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Subject: "Why do people look for more complic..."     Previous Topic | Next Topic
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Conferences The CTK Exchange Middle school Topic #43
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Liliya (Guest)
Jul-19-01, 11:01 PM (EST)
"Why do people look for more complicated ways of solving the same problem??"
   There are lots of ways of solving the same problems. Then why do Mathematicians still try to figure out some other ways?
Why do some people spend most of their time to do something that's already done?

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Charter Member
782 posts
Jul-19-01, 11:16 PM (EST)
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1. "RE: Why do people look for more complicated ways of solving the same problem??"
In response to message #0
   >There are lots of ways of
>solving the same problems.

>why do Mathematicians still try
>to figure out some other

If they did not, there would not be lots of ways of solving the same problems in the first place, right?

>Why do some people spend most
>of their time to do
>something that's already done?

I do not know about its being most of their time, but certainly many people spend a lot of time proving already established theorems.

Why? Because people seek enlightenment, not just truth. Often the first proof is too obscurred to shed light as to why something is true.

Some proofs are better than others - simpler, clearer, more transparent, more general, etc. Mathematicians seldom prove something just in order to prove, exercise their brain or build a reputation. What they prove is often a part of a more general scheme. Multiple proofs help them see a bigger picture.

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Michael Klipper
Jul-29-02, 08:31 AM (EST)
2. "Why complicated ways?"
In response to message #0
   I know what's annoying you, but there are often some good reasons for this.

The best reason I can think of is representation. A mathematical problem can be viewed through many different constructs, and often different constructs provide light to shed on problems which, although they didn't seem similar at first, end up being analogously solved.

I'll provide an example. What is the formula to 1 2 3... n, where n is a positive integer? It is n(n 1)/2 = (n^2) / 2 n / 2, commonly known as Little Gauss's Formula.
An easy way to prove this is to take the sum and write it twice; once forwards and once backwards:
1 2 3 ... n
n n-1 n-2 ... 1
(n 1) (n 1) (n 1) ...
This sum forms (n 1) n different times. Thus, when we doubled the sum, we got n(n 1), from which we easily get the formula.

However, using another method of proof, we get a way to generalize this process to ANY power, not just sums of numbers raised to the first power. This method will work to compute the sum of 1^2 2^2 3^2 ... n^2 as well as 1 2 3... n

We use summation properties (and the binomial theorem) to show this:
(i 1)^2 - i^2 = 2i 1
where my brackets stand for the summation operator.

However, since this sum is telescoping (i.e. (2^2 - 1^2) (3^2 - 2^2) will cancel the 2^2, which is very useful), we can write this too:
(i 1)^2 - i^2 = (n 1)^2 - 1.

Equating these forms, we get
(2i 1) = (n 1)^2 - 1 = n^2 2n
2( i) n = n^2 2n
2( i) = n^2 n
from which the answer follows.

Now, why go to all this trouble? Let me show that the same method works for the sum of squares! Note that (i 1)^3 expands to i^3 3i^2 3i 1, which you can check if you don't believe me.

(i 1)^3 - i^3 = (3i^2 3i 1)
but it also equals
(n 1)^3 - 1 = n^3 3n^2 3n
So, reusing Little Gauss's formula:
(3i^2 3i 1) = n^3 3n^2 3n
(3i^2 3i) = n^3 3n^2 2n
3( i^2) = n^3 3n^2 2n - (3n^2 / 2 3n / 2)
= n^3 3n^2 / 2 n / 2
i^2 = n^3 / 3 n^2 / 2 n / 6

Do you see how a change in approach and a representation change presents so much power? This method can be repeated for any positive integer power, not just 1 and 2. I hope this shows you some reasons why mathematicians revisit "oldies but goodies."

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