Question

Give the convergence interval for the expansions of the following functions f[x] in powers
of (x - a) for the given choices of 'a':

f[x] =1/(1-x) with a=4
f[x] = Sin[x] with a = pi (3.14....),

Answer 1

we can long divide and come up with a series for the first one

Answer 2

well i mean we know the expansion of 1/(1-x)

Answer 3

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

\[\frac{1}{1-x}=1+x+x^2+x^3+x^4+...\]
\[\frac{1}{1-x}=\sum_{n=0}^{inf} x^n\]
is that right?

Answer 4

So our radius of convergence is 1, and with that we should be able to determine the interval of convergence

Answer 5

so would it be 3 and 5?

Answer 6

for the first one

Answer 7

\[\lim_{n->inf}\left(\frac{x^n}{1}\frac{1}{x^{n-1}} \right)\]
\[\lim_{n->inf}\ x\]
\[|x| \lim_{n->inf}\ 1 \]
|x|<1 ; R=1

Answer 8

how do we do interval convergence? im gonna have to refresh the my memory lol

Answer 9

-1 < x < 1

if im reading this right

Answer 10

yes that is correct for a interval for a=0

Answer 11

then for x-a

-1 < x-a < 1

a-1 < x < a+1

maybe

Answer 12

yeah thats what i was thinking but i wasnt so sure

Answer 13

Sin[x] ; this ones from the derivatives built up I believe

for x=0

sin(0); 0
cos(0); 1
-sin(0); 0
-cos(0); -1

\[sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}...\]
right?

Answer 14

correct for a=0 yeah

Answer 15

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

Answer 16

\[\lim\frac{(-1)^n\ x^{(2n+1)}}{(2n+1)!}\frac{(2n-1)!}{(-1)^{(n-1)}\ x^{(2n-1)}}\]
\[\lim\frac{(-1)\ x^2}{(2n+1)!}\frac{(2n-1)!}{1}\]
\[|x^2|\ \lim\frac{(2n-1)!}{(2n+1)!}\]
now i just gotta determine what to do with that :)

Answer 17

\[(2n+1)!=(2n+1)(2n)(2n-1)!\]
maybe :)

\[|x^2|\ \lim\frac{1}{(2n+1)(2n)}\]
\[|x^2|\ \lim\frac{1}{4n^2+2n}\]
\[|x^2|\ \lim\frac{1}{2}\frac{1}{2n^2+n}\]
\[\frac{1}{2}|x^2|\ \lim\frac{1}{2n^2+n}\]
i think that goes to 0 tho

Answer 18

"Since L=0 <1 regardless of the value of x, this power series will converge for every x.

In these cases we say that the radius of convergence is R=inf and interval of convergence is:
-inf < x < inf

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx
#4