cant see it from here its cloudy bugger
Look at zee picturz n zee web my dear, quite stunning some of them.
...35, 56, 84, 120, 165...
Yes, well done. Now any idea why?
I'm so bored here that I worked out the formula for it n(n+1)(n+2)/6
and the formula for the sum of the first n squares n(n+1)(2n+1)/6
I'm just contemplating doing that for the first n cubes. You don't know the formula off hand do you to stop me? I'm doing it by solving simultaneous equations on paper and I've almost filled up my one sheet. I'll have to go and print something soon to get anoter sheet :)
Peace
Well, apart from giving you the ability to tell the number of cannonballs in a stack (triangular pyramid in the first case, square pyramid in the second) I really don't know. Are you expecting a siege? As for the sum of cubes, the only reason I can see for working it out is to pass the time while waiting for the siege to start. Do you have enough gunpowder?
Again you are spot on! I was eating cherry tomatoes in a tetrahedral pack and was wondering how many I'd need to make a perfect stack of each size. My you are on the ball there Kevin :)
You are rght, the sum of cubes is just to pass the time ... plus wondering if there were any similarities with the formlae for the squares and numbers themselves. I thnk I'll save the thril of that 'till anoher day :)
Yes, sad isn't it :) I got a Maths degree and I feel I have to exercise (exorcise?) my brain sometimes. How things fit together and stack is a very interesting subject and has applications in everyday things such as the design if package ... now that's fascinating isn't it? *grin*
You don't sound convinced there :) but that's OK, each to their own.
And I can't even balance my checkbook!!!!!
Doreen
Nor can I - I always have to add an entry "adjustment to reflect reality" and as long as it's not too large I leave it at that :)
(n(n+1)/2)^2 I think, for the sum of cubes. A perfect job for a difference engine - the fourth difference is a constant 6.
Yes, I've just worked that out ... admitedly using Excel as my paper and letting it do the subtractions but me doing the rest. I was phrasing it
n^2(n+1)^2 / 4
but it is the same as yours :) *this is soooo sad :) *
... and the sum of the fourth powers is n(n+1)(2n+1)(3n^2+3n-1)/30
I cannot see a patern coming here - must go to bed - I spent far to long trying to factorise that last equation before convincing myself it wasn't going to go.
Nite
Calculating sums of powers by difference engine: (Note that the initial example you gave, the sequence of sums of the triangular numbers does not fit this rule because it is not a sequence of sums of powers of n.)
Some definitions: sequence 0 is the sums of the constant sequence 1,1,1,1... sequence 1 is the sums of the whole numbers 1,2,3... sequence 2 is the sums of squares sequence 3 is the sums of cubes etc.
Each member of the sequence is equal to the previous member plus the preceding first difference. Each first difference is equal to the previous first difference plus the preceding second difference. Each second difference is equal to the previous second difference plus the preceding third difference. And so on, until the nth difference is zero.
Or algebraically: Let M(i,s) represent the ith member of sequence s. Let D(i,n,s) represent the ith member of the nth differences of sequence s. Then M(i,s)=M(i-1,s)+D(i-1,1,s) And D(i,n,s)=D(i-1,n,s)+D(i-1,n+1,s)
D(1,n,s) (the initial nth difference of sequence s) cannot be calculated from the above because D(0,n,s) does not exist, but can be calculated as follows.
To begin with, D(1,1,0)=1 Then where n<=s+1 D(1,n,s) = (n+1).D(n,s-1)+n.D(n-1,s-1) and where n>s+1 D(1,n,s)=0
I've tested this up to powers of six so far.
Argh! My head...my head! Make him stop!!!!!!!!! |