Find a power series representation for the function. (Center your power series representation at x = 0.)
f(x) =1/(7 + x)
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 +...
To find a power series representation for the function f(x) = 1/(7 + x), we can use the formula for the geometric series:
1/(1 - r) = 1 + r + r^2 + r^3 + ...
First, let's rewrite 1/(7 + x) as a fraction:
f(x) = 1/(7 + x) = 1/(7(1 + x/7))
Now we have a fraction in the form of 1/(1 - r) where r = -x/7.
Using the formula for the geometric series, we can rewrite f(x) as a power series:
f(x) = 1/(7(1 + x/7)) = 1/7 * 1/(1 - (-x/7))
Now let's substitute -x/7 into the formula:
f(x) = 1/7 * (1 + (-x/7) + (-x/7)^2 + (-x/7)^3 + ...)
To simplify further, we can expand the terms using the binomial theorem:
f(x) = 1/7 * (1 - x/7 + x^2/49 - x^3/343 + ...)
This is the power series representation for the function f(x) centered at x = 0.
To find a power series representation for the function f(x) = 1/(7 + x), we can use the concept of geometric series. A geometric series is a series in the form of ∑(n=0 to infinity) ar^n, where a is the first term and r is called the common ratio.
In this case, we can rewrite f(x) as 1/(7(1 + x/7)). Notice that 1/7 is just a constant factor and does not affect the general form of the power series since we will need to multiply the entire series by 1/7 later.
Now, we have f(x) = (1/7) * 1/(1 + (-x/7)), which is the same as f(x) = (1/7) * ∑(n=0 to infinity) (-x/7)^n.
However, it is important to ensure that the absolute value of x/7 is less than 1 for the power series to converge. In other words, |x/7| < 1.
Thus, we have the power series representation for f(x) centered at x = 0 as:
f(x) = (1/7) * ∑(n=0 to infinity) (-x/7)^n
This can be simplified further using the properties of exponents:
f(x) = (1/7) * (1 + (-x/7) + (-x/7)^2 + (-x/7)^3 + ...)
That's the power series representation for the function f(x) = 1/(7 + x), centered at x = 0.
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 - ...