a can only have two values, +-sqrt((1 + sqrt(17))/2). We can't put a minus between 1 and sqrt(17), because that would give us a complex value, and a has to be real. However, the other +- is correct, because both a and b can be positive or negative
and the +- at the beginning is kind of useless, because the two values are related to each other, if the real part is positive the imaginary part must be positive aswell. So they are either all two positive, or either all two negative, this implies the +- before the square root is more than enough. Actually the two +- at the beginning are kind of a mistake even, cause they mean you could choose one positive and one negative which is not correct, so they are useless and even wrong. This problem is actually poorly solved.
@ you're right, we should actually put one big +- before the whole complex number, but we don't need two solutions, we need one good solution, the positive one is better in my opinion
I forgot general algebra could be used to find the square root of a complex number, so when you said about evaluating I thought expected you to turn it into the exponential form and just halve the power... I might need more sleep
Wolfram Alpha agree with your "simplest" form, but can be coerced into giving 1/2 +/- (∜17)e^(i*atan(4)/2)/4. Which isn't very elegant as I've written it, but looks a lot better not written as linear text.
( paradox ) what ‘ s the set of all sets that ( property) don ‘ t contain themselves as sets . The sensibility that this need take a value , is a fallacy : a specious maxim .
Maybe I misunderstood something but since a is the real part don't we need to exclude t=(1-sqrt(17))/2 as it's a negative number and therefore taking the square root of it gives us complex number?
yes, that's what I also thought also he don't check for convergence, but if he did I think it would only converge to the solution with a minus sign. I haven't done the complete calculations though, I just thought visually about the complex plane
but in fact there are infinitely many solutions because you don't have to stop at just one "i". you can start subtracting from the second third and so on. you will get equations of the third and fourth degree, respectively, and so on to infinity.
Ok, but are we going to just accept that the sequence converges in the first place?:) The square root isn't even a well defined function in the complex plane because it would be multivalued(which is why pappa flammy got multiple answers for the limit, which shouldn't be the case if the sequence was convergent)
I got only one solution as follows: x_1 = -1/2 + 1/2 root4(17) e^(1/2 arctan(4)i) ≈ 0.3 + 0.62i I also got this: x_2 = -1/2 + 1/2 root4(17) e^((1/2 arctan(4)+pi)i) ≈ -1.3, -0.62i But when checking x_2 with the original equation sqrt( i - x ) = x, it does not work. So x_1 seems to be the only solution. Does anyone know if this is correct?
a can only have two values, +-sqrt((1 + sqrt(17))/2). We can't put a minus between 1 and sqrt(17), because that would give us a complex value, and a has to be real. However, the other +- is correct, because both a and b can be positive or negative
Reasoning looks sound
@ thanks, i was hoping he will realize that in the video
Exactly
and the +- at the beginning is kind of useless, because the two values are related to each other, if the real part is positive the imaginary part must be positive aswell. So they are either all two positive, or either all two negative, this implies the +- before the square root is more than enough.
Actually the two +- at the beginning are kind of a mistake even, cause they mean you could choose one positive and one negative which is not correct, so they are useless and even wrong. This problem is actually poorly solved.
@ you're right, we should actually put one big +- before the whole complex number, but we don't need two solutions, we need one good solution, the positive one is better in my opinion
I forgot general algebra could be used to find the square root of a complex number, so when you said about evaluating I thought expected you to turn it into the exponential form and just halve the power... I might need more sleep
Right lol like what the hell do you do with e^(i*arctan(4)/2)
Yay, so you didn't abandon the channel?
Please stay fam, we, your viewers, will support you.
Did you try a donation / subscription model?
Improvised Sessions are my favorites
Wolfram Alpha agree with your "simplest" form, but can be coerced into giving 1/2 +/- (∜17)e^(i*atan(4)/2)/4. Which isn't very elegant as I've written it, but looks a lot better not written as linear text.
You didn't get rid of negatives! You could argue, that sqrt is a multivalued function, but I'm still going to be cranky.
( paradox ) what ‘ s the set of all sets that
( property) don ‘ t contain themselves as sets .
The sensibility that this need take a value ,
is a fallacy : a specious maxim .
Maybe I misunderstood something but since a is the real part don't we need to exclude t=(1-sqrt(17))/2 as it's a negative number and therefore taking the square root of it gives us complex number?
yes, that's what I also thought
also he don't check for convergence, but if he did I think it would only converge to the solution with a minus sign. I haven't done the complete calculations though, I just thought visually about the complex plane
So what does this nested radicals converge to?
How can it converge to these +ve's and -ve's
but in fact there are infinitely many solutions because you don't have to stop at just one "i". you can start subtracting from the second third and so on. you will get equations of the third and fourth degree, respectively, and so on to infinity.
Cool but, can you use differential equations to solve 1+1?
Ok, but are we going to just accept that the sequence converges in the first place?:)
The square root isn't even a well defined function in the complex plane because it would be multivalued(which is why pappa flammy got multiple answers for the limit, which shouldn't be the case if the sequence was convergent)
b's value could have been rationalized.
whats the fourier and inverse transform of complex gamma function?
See I would just guess it between -1 and 1 with an i and move on
bruh, not gonna lie, it was a easy question
Indian students daily brakefast in class
I got only one solution as follows:
x_1 = -1/2 + 1/2 root4(17) e^(1/2 arctan(4)i) ≈ 0.3 + 0.62i
I also got this:
x_2 = -1/2 + 1/2 root4(17) e^((1/2 arctan(4)+pi)i) ≈ -1.3, -0.62i
But when checking x_2 with the original equation sqrt( i - x ) = x, it does not work. So x_1 seems to be the only solution.
Does anyone know if this is correct?
I Want Microsoft To Make Windows 12 Before GTA 6