Quote: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...
algebraic identity
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).
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
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
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
Quote: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.
Quote: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.
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.
This topic is closed to new replies.
Advertisement
Popular Topics
Advertisement
