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zysi [14]
3 years ago
10

Find a closed form for the generating function for each of these sequences. (assume a general form for the terms of the sequence

, using the most obvious choice of such a sequence.)
a.−1, −1, −1, −1, −1, −1, −1, 0, 0, 0, 0, 0, 0, ...
Mathematics
1 answer:
noname [10]3 years ago
4 0
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 n-th term in the sequence:

\{a_n\}=\{-1,\ldots,-1,0,\ldots,0,1,\ldots,1,2,\ldots,2,3,\ldots\}
\implies a_n=\left\lfloor\dfrac n7\right\rfloor-1

for n\ge0.

So the generating function for this sequence might be

G(a_n;x)=\displaystyle\sum_{n\ge0}\left(\left\lfloor\frac n7\right\rfloor-1\right)x^n

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?
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Rainbow [258]

Answer:

The solution to this equation could not be determined.

Step-by-step explanation:

Simplifying

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The solution to this equation could not be determined.

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2 years ago
HURRY PLEASE!!!
ozzi
1:04pm pretty sure i think that’s right
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Here, the differing digits are 1 and 0 in the thousandths place. The appropriate sign is ">"

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Answer:

D

Step-by-step explanation:

that what I think but not sure

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