Can equations such as
x^x^x^x^x = 2
or x↑↑5 = 2 if you prefer
be solved anylitically
*analytically
my bad.
>>7729321
no
you can't even solve x^x=2 without lambert's function
>>7729321
>↑↑
That's a single arrow dumbass
>>7729321
I could not help but notice your png was not optimized anon.
I have optimized your png.
Your png is now optimized.
>>7729380
Nope OP is right and you are today's faggot.
>>7729538
Knuth is a faggot and no one is going to use a single arrow for exponentiation
>>7729542
but a lot of calculators actually do that:
^ is the arrow ↑ without shaft.
>>7729611
>>7729470
Thank you based PNG optimizer.
>>7729470
smallest file I've ever seen on 4chan.
Thank you, based optimizer.
>>7729470
I love this meme
>>7729838
please
I know its between sprt(2) and 1.45
>>7730525
1.990... at 1.445
1.446602 at 2.0000031
>>7730513
shitty quality, compared to original
>>7729470
this is the best bot I have ever seen
>>7729380
>↑↑
That's a double arrow dumbass
i got the 5th root of 32
>>7730543
Looks the same to me
>>7730907
blame obama
>>7730907
delete the alpha channel
index the colors and/or gray scale
max png compression
>>7729321
What is the solution to this?
>>7729470
Thanks friend have a good day
1,43
no
i dont think that there is a analytical method.
maybe you can transform the equation to solve it somehow indirectly.
1,4326948055
excel:
x cell , 4 equivalent cells with same value,
functionvalue cell = 2, startvalue eg 1.1 for x,
now you can choose a target cell function, eg
[functionvalue - f(x)], for solution it has to be 0.
for result use manual interval or excel solver.
if not used earlier, solver needs to be activated.
>>7730884
You can compare with imagemagick (pic related. Use -metric AE), or zoom in and compare the images with an image viewer. That anon probably used some sort of color quantization, which can make an image more compressible but isn't lossless relative to the source.
>>7730907
-Check for and remove superfluous ancillary chunks (metadata), like text or comments.
-Check for and remove superfluous alpha channel.
-Change pixel data behind fully transparent pixels to something more compressible. This is technically lossy, but not perceptibly, as the pixel would be transparent anyway.
-Reduce bitdepth per pixel if possible, and viable. Very seldom reducing the bitdepth of an image makes it less compressible, but it can happen.
-Palette the image if less than 256 colors, sort that palette for best compressibility. Convert to grayscale if possible.
-Search for ideal scanline filters. This can be very processing intensive, so most programs just use a heuristic that tends to work decently.
-Run the resulting data through a deflate algorithm to generate the actual compressed stream.
-Improve the resulting blocks if possible.
There does exist an absolute maximum compression for a given image, but even with decent heuristics to avoid outright brute forcing filter combinations / deflate representation, it's still processing intensive. Adequately brute forcing any image that isn't very small, isn't even really possible. The computational investment and possibility of diminishing returns is part of why programs like paint or photoshop don't really bother, the other part is I doubt they care much. It could be done better, but their goal is to be fast and good enough. The user likely doesn't want to wait 9 or so seconds while their png compresses, much less 5-10+ minutes...
(In the case of photoshop, gimp, paint, whatever, it's usually better to save the image as the user created it and is expecting it to be saved.)
>>7731086
what does lossless means ?
>>7731164
It means that the compression function is reversible: whatever you feed into it, you can get it back from the compressed version. Lossy means that you only get an approximation of the input.
>>7731164
It means the original data can be perfectly reconstructed. So lossless compression is such that it's the same file that was compressed, when it's decompressed.
Let f(x) = x^x and the inverse function of it be g(x).
Then OP's equation is f(f(f(f(x))))=2. Hence the solution is g(g(g(g(2)))).
So how to get g(x)? The second poster is right. You gotta break out the special functions.
>>7731699
Messing around with it a little more I found the ODE:
[eqn] g'(x) = \frac{1}{\ln(g(x))+1}[/eqn]
So it's not surprising the solution requires special functions considering that the lambert special function satisfies:
[eqn] w'(x) = \frac{1}{e^{w(x)}+1} [/eqn]
>>7729542
>no one is going to use a single arrow for exponentiation
Lrn2algol fgt pls