Finding radius of convergence in a power series?

Question

a.) Suppose that (10x)/(13 + x) = (sum from n=0 to infinity) of c(sub-n)*(x^n).

b.) The function f(x)= (9)/(1+36x^2) is represented by the power series f(x)= (sum from n=0 to infinity) of c(sub-n)*x^n.

For both, find the first few coefficient, from 0 to 4, and find the radius of convergence. Also, what I mean by c(sub-n) is the letter c with a little n at its bottom.

Answer 1

Use the geometric series

1/(1 - t) = Σ(n=0 to ∞) t^n, convergent for |t| < 1.



a) 10x/(13 + x)

= 10 * [(x + 13) - 13] / (13 + x)

= 10 * [1 - 13/(13 + x)]

= 10 * [1 - 1/(1 + x/13)]

= 10 * [1 - 1/(1 - (-x/13))]

= 10 * [1 - Σ(n = 0 to ∞) (-x/13)^n], by the geometric series with t = -x/13

= 10 * [-Σ(n = 1 to ∞) (-x/13)^n]

= Σ(n = 1 to ∞) 10 (-1)^(n+1) * x^n /13^n

= Σ(n = 0 to ∞) 10 (-1)^n * x^(n+1) /13^(n+1), by reindexing the sum

= 10x/13 - 10x^2/169 + 10x^3 /13^3 - 10x^4 /13^4 + ...



This converges when |-x/13| < 1 <==> |x| < 13.

Hence, the radius of convergence is 13.

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b) f(x) = 9 * 1/(1 - (-36x^2))

= 9 * Σ(n=0 to ∞) (-36x^2)^n, by the geometric series with t = -36x^2

= Σ(n=0 to ∞) 9(-36)^n x^(2n).

= 9 + 0x - 9 * 36x^2 + 0x^3 + 9 * 36^2 x^4 + ...



This converges when |-36x^2| < 1 <==> |x| < 1/6.

So, the radius of convergence is 1/6.



I hope this helps!

Answer 2

to locate the era of convergence |x-a| < R, use the ratio or root try (often, ratio try will yield the consequent era). using the ratio try: a million. lim_n-inf._[ | [(x-8)^(n+a million) / 8^(n+a million)] * [8^n / (x-8)^(n)] | ]. 2. Simplify the expression [(x-8)^(n+a million) / 8^(n+a million)] * [8^n / (x-8)^(n)] interior the shrink: [(x-8)^n * (x-8)^a million * 8^n] / [(x-8)^n * 8^n * 8^a million]; (x-8)^n term and eight^n term cancel leaving: (x-8)/8. 3. The shrink simplifies to lim_n-inf._[ | (x-8)/8| ]. via fact the expression does not rely on 'n', you may manage it as a continuing. making use of the shrink-consistent rule, lim_n-inf._[ (x-8)/8| ] = |(x-8)/8|. 4. in accordance to the ratio try for the sum of a chain A_n from n = a million to infinity, if lim_n-inf._( |A_n+a million / A_n| ) < a million, then the sequence converges genuinely (If the sequence converges genuinely, there's a theorem that asserts that the sequence additionally converges). If lim_n-inf._( |A_n+a million / A_n |) > a million or equals infinity or DNE, then the sequence diverges. subsequently, the sequence with the sequence (x-8)^n / 8^n converges if |(x-8)/8| < a million and diverges otherwise. 5. locate the radius of convergence by ability of manipulating the inequality you ended with after taking the shrink which subsequently is (x-8)/8 < a million. a. |(x-8)/8| < a million b. - 8 < (x-8) < 8 ; subsequently for the sequence based around a = 8, the radius of convergence R is 8 or |x-a|

Answer 3

a) c_n = (-1)^n 10 / 13^(n+1) , | x | ≤ 1



b) c_2n = (-1)^n 9 [ 36^n] , | x | < 1/6 , c_(2n-1) = 0