Advanced methods in applied mathematics; lecture course by Richard Courant, Charles de Prima, John R. Knudsen

By Richard Courant, Charles de Prima, John R. Knudsen

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Note that g ◦ (λh) D = = = ( ◦ (λh)) ∗ Dg − g ∗ D( ◦ (λh)) ( ◦ (λh)) 2 ◦ (λh)) ∗ Dg − g ∗ (λDh) ∗ ( ◦ (λh)) ( ◦ (λh)) 2 ◦ (λh)) ∗ (Dg − λ(Dh) ∗ g) ( ◦ (λh)) 2 ( ( = 0, thus, g = β, ◦ (λh) for some scalar sequence β, or, g =β( ◦ (λh)) . 31). First note that q=( ◦ f) ∗ ( ◦ g) satisfies the equation Dq = (D(f + g)) ∗ q, since Dq = ( ◦ f ) ∗ D( =( ◦ g) + (D( ◦ f ) ∗ {(Dg) ∗ ( = q ∗ Dg + q ∗ Df ◦ f )) ∗ ( ◦ g) ◦ g)} + {(Df ) ∗ ( ◦ f )} ∗ ( = q ∗ D(f + g). By the uniqueness in the previous example, we see that ( ◦ f) ∗ ( ◦ g) = β ( ◦ (f + g)) for some constant β.

And ( ◦ c)n = 0, n ∈ Z+ . Thus ◦ c = ec . 22. Let a = {an }n∈N , b = {bn }n∈N ∈ lN such that b0 = 0. Then a ◦ b i is well defined. 11, bn = 0 for i > n. Thus ∞ n ai bni = a0 ai bni = i=0 n i=1 i=0 n=0 . 23. Let g = {gn } ∈ lN such that g0 = 0. , m − 1, and Hi = 1 for i ≥ m. Then H(m) ◦ g n ∞ = n (m) Hi (m) gni = Hi i=0 i=0 gni , n ∈ N. Hence, H(m) ◦ g n i n i=m gn = 0 n≥m . , m − 1 If we recall the convention that empty sums are equal to 0, then we may write (m) Hi 2 (m) Hi = gni 1 n! 24. Let max0≤i≤n |gi | .

5in ws-book975x65 Power Series Functions 51 The fact that ρ(|a|) = ρ(a) is clear from the previous result, that ρ(Da) = ρ(a) from lim sup |(Da)n |1/n = lim sup |nan |1/n = lim sup |an |1/n , n→∞ and that ρ n→∞ n→∞ a = ρ(a) from 1 an n+1 lim sup n→∞ 1/n = lim sup |an | 1/n . n→∞ In case each term ak in the sequence a ∈ l N is not zero, the ratio test for series also yields lim inf n→∞ an an ≤ ρ(a) ≤ lim sup . 4. Let a, b ∈ l N . For any α, β ∈ F, ρ(αa + βb), ρ(a ∗ b) ≥ min (ρ(a), ρ(b)) . Furthermore, (αa + βb)(λ) = αa(λ) + β b(λ) = αa(λ) + βb(λ) and a ∗ b(λ) = a(λ)b(λ) for |λ| < min (ρ(a), ρ(b)) .

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