Question #d43f9

Start by recalling what the general Taylor series polynomial is. The purpose is to find an approximation of the original function centered around some value #a#, using the following formula:

#f(x)=f(a)+f(a)^(1)(x-a)/(1!)+f(a)^(2)(x-a)^2/(2!)+f(a)^(3)(x-a)^3/(3!)+...#

The only difference with the Maclaurin series is that #a=0#, giving

#f(x)=f(0)+f(0)^(1)(x)/(1!)+f(0)^(2)(x)^2/(2!)+f(0)^(3)(x)^3/(3!)+...#

With that formula in hand, we can go about finding the first few derivatives

#f(x)=xcosh(2x)#

#f(x)^(1)=2xsinh(2x)+cosh(2x)#

#f(x)^(2)=4sinh(2x)+4xcosh(2x)#

#f(x)^(3)=8xsinh(2x)+12cosh(2x)#

#f(x)^(4)=32sinh(2x)+16xcosh(2x)#

#f(x)^(5)=32xsinh(2x)+80cosh(2x)#

Next, plug #x=0# into each of these derivatives

#f(0)=0#

#f(0)^(1)=1#

#f(0)^(2)=0#

#f(0)^(3)=12#

#f(0)^4=0#

#f(0)^5=80#

You can see the pattern that only the odd numbered derivatives have a non-zero solution. Finally, stick them in in the Maclaurin formula above

#f(x)=0+x/(1!)+0+(12x^3)/(3!)+0+(80x^5)/(5!)+...#

Removing the zeros and simplifying the factorials gives

#f(x)=x+2x^3+2/3x^5+...#

You can see that even these three terms (in red) follows the original function (in blue) for some distance.

I had to expand the window to #y=+-20# to even see where they start to split!