumed I did right because wolfram agrees but there have been times wolfram has been wrong

  14. freckles
    • one year ago
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    it seems the other way you have it also works

  15. Loser66
    • one year ago
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    I show you the book.

  16. Loser66
    • one year ago
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    theorem 1.3 and proposition 1.4

  17. Loser66
    • one year ago
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  18. Loser66
    • one year ago
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    To me, if I use limit comparison test, the result should be the same. But they are not, right? Ok, let check if use the proposition in the book \(lim|\dfrac{a_n}{a_{n+1}}|= lim\frac|{(-3i)^n}{n^3}*\dfrac{(n+1)^3}{(-3i)^{n+1}}|=|\dfrac{1}{(-3i)}lim\dfrac{n^3}{(n+1)^3}=\dfrac{1}{|-3i|}\)

  19. freckles
    • one year ago
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    \[|\frac{-1}{3iz}|<1 \\ \frac{|-1|}{|3iz|}<1 \\ \frac{1}{3|z|}<1 \text{ since } |iz|=|z| \\ \text{ multiply both sides by } |z| \\ \frac{1}{3}<|z| \\ \text{ hmm... I don't get why they put } a_{n+1} \text{ on the bottom } \\ \text{ \because this should be } |z|<\frac{1}{3} \text{ which says the radi is } \frac{1}{3}\]

  20. Loser66
    • one year ago
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    So the radius of convergence (if it works) is 1/3 while limit comparison test you post above is z > 1/3

  21. Loser66
    • one year ago
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    They are|dw:1444010175317:dw| completely different , right?

  22. freckles
    • one year ago
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    |z|>1/3 is definitely different from |z|<1/3 we should be getting |z|<1/3 which is what I obtained from doing |a_(n+1)/a_n|<1 but doing the book's way gives |z|>1/3 unless we are suppose to solve |a_n/a_(n+1)|>1 instead

  23. Loser66
    • one year ago
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    hey, the book's solution give me it is < 1/3

  24. freckles
    • one year ago
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    http://math.stackexchange.com/questions/378181/radius-of-convergence-of-power-series-of-complex-analysis here they also put the n+1th on top

  25. Loser66
    • one year ago
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    I know, but you tell me, should I follow my prof's and my book??? hehehe

  26. freckles
    • one year ago
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    if you got the book's way gives |z|<1/3 then did you solve |a_n/a_(n+1)|>1

  27. Loser66
    • one year ago
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    Nope, the concept is: if we calculate directly R by that method, that means on the radius R, the series converges (not including the boundary) |dw:1444010543836:dw|

  28. Loser66
    • one year ago
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    At the point on the boundary (the circle) the series may or may not converge.

  29. Loser66
    • one year ago
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    Because inside is convergent, outside is divergent, right on boundary : no information.

  30. freckles
    • one year ago
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    There are the exact ways I would have done the problems: we are doing n goes to infinity so I will probably not write the word limit... \[|\frac{a_{n+1}}{a_n}|<1 \\ |a_{n+1} \cdot \frac{1}{a_n}|=|\frac{(-3i)^{n+1}z^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(-3i)^nz^{n}}| =|(-3i)z \frac{n^3}{(n+1)^3}| \\=|-3iz| \cdot | (\frac{n}{n+1})^3|=|-1||3||iz| |(1)^3| \\ =1(3)|z|(1)=3|z| \\ \text{ so we have } |\frac{a_{n+1}}{a_n}|=3|z|<1 \\ \text{ solving for } |z| \text{ gives } |z|<\frac{1}{3} \text{ so radius is } \frac{1}{3}\] \[|a_n|^\frac{1}{n}<1 \\ |(\frac{-2 i n+1}{3n-2i})^nz^n|^\frac{1}{n}<1 \\ |\frac{-2i n+1}{3n-2i}z|<1 \\ |\frac{-2i n}{3n}z|<1 \\ |-1| |\frac{2}{3} |iz|<1 \\ \frac{2}{3}|z|<1 \\ |z|<\frac{3}{2} \\ \text{ which says the radius is } \frac{3}{2}\] I have no idea if this is what your teacher's way is or the book's but both seem to get me what wolfram also says as the answer

  31. Loser66
    • one year ago
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    For the second one 1) why do you let it <1 ? 2) How can you get \( from~~ |\frac{-2i n+1}{3n-2i}z|<1 \\to~~ |\frac{-2i n}{3n}z|<1 \)

  32. freckles
    • one year ago
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    1) http://tutorial.math.lamar.edu/Classes/CalcII/RootTest.aspx just using root test 2) n goes to infinity (you have the same degree on top and bottom; just take coefficients on top and bottom of term with highest exponent)

  33. freckles
    • one year ago
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    wait was 1) for both questions I did or for the second one only?

  34. freckles
    • one year ago
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    because I'm just using the standard root and ratio test

  35. freckles
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

  36. Loser66
    • one year ago
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    oh, you use root test!! not limsup as what the book says. Both them use |a_n|^(1/n) , and that confused me.

  37. Loser66
    • one year ago
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    But I still not understand how you simplify it nicely like that, please, put some calculation to help me understand it better.

  38. freckles
    • one year ago
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    which part are you still not convinced |iz|=|z| ?

  39. Loser66
    • one year ago
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    oh!! got it, you take limit so that it counts only the first numbers on both numerator and denominator. (the second one) (-2in / 3n)

  40. freckles
    • one year ago
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    right \[\lim_{n \rightarrow \infty} \frac{\color{red}{-2i } n +1}{\color{red}3n-2i}=\frac{-2i}{3}\]

  41. Loser66
    • one year ago
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    |dw:1444011753965:dw|

  42. freckles
    • one year ago
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    yeah I did put a disclaimer above all of my work

  43. freckles
    • one year ago
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    "we are doing n goes to infinity so I will probably not write the word limit... "

  44. Loser66
    • one year ago
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    Thank you so much. I do appreciate.

  45. freckles
    • one year ago
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    lol sorry I was feeling lazy didn't feel like writing it over and over

  46. freckles
    • one year ago
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    anyways I have to go peace

  47. Loser66
    • one year ago
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    Have a good night.

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