We know the definition for #sinh(x)#:
#sinh(x)=(e^x-e^-x)/2#
Since we know the Maclaurin series for #e^x#, we can use it to construct one for #sinh(x)#.
#e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/2+x^3/(3!)...#
We can find the series for #e^-x# by replacing #x# with #-x#:
#e^-x=sum_(n=0)^oo(-x)^n/(n!)=sum_(n=0)^oo(-1)^n/(n!)x^n=1-x+x^2/2-x^3/(3!)...#
We can subtract these two from each other to find the numerator of the #sinh# definition:
#color(white)(-e^-x.)e^x=color(white)(....)1+x+x^2/2+x^3/(3!)+x^4/(4!)+x^5/(5!)...#
#color(white)(e^x)-e^-x=-1+x-x^2/2+x^3/(3!)-x^4/(4!)+x^5/(5!)...#
#e^x-e^-x=color(white)(lllllllll)2xcolor(white)(lllllllll)+(2x^3)/(3!)color(white)(lllllll)+(2x^5)/(5!)...#
We can see that all the even terms cancel and all the odd terms double. We can represent this pattern like so:
#e^x-e^-x=sum_(n=0)^oo 2/((2n+1)!)x^(2n+1)#
To complete the #sinh(x)# series, we just need to divide this by #2#:
#(e^x-e^-x)/2=sinh(x)=sum_(n=0)^oo cancel2/(cancel2(2n+1)!)x^(2n+1)=#
#=sum_(n=0)^oo x^(2n+1)/((2n+1)!)=x+x^3/(3!)+x^5/(5!)...#
Now we want to calculate #f(1 \/ 2)# with an accuracy of at least #0.01#. We know this general form of the Lagrange error bound for an nth degree taylor polynomial about #x=c#:
#|R_n(x)|<=|M/((n+1)!)(x-c)^(n+1)|# where #M# is an upper bound on the nth derivative on the interval from #c# to #x#.
In our case, the expansion is a Maclaurin series, so #c=0# and #x=1 \/ 2#:
#|R_n(x)|<=|M/((n+1)!)(1/2)^(n+1)|#
The higher order derivatives of #sinh(x)# will either be #sinh(x)# or #cosh(x)#. If we consider the definitions for them, we see that #cosh(x)# will always be larger than #sinh(x)#, so we should work out the #M#-bound for #cosh(x)#
The hyperbolic cosine function is always increasing, so the largest value on the interval will be at #1 \/ 2#:
#sinh(1/2)=(e^(1/2)+e^(-1/2))/2=(sqrte+1/sqrte)/2=sqrte/2+1/(2sqrte)=M#
Now we plug this into the Lagrange error bound:
#|R_n(x)|<=(sqrte/2+1/(2sqrte))/((n+1)!)(1/2)^(n+1)#
We want #|R_n(x)|# to be smaller than #0.01#, so we try some #n# values until we get to that point (the lesser amount of terms in the polynomial, the better). We find that #n=3# is the first value that will give us an error bound smaller than #0.01#, so we need to use a 3rd degree taylor polynomial.
#sinh(1/2)~~sum_(n=0)^3(1/2)^(2n+1)/((2n+1)!)=336169/645120~~0.52#
