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

Jun 16, 2015

$1 - 2 \left(x + 1\right) + 5 {\left(x + 1\right)}^{2} - 4 {\left(x + 1\right)}^{3} + {\left(x + 1\right)}^{4}$

#### Explanation:

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, $f \left(x\right) = {x}^{4} - {x}^{2} + 1$ so $f ' \left(x\right) = 4 {x}^{3} - 2 x$, $f ' ' \left(x\right) = 12 {x}^{2} - 2$, $f ' ' ' \left(x\right) = 24 x$, $f ' ' ' ' \left(x\right) = 24$ and all higher-order derivatives are identically zero. Thus, $f \left(- 1\right) = 1 - 1 + 1 = 1$, $f ' \left(- 1\right) = - 4 + 2 = - 2$, $f ' ' \left(- 1\right) = 12 - 2 = 10$, $f ' ' ' \left(- 1\right) = - 24$, $f ' ' ' ' \left(- 1\right) = 24$ 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:

$1 - 2 \left(x + 1\right) + 5 {\left(x + 1\right)}^{2} - 4 {\left(x + 1\right)}^{3} + {\left(x + 1\right)}^{4}$

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

2) Use the substitution $u = x + 1$ to expand $f \left(x\right) = f \left(u - 1\right)$ 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:

$f \left(u - 1\right) = {\left(u - 1\right)}^{4} - {\left(u - 1\right)}^{2} + 1$

$= {u}^{4} - 4 {u}^{3} + 6 {u}^{2} - 4 u + 1 - \left({u}^{2} - 2 u + 1\right) + 1$

$= 1 - 2 u + 5 {u}^{2} - 4 {u}^{3} + {u}^{4}$

$= 1 - 2 \left(x + 1\right) + 5 {\left(x + 1\right)}^{2} - 4 {\left(x + 1\right)}^{3} + {\left(x + 1\right)}^{4}$, as before.