Simplify a summation and its product

[kleyton]

I have been working on representing the powers of numbers as a summation.

This is as far as I have gotten.

Power: 2
[itex]m^2 = \sum_{n=1}^m \left(2n -1\right)[/itex]

Power: 3
[itex]m^3 = \sum_{n=1}^m \left(3n^2 -3n +1\right)[/itex]

Power: 4
[itex]m^4 = \sum_{n=2}^m \left[6*(4n-6) * \left(\sum_{a=1}^{m-n+1} a\right)\right] + m^2[/itex]

I wanted to know if it is possible to simplify the equation for the 4th and 3rd power of a number so that the highest power in the equation in 1.

Thanks

 

[chiro]

Hey kleyton and welcome to the forums.

What kind of form are you looking to do? In your post you say you want to get your expressions down to linear functions of n like you did with your m^2 so I'll assume that this is your goal for all powers of m.

One way that might help you is to view things in a geometric way. We know that m^2 is basically a square full of individual dots and that you can you visualize the sum by seeing how each term of your summation is actually two sides that expand.

I'll do an example with finding m^2 where m = 3.

So I've got the following representation of 9 dots (they are x's in this post):

x x x
x x x
x x x

Here is each application of the sum

x ...|...x|...x
...|..x..x|...x
...|...|..x..x..x

[EDIT: I put dots to separate them because the formatting got screwed up]

The above shows the delta of each sum represented geometrically.

What I am trying to do is to get you to use that geometric idea for all of the other powers so that you can extend it to higher powers (even beyond 4).

There are probably other deltas that you can use, but I think this is a good one to start off with if you don't have anything else to go with.

 

[kleyton]

In your post you say you want to get your expressions down to linear functions of n like you did with your m^2 so I'll assume that this is your goal for all powers of m.

This is exactly what I was intending.

Your geometric method does seem interesting, but I do not quite seem to understand it.

The above is what you had tried to show in your post for 3^2. Could you explain what you meant by "delta". The image seem to show 1+3+5=9. That is what the summation does, but the arrangement is what I stuck at.

 

[chiro]

Hey kleyton.

What I meant by delta is the change with every summation term. So basically the first one has one 'x' corresponding to the term n=1, and then we have 3 'x's for n=2 and 5 'x's for n = 3 for m = 3.

Basically where I was going with this was to get to you to see how you can do this for say when you have cubic where you have a cube and so instead of having two lines you have three 'edges' as your n term increases. For a quartic (4th power) then you will have four 'cubes'.

The geometry is meant to help you think about it visually and then use this hint to go to higher powers in terms of algebra (i.e. the formula vs the geometric visual approach).

 

[Stephen Tashi]

kleyton,

Are you familiar with the mathematics called "The Calculus Of Finite Differences"?