Question

How do you find the Maclaurin series for f(x) using the definition of a Maclaurin series, of 4 sinh(4x)?

Answer 1

`4sinh(4x) = 16x + 128/3 x^3 + 512/15 x^5 + ... + (4^(2n+2))/((2n+1))x^(2n+1) + ...`

Explanation:

We derive a Maclaurin series from the infinite series

`f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ...`

We have:

`\ \ \ \ \ f(x) = 4sinh(4x)`
`:. f(0) = 4sinh(0) = 0`

Differentiate wrt to get the first derivative:

`\ \ \ \ \ f'(x) = 4cosh(4x) *4`
`:. f'(x) = 16cosh(4x)`
`:. f'(0) = 16cosh(0) = 16 =4^2`

Differentiate again wrt to get the second derivative:

`\ \ \ \ \ f''(x) = 16sinh(4x)*4`
`:. f''(x) = 64sinh(4x)`
`:. f''(0) = 0`

Differentiate again wrt to get the third derivative:

`\ \ \ \ \ f^((3))(x) = 64cosh(4x)*4`
`:. f^((3))(x) = 256cosh(4x)`
`:. f^((3))(0) = 256 =4^4`

Differentiate again wrt to get the fourth derivative:

`\ \ \ \ \ f^((4))(x) = 256sinh(4x)*4`
`:. f^((4))(x) = 1024sinh(4x)`
`:. f^((4))(0) = 1024sinh(0) = 0`

Differentiate again wrt to get the fifth derivative:

`\ \ \ \ \ f^((5))(x) = 1024cosh(4x)*4`
`:. f^((5))(x) = 4096cosh(4x)`
`:. f^((5))(0) = 4096 = 4^6`

`vdots`

And we can see a clear pattern forming, where

`f^((n))(0) = { (0, n " even"), (4*4^n, n " odd") :}`

So we can now form the Maclaurin series:

`f(x) = 0 + 16x + 0x^2 + 256/6 x^3 + 0x^4 +4096/120 x^5 + ...`
`\ \ \ \ \ \ \ = 16x + 128/3 x^3 + 512/15 x^5 + ... + (4*4^(2n+1))/((2n+1))x^(2n+1) + ...`
`\ \ \ \ \ \ \ = 16x + 128/3 x^3 + 512/15 x^5 + ... + (4^(2n+2))/((2n+1))x^(2n+1) + ...`

Answer 2

`4sinh(4x) = sum_(n=0)^oo 2^(4n+4)(x^(2n+1))/((2n+1)!)`

Explanation:

Take the MacLaurin series of `sinh t`:

`sinht = sum_(n=0)^oo (t^(2n+1))/((2n+1)!)`

substitute `t=4x` and multiply by `4`:

`4sinh(4x) = 4sum_(n=0)^oo ((4x)^(2n+1))/((2n+1)!)=sum_(n=0)^oo 2^2*2^(2(2n+1))(x^(2n+1))/((2n+1)!)=sum_(n=0)^oo 2^(4n+4)(x^(2n+1))/((2n+1)!)`