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

2 Answers
Apr 2, 2017

The Maclaurin series for #sin(x)# is given by:

#sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)=x-x^3/(3!)+x^5/(5!)+...#

Replacing #x# with #e^x#:

#sin(e^x)=sum_(n=0)^oo(-1)^n/((2n+1)!)(e^x)^(2n+1)=e^x-(e^x)^3/(3!)+(e^x)^5/(5!)+...#

So the first two nonzero terms are:

#f(x)=e^x-(e^x)^3/(3!)=e^x-e^(3x)/6#

So our approximation is:

#f(1/3)approxsin(e^(1/3))=e^(1/3)-e^(3(1/3))/6=e^(1/3)-e/6approx0.9426#

This compares to the actual value of #sin(e^(1/3))approx0.9847#.

Apr 2, 2017

#sin(e^(1/3)) ~~ 1#

Explanation:

The Maclaurin series for #f(x)# is given by:

#f(x) = sum_(n=0)^oo f^((n))(0)/(n!) x^n#

In our example:

#f^((0))(x) = sin(e^x)" "# so: #f^((0))(0) = sin(1)#

#f^((1))(x) = e^x cos(e^x)" "# so: #f^((1))(0) = cos(1)#

#f^((2))(x) = e^x cos(e^x) - e^(2x) sin(e^x)" "# so: #f^((2))(0) = cos(1)-sin(1)#

etc.

So the first few terms of the Maclaurin series for #sin(e^x)# are:

#(sin 1)+(cos 1)x+1/2(cos 1- sin 1)x^2+O(x^3)#

Using:

#sin 1 ~~ 0.8415#

#cos 1 ~~ 0.5403#

Then with #x=1/3# we have:

#sin(e^x) ~~ 0.8415+0.5403*1/3+1/2(0.5403-0.8415)*1/9+O(x^3)#

#color(white)(sin(e^x)) ~~ 0.8415+0.1801-0.0167+O(x^3)#

#color(white)(sin(e^x)) ~~ 1.0049#

Note that #sin# should be in the range #[-1, 1]# so we have not really used enough terms.

A calculator tells me:

#sin(e^(1/3)) ~~ 0.9846945#