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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
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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freckles
 one year ago
Best ResponseYou've already chosen the best response.1I haven't done radii of convergence with complex numbers before but I think it could be similar do you have the answers?

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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think if I were you I would try the root test on the second one

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that seem to work the key is knowing iz=z

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I am sorry, my computer is lagging.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I don't have the answers for them.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0In 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0And 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0As you stated above , the lim =1, hence we just consider \(\dfrac{1}{3iz} <1 \)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1for the first one I get z<1/3 which means the radius is 1/3

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0That throws me off also. We have R =\( lim \dfrac{a_n}{a_{n+1}}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=sum%28%283i%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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1it seems the other way you have it also works

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0theorem 1.3 and proposition 1.4

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0To 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 \(lim\dfrac{a_n}{a_{n+1}}= lim\frac{(3i)^n}{n^3}*\dfrac{(n+1)^3}{(3i)^{n+1}}=\dfrac{1}{(3i)}lim\dfrac{n^3}{(n+1)^3}=\dfrac{1}{3i}\)

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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0So the radius of convergence (if it works) is 1/3 while limit comparison test you post above is z > 1/3

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0They aredw:1444010175317:dw completely different , right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1z>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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0hey, the book's solution give me it is < 1/3

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://math.stackexchange.com/questions/378181/radiusofconvergenceofpowerseriesofcomplexanalysis here they also put the n+1th on top

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I know, but you tell me, should I follow my prof's and my book??? hehehe

freckles
 one year ago
Best ResponseYou've already chosen the best response.1if you got the book's way gives z<1/3 then did you solve a_n/a_(n+1)>1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Nope, 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0At the point on the boundary (the circle) the series may or may not converge.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Because inside is convergent, outside is divergent, right on boundary : no information.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1There 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... \[\frac{a_{n+1}}{a_n}<1 \\ a_{n+1} \cdot \frac{1}{a_n}=\frac{(3i)^{n+1}z^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(3i)^nz^{n}} =(3i)z \frac{n^3}{(n+1)^3} \\=3iz \cdot  (\frac{n}{n+1})^3=13iz (1)^3 \\ =1(3)z(1)=3z \\ \text{ so we have } \frac{a_{n+1}}{a_n}=3z<1 \\ \text{ solving for } z \text{ gives } z<\frac{1}{3} \text{ so radius is } \frac{1}{3}\] \[a_n^\frac{1}{n}<1 \\ (\frac{2 i n+1}{3n2i})^nz^n^\frac{1}{n}<1 \\ \frac{2i n+1}{3n2i}z<1 \\ \frac{2i n}{3n}z<1 \\ 1 \frac{2}{3} iz<1 \\ \frac{2}{3}z<1 \\ z<\frac{3}{2} \\ \text{ which says the radius is } \frac{3}{2}\] 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0For the second one 1) why do you let it <1 ? 2) How can you get \( from~~ \frac{2i n+1}{3n2i}z<1 \\to~~ \frac{2i n}{3n}z<1 \)

freckles
 one year ago
Best ResponseYou've already chosen the best response.11) 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)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1wait was 1) for both questions I did or for the second one only?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1because I'm just using the standard root and ratio test

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, you use root test!! not limsup as what the book says. Both them use a_n^(1/n) , and that confused me.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0But I still not understand how you simplify it nicely like that, please, put some calculation to help me understand it better.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1which part are you still not convinced iz=z ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh!! got it, you take limit so that it counts only the first numbers on both numerator and denominator. (the second one) (2in / 3n)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1right \[\lim_{n \rightarrow \infty} \frac{\color{red}{2i } n +1}{\color{red}3n2i}=\frac{2i}{3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yeah I did put a disclaimer above all of my work

freckles
 one year ago
Best ResponseYou've already chosen the best response.1"we are doing n goes to infinity so I will probably not write the word limit... "

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much. I do appreciate.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1lol sorry I was feeling lazy didn't feel like writing it over and over

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways I have to go peace
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