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#1
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Mathematical Equality
Prove that
N=[N/2] +[N/4] + [N/8] +...+[N/2^n] +... where [a/b] is the nearest integer function . |
#2
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Re: Mathematical Equality
[ QUOTE ]
Prove that N=[N/2] +[N/4] + [N/8] +...+[N/2^n] +... where [a/b] is the nearest integer function . [/ QUOTE ] For what value is [1/2] defined? |
#3
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Re: Mathematical Equality
[1/2]=1 ; [2.5]=3 etc .
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#4
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Re: Mathematical Equality
The conjecture is false. Counterexample: Let N = 1.1.
Did you mean to define the domain of the r.h.s. of the equality to be N = any integer? |
#5
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Re: Mathematical Equality
N is an arbitrary natural number .
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#6
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Re: Mathematical Equality
It's actually true for natural numbers .
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#7
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Re: Mathematical Equality
If you write N in binary it becomes somewhat clearer how this works. Say
N = 2^n a_n + ... + 2 a_1 + a_0. Then you can see that for each k: [N/2^k] = 2^{n-k} a_n + ... + 2 a_{k+1} + a_k + a_{k-1} Now collect all the terms multiplied by the same a_i in the sum [N/2] +[N/4] + [N/8] +...+[N/2^n] +... to find that this is equal to: a_0 + a_1 + a_1 + a_2 + a_2 + 2 a_2 + ... + a_n +Sum_{k=0}^{n-1} 2^k a_k = N. |
#8
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Re: Mathematical Equality
can't you just factor out N so that this summation becomes a geometric series that converges to 1?
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#9
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Re: Mathematical Equality
[ QUOTE ]
can't you just factor out N so that this summation becomes a geometric series that converges to 1? [/ QUOTE ] This brings us back to the hardest part of the problem, which is mathematically defining the 'nearest integer' function. I think it would be hard to rigorously show that you can factor out the numerator for this function considering 1/N does not necessarily equal [N/1]^-1. I had to use fourier series to define the nearest integer function as follows: NearestInt = x + 1/pi * sigma(k=1 => infinity)[sin(2*pi*k*x + 2*pi*k) From that the expansion of this problem becomes N = sigma(i=1 | infinity){N/(2^i) + 1/pi * sigma(k=1 | infinity)[sin(pi*k*N/(2^(i-1)) + pi*k)} The sin series converges to zero both with respect to i and k, because the harmonics cancel themselves out as well as each other, so you're left with the real number series N = sigma(i=1 | infinity){N/2^(i)} which clearly converges to N. |
#10
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Re: Mathematical Equality
didn't even read the part about nearest int func, never mind.
could you explain a little more how you used the fourier series to define the func? |
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