The Taylor series of a function is defined as:
#sum_(n=0)^oof^n(x_0)/(n!)(x-x_0)^n#
Where the #n# in only #f^n(x_0)# denotes the #n#th derivative of #f(x)# and not a power.
If we wanted to find, for example, the taylor series of #cosh(x)# around #x=0# then we set #x_0=0# and use the above definition. It is best to lay out two columns, one with the derivative and the other evaluating the value of #f^n(x_0)# at the point we wish to expand around.
- #f(x) = cosh(x)# #f(0) =1#
- #f'(x) = sinh(x)# #f'(0)=0#
- #f''(x) = cosh(x)# #f''(0)=1#
- #f'''(x) = sinh(x)# #f'''(0)=0#
- #f^(IV)(x)=cosh(x)# #f^(IV)(0)=1#
- #f^(V)(x) = sinh(x)# #f^(V)(0)=0#
- #f^(VI)(x)=cosh(x)# #f^(VI)(0) = 1#
We could continue these columns indefinitely but we should get a good approximation here.
We now have our values for #f^n(0)# so all that is left is to put them into the sum above and we get:
#1/(0!)(x-0)^0+0/(1!)(x-0)^1+1/(2!)(x-0)^2+0/(3!)(x-0)^3+1/(4!)(x-0)^4+0/(5!)(x-0)^5+1/(6!)(x-0)^6+...#
Simplifying this series gives us:
#1+x^2/2+x^4/24+x^6/720+...#
And thus we have the first four non - zero terms for #cosh(x)#, (but remember the series carries on infinitely). This will give us a fairly good approximation for values of #cosh(x)# near #0#. If you need more accuracy then you need to find more derivatives and continue building up the series.
Also, if you wish to expand the series around a value of #x_0# that is not 0 then it will not be as clean as this.
