Original post by eslene
Hiya...i want to know if (3(n+1))! is similar to 3!(n+1)n!
i need help...
i need help...
To test whether they are equivalent , then put n=4 (say) and use a calculator. IF the answers are the same, it might be coincidence , so repeat with another value for n.
Original post by steve2005
To use a slightly simpler example that illustrates the same point:
The reason is that, with 3(n!), you get 3*n*(n-1)*…*2*1 - but with (3n)!, you get (3n)*(3n-1)…*(n+2)*(n+1)*n*(n-1)*…*2*1. You can see that we've got all these extra terms (3n)…(n+1) in (3n)! which didn't appear in 3(n!) - and it's easy to check that (3n)(3n-1)…(n+2)(n+1) can never equal 3 as long as n is an integer greater than 0. (Non-essential outline of proof: by induction, the sequence whose nth term is (3n)(3n-1)…(n+2)(n+1) is increasing, and when n=1 we have (3n)…(n+1) = 3*2 = 6 > 3.)
The reason is that, with 3(n!), you get 3*n*(n-1)*…*2*1 - but with (3n)!, you get (3n)*(3n-1)…*(n+2)*(n+1)*n*(n-1)*…*2*1. You can see that we've got all these extra terms (3n)…(n+1) in (3n)! which didn't appear in 3(n!) - and it's easy to check that (3n)(3n-1)…(n+2)(n+1) can never equal 3 as long as n is an integer greater than 0. (Non-essential outline of proof: by induction, the sequence whose nth term is (3n)(3n-1)…(n+2)(n+1) is increasing, and when n=1 we have (3n)…(n+1) = 3*2 = 6 > 3.)
Original post by steve2005
To test whether they are equivalent , then put n=4 (say) and use a calculator. IF the answers are the same, it might be coincidence , so repeat with another value for n.
This forum is called "maths" not "physics".
Original post by JerzyDudek
This forum is called "maths" not "physics".
What's the point in producing a lengthy algebraic disproof when all you have to do is use a counter-example?
(3(n+1))! = (3n+3)! = (3n+3)(3n+2)(3n+1)...(1)
3!(n+1)n! = 6n!(n+1) = n(6n+6)*(n-1)(6n+6)...(1)
If your notation is right, this is what I got. They are not the same.
3!(n+1)n! = 6n!(n+1) = n(6n+6)*(n-1)(6n+6)...(1)
If your notation is right, this is what I got. They are not the same.
Original post by rawsex
What's the point in producing a lengthy algebraic disproof when all you have to do is use a counter-example?
Because this proof can be generalised and used for other cases while counter-example only exterminates one solution out of infinity.
Original post by JerzyDudek
(3(n+1))! = (3n+3)! = (3n+3)(3n+2)(3n+1)...(1)
3!(n+1)n! = 6n!(n+1) = n(6n+6)*(n-1)(6n+6)...(1)
If your notation is right, this is what I got. They are not the same.
3!(n+1)n! = 6n!(n+1) = n(6n+6)*(n-1)(6n+6)...(1)
If your notation is right, this is what I got. They are not the same.
3!(n+1)n! = 6n!(n+1) = n(6n+6)*(n-1)(6n+6)
Where did the 6 come from?
Original post by steve2005
3!(n+1)n! = 6n!(n+1) = n(6n+6)*(n-1)(6n+6)
Where did the 6 come from?
Where did the 6 come from?
3! = 6
Original post by eslene
need help on newton iteration
Oh.
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