Let ${a_n}$ be a sequence of real numbers. The backwards differences of this sequence are defined recursively:

Question

Let ${a_n}$ be a sequence of real numbers. The backwards differences of this sequence are defined recursively:

The first difference $∇a_n$ is an new sequence defined by:$∇a_{n} = a_{n} - a_{n-1} $

The $(k + 1)$st difference $∇^{k+1}a_n $ is $∇^{k+1}a_n = ∇^{k}a_{n} - ∇^{k}a_{n-1} $ Find $∇a_{n}$ and $∇^{2}a_{n}$ for $a_{n} = 2n$ and $a_{n} = n^2$

So far I have been able to determine (using the first difference formula) that $∇a_n$ for $a_n = 2n$ is as follows... $$∇a_n = 2n - (2(n-1))$$ $$∇a_n = 2$$ However I have not been able to figure out the $∇^2a_n$ case. As for the second case I have done the same in figuring out $∇a_n$ but have not been able to get $∇^2a_n$ $∇a_n$ for $a_n = n^2$ is as follows... $$∇a_n = n^2 - (n-1)^2$$ $$∇a_n = n^2 - n^2 - 2n + 1$$ $$∇a_n = -2n+1$$ Any help with the $∇^2a_n$ cases would be greatly appreciated. The main part I am having trouble understanding how we would get the $∇a_{n-1}$ case to use in the given formula.

Answer 1

Ok, for $a_n=n^2$ we have $∇a_n=a_n-a_{n-1}=n^2-(n-1)^2=n^2-n^2+2n-1=2n-1$. You had some signs wrong. As for the other, ${∇}^2a_n=∇(a_n-a_{n-1})=∇a_n-∇a_{n-1}=a_n-a_{n-1}-a_{n-1}+a_{n-2}=a_n-2a_{n-1}+a_{n-2}$. Applying this formula to $a_n=2n$, we get $∇^2a_n=2n-4(n-1)+2(n-2)=0$. This is the same as saying $∇ ∇a_n=∇2=0$, if this is the same as you mean by substituting the 2, then you were right, I misunderstood. You can do the other one in exactly this way.

Answer 2

For the first problem you know that $\nabla a_n=2$ for all $n$, so $\nabla a_n-\nabla a_{n-1}=2-2=0$.

In the second you made an algebra error: $\nabla a_n=n^2-(n-1)^2=n^2-n^2+2n-1=2n-1$, not $-2n+1$. Note that this is an error that you should have caught: $-2n+1$ is clearly negative for $n\ge 1$, but $n^2-(n-1)^2$ can’t possibly be negative for $n\ge 1$.

To see how to deal with $\nabla a_{n-1}$, it may be helpful temporarily to introduce an auxiliary sequence: let $b_n=\nabla a_n=a_n-a_{n-1}$. Then

$$\nabla a_{n-1}=b_{n-1}=a_{n-1}-a_{(n-1)-1}=a_{n-1}-a_{n-2}\;.$$

Or you can simply realize that $\nabla a_k=a_k-a_{k-1}$ no matter what $k$ is, and when $k=n-1$, this becomes $\nabla a_{n-1}=a_{n-1}-a_{n-2}$. At any rate, you now have

$$\nabla^2a_n=\nabla a_n-\nabla a_{n-1}=(2n-1)-\big(2(n-1)-1\big)\;,$$

and I’ll leave the cleaning up to you.