Math Is Fun Forum

Knight52

Guest

Sigma, Help!

I got given this question by my maths teacher to investigate the rules after x^3 plz help me!
These are the first 3 rules:

Σ(1,n)x=n(n+1)/2
Σ(1,n)x^2=n(n+1)(2n+1)/6
Σ(1,n)x^3=n^2.(n+1)^2/4

(n=the last or highest number)

How do i find the rule for to the power of four, or above?

I am in Yr 10 Extension

Thanks

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Sigma, Help!

I'll give you a hint

Take note that the sum of x³ is the square of the sum of x. What might you guess then about the relationship between the sum of x^4 and that of x²?

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Knight52

Guest

Re: Sigma, Help!

Thankyou very very much however your hint was slightly to obvious but i am still very grateful

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Sigma, Help!

uhm, but just squaring the x^2 formula doesnt work..?

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Sigma, Help!

I was sure that was it i remember doing this in FP1, what was the rule then? I know it was something similar to that atleast.

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The Beginning Of All Things To End.
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Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Sigma, Help!

I wonder if this is correct???

Last edited by Dragonshade (2008-04-24 04:03:14)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Sigma, Help!

Nice thinking! One thing - there's a term in the x^5 sum that isn't cancelled by the (x+1)^5 sum and so needs to be subtracted. If you're summing from 1 to n, then that term would be 1.
But otherwise, it looks correct and very elegant.

My way isn't nearly as nice, I just compared many layers of differences.

The first few terms of the sequence of sums are 1, 17, 98, 354, 979, 2275, 4676, 8772, ...
Now look at the differences between those terms, and then differences between the differences, and then differences between those...

1st differences: 16, 81, 256, 625, 1296, 2401, 4096, ...
2nd:                   65, 175, 369, 671, 1105, 1695, ...
3rd:                      110, 194, 302, 434, 590, ...
4th:                          84, 108, 132, 156, ...
5th:                             24,  24,  24,  ...

Oh look a pattern! The 5th level of differences is a constant 24. We're on the 5th level, so that helps us find the n^5 coefficient of the nth term in the sequence of terms. Being on the 5th level also means we need to divide by 5! = 120 (don't ask why ), and so the coefficient turns out to be 1/5. 

Now the fun part - we do it all again, but after taking away x^5/5 to adjust for the term we just found. Repeating the entire process with the adjusted sequence should give a constant term on the 4th level of differences (or earlier), which lets us find another coefficient for the nth term and then we do it all again, repeating until the entire sequence is found.

...Or we take a shortcut and make Excel do it.

Feeding the sequence of sums into Excel gives that the nth term is x^5/5 + x^4/2 + x^3/3 - x/30.
Factorising that gives:

Interestingly, this is the same as the sum of the first n squares, but with an extra term of

multiplied in.

But I much prefer Dragonshade's method to this one.

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Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Sigma, Help!

Oh yea, the 1 isnt canceled.   Wow, mathsyperson, your method is complicated, I couldnt figure out the difference thing. how did you even come up with that , awesome

Last edited by Dragonshade (2008-04-24 04:06:45)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Sigma, Help!

I came up with this thing ages ago (which could be why it's so inefficient ), as a way of finding nth terms that usually turned out to be quadratic (since they were Year 8 questions and so not too hard). For example, to find the sum of the first n integers:

The sequence is:                1, 3, 6, 10, 15, ...
The first differences are:       2, 3,  4,  5, ...
The second differences are:     1, 1, 1, ...

We're on the 2nd level here, so divide by 2! to get a term of x^2/2.

Now taking that away from the original gives 1/2, 1, 3/2, 2, 5/2, ...
First differences: 1/2, 1/2, 1/2, ...

And therefore there's an x/2 term there as well.
Taking that away gives 0, 0, 0, ... and so we're finished.

The nth term is x^2/2 + x/2, or more commonly x(x+1)/2.
So with that example, the number-crunching isn't quite as nasty.

There was a worksheet of with about 20 sequences where we needed to work out the nth terms (that were either linear or quadratic). I'd done most of them, and near the end I came across one that had a typo. I have no idea what it was now, but a similar example would be 2, 6, 12, 23, 38.
The 6 is actually meant to be a 5, and the teacher corrected it in the next lesson.

But I noticed that because whenever you go down a layer of differences, the length of the sequence goes down by 1, you could theoretically "cheat" by taking differences until there's only one number in the sequence. Then there would always be a "pattern" that you could use. A bit more investigation and I discovered the 'dividing by the factorial of the level' rule, and then it could be used for any sequence at all. I came up with a quartic function that correctly gave those 5 numbers. It didn't get marked though.

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Why did the vector cross the road?
It wanted to be normal.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Sigma, Help!

Hmm, maybe you need a general prove for the algorithm?  does this apply to other sequence?

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Sigma, Help!

interesting, although VERY long read.

http://members.aol.com/SciRealm/ForcedInduc.html

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The Beginning Of All Things To End.
The End Of All Things To Come.