Question

Find a power series representation for the function. (Center your power series representation at x = 0.)
f(x) =1/(7 + x)

Answer 1

what's the trouble? At x=0,
f(x) = 1/7
f'(x) = -1/(7+x)^2 = -1/49
f"(x) = 2/(7+x)^3 = 2/343
...
now just plug those into your usual series
f(x) = 1/7 - 1/49 x + 1/343 x^2 - ...

Answer 2

Maclaurin series of function f(x) is a Taylor series of function f(x) at: a = 0

f(x) = f(0) + [ f´(0) / 1! ] ∙ x + [ f´´(0) / 2! ] ∙ x² + [ f´´´(0) / 3! ] ∙ x³ + [ f⁽⁴⁾(0) / 4! ] ∙ x⁴ +...


f(0) = 1 / ( 7 + 0 ) = 1 / 7


Find derivatives of f(x ) = 1 / ( 7 + x ) at a = 0

f´(x) = - 1 / ( 7 + x )²

f´(0) = - 1 / ( 7 + 0 )² = - 1 / 7² = - 1 / 49


f´´(x) = 2 / ( 7 + x )³

f´´(0) = 2 / ( 7 + 0 )³ = 2 / 7³ = 2 / 343


f´´´(x) = - 6 / ( 7 + x )⁴

f´´´(0) = - 6 / ( 7 + 0 )⁴ = - 6 / 7⁴ = - 6 / 2401


f⁽⁴⁾(x) = 24 / ( 7 + x )⁵

f⁽⁴⁾(0) = 24 / ( 7 + 0 )⁵ = 24 / 7⁵ = 24 / 16807

So:

f(x) = f(0) + [ f´(0) / 1! ] ∙ x + [ f´´(0) / 2! ] ∙ x² + [ f´´´(0) / 3! ] ∙ x³ + [ f⁽⁴⁾(0) / 4! ] ∙ x⁴ +...

f(x) = 1 / 7+ [ - 1 / 49 / 1! ] ∙ x + [ 2 / 343 / 2! ] ∙ x² + [ - 6 / 2401 / 3! ] ∙ x³ + [ 24 / 16807/ 4! ] ∙ x⁴ +...

f(x) = 1 / 7+ [ - 1 / 49 / 1 ] ∙ x + [ 2 / 343 / 2 ] ∙ x² + [ - 6 / 2401 / 6 ] ∙ x³ + [ 24 / 16807/ 24 ] ∙ x⁴ +...

f(x) = 1 / 7 - x / 49 + x² / 343 - x³ / 2401 + x⁴ / 16807 +...