How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate #f(1/3)# given #f(x)=tanx#?

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Answer 1 · 2017-03-10T15:25:37

#tanx = x+x^3/3+o(x^3)#

#tan(1/3) ~= 28/81#

Start from the MacLaurin series for #sinx# and #cosx#:

#sinx = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!) = x-x^3/6+o(x^3)#

#cosx = sum_(n=0)^oo (-1)^n x^(2n)/((2n)!) = 1-x^2/2+ o(x^3)#

So that:

#tanx = sinx/cosx = (x-x^3/6+o(x^3)) /( 1-x^2/2+ o(x^3))#

Now consider:

#1/( 1-x^2/2+ o(x^3))#

If we put: #t = x^2/2+ o(x^3)# this expression can be seen as the sum of a geometric series:

#1/(1-t) = sum_(n=0)^oo t^n = 1+t+o(t^2)#

so:

#1/( 1-x^2/2+ o(x^3)) = 1+ x^2/2+ o(x^3) +o((x^2/2+ o(x^3))^2)#

The term:

#o((x^2/2+ o(x^3))^2) = o(x^4+x^2o(x^3)+o(x^6)) = o(x^4)+o(x^5)+o(x^6)#

is of higher order with respect to #o(x^3)# so we can ignore it, and we have:

#tanx = sinx * 1/( 1-x^2/2+ o(x^3)) = (x-x^3/6+o(x^3))(1+ x^2/2+ o(x^3))#

and ignoring again in the multiplications all factors of order higher than #x^3#

#tanx = x-x^3/6+x^3/2+o(x^3)#

#tanx = x+x^3/3+o(x^3)#

For #x=1/3# we have:

#tan(1/3)~= 1/3 +1/3*(1/3)^3= 1/3+1/81= 28/81#