Hi everyone, my name is Antonio, I live in Sicily and I love Maths, Physics and Statistics. I hope you will forgive my Italian accent and enjoy my videos. Ciao
im tired of explaining this things to "math teachers", who, as seemed, left the school in 10th grade, but im gonna do it again. 1. in R, exponentiation for a negative base DOES NOT EXIST BY DEFINITION. If you dont know why -explain it to yourself by simplifying (-8)^1/3, and then (-8)^2/6 (the same number in the exponent btw). 2. i (Imaginary unit) cannot be defined as sqrt(-1) because of this sequence: -1=i^2=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1. Congratulations, we have just proved that 1=-1.
Yea that rule only works with real numbers, you cant multiply sqrts of negative numbers like that, instead you have to write in terms of i and do exponts
@@piggydabest by assuming existance of i we letting roots of negative numbers have all properties of standart roots. If it wasnt true u wont be able to solve quadratic equations in complex world bc u wont be able to say that sqrt(A<0) = i*sqrt(abs(A))
In the second part, there's also the following proof: 2*ln(i) = i*pi => ln(i^2) = i*pi => ln(-1) = i*pi Replace -1 with Euler's identity: e^(i*pi)=-1 => ln[e^(i*pi)] = i*pi => i*pi = i*pi
x=y why : Taking natural log on both sides y ln(x)= x ln(y) 1/x ln(x) = 1/y ln(y) e^ln(1/x) ln(x) = e^ln(1/y) ln(y) Multiplying both sides by -1 e^ln(1/x) -ln(x) = e^ln(1/y) -ln(y) e^ln(1/x) ln(1/x) = e^ln(1/y) ln(1/y) Now taking Lambert's W function on both sides, which is the inverse of the function f(x)=xe^x We get ln(1/x)=ln(1/y) Taking exponent on both sides, 1/x = 1/y (x, y≠0) x=y
Since x and y are rational numbers they can be transform from x to y or y to x by multiplying by a constant. For example 5 can be transformed into 2 by multiplying with 2/5.
@@puremage0 Maybe that's true, but it wasn't explained. You would think a hyperintelligent AI robot would be kind enough to us mere humans to walk us through all the steps. Evidently they are too busy developing plans for overthrowing us from our current position of power.
I think there is not a predetermin d relation between X and y but by introducing such a relation we can more easily solve this problem. I initially thought about the case X=Y which is a true expression for any pair of the form (a,a)€R^2 But was a bit stumped about finding nontrivial solutions in a systematic way. And introducing a relation between X and Y helps determine the solution. I was thinking along the line of graphs so we consider Y fixed then we are looking for the intersections between polinomial x^y and exponential y^x for a given value of but it does not help algebraically
You could just factorise 4^x from both and equate the powers of 2. 4^x(1+1) = 1 4^x(2) = 1 2^2x(2) = 1 2^(2x+1)=1 2^(2x+1)=2^0 2x + 1 = 0 2x = -1 x = -0.5
les petits points de suspension ne forment pas une implication logique pour obtenir cette implication il faut réécrire la somme finie dans l’ordre décroissant sous la somme demandée comme Gauss est supposé l’avoir fait à l’âge de 7 ans
