algebraic identity

Do you mean how to proove these identities are true?
I believe you proove it with induction.
Or do you ask how did they discover it is true?
They probably did some examples and then "saw" the right equation with intuition.
They might have used talyor series, or all kind of methods.

Of course not.
I am asking how to construct a formula to find the sum of a sequence.
While M.I. is used for proving whether it is true or not.

A sum is usually very close to an integral. For instance, you know that the integral of n² is (1/3)n³, which means the partial sum will be equivalent, for large values of n, to this. So you know you will have to look for a polynomial of degree 3, the greatest coefficient of which is (1/3).

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Original post by TimChan I am asking how to construct a formula to find the sum of a sequence.

Apply axioms in any order.

To find a simple formula you have to apply them in the right order which is very hard find algorithmically.

The story I was told was that Gauss was assigned a punishment to calculate 1+2+3+...+n for some large n. He sat for a few minutes, then noticed that if he grouped the numbers like so,

(n + 1) + (n-1 + 2) + (n-2 + 3) + ...

eg.

1 + 2 + 3 + 4 = (1+4) + (2+3)

and noting that since each term in the above is equal to (n + 1), and that since each term is a grouping of two of the original n terms, there are a total of n/2 of them, thus the whole sequence is really n/2*(n+1). Or:

1+2+3+...+n = n*(n+1)/2

Of course if n is odd, the regrouped sequence has an extra term for the ((n+1)/2)th entry... eg.

1 + 2 + 3 + 4 + 5 = (1+5) + (2+4) + 3

which throws a wrench into things... though I guess the punishment Gauss got was an even number...

Or he noticed that there are (n-1)/2 terms of sum (n+1) and one term of (n+1)/2. So,

(n-1)*(n+1)/2 + (n+1)/2 = (n - 1 + 1)*(n+1)/2 = n*(n+1)/2

so it works out either way.

[Edited by - Geoff the Medio on December 4, 2005 6:09:50 AM]

Actually, his reasoning was:

1 + 2 + 3 + 4 = (1/2) * ((1+4) + (2+3) + (3+2) + (4+1)) = (1/2) * (4+1) * 4

Such things is discovered by thinking, I'd say. Since nobody knows how thinking really works (when it is something more or less complicated), there is no answer.
Automatic theorem provers do that similarly to how computer play chess: by exploring the tree of possibilities. Tree branching is defined by rules of chess or axioms of math and logic. For example at some position i can move pawn on b2 or pawn on c2 , similarly, in math when figuring out sum i can try to regroup terms one way or other way.
It is believed that brain is doing something like that. (so the speed and maximal depth of search depends to many factors such as genetics of said individual, nutrition, training, brain injuries if any, and so on. Training is probably most important)

but how about the other identities? how did mathematicans know that 1^2+2^2+3^2+...n^2 is equivalent to [n(n+1)(2n+1)]/6?

i'm reading a book which tells that those algebraic identities are derived from the combination formula, C(r,n) = C(r,n-1)+C(r-1,n-r). For example, put r be 2, we get [n(n+1)]/2 - [n(n-1)]/2 = n. But i don't know how it works.

Well, for most proofs you can know how it is done. You can even have explanations how one might arrive at certain identity (most often explanations has nothing to do with how it was done for first time)
But the discovery of something new is "creative" process and it is not quite known how brain does that (we don't know HOW we think. If we would, we'd make smart AI). Most likely brain does check really many possibilities "in background", somehow. Like chess, with enough training brain becomes capable of checking really many different ways of game.
(Heck, we don't even know how when you look at cat you recognize that it is cat. [smile])

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Original post by TimChan but how about the other identities? how did mathematicans know that 1^2+2^2+3^2+...n^2 is equivalent to [n(n+1)(2n+1)]/6?

I already answered that one...

In any case at some point mathematician who figured this out had to think of something new (or guess the formula, or make some general method).

Thanks for giving me a deeper knowledge of algebraic identity.

take a proofs class.i've actually proved that very identity..hell..one time i had to prove we have inifinitely many natural numbers...and a bunch of other crap i'd rather not prove....i think the first thing i do when it goes to proving does this look like it's true? after about twenty minutes i've come up with a decision and then i try to prove it.if i ever get stuck i then try to disprove it. if that doesn't work i try to prove it by disproving the opposite. and....rant end:-D

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Original post by Anonymous Poster In general proof to identities and theorems are discovered as a consequence of trying to find the solution of a larger problem. For example those generics expression for progressions of integer number are common on the design of digital filters. No one spent time say “you know I am going to find out what is the general expression for say 1 + 2 * 3 + 3/3 + 4 * 5 + 4 * 31 … = whatever” if it serve not purpose. Necessity is the mother of inventions

AP, please use your login when posting to the forums.

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Original post by TimChan how could the mathematicans construct algebraic identities from scratch? Some examples of algebraic identities are 1+2+3+...+n = [n(n+1)]/2 and, more complex, 1^2+2^2+3^2+...n^2 = [n(n+1)(2n+1)]/6.

There is a book called "how to solve it" by G. Polya. Polya was very interested in how mathematicians come up with their ideas and wrote several books on it. In "how to solve it" he tries to illustrates some of the ideas used to solve problems, like the ones you mention.

There are many methods to compute algebraic expressions, and I don't think there is a good way to describe how one discovers these things "in general". What you can do is learn many "tricks" that people have used before, and then you will be able to come up with your own "tricks". If a trick can be used twice, we call it a *method*.

Here's one way to figure out 1^2+2^2+3^2+...+n^2. Let's call it S_2 for short. Similarly, S_1 := 1+2+3+...+n = n(n+1)/2

Let's compute the difference between two consecutive cubes:
k^3 - (k-1)^3 = k^3 - (k^3 - 3k^2 + 3k - 1) = 3k^2 - 3k + 1

Now we can write all this expression for values of k between 1 and n:
n^3 - (n-1)^3 = 3n^2 - 3n + 1
(n-1)^3 - (n-2)^3 = 3(n-1)^2 - 3(n-1) + 1
(n-2)^3 - (n-3)^3 = 3(n-2)^2 - 3(n-2) + 1
...
3^3 - 2^3 = 3*3^2 - 3*3 + 1
2^3 - 1^3 = 3*2^2 - 3*2 + 1
1^3 - 0^3 = 3*1^2 - 3*1 + 1

Let's add all those identities together:
n^3 = 3*S_2 - 3*S_1 + 1

And now it's easy to make computations to figure out what S_2 is.

Alternatively you can use the "Hockey Stick Theorem" (a.k.a. "Christmas Stocking Theorem"), which is an identity with combinatorial numbers that will give you the answer right away. You can also use the Finite Difference method, and all sorts of other methods to compute S_2. There is a book by Knuth and others called "Discrete Mathematics", which explains many ways to solve this problem, and very powerful methods to compute many sums involving integers.