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Loser66

  • one year ago

Conjecture the formula for the nth term of {a_n} if the first ten terms of it is 1,0,0,1,0,0,0,0,1,0 Please, help

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  1. Loser66
    • one year ago
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    I want to use the floor function to put it in neat. Obviously, |dw:1440875326682:dw|

  2. Loser66
    • one year ago
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    But how?

  3. freckles
    • one year ago
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    what do you mean how if you say it is obvious?

  4. freckles
    • one year ago
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    is the piecewise part obvious but not how to write in the floor format? is that what you?

  5. freckles
    • one year ago
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    is that what you mean*?

  6. ikram002p
    • one year ago
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    oh that remind me of Legendre function

  7. Loser66
    • one year ago
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    When I say "obvious" , it means we can see it obviously.

  8. freckles
    • one year ago
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    ok then what do you need help on then

  9. Loser66
    • one year ago
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    I want the floor format.

  10. freckles
    • one year ago
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    oh okay

  11. Loser66
    • one year ago
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    \(\lfloor{\sqrt n}\rfloor\)

  12. Loser66
    • one year ago
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    How about this? \(1<\sqrt 2 < 2\), hence \(\lfloor{\sqrt 2}\rfloor =1\) and \(\lfloor{\dfrac{\lfloor{\sqrt 2}\rfloor}{\sqrt2}}\rfloor=0\)

  13. Loser66
    • one year ago
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    It works well for n =3,4, 5, 6... but how can I just list them out like this? I want a general logic to lead me there.

  14. ikram002p
    • one year ago
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    so sqrt(n^2) is integer, then [sqrt(n^2) ] would be same integer n else u would have an additional decimals there are two lovely things about floor function 1-[sqrt(n^2) ]<= sqrt(n^2) 2-[r]=0 if 0<r<1 and you already know [sqrt(n^2) ]<= sqrt(n^2) \(\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}<=1\) we have two cases case 1 \(\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}=1\) which mean n is an integer and n^2 is a square of integer also \([\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}] =1\) case 2 \(\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}<1\) from property 2 \([\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}] =0 \) which u already mean that n is not a square of integer

  15. ikram002p
    • one year ago
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    and you can use n =i^2 instead n^2 as the general term lol

  16. Loser66
    • one year ago
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    Hey, friend , I don't get why \(\sqrt n^2 \). Is it not that it is always = n and n >0 since it is the order of the term.

  17. Loser66
    • one year ago
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    If we consider \(\lfloor{\sqrt n^2}\rfloor = \lfloor{n}\rfloor\) and it is = n itself.

  18. Loser66
    • one year ago
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    since \(n\in \mathbb N\)

  19. ikram002p
    • one year ago
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    yes u are correct , its just a variable :D sorry to confuse u ;)

  20. ikram002p
    • one year ago
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    i just used it to explain but since u were asked to make general form of n then use n without n^2

  21. Loser66
    • one year ago
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    oh, I got you. Thank you so much.

  22. ikram002p
    • one year ago
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    you are the most wlc <3

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