Question

Little help here..

Answer 1

this seems to be a second part? is there information in the (i) part that might be useful?

as is; a_n is a rule that has to be determined; but the R is a mystery to me

Answer 2

Yes, It asks for the Expansion of tan^-1(x)

Answer 3

i beleive thats:
\[x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\frac{x^{13}}{13}... \]
if i remember that right

Answer 4

amistre64, that expansion is OK!

Answer 5

\[\int \frac{1}{1+x^2}dx=tan^{-1}\]

1-x^2 + x^4 .....
---------------
1+x^2 ) 1
(1+x^2)
--------
-x^2
(-x^2-x^4)
----------
x^4
(x^4+x^6)
----------
-x^6

\[\int (1-x^2 + x^4-x^6+x^8...)dx=tan^{-1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}...\]

yeah, had to make sure in my head :)

Answer 6

now we need to define a rule for an

(-1)^even to begin with

denom is (2n-1) for odds

and the top is just 1

and we got x^(2n-1)

and nstarts at 0 in the problem right?

\[a_n=(-1)^n\frac{x^{(2n+1)}}{2n+1}\]

Answer 7

almost had it in the right format:


\[\sum_{n=0}^{inf}a_nx^n\]

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

Answer 8

class is starting, and I got no idea what to do about an R yet :)

Answer 9

I think Still u didn come to the 2 nd part

Answer 10

\[g(x)=\frac{tan^{-1}(2x)}{1-3x}\]

\[tan(y)=2x\]
\[sec^2(y)\ y'=2\]
\[y'=\frac{2}{sec^2(y)}\]
\[y'=\frac{2}{tan^2(y)+1}\]
\[y'=\frac{2}{4x^2+1}=2-8x^2\]


2 -8x^2+ 32x^4 - 128x^6 + ...
-------------------
1+4x^2 ) 2
(2+8x^2)
---------
-8x^2
(-8x^2-32x^4)
-------------
32x^4
(32x^4+128x^6)
----------------
.....

the coeffs are a geometric seq with a common ratio of 4.

\( c_n = (-1)^n*2*4^n\) ; n = 0,1,2,3,4,...

\(2*{(2^2)}^n = 2^{(2n+1)}\)

these coeefs remain the same thru integration and get attached to our general tan-1 equation; therefore:

\[ tan^{-1}(2x)=2x-8\frac{x^3}{3}+32\frac{x^5}{5}-128\frac{x^7}{7}+512\frac{x^9}{9}...\]

and our summation becomes:
\[\large tan^{-1}(2x)=\sum^{inf}_{n=0}\ \frac{(-1)^n\ 2^{2n+1}}{(2n+1)}\ x^{(2n+1)} \]

when we put that into our original problem we get:
\[\large \sum^{inf}_{n=0}\ \frac{(-1)^n\ 2x^{2n+1}}{(2n+1)(1-3x)} \]

according to Pauls notes: http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

we would determine the convergence of this with one of the convergence tests; lets follow the lead here and do a ratio test.

Answer 11

flip it and where we see an n, replace it by n-1 and simplify

\[\lim_{n->inf}\ \left| \frac{(-1)^n\ (2x)^{2n+1}}{(2n+1)(1-3x)}*\frac{(2n-1)(1-3x)}{(-1)^{(n-1)}\ (2x)^{2n-1}} \right|\]

\[\lim_{n->inf}\ \left| \frac{(-1)(2x)^{2}(2n-1)}{(2n+1)}\right|\]
the x parts have no effect on us so factor them out
\[|(2x)^2|\ \lim_{n->inf}\ \left| \frac{2n-1}{2n+1}\right|\]
the lim of that goes to 1 and we are left with: |4x^2|

if |4x^2|<1 the series converges; if im reading this correctly

Answer 12

R = 1/2 maybe?

Answer 13

http://www.wolframalpha.com/input/?i=+sum+%28%28-1%29%5En+%282x%29%5E%282n%2B1%29%2F%28%281-3x%29%282n%2B1%29%29%29+from+0+to+infinity

i think the wolf agrees .