- anonymous

i need to express x/1-x as a power series

- katieb

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- anonymous

1 sec

- anonymous

ok

- anonymous

sorry bout that mate

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## More answers

- anonymous

?

- anonymous

so the series is \[\sum_{0}^{\infty} x / (1-x)\]

- anonymous

yeah

- anonymous

im sorry that is completely wrong.
1/(1-x) = \[\sum_{0}^{\infty} x^n\] im sorry that is completely wrong

- anonymous

where you are sending 0 --> infinity

- anonymous

what a series does is it models something else by using and infinite amount of terms. that is why you can equate what i stated above

- anonymous

and you sum allll of those terms

- anonymous

1/(1-x) is defined by
\[\sum_{n \rightarrow0}^{\infty}x^n\] = \[1 + x^2 + x^3 + x^4 + x^5 + ... + x^n\]
but this is not your equation.

- anonymous

right? you can find that series in any caluculus text book

- anonymous

that series has to be memorized

- anonymous

ya

- anonymous

so to get to your equation you must multiply everyyy term by x because
x * 1/(1-x) = \[x *\sum_{n=0}^{\infty} x^n\]

- anonymous

ok

- anonymous

multiplying by x your exponents are n and 1. when multiplying like bases you add exponents you get n + 1 so. \[x/(1-x) = \sum_{n=0}^{\infty} x^{n+1}\] = \[x + x^2 + x^3 + x^4 + x^5 + .... + x^{n+1}\]

- anonymous

cool right?

- anonymous

ok can i ask you one more question

- anonymous

yeah

- anonymous

so to find the interval of convergence how would i go about that

- anonymous

okay

- anonymous

have you learned the nth root test? or ratio test?

- anonymous

yeah

- anonymous

so you can take the ratio test. of the series. which states you must take the absolute value of the An+1 and divide by An you should get
| x^[ (n+1) +1 ] / x^(n+1) | = | x^(n+2) / x^(n+1) |

- anonymous

can you simplify that? dont froget the ABS and show me what you get

- anonymous

Is that just X^N?

- anonymous

|x^N|

- anonymous

you can look at it as multiples in the numerator and denominator

- anonymous

you have x^(n+2) = x^n * x * x
right? its all just boring exponent work

- anonymous

you get that?

- anonymous

in the numerator

- anonymous

yeah

- anonymous

you get that?

- anonymous

So is it just X^N+1?

- anonymous

dont forget you have a multiple of x^n in the denominator so they cancel each other
x^(n+2) / x^(n+1) = { x^n * x * x } / { x^n * x }

- anonymous

So just X then?

- anonymous

yeah. real quick work it out on a piece of paper while i type up the next part

- anonymous

ok

- anonymous

the ratio test states Xn+1 / Xn < \[\left| x_{n+1}/x _{n} \right|< \rho =1\]
Xn+1 / Xn = x

- anonymous

you can plug solve for x in the inequality
| x | < p = 1
| x | < 1
-1 > x > 1

- anonymous

so that is the IOC

- anonymous

ah ok got it

- anonymous

you can test the end points now
An = X^(n+1)

- anonymous

so to check for the end points you just plug into x

- anonymous

yeah

- anonymous

perfect

- anonymous

and for 1 it diverges because it doesnt converge
and -1 it diverges also right

- anonymous

but into the series. its kind of something a little technical
x^n doesnt equal series x^n

- anonymous

yeah because it is a divergent geometric series where r must be <1
r^n

- anonymous

or acutally its |r| < 1 to converge and your r's are -1 and 1. any x that is a fraction, for this series, would converge.

- anonymous

alright got it

- anonymous

the ratio test is just difficult with the exponents

- anonymous

ok thanks

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