How do you use the Maclaurin series sinx to find the maclaurin series for f(x) = ln(3+x)?

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Answer 1 · 2017-02-28T01:55:15

#ln(3+x) = ln3 - sum_(k=1)^oo (-1)^k/k x^k/3^k# for #x in (-3,3)#

We can determine the MacLaurin series of #f(x) = ln(3+x)# without using the MacLaurin series of #sinx#, by noting that:

#ln(3+x) = ln(3(1+x/3)) = ln(3) +ln(1+x/3) = ln3 + int_0^(x/3) (dt)/(1+t)#

The integrand function is the sum of a geometric series of ratio #(-t)#:

#1/(1+t) = sum_(n=0)^oo (-t)^n#

having radius of convergence #R=1#, so for #abs(x/3) < 1# we can integrate term by term:

#ln(3+x) = ln3 + int_0^(x/3) sum_(n=0)^oo (-t)^ndt = ln3 + sum_(n=0)^oo int_0^x(-t)^ndt = ln3 + sum_(n=0)^oo (-1)^n[t^(n+1)/(n+1)]_0^(x/3)#

#ln(3+x) = ln3 + sum_(n=0)^oo (-1)^n(x/3)^(n+1)/(n+1)#

Substitute #k=n+1#:

#ln(3+x) = ln3 - sum_(k=1)^oo (-1)^k/k x^k/3^k#