derive a formula for infinite series..

**carpark** · Feb 20th 2008, 06:39 PM

Derive a general formula for the coefficients bn, as defined by (1/(1-x)){summation to infinity with i=0}(ai)x^i = {summation to infinity with i=0}(bi)x^i

**mr fantastic** · Feb 20th 2008, 08:13 PM

Originally Posted by carpark

Derive a general formula for the coefficients bn, as defined by (1/(1-x)){summation to infinity with i=0}(ai)x^i = {summation to infinity with i=0}(bi)x^i

For $\, -1 < x < 1\,$ , $\, \frac{1}{1-x} = 1 + x + x^2 + x^3 + .......\,$ using the formula in reverse for an infinite geometric series.

For $\, x > 1\,$ or $\, x < -1\,$ , $\, \frac{1}{1-x} = -\frac{1}{x} \left( \frac{1}{1 - \frac{1}{x}}\right) = -\frac{1}{x} \left( 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ....\right) \,$ again using the formula in reverse for an infinite geometric series.

**mrbicker** · Feb 21st 2008, 12:06 AM

What if for each i>=0, ai=i, then what is bn?

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