Question

Question #d43f9

Answer 1

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

Explanation:

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!