Just a Single Variable Equation

"no"

That's not even a complicated problem, it would just take a while.

Okay buddy what steps would you take to solve it?

It's 2

What's the neat trick here?

I'm noticing a/b +- b/a patterns

whats the practical application of such an equation

OP, can you type all this in latex

Brute-forcing the problem at the moment. I believe there are 2 real solutions: one between 0.5 and 0.6, and one between 2.3 and 2.4.

OP, does this curve look familiar?

Type it again in polish notation so we can fucking parse it

OK I've narrowed it down
Approximate values of the solutions:

0.513638209
2.327004333

I don't know the exact values. Anyone who has autistically memorized preposterous square root identities, speak now.

>Brute forcing
>Just finding the zeros numerically
>Took 14 minutes
Calm down kid

Ask Cleo, she could probably help out

I've got a decent way of rewriting it. Let me see if I know how to mathpost.

Let [math] x_0 = y_0 = x [/math], let [math] x_n = y_{n-1}^{y_{n-1}} [/math], and let [math] y_n = x_{n-1}^{x_{n-1}} [/math].

Ah good. Then

[math]
\frac{x_5}{y_5} - \frac{x_4}{y_4} - \frac{x_3}{y_3} - \frac{x_2}{y_2} + \frac{x_1}{y_1} + \frac{y_1}{x_1} + \frac{y_2}{x_2} - \frac{y_3}{x_3} + \frac{y_4}{x_4} + \frac{y_5}{x_5} - x -2 = 0
[/math]

oh whoops, let [math] y_n = x_{n-1}^{\frac{1}{x_{n-1}}} [/math].

I don't see how this helps

It's the first time the equation has been written down outside of the OP's image.

Integrate the function by x.

>Someone else here knows who Cleo is
where the fuck am I

but OP, just find it numerically, unless you really need an analytical solution

Dude just square both sides lmao

Let [math]\log[/math] refer to [math]\log_x[/math] in the following:

[math] x_n =x^{\log{y_{n-1}^{y_{n-1}}} = x^{y_{n-1} \log{y_{n-1}}} [/math]

[math] y_n =x^{\log{x_{n-1}^{{\frac{1}{x_{n-1}}}} = x^{\frac{1}{x_{n-1}} \log{x_{n-1}}} [/math]

Thus,

[math]
x_n =x^{\product_{i=1}^n{z_{i-1, n-i}^{(-1)^{i-1}}})\log{x}} = x^{\product{a_i^{(-1)^i}}}
[/math]

And

[math]
y_n =x^{\product_{i=1}^n{z_{i, n-i}^{(-1)^i}})\log{x}} = x^{\product{a_i^{(-1)^i}}}
[/math]

Where [math]z_{i, j}= y_j[/math] when [math]i[/math] is even and [math]z_{i, j}= x_i[/math] when [math]i[/math] is odd.

Where and how does one learn a technique like this?

Let [math]log[/math] refer to [math]log_x[/math] in the following:

[math]
x_n =x^{\log{y_{n-1}^{y_{n-1}}}} = x^{y_{n-1} \log{y_{n-1}}}
[/math]

[math]
y_n =x^{\log{x_{n-1}^{\frac{1}{x_{n-1}}}}} = x^{\frac{1}{x_{n-1}} \log{x_{n-1}}}
[/math]

Thus,

[math]
x_n=(\prod_{i=1}^n{z_{i-1,n−i}^{(-1)^{i-1}}) \log x} = \prod_i=1^n{z_{i-1,n−i}^{(-1)^{i-1}}}
[/math]

And

[/math]
y_n=(\prod_i=1^n{z_{i,n−i}^{(-1)^i}) \log x} = \prod_i=1^n{z_{i,n−i}^{(-1)^i}}
[/math]

Where [math] z_{i,j} = y_j [/math] when [math] i [/math] is even and [math] z_{i,j} = x_j [/math]when [math] i [/math] is odd.

ok once more with feeling

Let [math]log[/math] refer to [math]log_x[/math] in the following:

[math]
x_n =x^{\log{y_{n-1}^{y_{n-1}}}} = x^{y_{n-1} \log{y_{n-1}}}
[/math]

[math]
y_n =x^{\log{x_{n-1}^{\frac{1}{x_{n-1}}}}} = x^{\frac{1}{x_{n-1}} \log{x_{n-1}}}
[/math]

Thus,

[math]
x_n=(\prod_{i=1}^n{z_{i-1,n−i}^{(-1)^{i-1}}) \log x} = \prod_{i=1}^n{z_{i-1,n−i}^{(-1)^{i-1}}}
[/math]

And

[math]
y_n=(\prod_i=1^n{z_{i,n−i}^{(-1)^i}) \log x} = \prod_{i=1}^n{z_{i,n−i}^{(-1)^i}}
[/math]

Where [math] z_{i,j} = y_j [/math] when [math] i [/math] is even and [math] z_{i,j} = x_j [/math]when [math] i [/math] is odd.

what technique

I just threw logs at the problem because the exponents were making me sad

To be blunt about it, im a calc 1 student trying to get better at math and I have no idea what you're doing.

The answer is x = -1

Yes, that's the one, you've got it. :3

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Nuuuu, you cracked my structure X3

Actually these would be super roots, but you've got them :3