*Find the general formula for where k can be any non-negative number.*

We know that a sum of polynomial of degree n would give polynomial of degree (n+1) by knowledge similar to integration, so we are not going to find the general formula through k, but through n for any fixed k.

For the first few k many of you should have already recited these:

1) , this is the series 1+1+1+...

2) , this is the summation of A.S..

3)

4)

For fixed k we only need to differentiate/integrate w.r.t. n, it seems easy:

.

Yes this is a diff. w.r.t. n (the summation formula is in terms of n, so we differentiate w.r.t. i in the summation sign.), but it looks like it's differentiate w.r.t k.

However, this would be a good news to us because we can link functions of different k together, while smaller k is trivially to compute.

We try to differentiate a few of them:

1)

2)

3)

Noticing the varying constant at the last, what happened?

Constant remind us about the constant results from integration, but why we get such a constant from integration, and are there any formula to compute that constant?

In elemental approach we can explain it like this by integration:

Now note that this C depends of the initial value of f(n), now we know that , so C is equal to zero because the function does not contain any other constant other than the integration concerning the polynomial.

However,

**the constant of is not necessarily related to the constant of , then we have to determine the constant again.**For constant with non-zero degree, i.e., , it's no longer vary along to initial value, so it could be fixed for further integration.

Now where C=0 for k=0. , by putting we have

However, it's extremely hard to generate the general formula of the constant because it's in a special recurrence form.

Let's try the formula for :

, putting n=1 gives . for k=4, though it will vary for larger k. Now we get . Try the first few term which is correct.

Problem to explore:

1) Show that and , hence find the formula for k=5,6,7 and try the first few terms.

2) Prove or disprove that is divisible by for all natrual n,k.

## No comments:

## Post a comment