We can derive the McLaurin series for sinh(x) from the one othe exponential function: as for every n:
[(d^n)/(dx^n) e^x ]_(x=0) = e^0=1
the Mc Laurin series for e^x is:
e^x=sum_(n=0)^oo x^n/(n!)
Now as:
sinhx = (e^x-e^(-x))/2
We have:
sinhx = 1/2[sum_(n=0)^oo x^n/(n!)-sum_(n=0)^oo (-x)^n/(n!)]
and it is easy to see that for n even the terms are the same and just cancel each other, so that just the odd order terms remain:
sinhx = 1/2[sum_(k=0)^oo x^(2k+1)/((2k+1)!)-sum_(k=0)^oo (-1)^(2k+1)x^(2k+1)/((2k+1)!)] = 1/2[sum_(k=0)^oo x^(2k+1)/((2k+1)!)+sum_(k=0)^oo x^(2k+1)/((2k+1)!)] = sum_(k=0)^oo x^(2k+1)/((2k+1)!)
We can reach the same conclusion directly, noting that:
d/(dx) sinhx = coshx
d^2/(dx^2) sinhx = d/(dx)coshx = sinhx
so that all derivatives of odd order equal coshx and all derivatives of even order equal sinhx
But sinh(0) = 0 and cosh(0) = 1 yielding the same result.
