Question

\[f(x)= \frac{3}{1-x^{4}}\]

Answer 1

ok so I need to find a power series rep'in the function

Answer 2

\[\sum c_{n}(x-a)^{n}\] ?

Answer 3

\[\sum_{n=0}^{\infty}\]

Answer 4

Just a suggestion, what do you think about doing the Taylor serie for that function? Then you see the different terms and you can make the power serie, uhm

Answer 5

Taylor and MacLaurin series is the next session...I haven't learned that yet

Answer 6

ok teach me, how would you use the Taylor series?

Answer 7

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/PowerSeries/series_as_functions.html

Answer 8

Using geometric series and the formula for the sum, tell me if this helpful

Answer 9

"Using geometric series and the formula for the sum..."
I"m not sure

Answer 10

neither do I lol I'm going to compute the serie using this method

Answer 11

lol...sounds good

Answer 12

Well I'm not sure, but \[a/1-r=\sum_{0}^{n} ar^n\]
in the function f(x)=3/1-x^4
a=3; r=x^4 so the series should be
\[\sum_{0}^{n} 3(x^4)^n\]
Anyway I don't know if this is correct or not, just a suggestion, ask someone else, sorry MathSofiya

Answer 13

does it say you need to to the Taylor series about x=0, or some other number?

Answer 14

no this pre-Taylor series...let me get my book

Answer 15

this section is titled power series

Answer 16

i'll write more details from this sections in a second, I'll be right back...sorry (10 mins)

Answer 17

You know, I was just brushing up on this stuff myself; I'm a little rusty on this way of finding power series

Perhaps if I can't parse it out for you right now these notes will help

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx

Answer 18

ok i'm back

Answer 19

\[\sum_{n=0}^{\infty} c_{n} x^{n}\] ok so a power series is of this form...lets see ...

Answer 20

I am beyond silly...wrong chapter...I'M SOO SORRY

Answer 21

ok the section is titled Representations of functions as power series
for f(x)=1/(1+x^2) we have \[\sum_{n=0}^{\infty}(-x^{2})^{n}\]

Answer 22

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

Answer 23

are you still there?

Answer 24

\[ \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k\] for\(-1<x<1\)
replace \(x\) by \(x^4\) radius of convergence does not change

Answer 25

similarly
\[\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}\] replace \(x\) by \(-x^2\) for your power series

Answer 26

gimmick is often trying to match up with a geometric series

Answer 27

\[f(x)= \frac{3}{1-x^{4}}\]
so here I would have
\[f(x)= \frac{3}{1-(x^{4})}\]

Answer 28

parentheses not really necessary here

Answer 29

\[\frac{1}{1-x^{4}}=\sum_{k=0}^{\infty}x^4\]

Answer 30

it is necessary here
\[\frac{1}{1+x}=\frac{1}{1-(-x)}\] to make it match up to the geometric series,

Answer 31

yes what you write is correct

Answer 32

what should we do with the three in the numerator?

Answer 33

slap it next to the x?

Answer 34

it is a constant, pull it out front

Answer 35

\[\frac{3}{1-x^{4}}=3\sum_{k=0}^{\infty}x^4\]

Answer 36

\[\frac{1}{1-x^{4}}=\sum_{k=0}^{\infty}x^4\]
\[\frac{3}{1-x^{4}}=3\sum_{k=0}^{\infty}x^4=\sum_{k=0}^{\infty}3x^4\]

Answer 37

where is your k?

Answer 38

oops that what i get for pasting

Answer 39

lolz...never copy and paste what I wrote...

Answer 40

\[\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k\]
\[\frac{1}{1-x^4}=\sum_{k=0}^{\infty}x^{4k}\]

Answer 41

I don't quite see how the left side of the equal sign equals the right side...I copied the first example from the book so it must be right...but could you explain why they're equal

Answer 42

Ok, so the first was right