Proof by induction, matrices | Mathematics | Page 1

exotic roostBOT Apr 9, 2023, 9:06 AM

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jessicajones7010

rotund hollow Apr 9, 2023, 9:54 AM

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This question is not simple to solve 💀 it's not powers of 3. I don't think I can do it right now

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I mean I have a formula i just can't write at the moment

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Maybe there's a simpler way. I doubt it

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Until then notice that your matrix is the sum of $I_2$ and a matrix with all 1's, and the latter one is simple enough to calculate the powers of it

exotic roostBOT Apr 9, 2023, 9:56 AM

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Pyrodynamic

rotund hollow Apr 9, 2023, 9:57 AM

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Where $I_2$ is the 2 by 2 identity matrix. Since any matrix commutes with the identity, we can use the binomial formula to compute $A^n$. However you need to be careful with it and you need to calculate a sum of binomial coefficients that is not trivial

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Refer to this if you can find it https://math.stackexchange.com/questions/1566506/sum-of-binomial-coefficients-multiplied-by-k2

Mathematics Stack Exchange

Sum of binomial coefficients multiplied by $k^2$

Show the method used to evaluate $\displaystyle\sum\limits_{k=1}^{12} {12\choose{k}}k^2$

The answer is $159744.$

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Hopefully I'll be back later to piece this together but I've given you the ends of the string

exotic roostBOT Apr 9, 2023, 9:58 AM

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Pyrodynamic

exotic roostBOT Apr 9, 2023, 1:11 PM

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jessicajones7010

fallow prawn Apr 9, 2023, 3:53 PM

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You can break the matrix A into a matrix with only 1's (which I will cal B) and the identity matrix

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So A^n = (B+I_2)^n

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You can expand this into

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Sigma i from 0 to n of B^i * (n choose i) ( the (I_2)^n-i doesn't matter as it is always I_2 )

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Now you just need to study the powers of the matrix B

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Which is not hard to prove using induction that B^n is a matrix with 2^(n-1) everywhere

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Oh

#Proof by induction, matrices