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.
