Sinh(x)/x

Question

Using maclaurin series

Answer 1

The terms of the infinite series sum: ∑f⁽ⁿ⁾(c)(x-c)ⁿ/n!, where c is a constant value to base the series on and integer n is such that 0≤n<∞. When n=0 the term is simply f(c). The next term is df/dx|_c(x-c)/1!. f⁽ⁿ⁾(c) is the nth derivative of f where f(x) in this case is sinh(x)/x.

We want a series for sinh(x)/x. sinh(x)=½(e^x-e^-x). When x=0 sinh(x)=0, and sinh(x)/x is undefined, therefore we need to find c≠0 such that sinh(c)/c is defined.

But there is another way. We already have a series for e^x=1+x+x²/2!+...+xⁿ/n!+... which can be derived if f(x)=e^x and c=0, because all derivatives of f(x) are e^x. This is one of the simplest Taylor or MacLaurin series.

Therefore e^-x=1-x+x²/2!-...+(-1)ⁿxⁿ/n!+...

From this we have a series for sinh(x):

½{(1+x+x²/2!+...+xⁿ/n!+...)-(1-x+x²/2!-...+(-1)ⁿxⁿ/n!+...)}=

½(2x+2x³/3!+2x⁵/5!+...2xⁿ/n!+...)=x+x³/3!+x⁵/5!+...xⁿ/n!+...

To find a series for sinh(x)/x we simply divide by x:

1+x²/3!+x⁴/5!+...x^n-1/n!+... where integer n>0.

Interestingly, plugging in x=0 gives f(0)=1, so sinh(x)/x=1 in the limit as x→0.