The way I thought about this limit, is that as we're taking the square root of a positive number very close to zero, the square root is bigger than the input, as we repeat this process infinite times, it has an compounding effect - making the limit 1. Surely, this logic does not neccesarily require it to approach one, but can we formalize this chain of thought to arrive at this answer? Or just use it as a vague intuitive explaination ?
So the way i think abt is like yk how 1/2+1/4+1/8…..1/2n = 1 It’s very similar but the therms get bigger instead of smaller like we started with 1/2n+……+1/4+1/2 =1
The thing is, sqrt(1+4x) > 1 for even the smallest value of x, since 1 = sqrt(1) and sqrt(1+4x) > sqrt(1) of any positive real x. Thus, the numerator will always by negative if you take the neg root as well.
@@filipeoliveira7001 You can take a negative square root and P will not be negative if x = 0. I think this step just skips over the solution that P still can be zero. Having x = 0 and P = 0 satisfies the polynomial equation
@@march3nkoWell, you see, its not just 0, its 0+, which is identical to zero in all but few niche cases. You can think of it as a positive number 0,0000...1 with as many 0 digits as you like, followed by 1 So like 1 + 0+ > 1 (by a tiny fraction, but still) You can think of such operations, like LHS is not exactly 1, but slightly overshoots. 0- is a same deal, but its negative and symbolizes undershoot.
How much time does it take to create a good manim animation? I tried it for a bit and couldn't really get a sense for how exactly things work, I did manage to create some stuff like a circle interpolating into a square but nothing too fancy. I really want to be able to make videos in this style even if just for my own fun
It depends. Most of my time is currently spent on playing the video in my mind. It is like I see the video in my mind virtually first, then I turn the scenes into codes. It seems that you are learning Manim right now. Just keep coding consistently for a while, you will be good :)
If you treat the expression (after sending n to infinity) as a function of x, then at least it won't be continuous at x=0 point as shown in the visual. Honestly I don't think it will be very trivial to prove its continuity elsewhere.
@@ron-math Thanks for clarifying! The visual and why we took the limit in the first place makes more sense now that I've watched it a few times. Great video!
Do you like my new end screen?
yeah 👍
Thanks bro!@@axbs4863
hey, what do you use to create these animations? just asking
Pretty good
Manim, pretty standard right now@@copter7013
Limit[(1+sqrt(1+4x))/2, x -> 0] = 1 -- This works regardless of the coefficient of x. The coefficient of x does affect the imaginary part however.
The way I thought about this limit, is that as we're taking the square root of a positive number very close to zero, the square root is bigger than the input, as we repeat this process infinite times, it has an compounding effect - making the limit 1. Surely, this logic does not neccesarily require it to approach one, but can we formalize this chain of thought to arrive at this answer? Or just use it as a vague intuitive explaination ?
Absolutely fantastic bro. It is good enough as an intuitive approach!
Yes, I think we can formalize it but it won't be very trivial.
So the way i think abt is like yk how 1/2+1/4+1/8…..1/2n = 1
It’s very similar but the therms get bigger instead of smaller like we started with 1/2n+……+1/4+1/2 =1
You can even put -1/2 inside
1:29 I dont get the argument. If x is less than or equal 1, there is no reason not to pick the negative square root
Square root implies the positive answer. In mathematics, when we inverse a square, we use ± square root.
The thing is, sqrt(1+4x) > 1 for even the smallest value of x, since 1 = sqrt(1) and sqrt(1+4x) > sqrt(1) of any positive real x. Thus, the numerator will always by negative if you take the neg root as well.
@@filipeoliveira7001 You can take a negative square root and P will not be negative if x = 0. I think this step just skips over the solution that P still can be zero. Having x = 0 and P = 0 satisfies the polynomial equation
@@march3nkoWell, you see, its not just 0, its 0+, which is identical to zero in all but few niche cases.
You can think of it as a positive number 0,0000...1 with as many 0 digits as you like, followed by 1
So like 1 + 0+ > 1 (by a tiny fraction, but still)
You can think of such operations, like LHS is not exactly 1, but slightly overshoots.
0- is a same deal, but its negative and symbolizes undershoot.
@@march3nko you’re right! Thank you pointing that out!
How much time does it take to create a good manim animation? I tried it for a bit and couldn't really get a sense for how exactly things work, I did manage to create some stuff like a circle interpolating into a square but nothing too fancy. I really want to be able to make videos in this style even if just for my own fun
It depends. Most of my time is currently spent on playing the video in my mind. It is like I see the video in my mind virtually first, then I turn the scenes into codes.
It seems that you are learning Manim right now. Just keep coding consistently for a while, you will be good :)
Would this nested radical be continuous?
Because x is positive, and the root is always positive
If you treat the expression (after sending n to infinity) as a function of x, then at least it won't be continuous at x=0 point as shown in the visual.
Honestly I don't think it will be very trivial to prove its continuity elsewhere.
@@ron-math Thanks for clarifying! The visual and why we took the limit in the first place makes more sense now that I've watched it a few times. Great video!
Thank you very much! Glad you figured it out!@@frendlyleaf6187
0^0