Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
1 answer:
You might be interested in
There really is no single "obvious" choice here...
Possibly the sequence is periodic, with seven copies of -1 followed by six copies of 0, or perhaps seven -1s and seven 0s. Or maybe seven -1s, followed by six 0s, then five 1s, and so on, but after a certain point it would seem we have to have negative copies of a number, which is meaningless.
Or maybe it's not periodic, and every seventh value in the sequence is incremented by 1? Who knows?
I'll go ahead and assume the latter case, that the sequence is not periodic, since that's technically somewhat easier to manage. We can assign the following rule to the
-th term in the sequence:
for
.
So the generating function for this sequence might be
As to what is meant by "closed form", I'm not sure. Would this answer be acceptable? Or do you need to find a possibly more tractable form for the coefficient not in terms of the floor function?
Possibly the sequence is periodic, with seven copies of -1 followed by six copies of 0, or perhaps seven -1s and seven 0s. Or maybe seven -1s, followed by six 0s, then five 1s, and so on, but after a certain point it would seem we have to have negative copies of a number, which is meaningless.
Or maybe it's not periodic, and every seventh value in the sequence is incremented by 1? Who knows?
I'll go ahead and assume the latter case, that the sequence is not periodic, since that's technically somewhat easier to manage. We can assign the following rule to the
for
So the generating function for this sequence might be
As to what is meant by "closed form", I'm not sure. Would this answer be acceptable? Or do you need to find a possibly more tractable form for the coefficient not in terms of the floor function?