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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

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?

**igloo** **myrtilles** **fourmis**

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

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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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

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?

**igloo** **myrtilles** **fourmis**

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

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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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

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.

**igloo** **myrtilles** **fourmis**

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

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

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

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.

**X'(y-Xβ)=0**

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

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

The Beginning Of All Things To End.

The End Of All Things To Come.

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

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.

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

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)*

**X'(y-Xβ)=0**

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

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

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

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!

Should I go to a math journal. Or was I doing some already-done work?

**X'(y-Xβ)=0**

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