Einstein Failed To Solve This | The HARDEST Problem On The Planet
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- Опубліковано 13 жов 2024
- In todays video, I will be teaching you how to solve an interesting problem. Make sure to like, subscribe, and also comment any questions or video ideas you may have relating to math!
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I guarantee Einstein didn’t struggle with this. Absolutely, unequivocally, incontrovertibly didn’t struggle in the slightest.
Neither did “E-U-L-E-R”.
Fake, Einstein never got this question so he didn't solve it
so this video is clickbait
@@1080GBAyeah
1^x ≠ 2
solved
disliked video
While the video is click bait there is an answer if x=ln(2)i/2kpi
(steps in my thinking about the problem).
STEP 1. There is no answer to your problem. That is "basic maths". Just because you can construct a problem, this does not mean that there is an answer to it.
STEP 2. However, I am very impressed by your answer at 6:20. Maybe I need to "forget everything I ever learnt about maths".
STEP3. WOW. Open ai says:
"Yes! Complex solutions for the equation 1^x = 2 do exist".
I can't post the equations they give, because it is awkwardly formatted when I clipped it from open ai.
But open ai's answer agrees with yours.
"This is a fascinating result that illustrates the rich behavior of complex exponents and logarithms! You're indeed exploring new territory beyond the constraints of real numbers".
Perhaps I misunderstand the solution to 1^x=2. I tried inserting 1^(-( i * ln(2) )/(2k * pi )) into wolfram alpha, and the result is always 1.
I then tried inserting the function using the eulers formula, ( e^(i * 2kpi ))^x=2 and the result for x does not appear to match your answer, "solution for variable x" => x = log(2) / (log(2pi) + log(e^i*k)). Even putting 1 to the power of that answer gives 1 every time.
what am I missing?
I don't understand why but putting 1 = 2^1/x gives this answer and putting the soln also gives 1 for even multiples of π (also it gives -1 for odd multiples of π which makes sense don't know why I mentioned it)
Don't know why though
Edit: Found this on an old bprp comment
Actually Wolfram-Alpha is correct. Too understand why we will need some function-theory/complex analysis (for example: Complex Analysis, Elias M. Stein S. 97-100). At first we will need a definition of z^w with w,z in C. For any z in C\(-∞,0] we can define a function
z^: C --> C by z^w:=exp(log(z)•w) where log is the principal branch of the logarithm (that means that log(1)=0). Of course you can choose another branch but in this case the definition does not match with the exponetialfunction with a real basis.
Using this definition we get:
1^x =exp(log (1)•x)=exp(0•x)=1 which states that the equation
1^x = 2 got no solution.
Now we take a look at the Question: "Can we finde a x in C such that 2^(1/x)=1?"
Using the definition we get
2^(1/x)=exp(log (2)•(1/x))
which is equal to 1 whenever log (2)/x=2πi•k, for any k in Z. This gives the solutions you are getting too.
I still don't understand it but maybe you do
top 10 clickbait
I've seen this type of question and attempts to justify it all too much here.
2:48
You didn't explain this at all. Theta = 0, so e^i0 = cos(0)+(i*sin(0)), so e = 1 right? i*0 = 0, so e^i0 = e^0 = 1, and cos(0) = 1, sin(0) = 0, so e = 1. So then you say "so I get theta = 2*k*pi." Where does that come from based on e^0=1?? Why would e^(i*2*k*pi) = 1 based on the previous example if we don't know what 2k is equal to?
Thank you this was the comment I was looking for. Literally making up numbers and variables to fit an idea that have no explanation or work to prove why they belong.
he didnt explain it properly, but the missing step is that k is some integer multiple of 2pi that satisfies the sin=0. so 0 works, 2pi works, 4pi works, etc.