anonymous
  • anonymous
i need to express x/1-x as a power series
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1 sec
anonymous
  • anonymous
ok
anonymous
  • anonymous
sorry bout that mate

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anonymous
  • anonymous
?
anonymous
  • anonymous
so the series is \[\sum_{0}^{\infty} x / (1-x)\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
im sorry that is completely wrong. 1/(1-x) = \[\sum_{0}^{\infty} x^n\] im sorry that is completely wrong
anonymous
  • anonymous
where you are sending 0 --> infinity
anonymous
  • 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
  • anonymous
and you sum allll of those terms
anonymous
  • 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
  • anonymous
right? you can find that series in any caluculus text book
anonymous
  • anonymous
that series has to be memorized
anonymous
  • anonymous
ya
anonymous
  • 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
  • anonymous
ok
anonymous
  • 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
  • anonymous
cool right?
anonymous
  • anonymous
ok can i ask you one more question
anonymous
  • anonymous
yeah
anonymous
  • anonymous
so to find the interval of convergence how would i go about that
anonymous
  • anonymous
okay
anonymous
  • anonymous
have you learned the nth root test? or ratio test?
anonymous
  • anonymous
yeah
anonymous
  • 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
  • anonymous
can you simplify that? dont froget the ABS and show me what you get
anonymous
  • anonymous
Is that just X^N?
anonymous
  • anonymous
|x^N|
anonymous
  • anonymous
you can look at it as multiples in the numerator and denominator
anonymous
  • anonymous
you have x^(n+2) = x^n * x * x right? its all just boring exponent work
anonymous
  • anonymous
you get that?
anonymous
  • anonymous
in the numerator
anonymous
  • anonymous
yeah
anonymous
  • anonymous
you get that?
anonymous
  • anonymous
So is it just X^N+1?
anonymous
  • 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
  • anonymous
So just X then?
anonymous
  • anonymous
yeah. real quick work it out on a piece of paper while i type up the next part
anonymous
  • anonymous
ok
anonymous
  • anonymous
the ratio test states Xn+1 / Xn < \[\left| x_{n+1}/x _{n} \right|< \rho =1\] Xn+1 / Xn = x
anonymous
  • anonymous
you can plug solve for x in the inequality | x | < p = 1 | x | < 1 -1 > x > 1
anonymous
  • anonymous
so that is the IOC
anonymous
  • anonymous
ah ok got it
anonymous
  • anonymous
you can test the end points now An = X^(n+1)
anonymous
  • anonymous
so to check for the end points you just plug into x
anonymous
  • anonymous
yeah
anonymous
  • anonymous
perfect
anonymous
  • anonymous
and for 1 it diverges because it doesnt converge and -1 it diverges also right
anonymous
  • anonymous
but into the series. its kind of something a little technical x^n doesnt equal series x^n
anonymous
  • anonymous
yeah because it is a divergent geometric series where r must be <1 r^n
anonymous
  • 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
  • anonymous
alright got it
anonymous
  • anonymous
the ratio test is just difficult with the exponents
anonymous
  • anonymous
ok thanks

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