Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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They are pretty tough. It is a big field all by itself. I can do a little bit of it.
Yes, I have used. I actually wanted to say I never derived any.
I will bet you have. Ever used Stirlings formula or the Hardy-Ramanujan formula?
Okay, but I have never used an asymptotic approximations.
Even better at times is an asymptotic answer. These require a bit more math but save a ton of computation.
Yes, that's true!
The method is useful at times. Sometimes even a package has trouble getting them. For instance: If you needed something like the coefficient of x^224 519, a recursion is convenient. Using expand is package abuse.
Hi bobbym,
Hi gAr;
Hi bobbym,
Okay, I'll remember.
Hi gAr;
Hi bobbym,
Hi Into a recurrence. One that would allow us to find the coefficients without expanding. The act of taking the log of g(x) and then differentiating it would result in linearizing the gf. I can no longer remember what my method was but I do still have the program that does it for me. The next best thing is a general formula for this type problem. Let us run a few test cases and in the best use of experimental math, see if we can spot a pattern. Proof comes later. yields the recurrence: yields the recurrence: yields the recurrence: yields the recurrence: Seems like a pattern has developed. yields the recurrence: So doing the simple: From the above formula: Filling in for the variables: |