How do you find the taylor series series for #f(x) = x^4 - x^2 + 1# at c=-1?

Question

Luke Alvarez · Accepted Answer

#1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4#

There are two methods:

1) Use #f(-1)+f'(-1)(x+1)+(f''(-1))/(2!)(x+1)^2+(f'''(-1))/(3!)(x+1)^3+(f''''(-1))/(4!)(x+1)^4+\cdots#

Here, #f(x)=x^4-x^2+1# so #f'(x)=4x^3-2x#, #f''(x)=12x^2-2#, #f'''(x)=24x#, #f''''(x)=24# and all higher-order derivatives are identically zero. Thus, #f(-1)=1-1+1=1#, #f'(-1)=-4+2=-2#, #f''(-1)=12-2=10#, #f'''(-1)=-24#, #f''''(-1)=24# and all higher-order derivatives at #x=-1# are zero.

Since #2! =2#, #3! =6#, and #4! =24#, this gives the answer above; the Taylor series is:

#1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4#

Since #f(x)# is a polynomial, #f(x)# equals its Taylor series for all #x# and we can write #f(x)=x^4-x^2+1=1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4# for all #x#.

2) Use the substitution #u=x+1# to expand #f(x)=f(u-1)# as a polynomial in powers of #u#. Then replace #u# by #x+1# at the end.

The binomial theorem (Pascal's triangle ) is helpful in doing the expansion:

#f(u-1)=(u-1)^4-(u-1)^2+1#

#=u^4-4u^3+6u^2-4u+1-(u^2-2u+1)+1#

#=1-2u+5u^2-4u^3+u^4#

#=1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4#, as before.

Christopher Agresta · Answer

undefined To find the Taylor series for $ f(x) = x^4 - x^2 + 1 $ at $ c = -1 $, we'll need to find the derivatives of $ f(x) $ at $ c = -1 $ and evaluate them, and then use these values to construct the Taylor series. Here are the steps:

1. Find the derivatives of $ f(x) $:

$$
   \begin{align*}
   f(x) & = x^4 - x^2 + 1 \
   f'(x) & = 4x^3 - 2x \
   f''(x) & = 12x^2 - 2 \
   f'''(x) & = 24x \
   f''''(x) & = 24
   \end{align*}
   $$

2. Evaluate the derivatives at $ c = -1 $:

$$
   \begin{align*}
   f(-1) & = (-1)^4 - (-1)^2 + 1 = 1 + 1 + 1 = 3 \
   f'(-1) & = 4(-1)^3 - 2(-1) = -4 - (-2) = -2 \
   f''(-1) & = 12(-1)^2 - 2 = 12 - 2 = 10 \
   f'''(-1) & = 24(-1) = -24 \
   f''''(-1) & = 24
   \end{align*}
   $$

3. Construct the Taylor series using these derivatives:

$$
   \begin{align*}
   f(x) & = f(-1) + f'(-1)(x + 1) + \frac{f''(-1)}{2!}(x + 1)^2 + \frac{f'''(-1)}{3!}(x + 1)^3 + \frac{f''''(-1)}{4!}(x + 1)^4 \
   & = 3 - 2(x + 1) + \frac{10}{2!}(x + 1)^2 - \frac{24}{3!}(x + 1)^3 + \frac{24}{4!}(x + 1)^4 \
   & = 3 - 2(x + 1) + 5(x + 1)^2 - 4(x + 1)^3 + (x + 1)^4
   \end{align*}
   $$

So, the Taylor series for $ f(x) = x^4 - x^2 + 1 $ at $ c = -1 $ is:

$$ 3 - 2(x + 1) + 5(x + 1)^2 - 4(x + 1)^3 + (x + 1)^4 $$