Thursday, April 12, 2018

integration - What is this sequence of polynomials?

NovaDenizen says the polynomial sequence i wanted to know about has these two recurrence relations



(1) $p_n(x+1) = \sum_{i=0}^{n} (x+1)^{n-i}p_i(x)$



(2) $p_{n+1}(x) = \sum_{i=1}^{x} ip_n(i)$



==




i was trying to calculate the probability of something and i came upon them. i needed to know what this was equal to:



$$p_n(x)=\sum_{k_n=k_{n-1}}^{x}....\sum_{k_3=k_2}^{x} \sum_{k_2=k_1}^{x} \sum_{k_1=1}^{x}k_1k_2...k_n$$



$k \in (1,2,...,x)$.



if you make it continuous and over the reals instead of over the natural numbers then its not too hard to see what that equals.



$$p_n(x) \approx \int_{k_n=0}^{x}...\int_{k_3=k_2}^{x}\int_{k_2=k_1}^{x}k_1k_2k_ndk_1dk_2...dk_n =\dfrac{(x)^{2n}}{2^n n!}$$




i computed some of these and got



$$p_1(x) \approx \frac{x^2}{2}, p_2(x) \approx \frac{x^4}{8}, p_3(x) \approx \frac{x^6}{48}, p_4(x) \approx \frac{x^8}{384}, p_5(x) \approx \frac{x^{10}}{3840},...$$
so im assuming that's the formula.



from the summation formula its easy to see that $$p_1(x)=\sum_{k=1}^x k=\frac{x(x+1)}{2}$$



i spent some time to compute $$p_2(x)=\sum_{k_2=k_1}^x\sum_{k_1=1}^x k_1k_2=x^4/8+(5 x^3)/12+(3 x^2)/8+x/12=x (3 x+1) (x+1) (x+2)/24$$




these agree with the approximations from integrating, which im guessing gives the first terms of $p_n(x)$.



also i think its might be fair to say that $\dfrac{(x)^{2n}}{2^n n!} < p_n(x) < \dfrac{(x+1)^{2n}}{2^n n!}$ if you can use the integral approximation to get lower and upper estimates of $p_n(x)$.



but im wondering what this sequence of polynomials is. i think i can just use the first terms of them to calculate the probabilities i wanted to know well enough, but it wouldn't hurt to know if this sequence of polynomials has a name. thanks.

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