How do you find the first five terms of the Taylor series for #f(x)=x^8+x^4+3# at #x=1#?

Question

Christopher Agresta · Accepted Answer

The Taylor series is a particular way to approximate a function with a polynomial in the neighbourhood of a generic point #(x_0,f(x_0))#.

#f(x)=f(x_0)+f'(x_0)(x-x_0)^1/(1!)+f''(x_0)(x-x_0)^2/(2!)+...#

or, with the series notation:

#f(x)=sum_(i=0)^(+oo)f^((i))(x_0)(x-x_0)^i/(i!)#.

So, let's calculate all the derivatives we need:

#f(1)=5#

#f^((1))(x)=8x^7+4x^3# and #f^((1))(1)=12#

#f^((2))(x)=56x^6+12x^2# and #f^((2))(1)=68#

#f^((3))(x)=336x^5+24x# and #f^((3))(1)=360#

#f^((4))(x)=1680x^4+24# and #f^((4))(1)=1704#

so:

#f(x)=5+12(x-1)^1/(1!)+68(x-1)^2/(2!)+360(x-1)^3/(3!)+1704(x-1)^4/(4!)#

It's easy to understand that if you would calculate the Taylor's series until the 8th order you will obtain the function itself, because it is a polynomial!

Christopher Allers · Answer

undefined To find the first five terms of the Taylor series for $ f(x) = x^8 + x^4 + 3 $ at $ x = 1 $, we need to find the derivatives of $ f(x) $ and evaluate them at $ x = 1 $. Then, we'll use these values to write out the Taylor series expansion.

First, let's find the derivatives of $ f(x) $:

$ f'(x) = 8x^7 + 4x^3 $

$ f''(x) = 56x^6 + 12x^2 $

$ f'''(x) = 336x^5 + 24x $

$ f''''(x) = 1680x^4 + 24 $

$ f'''''(x) = 6720x^3 $

Now, let's evaluate these derivatives at $ x = 1 $:

$ f(1) = 1^8 + 1^4 + 3 = 5 $

$ f'(1) = 8(1)^7 + 4(1)^3 = 8 + 4 = 12 $

$ f''(1) = 56(1)^6 + 12(1)^2 = 56 + 12 = 68 $

$ f'''(1) = 336(1)^5 + 24(1) = 336 + 24 = 360 $

$ f''''(1) = 1680(1)^4 + 24 = 1680 + 24 = 1704 $

Now, let's write out the first five terms of the Taylor series expansion:

$ f(x) \approx f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f''''(1)}{4!}(x-1)^4 $

$ = 5 + 12(x-1) + \frac{68}{2!}(x-1)^2 + \frac{360}{3!}(x-1)^3 + \frac{1704}{4!}(x-1)^4 $

Simplify if needed.