Thanks a ton. I found finite calculus very interesting and useful !
On Fri, Nov 6, 2009 at 11:35 AM, abhijith reddy <abhijith200...@gmail.com>wrote:
> Thank you so much ! :)
>
>
> On Fri, Nov 6, 2009 at 11:00 AM, Prunthaban Kanthakumar <
> pruntha...@gmail.com> wrote:
>
>> On a related note,
>> The solution I gave you is to find the nth element in the kth series.
>> If you want to sum the first 'n' elements of the kth series (call the
>> function s(n,k)), then it is easy to see that,
>>
>> *s(n,k) = f(n+1, k+1) - 1*
>>
>> where f(n+1, k+1) is the (n+1)th element in the (k+1)th series.
>> This can also be easily done using the summation operator of 'finite
>> calculus'.
>>
>>
>> On Fri, Nov 6, 2009 at 10:50 AM, Prunthaban Kanthakumar <
>> pruntha...@gmail.com> wrote:
>>
>>> This is a 'finite calculus' (differences & summations) problem.
>>> You can solve it using difference operator (actually its inverse which
>>> gives you the discrete integration which is nothing but summation).
>>> If you do not know finite calculus, Google for it (or refer Concrete
>>> Mathematics by Knuth).
>>>
>>> The solution for any k is.
>>>
>>> *f(n) = nC(k+1) + nC(k-1) + nC(k-3) + .... (all the way down to nC0 or
>>> nC1 depends on k is odd or even).*
>>>
>>> Here nCr is the binomial coefficient "n choose r".
>>>
>>> Eg: Let k = 3, n = 4
>>>
>>> f(4) = 4C4 + 4C2 + 4C0 = 1 + 6 + 1 = 8
>>>
>>> Another, k = 3 and n = 5
>>>
>>> f(5) = 5C4 + 5C2 + 5C0 = 5 + 10 + 1 = 16
>>>
>>>
>>> On Wed, Nov 4, 2009 at 11:23 AM, abhijith reddy <
>>> abhijith200...@gmail.com> wrote:
>>>
>>>> Is there a way to find the sum of the Kth series ( Given below)
>>>>
>>>> K=0 S={1,2,3,4,5,6,....}
>>>> K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
>>>> K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
>>>> K=1)
>>>> K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
>>>> K=2)
>>>>
>>>> Note that the common difference of Kth series is the (K-1) series
>>>>
>>>> Any ideas ??
>>>>
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>>>>
>>>
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