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# Little math thing i'm workin on

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posted on Sep, 30 2004 @ 09:48 PM
i'm not a math wizard or anything, and have only completed one year of calculus entering into college, but I've come up with this little problem which has required much thinking on my part. I'm not sure if anything like it has been done before (i'm sure it has) but I can't find anything on it. Anyways, here's my problem.

y = y -------This statement is always true.
y = 2y ------- This statement is only true when x = 0.
y = 2y + x -------- same as y = -x
y = 3y + x -------- same as y = -(1/2)x
y = 4y + x -------- same as y = -(1/3)x
and so on. It seems that the equivalent problem (if y = ay + x) equals y = (1/(a-1))x unless it is the first one.

Now here's where my brain pretty much exploded.

y = y^2 + x

When x = 1, y = ~-0.61803
and I can't figure out what comes next, my brain was just too jumbled and i haven't worked on it since... so I haven't figured out what to compare the above formula to.

Anyways, anyone have input on this?

posted on Sep, 30 2004 @ 09:51 PM
umm ill guess y=0 lol i dunno....this is hard math

posted on Sep, 30 2004 @ 10:48 PM
I'm not quite sure what you are trying to do, but it seems to me, that you want an equation for Y in terms of X.

So you have,
y=y^2+x
0=y^2-y+x

Now, recall the quadratic formula, which says for
a*y^2+b*y+c=0, y=(-b +- sqrt(b^2-4ac))/2a.
Note: sqrt stands for square rooot

Let a=1, b=-1, and c=x
So,
a*y^2+b*y+c = y^2-y+x = 0
So, we substitute our values of a, b, and c into the qudratic formula and get:
y=(1+sqrt(1-4*x))/2 or y=(1-sqrt(1-4*x))/2
There are two possible values of y for any given value of x.

Notice Y is only real, when x

posted on Sep, 30 2004 @ 10:57 PM
yeah, as far as i can tell, you want to find general eqns for y=y^(n+1) + x? is that right? its easy enough... i guess. just try things out. the sqrt will mess things up.

posted on Sep, 30 2004 @ 10:57 PM
.
are you just trying to do the algebra?

The first set of equations is linear (degree one)
This is a parabolic (second degree ) equation
y = y^2 + x
y - y^2 = x
y ( 1 - y ) = x

are you treating this as though x were a function of y?
f(y)?
if y = 0 ; x = 0

I believe this is a sideways parabola with apogee/vertex (term?) at (x,y) = (1/4,1/2)
I don't believe you can have any values of x greater than 1/4,
The tails of the parabola shoot off from there to the negative x side of the plane.

how you got the x = 1 value, I am not clear on.

. . . . y
____ .
. . . . \
x . . . | (0,0) . . x . . ; (this sure is one crude graphic)
____/.
. . . . y
.

posted on Sep, 30 2004 @ 11:35 PM

y = y -------This statement is always true.
y = 2y ------- This statement is only true when x = 0.
y = 2y + x -------- same as y = -x
y = 3y + x -------- same as y = -(1/2)x
y = 4y + x -------- same as y = -(1/3)x
and so on. It seems that the equivalent problem (if y = ay + x) equals y = (1/(a-1))x unless it is the first one.
Now here's where my brain pretty much exploded.
y = y^2 + x
When x = 1, y = ~-0.61803

y - ay = x
y*(1-a) = x
y = x / (1-a)
y = (1/(1-a))*x

Or take out a negative:
y = -(1/(-1--a))*x = -(1/(-1+a))*x = -(1/(a-1))*x

If x =1, then y = -(1/(a-1))*1, so what's the value for a?

If y = y^2 + x, then a = 1.

y^2 = ( -(1/(a-1))*x )^2

At this point you have to take into account that squaring something will cause it to always become positive, so you need to take the +/- of the values, unless you're only interested in the postive ones.

y = (ay)^2 + x OR y = a*(y)^2 + x ?

y = (ay)^2 + x = a^2*y^2 + x

Does this help?

posted on Oct, 1 2004 @ 10:03 PM
As far as I can tell, you're trying to find a general formula y = ay + x, and are having difficulties when a = y. What you have to understand, is that when a = y, a is varying with y, so you get a quadratic equation.
x = y-ay
x = y(1 - a)

You'll find this holds for all a including y.
Only other thing that I can think of is that you're trying to solve the differential equation y' = ay + x, with y as a fucntion of x... but I doubt it.

posted on Oct, 1 2004 @ 10:12 PM
I believe .61803 is the golden ratio or PHI prounounced FEE , maybe it's a trick question and nothing comes next....?

posted on Oct, 2 2004 @ 02:21 AM
[edit on 2-10-2004 by nschmoyer]

posted on Oct, 2 2004 @ 05:53 PM
Well I am no math major so I could be just rambling, but here's what I'm thinking.

For the solution Y = Y^2 + X, the simple solution would be to have one side equal zero, making it Y^2 - Y + X = 0 and applying the quadratic formula -- makes perfect sense. BUT, when I dust off my ole calculator and start plugging numbers in, I find that (.61803)^2 + 1 = .61803 (or rather very close, as the decimal continues and It gets irritating finding decimal values)

EDIT The .61803 should be replaced with a -.61803, my mistake.... that just made the entire equation not make sense.. sorry..

[edit on 2-10-2004 by nschmoyer]

posted on Oct, 2 2004 @ 09:08 PM
And my entire theory has been debunked again by myself..

OKAy you guys are correct.

When using -.61803, one side of the equation is positive and one is negative... ALTHOUGH the absolute values of the numbers mirror each other.. bleh

posted on Oct, 4 2004 @ 01:33 AM
Whats this thingy? "^" its a cool sign

posted on Oct, 5 2004 @ 08:00 PM
It is an exponent sign.

5^2 means five to the second power, or 5 X 5 = 25.

2^4 is two to the fourth powr or 2 X 2 X 2 X 2 = 16.

[edit on 5-10-2004 by Off_The_Street]

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