anonymous
  • anonymous
Find a Maclaurin series for f(x): ln(1+x)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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).
anonymous
  • anonymous
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
anonymous
  • anonymous
\[\ln(1-x)\] \[\frac{1}{1-x}=(1-x)^{-1}\] \[-1(1-x)^{-2}\] \[2(1-x)^{-3}\] \[-6(1-x)^{-4}\] etc general term being \[(-1)^{k+1}k!(1-x)^{-k}\]

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amistre64
  • amistre64
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
anonymous
  • anonymous
replace x by zero and get \[(-1)^{k+1}k!\]
amistre64
  • amistre64
Mickie is with (x-a) tho right?
anonymous
  • anonymous
except for some reason i used \[\ln(1-x)\] instead of \[\ln(1+x)\]!!!!!
anonymous
  • anonymous
no soup for me
amistre64
  • amistre64
back to ye olde dungeon with ye ;)
amistre64
  • amistre64
i think pavan and I got the same notion going on :)

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