Let g (t) = 1/1+4t2, and let be the Taylor series of g about 0. Then: a2n = for n = 0, 1, 2, . . . A2n+1 = for n = 0, 1, 2, . . . The radius of convergence for the series is R = Hint: g is the sum of a geometric series.

Answers

Answer 1

The Taylor series of g about 0 is given by 1 - 4t^2 + 16t^4 - 64t^6 + ... The coefficients a2n and a2n+1 are given by a2n = (-1)^n * 4^n/(2n+1) and a2n+1 = 0. The radius of convergence for the series is R = 1/2sqrt(2).

The Taylor series of g about 0 is given by:

g(t) = ∑[n=0 to infinity] ((-1)^n * 4^n * t^(2n))/(2n+1)

That this is the sum of a geometric series with first term a=1 and common ratio r=-4t^2. Therefore, we can use the formula for the sum of an infinite geometric series to get the Taylor series of g. The formula is:

S = a/(1-r)

Plugging in our values, we get:

g(t) = 1/(1+4t^2) = 1 - 4t^2 + 16t^4 - 64t^6 + ...

To find the coefficients a2n and a2n+1, we just need to look at the terms that have even and odd powers of t:

a2n = (-1)^n * 4^n/(2n+1)

a2n+1 = 0

The radius of convergence for the series is R = 1/2sqrt(2). We can see this by using the ratio test:

lim[n→∞] |a_n+1/a_n| = 4t^2/(2n+3) → 1 as n → ∞

Therefore, the series converges for |t| < 1/2sqrt(2).

To know more about geometric series, visit;

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Step-by-step explanation:

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