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Apr
10
comment What particular solution should I guess for this relation?
If, hypothetically, there was yet another shared term between my guess and the $f(n)$, I would have to multiply the guess by $n$ again, right? In that case the guess would be $(an^3+bn^2)2^n$?
Apr
10
revised What particular solution should I guess for this relation?
deleted 1 characters in body
Apr
10
comment What particular solution should I guess for this relation?
@user88595: Oh no, actually, I saw your comment and said "yeah I should invert the sign" so I went and inverted it without realising you already had inverted so it returned to its original form lol.
Apr
10
revised What particular solution should I guess for this relation?
added 3 characters in body
Apr
10
comment What particular solution should I guess for this relation?
How did you rearrange $f(n) = 2f(n-1) + n2^n$ to $f(n)-f(n-1) = n2^n$? I ask because of $2f(n-1)$, how did it become just $f(n-1)$?
Apr
10
comment What particular solution should I guess for this relation?
@ajotatxe: This stuff confuses me. If we chose $b_0$ to replace $n$ there, why aren't we replacing the exponential $n$ too?
Apr
10
asked What particular solution should I guess for this relation?
Apr
10
comment Does the order I multiply the characteristic equation's factors in the homogeneous solution matter?
@achillehui: But suppose that $b_0 = 1$ and $b_1 = 2$ and $n = 1$. Clearly $$1\cdot2+2\cdot 5 \neq 2\cdot2+ 1 \cdot 5$$ I'm asking about product distribution, not addition.
Apr
10
asked Does the order I multiply the characteristic equation's factors in the homogeneous solution matter?
Apr
10
comment About the particular solution given an homogeneous solution in a recurrence relation.
Oh, very interesting. This explains $d_0$ on the third one too I guess. But then, what about $d_1n$?
Apr
10
asked About the particular solution given an homogeneous solution in a recurrence relation.
Apr
10
comment How to solve non-homogeneous recurrence relations?
I'd have $$f(n) = f(n-1) + 2f(n-2) + n + 2^n + K + an + b + cn2^n$$ and still wouldn't have the fainted idea what's going on.
Apr
10
comment How to solve non-homogeneous recurrence relations?
Into what equation? The whole $f(n) = f(n-1) + 2f(n-2) + n + 2^n + K$?
Apr
10
asked How to solve non-homogeneous recurrence relations?
Apr
9
accepted How to calculate the determinant when the diagonal is in terms of $k$?
Apr
9
comment How to calculate the determinant when the diagonal is in terms of $k$?
$-1/2$ makes them linearly dependent, I think? But I guess there could be other values too.
Apr
9
asked How to calculate the determinant when the diagonal is in terms of $k$?
Apr
9
accepted How to determine column dependency without calculating the determinant?
Apr
9
comment How to determine column dependency without calculating the determinant?
Yeah, it makes sense. Although the question says "without any additional calculations" I suppose this is the right way (I don't think there's a way to determine this without doing any calculations).
Apr
9
comment How to determine column dependency without calculating the determinant?
You mean a combination of the form $aC_1 + bC_2 = C_3$?