Ok can someone please explain why this is.
I've a series x^3/3 + x^6/6 +x^9/9 + ……
which can be represented as Sigma {n=1 to infinity} x^3n/n
Question is find a simple formula. the answer gives the following:
first step is to differentiate - f’(x)=x^2 + x^5 +x^8 + …. -- Why?????
Provided |x|<1 then this is a sum of an infinite number of terms
Sum = a/(1-r) = x^2/(1 – x^3)
"a" first term = x^2 ---- Why not use 1/3 as per original equation?
"r" the common ratio ---x^3 - this makes sense is in first equation
Therefore f’(x) = x^2/(1 – x^3)
Integrating yields f(x) = -1/3.log(1-x^3) + c - Why integrate
I've a series x^3/3 + x^6/6 +x^9/9 + ……
which can be represented as Sigma {n=1 to infinity} x^3n/n
Question is find a simple formula. the answer gives the following:
first step is to differentiate - f’(x)=x^2 + x^5 +x^8 + …. -- Why?????
Provided |x|<1 then this is a sum of an infinite number of terms
Sum = a/(1-r) = x^2/(1 – x^3)
"a" first term = x^2 ---- Why not use 1/3 as per original equation?
"r" the common ratio ---x^3 - this makes sense is in first equation
Therefore f’(x) = x^2/(1 – x^3)
Integrating yields f(x) = -1/3.log(1-x^3) + c - Why integrate
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By differentiating, you get rid of the denominator and end up with the sequence being the sum of a geometric series with common ration x^3. There is no common ratio in the original sequence--the ratios are x^3/6, 6/9 x^3, 9/12 x^3 etc.
Then you can use the usual formula for the sum of a geometric series to change it to a closed formula. Then integrating gets you back to the original formula, +/- a constant.
This technique would only work with some sequences, not with just any. What's special about this one is that differentiating leaves you with a geometric sequence.
Then you can use the usual formula for the sum of a geometric series to change it to a closed formula. Then integrating gets you back to the original formula, +/- a constant.
This technique would only work with some sequences, not with just any. What's special about this one is that differentiating leaves you with a geometric sequence.