Question

integral sin(x)/3x

find the first five non zero term centered around x=0. what is radius of convergence?

Answer 1

it is sin(x)/3x and I need to find radius of convergence. I would really appreciate if anyone helps me out with this one.

Answer 2

let's try this
http://upload.wikimedia.org/wikipedia/en/math/e/5/9/e59ac417d852ff5c18fe14f655501bab.png

Answer 3

lin n->inf 1/(1/(2i+1)!)^(1/n)

Answer 4

do you mean lim ?

Answer 5

limit

Answer 6

wouldnt answer be just zero?

Answer 7

i found excellent example

Answer 8

or do I have to use ratio or root test? The function we are taking limit is derived from taylor series??

Answer 9

I was trying to list taylor series and try to convert the series into sigma of a function and take ratio or root test to find radius.

Answer 10

ratio test i guess

Answer 11

lin n->inf 1/(1/(2i+1)!)^(1/n) just to ratio test for this one

Answer 12

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx
example 1 below ... BRB

Answer 13

Thank you so much. I think my concern was how you got inf 1/(1/(2i+1)!)^(1/n) from sinx/3x

Answer 14

expand sin using taylor seris
http://blogs.ubc.ca/infiniteseriesmodule/units/unit-3-power-series/taylor-series/maclaurin-expansion-of-sinx/

Answer 15

how about 3x?

Answer 16

divide x's by x ... 1/3 is just a constant.

Answer 17

and therefore we get from sinx/3x to 1/(1/(2i+1)!)^(1/n)

Answer 18

no ... that would be the nth root of of coefficint. plus 1/3 is missing.

Answer 19

sorry but do you mean? then (1/(1/(2i+1)!)^(1/n)) +(1/3)

Answer 20

radius of convergence is also expained here.
http://blogs.ubc.ca/infiniteseriesmodule/units/unit-3-power-series/taylor-series/maclaurin-expansion-of-sinx/
you just have to write 2k for 2k+1

an have 1/3 out

Answer 21

I'll be back in an hour.

Answer 22

okay thank you

Answer 23

I think I got stuck in the process. Kind of lost.

Answer 24

can anyone help me out?

Answer 25

i guess it would be same ... 0

Answer 26

what do you mean??

Answer 27

Here are the first few terms
\[-\frac{x^{11}}{1317254400}+\frac
{x^9}{9797760}-\frac{x^7}{105
840}+\frac{x^5}{1800}-\frac{x
^3}{54}+\frac{x}{3}

\]
Read it backwards

Answer 28

\[ \sin(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac { x^{2n+1}}{(2n+1)!}
\]

\[ \frac {\sin(x)}{3x} = \sum_{n=0}^{\infty} (-1)^{n+1}\frac { x^{2n}}{3(2n+1)!}
\]

\[ \int \frac {\sin(x)dx}{3x} = \sum_{n=0}^{\infty} (-1)^{n+1}\int \frac { x^{2n+1}dx}{3(2n+1)!}= \sum_{n=0}^{\infty} (-1)^{n+1 } \frac { x^{2n}dx}{3(2n+1)(2n+1)!}
\]
Apply the ratio test with
\[ a_n = \frac { x^{2n}}{3(2n+1)(2n+1)!}
\]
to deduce that
\[ R = \infty\]

Answer 29

thank you !!