##sinhx =sum_(k=0)^oo x^(2k+1)/((2k+1)!)##

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.