Kainui
  • Kainui
I keep finding myself interested in this question.
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Kainui
  • Kainui
This is what's called the finite difference, \(\Delta f(n) = f(n+1)-f(n)\) So that means: \(\Delta n = 1\) \(\Delta n^2 = 2n+1\) \(\Delta n^3 = 3n^2+3n+1\) \(\cdots\) \(\Delta n^k =(n+1)^k-n^k = \sum_{i=0}^{k-1} \binom{k-1}{i} n^i\) Now my question is, what's the general form for the reverse relations: \(n = \frac{1}{2} \Delta n^2 - \frac{1}{2} \Delta n\) \(n^2 = \frac{1}{3} \Delta n^3 +\frac{1}{2} \Delta n^2 + \frac{1}{6} \Delta n\) \(n^3 = \frac{1}{4} \Delta n^4 + \frac{1}{2} \Delta n^3 + \frac{1}{4} \Delta n^2 \) \(\cdots\) \(n^k = ?\)
isaacsgames
  • isaacsgames
hi
ParthKohli
  • ParthKohli
interesting - from what you've written so far I could only gather that\[\frac{1}4 = \frac{1}2 - \frac{1}4\]\[\frac{1}6= \frac{1}2-\frac{1}3\]My brain may be constructing stupid patterns from this limited information though, lol

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ParthKohli
  • ParthKohli
and also that the first term is \(\frac{1}{k+1}\Delta n^{k+1}\) but all of these seem pretty obvious
anonymous
  • anonymous
I think there's a slight typo, shouldn't it be \[\Delta n^k=\sum_{i=0}^{k-1}\binom ki n^i~~?\]The way you currently have it would make the leading term's coefficient \(1\), though it should be \(\dbinom k{k-1}=\dbinom k1=k\). Anyway, interesting question.
Kainui
  • Kainui
Ah yeah, should just be \(\binom{k}{i}\) my bad. Also, in case you guys were wondering what possible uses this has, you have already seen this before: Since the finite difference operator is a linear operator, I can push the constants in and factor it out of both of them like this: \(n = \frac{1}{2} \Delta n^2 - \frac{1}{2} \Delta n = \Delta \left( \frac{1}{2} n^2 - \frac{1}{2} n \right) = \Delta \left( \frac{n(n-1)}{2} \right)\) So, importantly: \(n= \Delta \left( \frac{n(n-1)}{2} \right)\) Sum over both sides, the right hand side is a telescoping series (I like to think fundamental theorem of calculus): \[\sum_{n=1}^{k-1} n = \sum_{n=1}^{k-1} \Delta \left( \frac{n(n-1)}{2} \right) = \frac{k(k-1)}{2}\] Yay.
ganeshie8
  • ganeshie8
this somehow reminds of the good old telescoping thingy : \[\sum\limits_{i=1}^n (i+1)^k- i^k = (n+1)^k-1\]
anonymous
  • anonymous
You're looking to write \(n^k\) as a linear combination of the \(\Delta n^i\)s, right? Something tells me \[n^k=\frac{\Delta n^{k+1}-\displaystyle\sum_{i=0}^{k-1}\binom {k+1}in^i}{\dbinom{k+1}k}\]isn't satisfactory.

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