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matemusic
Member since Jul-30-02
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Aug-01-02, 07:52 AM (EST) |
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2. "RE: induction and function"
In response to message #1
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>Try simple cases. What do you get plugging in m = n =
>0? What do you get if only one of them is 0?
Of course it'easy to prove that f(0)=0 and f(n²)=(f(n))², even it'rather easy to calculate f(1),f(2), then f(4),f(8) , then f(5) (plug m=2,n=1), then f(3) (plug m=3,n=4) ,etc...
In fact i have calculate f(m) for m=0 to 20 and i stop to 20 because I wanted a general proof of the "obvious" fact that f(m) = m by induction, but it does'nt seem to be quite so easy
Thanks a lot if you could help me
sincerely
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sanjay
Member since Aug-18-02
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Aug-18-02, 12:07 PM (EST) |
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3. "RE: induction and function"
In response to message #0
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Induction generally can only be used on integer
arguments because the proof depends on a process
that hits every number. Keeping this in mind, here
is a general guideline that should assist you in
solving the induction problem above:Firstly, make sure you know where you are
starting from. It is usually a good idea to
include both 0 and 1 in the base cases. (They
can be calculated by hand and then assumed in the
rest of the induction.) f(0) = 0
f(1) = 1 In induction, we assume that we have already proved
that f(m) = m for all values of m <= k, and then
we prove that f(k+1) = k+1. This should help you
get started on your inductive proof. Sanjay
Sanjay |
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Michael Klipper
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Aug-19-02, 06:42 AM (EST) |
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4. "RE: induction and function"
In response to message #3
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One thing that might be useful to know is that the summation property of this function does not have to be restricted to two terms. For example, f(a^2 + b^2 + c^2) = (f(a))^2 + (f(b))^2 + (f(c))^2, which is easily proven through iteration of the sum property. The reason why this property is useful is that, although not every number is a sum of two perfect squares, every number must be a sum of some finite number of perfect squares (at worst, n is the sum of n ones). |
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