How do you find the Taylor's formula for #f(x)=e^x# for x-2?

Question

Luke Alvarez · Accepted Answer

#f(x) = e^2( 1 + (x-2) + (x-2)^2/2 + (x-2)^3/6 +...)#

Did you mean find the Taylor series of #f(x)# about #x = 2# ?

The Taylor expansion of a function about a point a is defined as

#f(x) = f(a) + f'(a) (x-a) + (f''(a))/(2!) (x-a)^2 + (f'''(a))/(3!) (x-a)^3 + ... = sum_(n=0)^(oo) (f^((n))(a))/(n!) (x - a)^n#

Taking #a = 2# we have that

#f(x) = e^2( 1 + (x-2) + (x-2)^2/2 + (x-2)^3/6 +...)#

as the derivative of #e^x# is simply #e^x#.

Elijah Abram · Answer

undefined To find Taylor's formula for $f(x) = e^x$ centered at $x = 2$, we first need to find the derivatives of $f(x)$ at $x = 2$. Then, we'll evaluate those derivatives at $x = 2$ and plug them into Taylor's formula.

1. Find the derivatives of $f(x) = e^x$:
$$f'(x) = e^x$$
$$f''(x) = e^x$$
$$f'''(x) = e^x$$
and so on. Notice that all derivatives of $e^x$ are $e^x$.

2. Evaluate the derivatives at $x = 2$:
$$f(2) = e^2$$
$$f'(2) = e^2$$
$$f''(2) = e^2$$
and so on.

3. Taylor's formula for $f(x)$ centered at $x = 2$ is:
$$f(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 + \frac{f'''(2)}{3!}(x - 2)^3 + \cdots$$

Substituting the values we found:
$$f(x) = e^2 + e^2(x - 2) + \frac{e^2}{2!}(x - 2)^2 + \frac{e^2}{3!}(x - 2)^3 + \cdots$$