Question

Find the first five non-zero terms of power series representation centered at x=0 for the function below.
f(x)=ln(6−x)

Answer 1

it might be useful to know the power series for ln(1-x) to start with.

Answer 2

Oh I got the answer
ln(6 - x) = ln 6 - Σ(n = 0 to ∞) x^(n+1)/[6^(n+1) * (n+1)], which has R = 6

Answer 3

thats fair

Answer 4

I'm just confused on finding the first five terms?

Answer 5

ln(6-x)

the derivative is -1/(6-x)

which can be long divided to produce the derivatives power series


-1/6 -x/6^2
--------------
6-x | -1
(-1+x/6)
-------
-x/6
(-x/6+x^2/6^2)
--------------
-x^2/6^2

i see a pattern

\[\frac{-1}{6-x}=\sum_0\frac{-1}{6^{n-1}}x^n\]
\[\int \frac{-1}{6-x}=\int\sum_0\frac{-1}{6^{n-1}}x^n\]
\[ln(6-x)=\sum_0\frac{-1}{6^{n-1}(n+1)}x^{n+1}\]

Answer 6

pull out the first 6 terms; n=0 to 5

Answer 7

hmm, first term seems of course is ln(6-0) for the constant

Answer 8

the second term is -x^2 / 2

Answer 9

1/6^(n+1) slipped of the finger

Answer 10

no

\[ln(6-x)=ln(6)+\sum_0\frac{-1}{6^{n+1}(n+1)}x^{n+1}\]
\[ln(6-x)=ln(6)+\frac{-1}{6^{0+1}(0+1)}x^{0+1}+\sum_1\frac{-1}{6^{n+1}(n+1)}x^{n+1}\]
\[ln(6-x)=ln(6)-\frac{1}{6}x+\frac{-1}{6^{1+1}(1+1)}x^{1+1}+\sum_2\frac{-1}{6^{n+1}(n+1)}x^{n+1}\]
etc ...



ln(6) - x/6

Answer 11

just pull out terms

Answer 12

ln(6) - x/6 -x^2/2(6^2) - x^3/3(6^3) .... see a pattern?

Answer 13

oh ok

Answer 14

@amistre64 when ur done i need u

Answer 15

i believe im done, but please do not post for help i another persons question, its considered rude.