A tricky question that confused a lot of students | Complex numbers
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- เผยแพร่เมื่อ 8 ต.ค. 2024
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Let's be clear here. 1^x=1, 1^x does not equal -4 under any conditions. The author of the video found the solution to x for e^(i*2*π*k*x)=-4, not 1^x=-4. Any solution for e^(i*2*π*k*x) = -4 would involve complex numbers, but the original equation 1^x = -4 is a contradiction for any real or complex x. The expression 1^x always equals 1 for any x, so there is no solution to 1^x = -4 in the real or complex number systems. Substituting 1 with e^(i*2*π*k) leads to complex solutions that do not satisfy the original equation, as 1^x remains 1 regardless of x.
To put it more simply (I have a simple mind), in the original equation 1^x=-4, you can't change fundamental laws of math: 1 raised to any power is 1 and if that is true, then 1^x is NOT equal to -4 in any case, you matter how you try to manipulate the math.
Hi, I really like all of your videos. You do a GREAT job of explaining your steps! I learn so much every time I watch your videos. I would love to see you work through Schrödinger's Equation if you ever want to tackle it?
I think the k = 3 and 4 parts are unnecessary -- you don't really need to establish the pattern this far. If you must, use k in {1,2}, but that's all. Going to 3, 4, 1000 is unnecessary.
Mister new Euler, just one example of one of your errors and its consequences.
In in minute 10:00 you have:
Ln(e^(i2kpi*x) = Ln(-4), (note the capital L for the complex log.)
Applying the proper formula for the logarithm of complex number you will have:
ln(1)+ i(2kpi*x + 2mpi) = Ln(-4)
Do not apply ln(1) = 0. This is no longer true according to your algebra. The very 1st line of your presentation is 1^x = -4, after applying ln, and dividing by x, you have ln(1) = ln(-4)/x, and this is your algebra, erroneous, but you are bound to use this value of ln(1) all the time in your calculations. Also remember that in your algebra 1*1 is not equal 1, and 1*x is not equal x, and probably 2*x is not equal x + x.
After substituting for ln(1) according to your algebra you have following to solve:
ln(-4)/x+ i(2kpi*x + 2mpi) = Ln(-4)
Also you can't remove the principal value of the complex number removing k = m = 0. There is no complex number w/o its principal value.
"Mister new Euler" ©
Ahah😂 Golden!
Complex numbers are not complex but calculations with complex numbers definitely tend to get complex. A lot of writing gets involved and you easily make a sloppy mistake.
1^x = −4
x · ln(1) = ln(−4)
In real number, any logarithm of 1 is 0. But here we can cleverly use the identity ln(−1) = π · i:
x · ln((−1)²) = ln(−1) + ln(4)
2x · ln(−1) = ln(−1) + ln(4)
x = [ln(−1) + ln(4)] / [2 · ln(−1)]
x = 1/2 + ln(4) / (2π · i).
well... if someone asks, what is
1^(1/2 - i/pi ln(2))
would you really says it's -4? Because 1^x = 1 for all complex x, not only for real x.
only of you add branches from 1^x=exp(x ln(1))
you CAN get a multivalued function - without any branch cuts!!. but usually 1^x = 1 and nothing else
WolframAlpha believes that if you raise 1 to the power you got in your answer, you get exactly 1. Not -4
as would be expected for any value of x.
Wrong! But it will certainly fool a lot of people. I shouldn't even be commenting, because I know that the intention of this video is to attract people (particularly those who know that the so-called "solution" doesn't make any sense). But, I repeat, the so-called "solution" is wrong! The equation presented has no solution, either in real or complex numbers.
No need for variable "m".
Substitute left side with e^ (i2kπx), substitute right side with: 4e^(i(π+2kπ)).
well the solution has two degrees of freedom, no? so you do need two variables
wow. You have certainly caught my attention. I noticed the other day from looking at these maths questions, that log(-2) has an infinity of complex values. "I don't remember learning that at school". Maybe I misremembered what I read the other day. but whatever it said, it has prepared me have a look at your post for this question.
Now I'm stuck on 1^x = -4
take logs of both sides
x log 1 = log (-4).
I was prepared from what I read last week, to accept that log(-4) has exotic values.
But how can x log 1 be anything else but 0??
I will watch your post anyway to see if it helps.
I know I often make mistakes, but open ai agrees with me.
Even open ai sometimes makes mistakes, and it often makes mistakes on difficult problems.
(1)^2=1(x ➖ 1x+1 ). 2^2 (x ➖ 2x+2).
Somehow I never learned about logarithms between high school and college. I figured that was part of this, but I don't know how to use them.
Do not warry, the author of presentation don't know how to use logarithms of complex numbers.
At 5:43 I pointedly ask why anyone would not expect the thame anther for all k.
OMG ...... endless loops of complex numbers
Your videos are very informative! Thank you for making them. Also, what is your accent?
My advice: DO NOT REMAIN MISINFORMED by this video
@@hajstra1307 I just was asking your nationality. You have a very unique accent.
@@epd807 I an not the author of this video
Hello i am the first.
A inveja é uma merda