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#1 2006-05-31 01:52:13

John E. Franklin
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Registered: 2005-08-29
Posts: 3,582

graphs slid around with subtraction

I want to know WHY this works.
Take most any 2D graph with y and x in the equation, and notice that you can slide the graph up, down, left, and right by subtracting from x to move right, and subtracting from y to move up.
Adding to y moves the whole graph down, and adding to x moves the whole graph left.
For example.
y = x^3 + x^1.8
This graph has a certain shape.
To move the whole graph down 5.5 units and right 33 units, then we get this:
y + 5.5 = (x - 33)^3 + (x - 33)^1.8
I don't know if I was taught this or if I just saw it through the years and now it seems to be true.
But I'd like to know WHY it works.

Here is another example:
Start with    y = 5
Move the whole graph left by 13 which essentially does nothing since it is still a horizontal line.
But where is the x?  y = 0(x + 13) + 5
y = 5 still.

Here is another one:
y = sin(x), where sine is in degrees.
Move whole graph left by 90 degrees:  y = sin(x + 90), which by the way is probably y = cos(x), but that's off the subject.

Anyone know why this simple substitution for y and x appears to move the graph, or does it move the axis the other way?


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#2 2006-05-31 13:46:20

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: graphs slid around with subtraction

The easiest way to see why this happens is to make a table.  In general we have:

f(x) be some function
g(x) = f(x - a)

x  |  x - a
-----------
0  |  -a
1  |  1 - a
2  |  2 - a
3  |  3 - a

So if we are looking at f(x) and g(x):

f(0) and g(a) = f(a - a) = f(0)
f(1) and g(a + 1) = f(a + 1 - a) = f(1)
f(2) and g(a + 2) = f(a + 2 - a) = f(2)

and so on.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-06-01 08:01:01

John E. Franklin
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Registered: 2005-08-29
Posts: 3,582

Re: graphs slid around with subtraction

Hey thanks Ricky, I like to see how other people can explain things.
Does this show why it goes in the opposite direction as the sign?
Is this example horizontal or vertical movement?


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#4 2006-06-01 08:45:48

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: graphs slid around with subtraction

This is just horizontal movements.

And it does explain why the sign is reversed.  If we plug in x-a like I did for g:

g(x) = f(x - a)

If x = 0, then g(0) = f(-a).  But to put it another way, g(x + a) = f(x - a + a), and so if x = 0, then g(a) = f(0).  And like I showed before, the same is true for 1, 2, 3... etc.  So even though your pugging in a negative a, it shifts it over a positions.

For verticle, it's basically the same concept:

g(x) = f(x) + b


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-06-01 11:17:07

John E. Franklin
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Registered: 2005-08-29
Posts: 3,582

Re: graphs slid around with subtraction

Yes but notice that if the relationship is a circle, you go up by altering the y with subtraction to move it up before the y is squared.    So g(x)=f(x) + b may not work for circles.   I originally said you get the -a right in with the y to go up by a, so if the y is not solved for and is in a mess, the vertical constant is right in tight with the y.
For example  25 = (y-6)^2 + x^2  would move the circle up vertically by 6 I think.  I might be wrong, but that's how I think it is.


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#6 2006-06-01 12:21:51

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: graphs slid around with subtraction

But a circle isn't a graph!  It isn't well defined.  In other words, it doesn't pass the verticle line test.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#7 2006-06-07 02:59:58

George,Y
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Registered: 2006-03-12
Posts: 1,306

Re: graphs slid around with subtraction

that doesn't matter, the shifting rule is not limited to functions.

for post 4, Ricky, I recommand you  using x[sub]1[/sub] and x[sub]2[/sub] instead of both xs, to make it clearer.

post 2 is very clear.

I think the shifting rule also applies to x²+y²≤4 and (x-.5)²+(y+1.5)²≤4 as well.


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#8 2006-06-07 04:13:00

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: graphs slid around with subtraction

Ricky wrote:

But a circle isn't a graph!

x^2 + y^2 = r^2 can be rewritten y = +/- sqrt{r^2-x^2}

which if you plot with the +/- will give you a circle

(and if you plot complex numbers aswell, you get a verticaly standing circle, and then two horizontal square root patterns)

take the x axis, as x
the y axis as the real part of the circle function
and the z axis as the imaginary part of the circle function

obviously the circle has all z = 0
and the outper parts have all y = 0

View Image: circle.JPG

The Beginning Of All Things To End.
The End Of All Things To Come.

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#9 2006-06-07 08:32:54

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: graphs slid around with subtraction

x^2 + y^2 = r^2 can be rewritten y = +/- sqrt{r^2-x^2}

which if you plot with the +/- will give you a circle

If you have two functions:

y1 = sqrt(r^2 - x^2)
y2 = sqrt(r^2 - x^2)

And you try to apply the shifting rule to both those functions, it will work.  But it only works when you can put a function in the form of:

y = f(x)

Which is not always possible.

Edit: And I made a minor change to my post #2 which I think makes it a bit clearer.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#10 2006-06-07 14:50:32

George,Y
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Registered: 2006-03-12
Posts: 1,306

Re: graphs slid around with subtraction

shifting rule can also be applied to
f(x,y)=0

Because:
Break the curve into very small pieces. Locally, you will solve out {y=g[sub]i[/sub](x)}, get one to one functions, and then you just need to connect them together.
Now the altogether f(x,y)=0 is ready for shifting rule.

Accually, that's how the theorem about implicit differential works.

Last edited by George,Y (2006-06-07 14:53:21)


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#11 2006-06-07 14:56:13

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: graphs slid around with subtraction

basicly the reason the shifting occurs is it puts it a few steps ahead, say for instance f(x) = 2x. To get a 4 you'd have to inset an x value of 2. But if f(x) = 2x + 1 then x can be one less and produce the same value. So the whole graph is shifted horizontally one unit. It basicly gives it a headstart.


A logarithm is just a misspelled algorithm.

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#12 2006-06-07 15:20:33

George,Y
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Registered: 2006-03-12
Posts: 1,306

Re: graphs slid around with subtraction

Another proof about g(x,y)=0 qualified for shifting rule

first about f(x)=y
f(x[sub]0[/sub])=y[sub]0[/sub]
then f(?-a)=y[sub]0[/sub]
The question here may seem to be solving an equation out. But not so.
First, x[sub]0[/sub]+a is qualified,
for f(x[sub]0[/sub]+a-a)=y[sub]0[/sub]
Second, every x satisfying f(x)=y[sub]0[/sub] can be grouped into {x[sub]i[/sub]}
then for each i, or each x[sub]i[/sub]
f(x[sub]i[/sub]+a-a)=y[sub]0[/sub]
here we proved x[sub]i[/sub]+a are qualified solutions for f(x-a)=y[sub]0[/sub], but are they all solutions? Is there a solution out of {x[sub]i[/sub]+a}? If there is one, the shifted function may have one more point at the altitude y=y[sub]0[/sub]

IF y[sub]0[/sub]=f(x) and y[sub]0[/sub]=f(x-a) has the same number of solutions, {x[sub]i[/sub]+a} is the complete solution set for f(x-a)=y[sub]0[/sub].

The assumption is what I cannot prove. But if anyone prove it, or has proved it, the proof is complete. Or to put it inanother way, now that f(x)=y can be applied by shifting rule, the assumption should be true.

The proof for g(x,y)=0 is the same. the same projection. the difference is minor.
for any x[sub]i[/sub] satisfying g(x,y[sub]0[/sub])=0
x[sub]i[/sub]+a satisfy g(x-a,y[sub]0[/sub])=0.

IF g(x,y[sub]0[/sub])=0 and g(x-a,y[sub]0[/sub])=0 have the same amount of solutions, the shifting rule applies.

I think the shifting rule can also be applied to inequality like g(x,y)≤0 because equalities and inequalities have tight connections. However, strict proof need rigorous knowledge on inequality theorems. So I assume it.

Last edited by George,Y (2006-06-07 15:21:52)


X'(y-Xβ)=0

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#13 2006-06-07 15:38:47

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: graphs slid around with subtraction

Proof of the assumption
--I discovered no knowledge about solving a polynomial solution is needed

the proof before assumpition IMPLIES that f(x-a)=y[sub]0[/sub] has no less solutions than f(x)=y[sub]0[/sub], for there is a solution for f(x)=y[sub]0[/sub], there exists a corresponding one for f(x-a)=y[sub]0[/sub].

This is a corollary from the x[sub]i[/sub] and x[sub]i[/sub]+a theorem.

Note this corollary applies to any a and any function.
Name f(x-a)=g(x)
t=x-a
f(t)=g(t+a)
t=x
f(x)=g(x+a)=g(x-(-a))
Using the corollary,
g(x-(-a))=y[sub]0[/sub] has no less solutions than g(x)=y[sub]0[/sub]
that's analogous to say f(x)=y[sub]0[/sub] has no less solutions than f(x-a)=y[sub]0[/sub]

Together, f(x)=y[sub]0[/sub] and f(x-a)=y[sub]0[/sub] has the same amount of solutions, assumption PROVEN!
smile smile smile
Should I go to  a math journal. Or was I doing some already-done work?


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