By far the most complicated part is the fact that the ratio of successive terms in the Fibonacci sequence approaches a specific number (which happens to be the golden ratio, which happens to be close to the ratio of km/mi).
Fun fact: If you have a scientific calculator (literal or app) but no other conversion tool available, the conversion factor between miles and kilometres is almost exactly ln 5. Disturbingly close in fact.
That's fewer keypresses than generating the Golden ratio or working out Fibonacci numbers. But if all you have is your head then, yeah, the Fibonacci trick is good enough in a pinch.
There are many such ways to memorize conversion ratios. Admittedly, this one is particularly cool, since you can construct it from the fairly trivial fibonacci series. But I still feel, it's no replacement for the actual solution; get rid of imperial and adopt metric.
I don't believe you can just factor our e^x in the second step like that. That seems in incorrect to me. Then again I've been out of calculus for years now.
I know that, but @Urist said that most of us here are bad at basic algebra. Not that I'm great at it (I had an 8 average in all 4 maths in uni), but I do believe that most of here were correct. This is unsolvable using classic algebra.
Just to be clear: the comment was meant in a lighthearted way and not meant to be taken personally, especially toward those who are confident with their math skills.
Would it be a rabbit hole to try and find any merit in this solution when interpreting it as: "if x is in a superposition of 2 and -2, the x + 2 = x - 2 would be true in 1/4 of the observations", or something like that?
It is the closest thing to a "solution" that I can imagine, but doesn't fit any laws that I know of or understand, and would probably break down on any scrutiny, but it feels like something is there.
x cant be both values at the same time, not under what most people consider to be math. Feel free to write your own logical system and see where that takes you, though.
I think the only “solution” that works is addition/subtraction under mod 4 (or mod 2 I suppose) like another poster suggested. Then we’d have:
Correct, not solvable with rational numbers. I should have been more clear. When we’re doing arithmetic modulo x, it’s assumed to be with integers.
To be clear, this is a solution only in Z_4 which is not what most people mean when they’re look for answers to algebra problems. And it would be a solution for all x in Z_4 (which are integers, see this page that I assume is a good summary)
Yes, that is correct, this is solvable in modular algebra... but, in that case, the tripple horizontal line equal sign should have been used, not the double horizontal line one, which of course indicates classic algebra.
Haha I got that :) @Urist is right, I was halfheartedly looking for a logic system in which it could make sense. Still, I would have major issues with the first step as it is shown, but I am wondering about systems where, say, each x <- {..}, then what would be the set, and the probability of the correct solution.
Something I need to be more awake for, and it may be easier to solve without resorting to powers and roots, haha.
Except the first assumption that e^x = its own integral, everything else actually makes sense (except the DX are in the wrong powers). You simply treat the "1" and "integral dx" as operators, formally functions from R^R into R^R and "(0)" as calculating the value of the operator on a constant-valued function 0.
EDIT: the step 1/(1-integral) = the limit of a certain series is slightly dubious, but I believe it can be formally proven as well.
EDIT 2: I was proven wrong, read the comments
my favourite part is when you do by parts and it results in integral(say equation 1), then you do another and it results in original integral you started with(say equation 2).
so, you go clever and add both and find the result. it's clever and satisfying.
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