[Guest]

if N = the set of real numbers

find X

X^N = X^(N-1) + X^(N-2}

I'm pretty sure this works for any value of N and only one value for X

Please give reason for the answer and the method used(this will be very important to me because I got it by observation)

[Guest]

I have an answer to the problem, but not yet worked out for all powers, although it does work for N=1 and N=2

Putting N=2 gives X^2 = X + 1

Solving for X gives X = (1+ sqrt 5)/2

Interestingly, this is the limit of the ratio of consecutive terms in the Fibonacci sequence, the golden ratio.

[Guest]

By iteration,

if X^2 = X^1 +X^0, and we multiply both sides by X, then we get

X^3 = X^2 + X^1, and so it is true for any N X^N = X^(N-1) + X^ (N-2)

:)

[Guest]

If you divide through by X^(N-2) you get

X^2 = X + 1

Which is a simple quadratic with roots of (1+SQRT(5))/2 and (1-SQRT(5))/2 and doesn't depend on N

Edited June 6, 2010 by dannchen

[Guest]

Whoops, I missed a minus sign in my first reply. Dannchen is right

[Guest]

Try this one.

[Guest]

X^(N) = X^ (N-1) + X^(N-2)

N Log X = (N-1) Log X + (N-2) Log X

N = N - 1 + N - 2

N = 2N - 3

N = 3

[Guest]

For positive N,

X^N = X^(N-1)+X^(N-2)

X^(N+2) = X^(N+1) + X^N

X^N(X^2+X+1)=0

Either X = 0 or X = (1+-Sqrt(5))/2, by the quadratic equation.

For negative N,

X = (1+-Sqrt(5))/2, only.

[random7]

if N = the set of real numbers

find X

X^N = X^(N-1) + X^(N-2}

I'm pretty sure this works for any value of N and only one value for X

Please give reason for the answer and the method used(this will be very important to me because I got it by observation)

X^N = X^N-1 + X^N-2

X^N - X^N-1 - X^N-2 = 0

X^N-2 (X² - X - 1) = 0

Now if X² - X - 1 = 0 then:

X = (1 +- sqrt(5))/2 for all N an element of the reals.

If X^N-2=0 then:

X = 0 for all N an element of the reals except N = 0, 1, 2.

This proves that there are three potential answers for X, with X = 0 only if N != 0, 1, 2.

EDIT: After posting this at the same time as marsupialsoup, I realised from their response that N cannot be less than or equal to 2 for the X = 0 solution. That is, if X = 0, N > 2.

Edited June 7, 2010 by random7

[Guest]

Just substitute X^N-2 for t and you'll get:

t³=t²+t

One answer is 0 but that only works for N>2

Other answer is like everyone else answered:

t²-t-1=0

(1±sqrt(5))/2

Which I believe is the Golden Ratio.

http://en.wikipedia.org/wiki/Golden_ratio

[Guest]

For positive N,

X^N = X^(N-1)+X^(N-2)

X^(N+2) = X^(N+1) + X^N

X^N(X^2+X+1)=0

Either X = 0 or X = (1+-Sqrt(5))/2, by the quadratic equation.

For negative N,

X = (1+-Sqrt(5))/2, only.

My mistake, I multiplied both sides of the equation by 0.

The two cases are N>2 and N<=2, for N being real, X should be positive. This is a problem with polynomials over the real field; sometimes there are no roots...

[Quantum.Mechanic]

X^(N) = X^ (N-1) + X^(N-2)

N Log X = (N-1) Log X + (N-2) Log X

N = N - 1 + N - 2

N = 2N - 3

N = 3

The log of a sum is not the sum of the logs.

The problem stated that this worked for all N, but only 1 X. Where's your X?

[Guest]

^ You mean it should have been in step 2 instead of:

N Log X = (N-1) Log X + (N-2) Log X

Should have been

N Log X = Log(X^ (N-1) + X^(N-2))

Just in case anyone like me didn't understand...

[Guest]

Here's a thought:

I strikes me that there is a simple solution based on the concept that there is one real number that stays the same whether you add it to itself or multiply itself by itself... You could project that through simple identity definitions, but I may be underthinking this.