Proving asymptotics to sequences

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The discussion focuses on proving asymptotic behavior for a sequence defined by the recurrence relation a_n = ∑_{k=1}^n f(k)·a_{n-k}, where f(k) is a known function. The user is exploring methods to estimate a_n for large n, considering the use of first and second differences. A proposed approach involves transforming the recurrence into an integral form and applying Laplace transforms to derive asymptotic expressions. The discussion emphasizes the importance of convolution properties in this context.

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  • Understanding of recurrence relations and sequences
  • Familiarity with Laplace transforms
  • Knowledge of asymptotic analysis
  • Basic concepts of convolution in mathematics
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  • Study the properties of convolution in detail
  • Learn about Laplace transforms and their applications in asymptotic analysis
  • Explore methods for estimating sequences, including first and second differences
  • Investigate asymptotic notation and its usage in mathematical proofs
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Mathematicians, researchers in combinatorics, and anyone involved in analyzing sequences and asymptotic behavior in mathematical functions.

CRGreathouse
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Suppose I have a sequence

[tex]a_0 = 1[/tex]

[tex]a_n = \sum_{k=1}^n f(k)\cdot a_{n-k}[/tex]

where f(n) is a known function (in binomial coefficients, powers, and the like).

In general, how would I go about proving that [itex]a_n\sim g(n)[/itex]? I'm working on more closely estimating the function by calculating its value for large n, along with first and second differences. (Suggestions on better methods are welcome, though I think I'm nearly there -- the second differences look suspiciously hyperbolic.)

Any help would be appreciated.
 
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umm..if we had the sequence :

[tex]a_n +b_n = \sum_{k=1}^n f(k)\cdot a_{n-k}[/tex]

replace the sum by an integral so:

[tex]a(n)+b(n) = \int_{k=1}^n dkf(k)a(n-k)[/tex]

if you obtain A(s) F(s) and B(s) as the Laplace transform of a(n) b(n) and f(k)

[tex]A(s)+B(s)=F(s)A(s)[/tex] (using the properties of "convolution" )

invertng you could get an asymptotic expression for a(n)
 

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