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
