a) Find the Taylor series associated to f(x) = x^-2 at a = 1.
Be sure to show the general term of the series.
b) Find the radius of convergence of the series.
c)Use Lagrange's Remainder Theorem to prove that for x
in the interval of convergence with x > 1; the power series converges
to f(x).
a) To find the Taylor series associated with f(x) = x^(-2) at a = 1, we need to first find the derivatives of f(x) at x = 1.
f(x) = x^(-2)
f'(x) = -2x^(-3)
f''(x) = 6x^(-4)
f'''(x) = -24x^(-5)
...
We can observe a pattern in the derivatives:
n-th derivative of f(x) = (-1)^(n+1) * n! * x^(-n-1)
Next, we evaluate these derivatives at x = 1:
f(1) = 1^(-2) = 1
f'(1) = -2(1)^(-3) = -2
f''(1) = 6(1)^(-4) = 6
f'''(1) = -24(1)^(-5) = -24
...
Again, we observe a pattern:
n-th derivative of f(x) at x = 1 = (-1)^(n+1) * n!
Now, we can express the Taylor series for f(x) = x^(-2) at x = 1:
f(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2! + f'''(1)(x-1)^3/3! + ...
Using the pattern we found, we can write the general term of the series:
f^(n)(1)(x-1)^n/n!
where f^(n)(1) represents the n-th derivative of f(x) evaluated at x = 1.
b) To find the radius of convergence of the series, we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is L as n approaches infinity, then the series converges if L < 1 and diverges if L > 1.
In our case, we want to determine the radius of convergence, which is the value of r such that the series converges for |x-1| < r. Let's apply the ratio test to our series:
|f^(n+1)(1)(x-1)^(n+1)/(n+1)!| / |f^(n)(1)(x-1)^n/n!|
Simplifying, we get:
|f^(n+1)(1)(x-1)/(n+1)|
In order for the series to converge, the limit of this expression as n approaches infinity should be less than 1:
lim(n->inf)|f^(n+1)(1)(x-1)/(n+1)| < 1
The radius of convergence, denoted by R, is given by the reciprocal of this limit:
R = 1 / lim(n->inf)|f^(n+1)(1)(x-1)/(n+1)|
c) Lagrange's Remainder Theorem states that if a power series converges to a function f(x) on an interval (-R, R), then the error term, denoted as R_n(x), which is the difference between the function and the nth degree Taylor polynomial, can be bounded by:
|R_n(x)| <= M |x-a|^(n+1)/(n+1)!
where M is some constant.
In our case, f(x) = x^(-2) and a = 1.
To prove that the power series converges to f(x) for x > 1, we need to show that the remainder term R_n(x) approaches zero as n approaches infinity.
|R_n(x)| = |f(x) - P_n(x)| = |x^(-2) - P_n(x)|
where P_n(x) is the nth degree Taylor polynomial of f(x).
By applying Lagrange's Remainder Theorem, we have:
|R_n(x)| <= M |x-a|^(n+1)/(n+1)!
Since x > 1 and a = 1, |x-a| = x-1 and |x-a|^(n+1) = (x-1)^(n+1).
Therefore, the inequality becomes:
|R_n(x)| <= M (x-1)^(n+1)/(n+1)!
As n approaches infinity, both (x-1)^(n+1) and (n+1)! grow without bound. However, (x-1)^(n+1)/(n+1)! converges to zero as n approaches infinity.
Thus, |R_n(x)| approaches zero, which means the power series converges to f(x) for x > 1.