I am trying to identify the mathematical properties of this polynomial $(4x^3+6x^2+4x+1)^k$. Most of all I’ve been attempting to describe an actual relationship to binomial coefficients, such as binomial$(4,n)^k$, and it’s very nature itself. How would this polynomial be written in a ‘concise manner’ resembling the binomial properties instead of just terms? Such as the equivalent of writing binomial$(n,k)=n!/(k!(n-k)!)$...? On Wolfram the gamma function shows up many times, it’s use seems unsatisfactory, and too mathematically rigorous. I’ve tried several methods of finding a useful method for years; multinomials, examining Stirling numbers, Catalan numbers, and so on, and nothing seems to work. I’m not worried about the complexity, it just a matter of finding a better more elegant representation.
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1Note that you can rewrite that as $((1+x)^4-x^4)^k$ – Thomas Andrews Nov 13 '17 at 02:33
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There's the multinomial formula (it's not as bad as it looks): https://en.wikipedia.org/wiki/Multinomial_theorem#Theorem – Theo Bendit Nov 13 '17 at 02:34
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I’ve noticed over the years how pascal’s triangle and the limiting behavior it represents to the polynomial (4x^3+6x^2+4x+1)^k, but I have never seen any mathematics resembling this approach. – Mike Moye Nov 13 '17 at 02:37
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Sure. You can use binomial coefficients as the thing is $(x+1)^4 - x^4.$ If you know how to write $(a-b)^k,$ you have a way to write $\left((x+1)^4 - x^4 \right)^k.$
Will Jagy
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I have haven’t seen that before, I was wrote (x^4+4x^3+6x^2+4x+1)-(4x^3+6x^2+4x+1)=x^4 once. I like your way better, but do you know how I can make it into more of a ‘binomial(n,k)’ kind of formula. – Mike Moye Nov 13 '17 at 02:46
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In further examination, an easier way to write (x^4+4x^3+6x^2+4x+1)^k-(4x^3+6x^2+4x+1)^k-(6x^2+4x+1)^k-(4x+1)^k-1? – Mike Moye Nov 13 '17 at 02:55
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Another note: if x=1, k=1? {16^n-15^n-11^n-5^n-1^n}. Trying to understand these properties compared to just 16^n, compared to binomial(n,k). – Mike Moye Nov 13 '17 at 03:00
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Sorry, I’m not sure? Something similar to binomial({3,5},n)? It eludes me. – Mike Moye Nov 13 '17 at 03:11
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@MikeMoye $(a-b)^3 = a^3 - 3 a^2 b + 3 a b^2 - b^3. $ Can you expand $(a-b)^5 ; ; ?$ – Will Jagy Nov 13 '17 at 03:30
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