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Did anyone else have a look at the Venus transit? I must admit I… - Peter Sheil [entries|archive|friends|userinfo]
Peter Sheil

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[Jun. 8th, 2004|01:30 pm]
Peter Sheil
Did anyone else have a look at the Venus transit? I must admit I didn't bother trying to see it myself as I thought the image would be too small to see clearly. I had a putter around the web and the best images I found where at the Institute for Solar Physics, Stockholm (that's Sweden for my American readers ... part of the spikey bit pointing down just left of Russia :P ) Their main page is here and they have a gallery of images from the transit. Have you found any better ones? If so then please post a link to share them.

And now for something comletely different, as a well known TV series once said.
What is the next number ib this series?

1, 4, 10, 20, ?

And for the super clever of you there is an additional question (for extra points). Why did I think of this as I was having my lunch? Need a clue? Well ... I'll tell you what I had if you like.

Chicken and Bacon sandwich
2 snack pork pies
a small packet of cherry tomatoes
a banana

There, that's much clearer now isn't it? *evil laugh*

I'm off now to finish eating my fruit. I look forward to your comments.
Peace
peter
LinkReply

Comments:
From: fruitbatz
2004-06-08 06:13 am (UTC)
cant see it from here its cloudy
bugger
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[User Picture]From: petersheil
2004-06-08 07:57 am (UTC)
Look at zee picturz n zee web my dear, quite stunning some of them.
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From: fruitbatz
2004-06-08 08:17 am (UTC)
ok smartarse..lol
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[User Picture]From: kevinrtaylor
2004-06-08 06:37 am (UTC)
...35, 56, 84, 120, 165...
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[User Picture]From: petersheil
2004-06-08 07:54 am (UTC)
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
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[User Picture]From: kevinrtaylor
2004-06-08 08:30 am (UTC)
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?
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[User Picture]From: petersheil
2004-06-08 08:38 am (UTC)
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 :)
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[User Picture]From: feath
2004-06-08 08:54 am (UTC)
You do this for fun? O_o
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[User Picture]From: petersheil
2004-06-08 08:59 am (UTC)
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*
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[User Picture]From: feath
2004-06-08 09:29 am (UTC)
*cautious nod*
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[User Picture]From: petersheil
2004-06-08 09:44 am (UTC)
You don't sound convinced there :) but that's OK, each to their own.
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[User Picture]From: peaceful_fox
2004-06-08 09:35 am (UTC)
And I can't even balance my checkbook!!!!!

Doreen
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[User Picture]From: petersheil
2004-06-08 09:42 am (UTC)
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 :)
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[User Picture]From: kevinrtaylor
2004-06-08 09:10 am (UTC)
(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.
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[User Picture]From: petersheil
2004-06-08 09:40 am (UTC)
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 :) *
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[User Picture]From: petersheil
2004-06-08 05:41 pm (UTC)
... 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
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[User Picture]From: kevinrtaylor
2004-06-09 04:19 am (UTC)
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.
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[User Picture]From: peaceful_fox
2004-06-10 06:00 am (UTC)
Argh! My head...my head! Make him stop!!!!!!!!!
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