Question

The function f(x)=ln(6−x) is represented as a power series. Find the first few coefficients in the power series.

Answer 1

Yea because I have been stuck on this for a while. And could use any help.

Answer 2

to me, I consider the term (6-x)
\[\frac{ 1 }{ 6(1- \frac{ x }{ 6 }) }= \frac{ 1 }{ 6 }\sum_{0}^{\infty} 1 + (\frac{ x }{ 6 })+(\frac{ x }{ 6 })^2+ (\frac{ x }{ 6 })^3+.......\]
then take integral both sides I let 1/6 in the front aside, I calculate the next part.
integral LHS = ln (6-x) and integral RHS you can take int few of them and stop at wherever you like

Answer 3

that's what I think, or you can multiple 1/6 into each term after the sum notation, and then take integral.

Answer 4

Thank you that should help.

Answer 5

@sportymort let hear other's opinion

Answer 6

@Spacelimbus

Answer 7

he's off line, how ???

Answer 8

Depends on what kind of topic this is based on, like usually this is an application of substitution. If we consider a Maclaurin Series then:

\[\ln(1+x)= \sum_1^\infty \frac{ (-1)^\left( n+1 \right)x^n }{ n }\]
Usually such kind of problems are faced first, nevertheless you can go from a substutiton from here.

\[\ln(1+5-x)=\ln(6-x)=\sum_1^\infty \frac{ (-1)^\left(n+1 \right)(5-x)^n }{ n }\]

Answer 9

yep, much more neatly than mine.

Answer 10

anyway, he 's gone. I just want to study more from his problem.

Answer 11

Yours was great @Loser66, that's the way how you would approach it manually, just make sure that you compute each successful derivative right. The results should remain the same though.