Another Factorial Equation

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Yeah, move the 1 to the RHS, factor n^2-1 as (n+1)(n-1) and divide both sides by (n-1). This leaves (n-2)! = (n+1). The LHS obviously grows faster than the RHS, so you can just test low values of n until either you hit the solution or the LHS is larger. Which is n=5.

rickdesper
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The gamma function is an extension of the factorial function. (With a shift.) If you use factorial notation, you're referring to the factorial function, not the gamma function.

rickdesper
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( n -1 ) ! = n^2 -1 = ( n -1 )( n + 1 )
( n -2 )! = n + 1 ( n ≠ 1 )
n -2 = t , n = t +2
t ! = t + 3
t !- t = 3
t ( ( t -1 ) ! -1 ) = 3 = 3・1
t ≠ 1 , t = 3
n = t +2 = 3+2 = 5

ひろ-nzk
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Trying out in excel, I got the answer 5 too. But in the end n=sqrt(2) was valid as well (by narrowing down...). Why???

mystychief
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There is something wrong with the graph. It does not give the solution n = 5.

lhdill
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This was fairly straightforward. A little algebraic manipulation leads to: n = (n-2)! -1. Now LHS is a constant function, whereas RHS is an increasing function, hence, they will intersect at only 1 point which would be the solution. also: n>2 for practical purposes (unless we invoke the gamma function). this leads to n=5.

InnocentNeuron
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Ah, nice! The same reason you can’t just go ahead and cancel - divide through by (n-1) - if (n-1) is zero, is the same reason 0! Does not exist!

BlaqRaq
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You really insist on doing everything algebraically. It's OK to enumerate possible solutions to find the solution. When you get to k! = k+3, it's pretty obvious that k=3.

rickdesper
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This is confusing! Why n=1 is NOT a solution? We do simple adding substarcring and groupung to get (n-1)(something else)=0 from the original equation. Therefore we must certainly have all the roots from (n-1)(something else)=0 to be roots of the original equation. But n=1 is NOT a root of the original.equation. Why? May be rhe answer is that (n-1)! = (n-1)(n-2)! does NOT hold for n=1.

dimitardimitrakov
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I hate factorials. They're so lame. Perchance.

omerzaferdundar