Question

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] f(x) = 10 x , a = −2

Answer 1

Answer:

$\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n$

Step-by-step explanation:

Rn(x) →0

f(x) = 10/x

a = -2

Taylor series for the function f at the number a is:

$f(x) = \sum^\infty_{n=0} \frac{f^{(n)}(a)}{n!} (x - a)^n$

$f(x) = f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f"(a)}{2!} (x-a)^2 + ...$ ............ equation (1)

Now we will find the function f and all derivatives of the function f at a = -2

f(x) = 10/x            f(-2) = 10/-2

f'(x) = -10/x²         f'(-2) = -10/(-2)²

f"(x) = -10.2/x³      f"(-2) = -10.2/(-2)³

f"'(x) = -10.2.3/x⁴     f'"(-2) = -10.2.3/(-2)⁴

f""(x) = -10.2.3.4/x⁵    f""(-2) = -10.2.3.4/(-2)⁵

∴ The Taylor series for the function f at a = -4 means that we substitute the value of each function into equation (1)

So, we get $\sum^\infty_{n=0} - \frac{10(x+2)^n}{2^{n+1}}$ Or $\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n$