/sci/ - Science & Math

Just take the derivative bro

>>14553840
Show a picture of the graph

>>14553870

I've also done the derivatives for the main function (no =x) and it doesn't really help me understand why the graph is the way it is.

>>14553870
FYI, it's hard to tell in that graph what's going on, but if you look at the array of coordinates you can see that it behaves oddly. First it decreases rapidly to y=32 (x= {8,9,10} then y starts rising again but the rise itself is slowing.

>>14553844
I appreciate that, but didn't really help me understand.

To be clear, I'm just calculating the difference between inverse exponents where that difference reaches parity with x. In a programmatic sense, think of x as a base and y as an exponent and I'm wanting to find the point at which the difference between x^y and y^x has grown at a rate that is equal to x.

>>14553883
Looks like it just converges to Y=X, but with a vertical asymptote at X=1?

>>14553917
and -1 but the rate of convergence makes no sense, it's not only inconsistent, but it fluctuates back and forth

>>14553917

Here's the example for the first integer (obviously not solution for 1). It first becomes equivalent where y = 65 and x=2. That path to convergence get's shorter until it hits 32 for x = 8,9,10, then it rises to 33 and eventually 34 and so on, but that rising itself decays, it's just strange.

>>14553917
I guess what I'm trying to really understand here is the shift in each derivative. I get the asymptotes because there will never be a inverse exponential parity for base 1, but why, for all other integers, does that parity descend, then plateau, then ascend to a much longer plateau then descend, etc (you can't see that in these derivative graphs but it should based on the second derivative).

>>14553935
Yes, so it's strange that when considering your initial OP images contains no numbers what so ever, that when graphed any number would get such a specific and isolated relavancy

>>14553896
Now set it to zero.

>>14553952
There is no solution for that because you can't compute the denominator.

>>14553953
>>14553957

To be clear I'm saying you can't compute the denominator because setting x = 0 requires the denominator to compute x^-1 since in the numerator y would necessarily need to be 0 (to get 1-1 to resolve x = 0 in the final assignment)

>>14553840
Well it is a very nice, balenced function, with everything off kiltering each other; everything being opposite:

There are 4 terms (roughly),
x^y
y^x
x^y-1
y-1^x

I would also be interested in other such balences (like x^y-1 - y^x-1 .. or is that not intersting?)

Anyway there are those 4 terms;

2 of them are divided by 2 of them.

The top 2 are minusing one of the other
The bottom 2 are minusing one of the other

This leaves 1 term on top and 1 term on the bottom.

That's a total of 2 terms.

So that's 2 terms, from 4 terms.

When considering the exponentials, that's 8 ideas.

2, 4, and 8 are all in the family of 32.

8 times 4 is 32.

>>14553963
You've lost me.

Here, you can see it partially in what desmos can graph:

https://www.desmos.com/calculator/cizzuxtkqt

>>14553964
The first part is unrelated possibly seeking, to find any possible relation to 32 I could, and I did somewhat;

There is some relation to 32 in some relation of quantity of terms and times you are asking things to be done, and number of results from seperate asking of the terms to carry out interactions with variables

>>14554064
>>14553963

>>14553840
Have you tried simplifying it?
You should end up with [math]\frac{ln(x)}{x} = ln\left(\frac{y}{y-1}\right)[/math]

Don't graph exactly, otherwise ur gonna get programming errors. Shit like 10^32-32^10 is gonna end up looking weird

>>14554273
Good start anon. But it's not actually simplified. The simplification is

[math] y = \frac{x^{1/x}}{x^{1/x} - 1} [/math]

We can compute the derivative

[math] y' = (\log x - 1)\frac{x^{1/x} - 1}{x^2 (x^{1/x} - 1)} [/math]

This has a minimum value at x = e, which yields

[math]y' \approx 3.25 [/math]

>>14554273
>>14554603
Why and how does the number 32 come into play and get involved?

And that second anon posts 3.25
Move the decimal place over and that's 32.5
A 32 with the inspiration of a half, a motivation upwards

>>14553840
Of this implicit equation define a scalar field.
And calculate its gradient
Search for the point where the gradient is the null vector

>>14554705
Guy, graph the simplified equation