I Need Help Finding Inverse Functions.

[GlassKnees]

I am trying to solve a problem in an old textbook. I am given a function of two variables that map ordered pairs of positive integers to the set of positive integers:

[math]F(x,y) = (x+y-1)(x+y-2)/2+y[/math]
I am asked to find inverse functions K and L with the property K(F(x,y)) = x and L(F(x,y)) = y.

To help myself, I created a table with x's across the top and y's down the side and filled in values for F(x,y) and here's what I'm seeing:

K(1) = 1, K(2) = 1, K(3) = 2, K(4) = 1, K(5) = 2, K(6) = 3, K(7) = 1, K(8) = 2, K(9) = 3, K10) = 4, K(11) = 1, K(12) = 2, K(13) = 3

And L(1) = 1, L(2) = 2, L(3) = 1, L(4) = 3, L(5) = 2, L(6) = 1, L(7) = 4, L(8) = 3, L(9) = 2, L(10) = 1, L(11) = 5, L(12) = 4, L(13) = 3

See attached file. Can anyone help me?

 

[topsquark]

I am trying to solve a problem in an old textbook. I am given a function of two variables that map ordered pairs of positive integers to the set of positive integers:

[math]F(x,y) = (x+y-1)(x+y-2)/2+y[/math]
I am asked to find inverse functions K and L with the property K(F(x,y)) = x and L(F(x,y)) = y.

To help myself, I created a table with x's across the top and y's down the side and filled in values for F(x,y) and here's what I'm seeing:

K(1) = 1, K(2) = 1, K(3) = 2, K(4) = 1, K(5) = 2, K(6) = 3, K(7) = 1, K(8) = 2, K(9) = 3, K10) = 4, K(11) = 1, K(12) = 2, K(13) = 3

And L(1) = 1, L(2) = 2, L(3) = 1, L(4) = 3, L(5) = 2, L(6) = 1, L(7) = 4, L(8) = 3, L(9) = 2, L(10) = 1, L(11) = 5, L(12) = 4, L(13) = 3

See attached file. Can anyone help me?

First, is your function
F(x,y) = (x+y-1)(x+y-2)/2+y --> [imath]F(x,y) = \dfrac{(x+y-1)(x+y-2)}{2}+y[/imath]

or
F(x,y) = (x+y-1)(x+y-2)/(2+y) --> [imath]F(x,y) = \dfrac{(x+y-1)(x+y-2)}{2+y}[/imath]

If the second, parenthesis are needed!

The general idea would be to, say, set z = F(x, y) and to find K(F(x, y)) = x set x = F(z, y) and solve for z, where y is taken to be constant.

So, for example, if F(x, y) is of the second form:
[imath]x = \dfrac{(z + y - 1)(z + y - 2)}{2 + y}[/imath]

[imath]x(2 + y) = z^2 + (2y - 3)z + y^2 - 3y + 2[/imath]
which is a quadratic in z.

Give it a try.

-Dan

 

[blamocur]

I believe your table is incorrect in at least one place : [imath]F(1,3) \neq 4[/imath]

I'd use the property [imath]\forall x\geq 2: F(x+1,y) - F(x,y) > y[/imath]

 

[Dr.Peterson]

I am trying to solve a problem in an old textbook. I am given a function of two variables that map ordered pairs of positive integers to the set of positive integers:

[math]F(x,y) = (x+y-1)(x+y-2)/2+y[/math]
I am asked to find inverse functions K and L with the property K(F(x,y)) = x and L(F(x,y)) = y.

To help myself, I created a table with x's across the top and y's down the side and filled in values for F(x,y) and here's what I'm seeing:

K(1) = 1, K(2) = 1, K(3) = 2, K(4) = 1, K(5) = 2, K(6) = 3, K(7) = 1, K(8) = 2, K(9) = 3, K10) = 4, K(11) = 1, K(12) = 2, K(13) = 3

And L(1) = 1, L(2) = 2, L(3) = 1, L(4) = 3, L(5) = 2, L(6) = 1, L(7) = 4, L(8) = 3, L(9) = 2, L(10) = 1, L(11) = 5, L(12) = 4, L(13) = 3

See attached file. Can anyone help me?

You mislabeled x and y in your table. Row numbers are x, column numbers are y. The function is as you wrote it, [math]F(x,y) = \dfrac{(x+y-1)(x+y-2)}{2}+y[/math]
It may be possible to solve directly as a quadratic, but my experience with this sort of function is that it works nicely to first work out which diagonal (1; 2,3; 4,5,6; 7,8,9,10; ...) a number belongs to, which can be done by solving a simpler quadratic equation and taking the greatest integer, and then construct the answer from that.

 

[GlassKnees]

First, is your function
F(x,y) = (x+y-1)(x+y-2)/2+y --> [imath]F(x,y) = \dfrac{(x+y-1)(x+y-2)}{2}+y[/imath]

or
F(x,y) = (x+y-1)(x+y-2)/(2+y) --> [imath]F(x,y) = \dfrac{(x+y-1)(x+y-2)}{2+y}[/imath]

If the second, parenthesis are needed!

The general idea would be to, say, set z = F(x, y) and to find K(F(x, y)) = x set x = F(z, y) and solve for z, where y is taken to be constant.

So, for example, if F(x, y) is of the second form:
[imath]x = \dfrac{(z + y - 1)(z + y - 2)}{2 + y}[/imath]

[imath]x(2 + y) = z^2 + (2y - 3)z + y^2 - 3y + 2[/imath]
which is a quadratic in z.

Give it a try.

-Dan

Dan, thank you for your response. I will give it a try. I'm not experienced with typing in math equations in this forum. At any rate, I will give it a try tomorrow. Thanks again for your help!

 

[GlassKnees]

I believe your table is incorrect in at least one place : [imath]F(1,3) \neq 4[/imath]

I'd use the property [imath]\forall x\geq 2: F(x+1,y) - F(x,y) > y[/imath]

Actually, it is correct. F(1,3) = (1+3-1)(1+3 -2) divided by two and add 1. If you examine the table I made as shown in the attachment, there is a nice pattern, but I Was unable to come up with the inverse functions given this insight. But I've gotten some responses and will try their suggestions.

 

[GlassKnees]

You mislabeled x and y in your table. Row numbers are x, column numbers are y. The function is as you wrote it, [math]F(x,y) = \dfrac{(x+y-1)(x+y-2)}{2}+y[/math]
It may be possible to solve directly as a quadratic, but my experience with this sort of function is that it works nicely to first work out which diagonal (1; 2,3; 4,5,6; 7,8,9,10; ...) a number belongs to, which can be done by solving a simpler quadratic equation and taking the greatest integer, and then construct the answer from that.

 

[GlassKnees]

Dr. Peterson, thank you for your response. I'm not quite sure I followed your suggestion, but I will look at it more carefully tomorrow. If I'm still confused, I may ask for your help again (warning!)

Thanks again!

 

[blamocur]

Actually, it is correct. F(1,3) = (1+3-1)(1+3 -2) divided by two and add 1. If you examine the table I made as shown in the attachment, there is a nice pattern, but I Was unable to come up with the inverse functions given this insight. But I've gotten some responses and will try their suggestions.

What is the value of [imath]y[/imath] when you compute [imath]F(1,3)[/imath]?

 

[GlassKnees]

What is the value of [imath]y[/imath] when you compute [imath]F(1,3)[/imath]?

The function is F(x,y) = ((x+y-1)(y+y-2))/2 + y

For F(1,3), y is 3; F(1,3) = 4

 

[Dr.Peterson]

The function is F(x,y) = ((x+y-1)(y+y-2))/2 + y

For F(1,3), y is 3; F(1,3) = 4

No, you mean F(x,y) = ((x+y-1)(x+y-2))/2 + y.

And F(1,3) = ((1+3-1)(1+3-2))/2 + 3 = ((3)(2))/2 + 3 = 3 + 3 = 6.

That disagrees with your table, as labeled:

​

The red circle is where 6 is located, which according to the table would be F(3, 1). The green circle is what the table says F(1,3) would be.

The table should have been labeled like this:

​

Do you understand this?

Now you need to find the inverse ...

Here are the diagonals I referred to in my hint:

​

Given a value of F, my suggestion is to find the value at the lower left of the diagonal it's on, and then use that to find x and y. You can start by finding a formula for just that first column (with y = 1) and inverting that.

But I'm sure there are other ways.