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anonymous

  • 5 years ago

I am learning how to find the limit of a sequence and this is the problem. (n!)/(n^n). So the first step is to expand the numerator, and in the video the numerator expands as follows: (n) (n-1) (n-2) .....(2)(1). I and confused about how the person carried out the expansion. how do they get (2) (1) as the last terms? and if I did the expansion, wouldn't it just continue to (n-1)(n-2)(n-3) (n-4)?

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  1. anonymous
    • 5 years ago
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    I am*

  2. amistre64
    • 5 years ago
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    hmm....n! means 2*3*4*.....*(n-2)*(n-1)*(n) right?

  3. amistre64
    • 5 years ago
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    if n=3, to keep things simple: we got: (3)(3-1)(3-2) if we go any further we end up wit a (3-3)=0 and that messes up the factorial right?

  4. anonymous
    • 5 years ago
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    yes. but in this case we dont know what n is. so I thought it would just go onto infinity. how did they get (2)(1) at the end?

  5. amistre64
    • 5 years ago
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    n doesnt mean infinity it just means "a number that we can choose" n=3 is perfectly legit. the logic is that when you begin with a number, lets call it "n" that the factorial of the number will be: (n)(n-1)(n-2)(n-3) .... all the way down to.....(n-n+3)(n-n+2)(n-n+1) n-n+3 = 3 right? n-n+2 = 2 right? you see the logic?

  6. amistre64
    • 5 years ago
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    regardless of the value of "n"; when it gets close to 0 as it moves along down the number line

  7. anonymous
    • 5 years ago
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    kk let me read your reply. one minute..

  8. anonymous
    • 5 years ago
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    okay I think I am seeing the logic. please continue, when n gets close to 0....

  9. amistre64
    • 5 years ago
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    that was a typo...that when wasnt supposed to be there :) we can start at "1" and move all the way to "n" and it will be the same value right? so lets turn this factorial around and start with 1...............n (1)(2)(3)(4) ... (n-3)(n-2)(n-1)(n) do you see how gettting close to "n" is just (n-3) and those other notations?

  10. amistre64
    • 5 years ago
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    think of it like this: you really want to get to the fridge to get a soda.... but someplace between here and the fridge, you stop. look behind you and you will see how far you have come: (1step)(2steps)(3steps). now look ahead of you towards the fridge, how far to you have to go?: (fridge-3steps)(fridge-2steps)(fridge-1steps)(fridge,your there)

  11. anonymous
    • 5 years ago
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    oh wow!!!!! great analogy! I am trying to apply that to the n!. so I get to the fridge (n) and want to retrace my steps. it would be (n-1) (n-2) (n-3) (n-4)........

  12. amistre64
    • 5 years ago
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    yes..... exactly

  13. amistre64
    • 5 years ago
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    your between the fridge and the sofa after getting your soda, the fridge behind you is: (n)(n-1)(n-2)(n-3)(n-4) ...you are here now..(3)(2)(1, safely back at the sofa)

  14. amistre64
    • 5 years ago
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    your just counting down the steps from your first trip.....

  15. anonymous
    • 5 years ago
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    and is the midpoint I am standing at considered 0 on the number line? because the values become postive?

  16. amistre64
    • 5 years ago
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    not really, we are measureing a distance, and distance is always a positive value; but if we were to use the number line it would not matter..... but in these exansions you really want to start at an origin of 0 and measure a distance of "n" steps.

  17. amistre64
    • 5 years ago
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    les say we want to go 10 steps..... 1,2,3,4,5 were are here, 10-4,10-3,10-2,10-1,10. 10-4 = 6 10-3=7 10-2=8 10-1=9 10=10

  18. anonymous
    • 5 years ago
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    Thank you so much for the thorough explanations with great examples. I think I am getting a grasp of this concept now. I really appreciate your help! :) :) :)

  19. amistre64
    • 5 years ago
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    youre welcome...... math jargon can be rather complicated. these mathmatickers just aint got nothing better to do i think :)

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