Question

Find a Maclaurin series for f(x):

ln(1+x)

Answer 1

The Maclaurin series is the Taylor series around x=0.

Now I assume you know the following:

If |x| < 1 then 1/(1-x) = 1 + x + x^2 + x^3 + ...

Then replacing x by -x:

1/(1+x) = 1 - x + x^2 - x^3 + ...

Integrating on both sides:

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... if |x|< 1

Therefore ln(1+x) = sum ( (-1)^(k-1) x^k / k , k=1...infinity) for all x in (-1,1).

Answer 2

that is indeed the snappy way to do it, provided you know that you are allowed to integrate a power series term by term. you could also just grind it out, it is not that bad for this one

Answer 3

\[\ln(1-x)\]
\[\frac{1}{1-x}=(1-x)^{-1}\]
\[-1(1-x)^{-2}\]
\[2(1-x)^{-3}\]
\[-6(1-x)^{-4}\] etc general term being
\[(-1)^{k+1}k!(1-x)^{-k}\]

Answer 4

D(ln(1+x)) = 1/(1+x)


1-x+x^2-x^3+x^4-x^5 ...
------------------
1+x ) 1
(1+x)
-----
-x
(-x-x^2)
------
+x^2

\[\int 1-x+x^2-x^3+x^4-x^5 ...dx\]\[\hspace{10em}=ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}...\]

which is also the basic power series

Answer 5

replace x by zero and get
\[(-1)^{k+1}k!\]

Answer 6

Mickie is with (x-a) tho right?

Answer 7

except for some reason i used
\[\ln(1-x)\] instead of
\[\ln(1+x)\]!!!!!

Answer 8

no soup for me

Answer 9

back to ye olde dungeon with ye ;)

Answer 10

i think pavan and I got the same notion going on :)