Question

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

Answer 1

1 sec

Answer 2

ok

Answer 3

sorry bout that mate

Answer 4

?

Answer 5

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

Answer 6

yeah

Answer 7

im sorry that is completely wrong.

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

Answer 8

where you are sending 0 --> infinity

Answer 9

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

Answer 10

and you sum allll of those terms

Answer 11

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.

Answer 12

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

Answer 13

that series has to be memorized

Answer 14

ya

Answer 15

so to get to your equation you must multiply everyyy term by x because

x * 1/(1-x) = \[x *\sum_{n=0}^{\infty} x^n\]

Answer 16

ok

Answer 17

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

Answer 18

cool right?

Answer 19

ok can i ask you one more question

Answer 20

yeah

Answer 21

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

Answer 22

okay

Answer 23

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

Answer 24

yeah

Answer 25

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) |

Answer 26

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

Answer 27

Is that just X^N?

Answer 28

|x^N|

Answer 29

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

Answer 30

you have x^(n+2) = x^n * x * x

right? its all just boring exponent work

Answer 31

you get that?

Answer 32

in the numerator

Answer 33

yeah

Answer 34

you get that?

Answer 35

So is it just X^N+1?

Answer 36

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 }

Answer 37

So just X then?

Answer 38

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

Answer 39

ok

Answer 40

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

Xn+1 / Xn = x

Answer 41

you can plug solve for x in the inequality

| x | < p = 1

| x | < 1

-1 > x > 1

Answer 42

so that is the IOC

Answer 43

ah ok got it

Answer 44

you can test the end points now

An = X^(n+1)

Answer 45

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

Answer 46

yeah

Answer 47

perfect

Answer 48

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

Answer 49

but into the series. its kind of something a little technical

x^n doesnt equal series x^n

Answer 50

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

r^n

Answer 51

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.

Answer 52

alright got it

Answer 53

the ratio test is just difficult with the exponents

Answer 54

ok thanks