2^n = 2*n!
- Thread starter hlreed
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HallsofIvy
Registered Senior Member
They WAY to solve 2^n= 2 n! is to plug in values of n and see what happens!
If n=0, then 2^n= 1 and 2(0!)= 2 so that's not true
If n= 1, then 2^n= 2 and 2(1!)= 2 Yes, a solution!
If n= 2, then 2^n= 4 and 2(2!)= 4 Yes, a solution!
If n= 3, then 2^n= 8 and 2(3!)= 12 so that's not true.
If n= 4, then 2^n= 16 and 2(4!)= 48 so that's not true.
In fact, it should be clear that 2(n!) will increase much faster than 2^n (because n! increases faster than 2^(n-1)) so the only solutions are n= 1 and n= 2.
If n=0, then 2^n= 1 and 2(0!)= 2 so that's not true
If n= 1, then 2^n= 2 and 2(1!)= 2 Yes, a solution!
If n= 2, then 2^n= 4 and 2(2!)= 4 Yes, a solution!
If n= 3, then 2^n= 8 and 2(3!)= 12 so that's not true.
If n= 4, then 2^n= 16 and 2(4!)= 48 so that's not true.
In fact, it should be clear that 2(n!) will increase much faster than 2^n (because n! increases faster than 2^(n-1)) so the only solutions are n= 1 and n= 2.
What I need is a symbolic way to show that n cannot be greater than 2.
x^n = x*n! is the more general term. Need n = f(x).
All this is in integers.
What this came from was z^n = x^n + y^n true when n < 3
dn-1(z^n) = z*n! likewise for x, and y so that
z = x + y since n! is common to all. This says z must be 2
dn(z^n) = n! likewise for x and y
But we are asking about z^n = 2*n!
So z = 2
so 2^n = 2*n! (only for n = 1 or n = 2.)
So is that Fermat's short proof?
x^n = x*n! is the more general term. Need n = f(x).
All this is in integers.
What this came from was z^n = x^n + y^n true when n < 3
dn-1(z^n) = z*n! likewise for x, and y so that
z = x + y since n! is common to all. This says z must be 2
dn(z^n) = n! likewise for x and y
But we are asking about z^n = 2*n!
So z = 2
so 2^n = 2*n! (only for n = 1 or n = 2.)
So is that Fermat's short proof?
HallsofIvy
Registered Senior Member
Would you mind explaining what in the world you are talking about? In the first place, what is "d" in "dn-1(z^n) = z*n! " and how does it follow?
Why would it be true that z= x+y? (In general if z^n= x^n+ y^n for ANY n other than 1 (even n=2) z can't equal x+y.)
And if it were true why would it then follow that z= 2? There is surely no reason to think that x and y are both 1.
Why would it be true that z= x+y? (In general if z^n= x^n+ y^n for ANY n other than 1 (even n=2) z can't equal x+y.)
And if it were true why would it then follow that z= 2? There is surely no reason to think that x and y are both 1.
Sorry,
dn(x) is the nth derivative of x.
If you differentiate a power, x^n n times you always get n!.
Do it n-1 times and you get x*n!
So dn-1(z^n = x^n + y^n) = z*n! = x*n! + y*n!
Divide out the n! and you get z = x + y
The least value x and y can have is 1. So z = 2 if that is so.
x^4
4*x^3 = d1 = d(x^4)
4*3*x^2 = d2 = d(d1)
4*3*2*x = d3 = d(d2) last x value is x*4!
4*3*2*1 = d4 = d(d3) now 4!
I thought this was well known.
The derivatives are my own notation, which I am trying to sell. They work not on coordinates but on next number and this number in a time stream. d(x) = x[now] - x[next] dt is 1 in this system so differentiation and integration are functions.
The general Fermat system is:
k^n = k*n! ; k is the number of sum variables. For z^n=x^n+y^n
k = 2
If n = 1 then k = k meaning every sum is ok.
If n = 2 then k^2 = k*2! ; or k = 2! or k = 2
If n = 3 then k^2 = 3! or k = sqrt(6)
There are no z^n = u^n+x^n+y^n either.
dn(x) is the nth derivative of x.
If you differentiate a power, x^n n times you always get n!.
Do it n-1 times and you get x*n!
So dn-1(z^n = x^n + y^n) = z*n! = x*n! + y*n!
Divide out the n! and you get z = x + y
The least value x and y can have is 1. So z = 2 if that is so.
x^4
4*x^3 = d1 = d(x^4)
4*3*x^2 = d2 = d(d1)
4*3*2*x = d3 = d(d2) last x value is x*4!
4*3*2*1 = d4 = d(d3) now 4!
I thought this was well known.
The derivatives are my own notation, which I am trying to sell. They work not on coordinates but on next number and this number in a time stream. d(x) = x[now] - x[next] dt is 1 in this system so differentiation and integration are functions.
The general Fermat system is:
k^n = k*n! ; k is the number of sum variables. For z^n=x^n+y^n
k = 2
If n = 1 then k = k meaning every sum is ok.
If n = 2 then k^2 = k*2! ; or k = 2! or k = 2
If n = 3 then k^2 = 3! or k = sqrt(6)
There are no z^n = u^n+x^n+y^n either.
HallsofIvy
Registered Senior Member
Yes, it is well known that the n-1 derivative of x^n i(with respect to x) is n! x.
And I have no problem with you inventing notation (especially on the internet where we can't use regular notation. It would be a good idea to explain it first however! On difficulty with your not
However, it is NOT true that if x^n+ y^n= z^n then
"differentiating n-1 times" gives x n!+ y n!= z n! because you can only differentiate with respect to one variable.
It is also true that since "Fourier's last theorem" requires that x, y, and z be integer valued, the derivative isn't going to tell us much about that.
Your "version" appears to be "finite differences" which goes back to Boole in the nineteenth century. In finite differences it is NOT true that "d(x^n)= nx^(n-1). In finite differences,
d(x^(n))= nx^((n-1)) (Where x^(n) is x*(x-1)*(x-2)...(x-n+1).)
And I have no problem with you inventing notation (especially on the internet where we can't use regular notation. It would be a good idea to explain it first however! On difficulty with your not
However, it is NOT true that if x^n+ y^n= z^n then
"differentiating n-1 times" gives x n!+ y n!= z n! because you can only differentiate with respect to one variable.
It is also true that since "Fourier's last theorem" requires that x, y, and z be integer valued, the derivative isn't going to tell us much about that.
Your "version" appears to be "finite differences" which goes back to Boole in the nineteenth century. In finite differences it is NOT true that "d(x^n)= nx^(n-1). In finite differences,
d(x^(n))= nx^((n-1)) (Where x^(n) is x*(x-1)*(x-2)...(x-n+1).)
Differentiation is the same either finite or symbolic. Ie. you get the same answers
Take the same sample x^4. Simply let x count so we have a stream.
0^4 = 0
1^4 = 1 d1,1 = 1
2^4 = 16 d1,2 = 15 d2,2 = 14
3^4 = 81 d1,3 = 65 d2,3 = 50 d3,3 = 36
4^4 = 256 d1,4 =175 d2,4 = 110 d3,4 = 60 d4,4 = 24
5^4 = 625 d1,5 = 369 d2,5 = 194 d3,5 = 84 d4,5 = 24
6^4 =1296d1,6 = 671 d2,6 = 302 d3,6 = 108 d4,6 = 24
of course 4! = 24
Much harder this way, but this can be mechanized.
Now i4(24) = x^4 with proper constants.
36 50 65 81
24 60 110 175 256
24 84 194 369 625
24 108 302 671 1296
This also is mechanizable
Take the same sample x^4. Simply let x count so we have a stream.
0^4 = 0
1^4 = 1 d1,1 = 1
2^4 = 16 d1,2 = 15 d2,2 = 14
3^4 = 81 d1,3 = 65 d2,3 = 50 d3,3 = 36
4^4 = 256 d1,4 =175 d2,4 = 110 d3,4 = 60 d4,4 = 24
5^4 = 625 d1,5 = 369 d2,5 = 194 d3,5 = 84 d4,5 = 24
6^4 =1296d1,6 = 671 d2,6 = 302 d3,6 = 108 d4,6 = 24
of course 4! = 24
Much harder this way, but this can be mechanized.
Now i4(24) = x^4 with proper constants.
36 50 65 81
24 60 110 175 256
24 84 194 369 625
24 108 302 671 1296
This also is mechanizable
HallsofIvy
Registered Senior Member
What's hard about that?
If k= 2 then the only solutions are n= 1 and n= 2.
If k>2 then the only solution is n= 1.
If k= 2 then the only solutions are n= 1 and n= 2.
If k>2 then the only solution is n= 1.
Why?
Oh well. From memory. Fermat was reading a book on integer algebra which had an equation z^n = z^n + y^n and Fermat made a note in the margins that he had a short proof that n could not be > 2 but it was too long for the margin. Which drove people crazy, of course.
Andrew Wiles made a general proof of this conjecture but it took a long trail that I, for one, cannot follow. I still think Fermat had a short proof.
While messing around with numbers some time ago, I discovered, numberically, that the nth deriviative of a n power number was n!
Later I found out everyone knew that already.
Let dn(x) be the nth derivative of x.
Then dn(x^n) = n!
Numerically, dn-1(x^n) = x*n! ;(...3*2*x)
So if you see x^n you can see n! and vise versa.
Also, given n! you can create x^n
The Fermat conjecture becomes a question of whether any power can be a sum of same powers.
Back to the equation: z^n = x^n + y^n
Differentiate both sides n-1 times, getting z*n! = x*n! + y*n!
or z = x + y
The minimum value of the integer z is 2.
Now we ask the question: can 2^n = 2*n! ;We must have a power as a result of the addition. The elements added must represent powers. 2*n! = 1^n + 1^n is an alternate question.
So 2^n = 2 and n = 1
2^n = 2*n! and n = 2
There are no other possibilities.
The real short proof is that 2 is the only number whose addition and multiplication are the same. 2^2 = 2 + 2
Ok?
Oh well. From memory. Fermat was reading a book on integer algebra which had an equation z^n = z^n + y^n and Fermat made a note in the margins that he had a short proof that n could not be > 2 but it was too long for the margin. Which drove people crazy, of course.
Andrew Wiles made a general proof of this conjecture but it took a long trail that I, for one, cannot follow. I still think Fermat had a short proof.
While messing around with numbers some time ago, I discovered, numberically, that the nth deriviative of a n power number was n!
Later I found out everyone knew that already.
Let dn(x) be the nth derivative of x.
Then dn(x^n) = n!
Numerically, dn-1(x^n) = x*n! ;(...3*2*x)
So if you see x^n you can see n! and vise versa.
Also, given n! you can create x^n
The Fermat conjecture becomes a question of whether any power can be a sum of same powers.
Back to the equation: z^n = x^n + y^n
Differentiate both sides n-1 times, getting z*n! = x*n! + y*n!
or z = x + y
The minimum value of the integer z is 2.
Now we ask the question: can 2^n = 2*n! ;We must have a power as a result of the addition. The elements added must represent powers. 2*n! = 1^n + 1^n is an alternate question.
So 2^n = 2 and n = 1
2^n = 2*n! and n = 2
There are no other possibilities.
The real short proof is that 2 is the only number whose addition and multiplication are the same. 2^2 = 2 + 2
Ok?
Back to the equation: z^n = x^n + y^n
Differentiate both sides n-1 times, getting z*n! = x*n! + y*n!
or z = x + y
Sigh.
Take away the coordinates x,y,z and instead create three data streams x, y and z. (Data streams are a system of curve, index, emitters, nodes, motors.)
1. S = curve
2. j = index
3. S[n] = emitter ; sensor
4. S[j] = node ; any function, point on the curve
5. S[0] = motor ; data stream sink
d(S) = S[j] - S[j-1] ;for any S
(DS^2 = dx^2 + dy^2 in x,y coordinates)
Sensors compute S1, S2, S3.
d(S1) reads S1, computes d(S1) ;(no x or y here)
d1 reads d(S1), compute d1(S1)
Duplicate for S2 and S3
Repeat this for n times.
dn-1(Sk) = Sk*n! and dn(Sk) = n!
Rename S1 to x
Rename S2 to y
Rename S3 to z
z = x + y ;let x=y = 1
z = 2
Sum is 2*n!
2^n = 2*n!
or
k^n = k*n!
Differentiate both sides n-1 times, getting z*n! = x*n! + y*n!
or z = x + y
Sigh.
Take away the coordinates x,y,z and instead create three data streams x, y and z. (Data streams are a system of curve, index, emitters, nodes, motors.)
1. S = curve
2. j = index
3. S[n] = emitter ; sensor
4. S[j] = node ; any function, point on the curve
5. S[0] = motor ; data stream sink
d(S) = S[j] - S[j-1] ;for any S
(DS^2 = dx^2 + dy^2 in x,y coordinates)
Sensors compute S1, S2, S3.
d(S1) reads S1, computes d(S1) ;(no x or y here)
d1 reads d(S1), compute d1(S1)
Duplicate for S2 and S3
Repeat this for n times.
dn-1(Sk) = Sk*n! and dn(Sk) = n!
Rename S1 to x
Rename S2 to y
Rename S3 to z
z = x + y ;let x=y = 1
z = 2
Sum is 2*n!
2^n = 2*n!
or
k^n = k*n!