If the equation " x^2 = x + 1" has two solutions, why does only the positive one make sense in this context? Could the negative root be valid for an infinite sequence of square roots like this? (really great video btw
First, we need to properly define the problem. The most logical one is to say: u_0 = 1, u_n = sqrt(1+u_(n-1) ) Which defines a sequence which nests more and more square roots in this way. Then, by definition of the square root, the result is always positive. Moreover, one can show that this does in fact converge to the positive solution, by showing that this solution is "stable" - that is, going to the next term always brings you closer to the positive root. This is done by analysing the recursion with tools such as pertubation analysis.
Hey! I haven't watched the video, and I'm not planning on doing so either, but I wanted to tell you that the approximation in the title has a wrong digit. The first seven should be an eight.
Hey! Just stumbled on your channel and found your style really interesting, what do you use to write in these videos? as in device, platform etc. All the love and support 🎉❤
I disagree about X having to be greater than 0. E.g., x = sqrt(25).. x can be positive or negative 5.. it is true that the number inside the square root has to be positive, but that is the case, even with (1-sqrt(5))/2. that is equal to approsximately -0.618, and if you add that value to 1, then we always have a positive number inside the square root. you can confirm this by plugging (1-sqrt(5))/2 into the equation for sqrt(1+x) = x... when you do, you get 0.618 on both sides of the equation the negative solution appears to be valid, and I don't understand your basis for denying it
Every positive number x has two square roots: +- sqrt(x) Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. [1] 1. en.wikipedia.org/wiki/Square_root
@ the video presenter states that “because this is a square root, you can’t have a negative result”. What you’re saying might be the correct explanation of why we disregard it. But if you substitute (1-sqrt(5))/2, for x, you get the correct result. The dude just throws it out like it’s unimportant and then uses some hand wavey and ultimate incorrect explanation of why
That's a shocking and interesting problem
If the equation " x^2 = x + 1" has two solutions, why does only the positive one make sense in this context? Could the negative root be valid for an infinite sequence of square roots like this? (really great video btw
First, we need to properly define the problem. The most logical one is to say:
u_0 = 1, u_n = sqrt(1+u_(n-1) )
Which defines a sequence which nests more and more square roots in this way. Then, by definition of the square root, the result is always positive. Moreover, one can show that this does in fact converge to the positive solution, by showing that this solution is "stable" - that is, going to the next term always brings you closer to the positive root. This is done by analysing the recursion with tools such as pertubation analysis.
Hey! I haven't watched the video, and I'm not planning on doing so either, but I wanted to tell you that the approximation in the title has a wrong digit. The first seven should be an eight.
Hey! Just stumbled on your channel and found your style really interesting, what do you use to write in these videos? as in device, platform etc.
All the love and support 🎉❤
@@MAHMOUDstar3075 Hi! Glad you like the videos :) I use an iPad, with a stencil and an app called “Notability” - it’s quite intuitive, hope this helps
@@amaarculusalright thanks alot, keep up the good work sir 🎉❤
I disagree about X having to be greater than 0. E.g., x = sqrt(25).. x can be positive or negative 5..
it is true that the number inside the square root has to be positive, but that is the case, even with (1-sqrt(5))/2. that is equal to approsximately -0.618, and if you add that value to 1, then we always have a positive number inside the square root.
you can confirm this by plugging (1-sqrt(5))/2 into the equation for sqrt(1+x) = x... when you do, you get 0.618 on both sides of the equation
the negative solution appears to be valid, and I don't understand your basis for denying it
the square root of function is defined as being the principle root, i.e. always positive in reals
Every positive number x has two square roots: +- sqrt(x)
Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. [1]
1. en.wikipedia.org/wiki/Square_root
@ the video presenter states that “because this is a square root, you can’t have a negative result”. What you’re saying might be the correct explanation of why we disregard it.
But if you substitute (1-sqrt(5))/2, for x, you get the correct result. The dude just throws it out like it’s unimportant and then uses some hand wavey and ultimate incorrect explanation of why
Erm actually when I plugged in (1+sqrt(5))/2 into the calculator i got 1.6180339887498948 so the title is wrong.... Reported for "Misinformation"