cheater1 said:ohh i see it now, thanks you. My class just started this section and I'm kinda new to this. Is there a shortcut of doing this or I have to write out the terms to find power series?
No, if you use the technique I suggested, you get a first term of +1, not -1. Dick's approach and mine yield the same results.seeker247 said:I'm very thankful for this tread, as I ran into the same problem as cheater1.
I would like to point out, however, that while with Dick's method the 1st term of the series is 1, with Mark44's the first term appears to be -1. Any help in understanding why this is so would be greatly appreciated.
Mark44 said:You are doing the division incorrection. When you divide 1 + x by 1 - x, the first term you get is 1, not -1.
A power series is an infinite series of the form ∑n=0∞ cnxn, where cn are constants and x is the variable.
To represent (1+x)/(1-x) as a power series, we can use the geometric series formula:
1 + x + x2 + x3 + ... = 1/(1-x).
We can then substitute -x for x in the formula to get:
1 - x + x2 - x3 + ... = 1/(1+x).
Finally, multiplying both sides by x gives us the desired form (1 + x)/(1 - x) = 1 + 2x + 2x2 + 2x3 + ... = ∑n=0∞ 2xn.
The interval of convergence for the power series (1+x)/(1-x) is -1 < x < 1. This means that the series will converge for all values of x within this interval and diverge for all values of x outside of this interval.
Yes, the power series representation of (1+x)/(1-x) can be used to approximate the function for all values of x within the interval of convergence. However, it may not be as accurate for values of x closer to the endpoints of the interval.
Representing a function as a power series allows us to approximate the function easily and accurately, especially for values of x close to the center of the interval of convergence. It also allows us to manipulate the function algebraically, making it easier to solve complex problems and find patterns in the function's behavior.