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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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
#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#
