Where is this converging to on $y=x$?

Question

Well I was playing with graphs and I started plotting equations as the following:

$$\underbrace{x+y}_{degree=1}=1 \tag{1}$$

$$\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{2}$$

$$\underbrace{x^3+y^3+x^2y+xy^2}_{degree=3}+\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{3}$$

$$\underbrace{x^4+x^3y+x^2y^2+xy^3+y^4}_{degree=4}+\underbrace{x^3+y^3+x^2y+xy^2}_{degree=3}+\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{4}$$

and so on ...

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And here are the plots : (Click on them to get a better pic)

Zoom:1

Zoom:2

Zoom:3

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It seems like these graphs are converging to some value on the (red dashed line) $y=x$.

What is this value?

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My Attempt:

Since the graphs are converging to $y=x$, hence we solve the two equations as :

(Let's take degree 2)

$$\Rightarrow x^2+y^2+xy+x+y=1 \space\space\space and \space\space\space y=x$$

$$\Rightarrow x^2+x^2+x^2+x+x=1$$

$$\Rightarrow 3x^2+2x=1$$

Hence if we take $degree=n$:

The equation becomes : $\sum\limits_{n=1}^{\infty} (n+1)x^n=1$

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1. Am I right?
2. How do you solve this?
3. Any more comments on this question?
4. How to represent equations $(1),(2),(3),(4),\cdots$ in a more general way?

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WolframAlpha showed that when solved (IDK how?) $x=1\pm \frac{1}{\sqrt{2}}$ and via plotting it looks like $x=1 - \frac{1}{\sqrt{2}}$.

How?

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Thanks!!

Answer 1

If you think of the both sides of that equation as functions of $x$, you're looking for a value $x$ with $$ f(x) = g(x) $$ where $f$ is a sum and $g$ is a constant function. The sum happens to also be $$ F'(x) $$ where \begin{align} F(x) &= \sum_{n=1}^\infty x^{n+1}\\ &= \sum_{n=2}^\infty x^{n}\\ &= -1-x + \sum_{n=0}^\infty x^{n}\\ &= -1-x + \frac{1}{1-x} \end{align} where this last step is from the geometric series formula. Hence $$ F'(x) = -1 + \frac{1}{(1-x)^2} $$ Setting this to $1$, we get \begin{align} 1 &= -1 + \frac{1}{(1-x)^2}\\ 2 &= \frac{1}{(1-x)^2}\\ 2(1-x)^2 &= 1 (1-x)^2 &= \frac{1}{2}\\ 1-x &= \pm\frac{1}{\sqrt{2}}\\ -x &= -1 + \pm\frac{1}{\sqrt{2}}\\ x &= 1 + \pm\frac{1}{\sqrt{2}} \end{align}

Pretty nifty!

A partial answer to question 4:

Suppose you add a new variable, $z$, and include in your formulas enough copies of $z$ in each term to make the terms all have the same degree. So you get

\begin{align} F_1 (x, y, z) &= x + y\\ F_2(x,y, z) &= x^2 + xy + y^2 + xz + yz \\ F_3(x, y, z) &= x^3 + x^2y + xy^2 + y^3 + x^2z + xyz + y^2 z + xz^2 + yz^2 \ldots \end{align} Then each $F_k$ is almost a sum of every $k$th power monomial in the three variables: \begin{align} G_1 (x, y, z) &= x + y + z\\ G_2(x,y, z) &= x^2 + xy + y^2 + xz + yz + z^2 \\ G_3(x, y, z) &= x^3 + x^2y + xy^2 + y^3 + x^2z + xyz + y^2 z + xz^2 + yz^2 + z^3 \ldots \end{align} the only difference being the addition of the $z, z^2, z^3$ at the end of each one.

The equations you're solving are \begin{align} F_1 (x, y, z) &= 1\\ F_2(x,y, z) &= 1 \\ F_3(x, y, z) &= 1\\ \ldots \end{align} which I propose to rewrite as \begin{align} G_1 (x, y, z) &= 0 \text{ and } z = 1\\ G_2(x,y, z) &= 0 \text{ and } z = 1\\ G_3(x, y, z) &= 0 \text{ and } z = 1\\ \ldots \end{align} which shows that the things you've drawn are the $z = 1$ slices of certain very symmetric surfaces in 3-space, ones defined (for $k$) as the 0-level sets of the sum of all homogeneous symmetric monomials of total degree $k$.

Is that the kind of "general" statement you were looking for?

BTW: cool question!