Proving the Arithmetic Series Property for x and y When x=! -1, y=! -1, x=! -y

[Nikitin]

Homework Statement

When x=! -1, y=! -1, x=! -y then x and y are two numbers so that 1/(x+1) + 1/(x+y) + 1/(y+1)... is an arithmetic serie. Show that then also x² + 1 + y²... must be an arithmetic serie.

The Attempt at a Solution

I tried to find the differentials between each number in the first line, then make them equal each other but I really didn't get anywhere. I figured out though that if x=y=1 then both of the series are arithmetic.

Anyways can I get some help pls?

 

[Dick]

a+b+c+... an arithmetic series if b-a=c-b, right? Apply that to both your series and show they lead to the same condition on x and y.

 

[Nikitin]

Yes, I wrote in the OP that I tried that but I didn't get any smarter.. can you maybe please solve this problem and show me how you did it?

BTW: What do you mean by condition? You mean that x is the same number in both the series, and y is the same number in both the series? I got y=0.216 and Y=1, and x=0.37 and x=1, if I remember correct. That was wrong.

 

[Dick]

Yes, I wrote in the OP that I tried that but I didn't get any smarter.. can you maybe please solve this problem and show me how you did it?

BTW: What do you mean by condition? You mean that x is the same number in both the series, and y is the same number in both the series? I got y=0.216 and Y=1, and x=0.37 and x=1, if I remember correct. That was wrong.

Look, if x^2+1+y^2+... is arithmetic then 1-x^2=y^2-1. Yes? That leads to the 'condition' that 2=x^2+y^2. There are lots of solutions to that. Like x=0 and y=sqrt(2). All you have to do is do some algebra to show that the other series gives you the same equation in x and y.

 

[Nikitin]

Well sure x=y=1 is a solution to both but that's not showing anything is it? I already did that difference between each joint stuff a while ago, and I got

2=y^2 + x^2 in the second series and
2= (1/y) + (1/x) in the first.

that doesn't really tell me anything.. And are you suggesting that the difference in the first serie and the second are the same? Whut?

 

[Dick]

Well sure x=y=1 is a solution to both but that's not showing anything is it? I already did that difference between each joint stuff a while ago, and I got

2=y^2 + x^2 in the second series and
2= (1/y) + (1/x) in the first.

that doesn't really tell me anything.. And are you suggesting that the difference in the first serie and the second are the same? Whut?

No, the differences aren't the same between the two series. What is true is that b-a=c-b for each series. I think you did the algebra wrong for the second series. Can you show how you got 2= (1/y) + (1/x)?

 

[Nikitin]

b-a= (1/y) - 1, c-b= 1 - 1/x. c-b=b-a. (1/y) - 1 = 1 - 1/x => 2 = 1/y + 1/x.

But I don't get it, what am I supposed to do with this? can somebody help? How do I prove that as a result of the first serie, the second one must also be arithmetic??

 

[Dick]

b-a= (1/y) - 1, c-b= 1 - 1/x. c-b=b-a. (1/y) - 1 = 1 - 1/x => 2 = 1/y + 1/x.

But I don't get it, what am I supposed to do with this? can somebody help? How do I prove that as a result of the first serie, the second one must also be arithmetic??

You can't simplify 1/(x+y)-1/(x+1) to 1/y-1 by magically canceling the x. That's wrong algebra. You need to find a common denominator if you are going to combine them.

 

[Nikitin]

ah damn, you're right, I've been doing lots of mistakes of this type lately (sleep deprivation I guess).

OK that was the root of the entire problem, I mixed up the fractions-math. Sorry. thanks for the help, I get it now.