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 Post subject: Polynomial
PostPosted: Thu, 5 Aug 2010 08:22:52 UTC 
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A polynomial f is given. All we know about it is that all its coefficients are non-negative integers, f(1) = 6 and F(7)=3438. What is the value of f(3)?


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 Post subject: Re: Polynomial
PostPosted: Thu, 5 Aug 2010 09:27:50 UTC 
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CrowsZero wrote:
A polynomial f is given. All we know about it is that all its coefficients are non-negative integers, f(1) = 6 and F(7)=3438. What is the value of f(3)?


Think you mean f(7)=3438, not F(7)=3438.

Just bound, and fit.

Spoiler:
From f(7)=3438 you know f is at most a quartic. Also, since f(1)=6<7, you know you only need to covert f(7) to base 7 and read off the answer. Now 3438_{10}=13011_{7}, so f(x)=x^4+3x^3+x+1, hence f(3)=166.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Tue, 10 Aug 2010 02:14:05 UTC 
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I'm sorry I don't understand the reply given. Could you be more explicit? Thanks.


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 Post subject:
PostPosted: Tue, 10 Aug 2010 03:01:08 UTC 
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f(x) = a+bx+cx^2+dx^3+ex^4+... where all coefficients are non-negative integers.

The smallest possible quintic would be f(x)=1x^5, but that makes f(7) = 16807, which is too high, so f(x) must be of order 4 or lower.

f(1)=6, so a+b+c+d+e=6, so no coefficient exceeds 6.

f(7)=3438, so a+7b+7^2c+7^3d+7^4e=3438. Because all coefficients are in {0,1,2,3,4,5,6}, solving this is equivalent to finding 3438 in base 7.


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PostPosted: Tue, 10 Aug 2010 03:13:53 UTC 
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Thanks, Aswoods. Rafick


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