No, I got that part, but I don't think I understand the significance of the indexed y values and their relationships to the indexed x values. The criterion seems to suggest that P3(xn)=yn for each, but that strikes me as something that is defined as a constraint rather than something that is to be proved. Also, I woke up then and now so that might be playing a factor in my confusion.
OK, you got it then, I believe.
P3 is specifically built so that P3(xn)=yn for n from 1 to 4. The proof lies in its construction. I guess the sentence can be understood as "we know it works because we built it like that, however you may verify it yourself"
I feel like the sentence also means "it's kinda obvious when you think about it, so we won't explain, but it's actually important, so you probably should make sure you agree".
The function should be cubic, so you should be able to write it in the form "f(x) = ax^3 + bx^2 + cx + d". You could work out the entire thing to put it in that form, but you don't need to.
Since there are no weird operations, roots, divisions by x, or anything like that, you can just count how many times x might get multiplied with itself. At the top of each division, there are 3 terms with x, so you can quite easily see that the maximum will be x^3.
It's useful to know what the values x_i and x_y are though. They describe the 3 points through which the function should go: (x_1, y_1) to (x_3, y_3).
That also makes the second part of the statement ready to check. Take (x_1, y_1) for example. You want to be sure that f(x_1) = y_1. If you replace all of the "x" in the formula by x_1, you'll see that everything starts cancelling each other out. Eventually you'll get "1 * y_1 + 0 * y_2 + 0 * y_3", thus f(x_1) is indeed y_1.
They could have explained this a bit better in the book, it also took me a little while to figure it out.
It's not too bad, once you consider that everything in each term is a constant value, except for "x" itself.
So the numerator of each term is the product of three linear factors, like (x-4)(x-2)(x-6), which should produce a cubic, like x³ - 12x² + 44x - 48. Then the denominator of each term is a pure constant, so it would be like taking that cubic and dividing it by 4, getting 0.25x³ - 3x² + 11x - 12. Then the yₙ terms are also constants, so no different than doing something like multiplying by 2, getting you something like 0.5x³ - 6x² + 22x - 24, if I take that example a bit too far. And at that point it's just the sum of four cubics, which will be cubic as long as the x³ terms don't perfectly cancel out - which I believe would only happen if the four pairs of points used to make the function were all perfectly laying on the same line or parabola.
The construction's also pretty clever: OP said the point was to fit the function to the four points (x₁, y₁), (x₂, y₂), (x₃, y₃), (x₄, y₄). Let's say we set x = x₂, then. Because (x-x₂) appears in the numerator of every term but the second term, every term but the second term will have a 0 in its numerator and cancel out - so we only need to consider the second term. Its numerator is then (x₂-x₁)(x₂-x₃)(x₂-x₄) - exactly the same as its denominator. So they both cancel out, leaving only y₂ - meaning we get P₃(x₂) = y₂, as desired.
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.
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
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.
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