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While analyzing square and cube functions, i found the following:

for y=x^2

x=1, y=1
          +3
x=2, y=4      +2
          +5
x=3, y=9      +2
          +7
x=4, y=16     +2
          +9
x=5, y=25

increase of increase (well, how else should i say this) is +2.

What does this signify?

Same pattern for y=x^3, only number in question being +6 and appearing after another round of measuring increase:

x=1, y=1
          +7
x=2, y=8      +12
          +19     +6
x=3, y=27     +18
          +37     +6
x=4, y=64     +24
          +61
x=5, y=125
Jnsx
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    I made the tables for $x^4$ and $x^5$ and obtained a similar pattern: the increasing of the $\cdots$ increasing of $x^k$ is $k!$ (the word "increasing" is written $k$ times). To put this formally, we do the following: Let $k\in\mathbb{N}$ be fixed. Define the sequence $y^k_n=n^k$ (here $n=1,2,\ldots$). Define $y^k_{n,0}=y^k_n$ and for $i\geq 0$, define $y^k_{n,{i+1}}=y^k_{n+1,i}-y^k_{n,i}$ (so that $y^k_{n,i}$ is the "increasing-$i$ times"). We want to know if: For any choice of $k$, we have $y^k_{n,k}=k!$ for every $n$. – Luiz Cordeiro Sep 06 '14 at 03:16
  • Trying to put $y^{k+1}{n,k+1}$ in terms of $y^{k+1}_n$, apparently the following formula holds for every $n,s,p,l$: $y^s{n,p}=\sum_{i=0}^l\binom{l}{i}(-1)^iy_{n+l-i,p-l}^s$. Maybe (after proving this) if we use $s=p=l$ all equal to $k$, we can obtain the result.... – Luiz Cordeiro Sep 06 '14 at 03:32
  • Maybe this can help you

    $y_n = (1+n)^2 = 1 + 2n + n^2 = (1+2n) + n^2$ (1+2n is odd)

    $y_{n+1} = (1+(n+1))^2 = 1 + 2(n+1) + y_n$ (2(n+1)+1 is the next odd number)

     – Aldo Sep 06 '14 at 03:47
  • I think applying n-times the mean value theorem you could get your result since $\frac{d^n (x^n)}{dx^n} = n! $ – Héctor Sep 06 '14 at 04:54
  • I agree that it seems to have some relation to the $n^{th}$-derivative of $x^n$ being $n!$. But simply aplying the mean value theorem gives us some weird points in $y_n\in (n,n+1)$, and the distance of $y_n$ and $y_{n+1}$ is not $1$, so applying the mean value theorem repeatedly gets more and more difficult. Also, why choosing the function $x^n$ and not $x^n+\sin(2\pi x)$? – Luiz Cordeiro Sep 06 '14 at 06:15

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I think that you are just showing that, whatever $x$ could be, the $n^{th}$ derivative of $x^n$ is a constant $$\frac{d^2}{dx^2}(x^2)=2$$ $$\frac{d^3}{dx^3}(x^3)=6$$ This corresponds to the number of steps required to arrive to your constant term. With $x^4$, one more round would give you $24$ as constant. In fact $$\frac{d^n}{dx^n}(x^n)=n!$$ what you would get after $n$ rounds.

Claude Leibovici
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