Interesting stuff..

[thinktank2]

Playing with factorials I found the following

x^2 - x = x! (or) x(x-1) = x!

has unique Solution: x = 2

x^3 - 3x^2 + 2x = x! (or) x(x-1)(x-2) = x!

has unique Solution: x = 3

Similarly,

x^4 - 6x^3 + 11x^2 - 6x = x! has the unique solution x=4
and

x^5 - 10x^4 + 35x^3 - 50x^2 + 24x = x !

has the unique solution x=5

You can notice that solution for all these equation is the highest power of x in the equation. I think this is unique

because other than the equation x = 1 ...

I haven't heard of a polynomial equation that has a unique solution with the solution equal to the highest power of x in the equation.

The coefficents of these equation seems to follow a pattern.
I would like to know how to generate the coefficents for the nth degree equation.

The only cubic equation that can be equal to x! is

x^3 - 3x^2 + 2x

So... even the polynomial eqs are unique for a factorial.

[NotLKH]

Originally posted by thinktank2
Playing with factorials I found the following

x^2 - x = x! (or) x(x-1) = x!

has unique Solution: x = 2
.................

Hmm, 3^2 - 3 = 6 = 3!

[NotLKH]

The above can be solved like this:
x(x-1)=x(x-1)(x-2)
x(x-1)(1-x+2)=0
yielding
x = 0
x = 1
x = 3

[NotLKH]

So, Certainly
x(x-1)(x-2)...(x-n) always equals (n+1)!, when x = n+1
{test::if x = 4: 4(4-1)(4-2)(4-3) = (3+1)!, yepp!}

AND, if My math is correct:

x(x-1)(x-2)....(x-n) = x(x-1)(x-2)....(x-n)(x - n - 1)
or:
x(x-1)(x-2)....(x-n) (1-x+n+1)=0
meaning
x(x-1)(x-2)....(x-n) = x(x-1)(x-2)....(x-n)(x - n - 1) always has the solution:

x = n+2
or
x(x-1)(x-2)...(x-n) = (n+2)!

lets test.
if x = 7, then n = 5
first, place n into the equation
x(x-1)(x-2)(x-3)(x-4)(x-5) = x(x-1)(x-2)(x-3)(x-4)(x-5)(x-5-1)
next place x into it:

7(7-1)(7-2)(7-3)(7-4)(7-5) = 7*6*5*4*3*2*1
or
7*6*5*4*3*2 = 7*6*5*4*3*2*1
yepp.

also
6*5*4*3*2*1 = 6*

So, for the equation:
x(x-1)(x-2)(x-3)(x-4)(x-5) = x!
there are 2 answers:
x=6, k = 6
and
x = 7, k = 7

Now, this does not prevent other solutions to exist.

The real form of the equation is:
x(x-1)(x-2)(x-3)...(x-n) = x(x-1)(x-2)(x-3)...(x-n)(x-n-1)(x-n-2)(x-n-3)....(x-n-b).
where x-n-b = 1.

And I've shown that x can have 2 soulutions, x = (n+1) {ie... b=0}
or x = (n+2) {ie...b=1}

What would be most impressive is if you could prove that there are no other values for b that produces Integer solutions for x, where n > 0, x > n, x > b, b >= 0; x,n,b are integral values.

[thinktank2]

Sorry, that was a blunder. I Was checking like n! and (n-1)! but failed to check for (n+1)!

Now , I can say there are only two solutions for an equation(Let's call it the TT equation ) of any degree.

Originally posted by NotLKH
The above can be solved like this:
x(x-1)=x(x-1)(x-2)
x(x-1)(1-x+2)=0
yielding
x = 0
x = 1
x = 3

This is not the correct way because we cannot replace n! by something like n(n-1) or n(n-1)(n-2)

n! represent the exact value.

x(x-1) = x(x-1)(x-2) ---- Eq1 is a different equation altogether.
solving u get 0,1,3

But only 3 satifies the equation
x(x-1) = x! ---- Eq 2

and an Important value 2 is not found by equation 1

[thinktank2]

Well...it's boring to develop a proof for it. But you can see that

x(x-1)(x-2)(x-3)....(x-(n-1)) = x!

has the solutions

x = n and
x = n+1

and

For x=n,

you can write 2 equations

x(x-1)(x-2).........(x-(n-1)) = x!

and

x(x-1)(x-2)....(x-(n-2)) = x!

[NotLKH]

Originally posted by NotLKH
....Now, this does not prevent other solutions to exist.

The real form of the equation is:
x(x-1)(x-2)(x-3)...(x-n) = x(x-1)(x-2)(x-3)...(x-n)(x-n-1)(x-n-2)(x-n-3)....(x-n-b).
where x-n-b = 1.
... and where n > 0, x > n, x > b, b >= 0; x,n,b are integral values.

And I've shown that x can have 2 soulutions, x = (n+1) {ie... b=0}
or x = (n+2) {ie...b=1}

....

Actually, if the equation is altered to:
x(x-1)(x-2)(x-3)...(x-n) = abs(x)(abs(x)-1)(abs(x)-2)(abs(x)-3)...(abs(x)-n)(abs(x)-n-1)(abs(x)-n-2)(abs(x)-n-3)....(abs(x)-n-b),
where abs(x)-n-b = 1, and x <> 0, then we see that your original eq:

x(x-1) = a factorial, then it really has 4 solutions,
x = 2, x = 3, x = -1, and x = -2:

2*(2-1) = 2!
3*(3-1) = 3!
(-1)*(-1-1) = 2!
(-2)*(-2-1) = 3!

or:
x(x-1) = (abs(x-1))!, x = {3, 2, -1, -2}

[thinktank2]

Originally posted by NotLKH

x(x-1) = a factorial, then it really has 4 solutions,
x = 2, x = 3, x = -1, and x = -2:

2*(2-1) = 2!
3*(3-1) = 3!
(-1)*(-1-1) = 2!
(-2)*(-2-1) = 3!

No , You have misunderstood the equation

For example The TT second degree equation is not

x(x-1) = some factorial

it is

x(x-1) = x! (get the difference ??... though they seem to be the same)

x(x-1) = (abs(x-1))!, x = {3, 2, -1, -2}

May be we should call that NotLKH's Second degree equation

but you see...it's not that elegant

[thinktank2]

Sad thing is.. mathematica cannot directly solve ThinkTank equations

Something like

Solve[x(x-1) == x!, x ]

doesn't work, { It says it can't solve transcendental expressions }

I plotted the curves for

Abs(x(x-1)-x!) and they cut the x-axis at two points.

But again the graph is kind of spoiled since, x is stepped up in non integer amount. I am trying to figure out how to step up x in integers...

[NotLKH]

Thinktank2,
Perhaps, instead of feeding it x!, you could feed it the expansion:
{from your link: http://mathworld.wolfram.com/Factorial.html}



Mathematica might be able to use it that way.
BTW, It works on Non-Integer Values, so if you ever wondered what 2.1! was, it'll give it to you.

[DerFarm]

Look at it this way, TT, have you ever been considered
transcendental before?

[thinktank2]

OMG ...edit edit edit...

Solve[x(x-1) == x!, x ]

doesn't work, { It says it can't solve transcendental expressions }

I plotted the curves for ....