solutions for b

2 2 alternative answers for 2a 2b since 2 2

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Unformatted text preview: ⋆ 1 6 ·4 ·2 ·6 ·4 ·2 − = 1 4 ·2 ·4 ·2 + 1 1 2 ·2 − = − + = 2 !! 2 x2 x4 1 x2 + x8 1 x6 + 8 !! 2 x8 − · · · x2 ℓ . 2 2 ℓ !! Alternative Answers for (2a), (2b) : Since 2 2 ℓ !! 6 !! 2 1 ℓ x10 + · · · 1 x4 − 2 4 !! −1 x6 = ℓ! 2 · 22 ℓ , you may write your answers alternatively as (2a) Jx = ∞ ℓ=0 (2b) J x = 1 − 1 1! 2 x 2 1 ℓ −1 2 + ℓ! 1 2! 2 x 2 3 2 4 − 1 3! 2ℓ 1 2 · 2 · x2 ℓ . x 2 6 + 1 4! 2 x 2 8 − ··· . (3) ∞ The radius of convergence for 1 ℓ −1 ℓ=0 · x2 ℓ 2 2 ℓ !! R = +∞. Work : Consider the series with variable u: ∞ ℓ=0 aℓ = 2 · uℓ . 2 ℓ !! 1 ℓ Set 1 ℓ −1 −1 Then . 2 2 ℓ !! aℓ 1 = 2 , 2 ℓ !! aℓ+1 1 = 2 2 ℓ + 2 !! . 1 2 2 ℓ !! =⇒ lim ℓ− ∞ → aℓ aℓ+1 = lim ℓ− ∞ → 1 2 2 ℓ + 2 !! 4 is = lim ℓ− ∞ → 2 2 ℓ + 2 !! 2 ℓ !! 2 = lim ℓ− ∞ → 2ℓ + 2 = + ∞. Hence the series ( ∗) ∞ ℓ=0 1 ℓ −1 2 · uℓ 2 ℓ !! with variable u has radius of convergence + ∞. Then the original series ∞ ℓ=0 1 ℓ −1 2 · x2 ℓ 2 ℓ !! with variable x must have radius of convergence + ∞. Indeed, the original series with variable x is merely the outcome of the substitution of u = x2 in the series (∗) above with variable u . We have just proved that the series (∗) is convergent for an arbitrary real number u. It is then clear that the original series...
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This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

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