derivative of sqrt(x+sqrt(x+sqrt(x+…))),

implicit differentiation,

blackpenredpen,

math for fun,

source

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28

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derivative of sqrt(x+sqrt(x+sqrt(x+…))),

implicit differentiation,

blackpenredpen,

math for fun,

source

Copyright ©2024 EduGorilla Community Pvt. Ltd.

Hmm so you say y = sqrt(x+sqrt(x+…)

But you can rewrite this in infinitely many ways as this weird sum of square roots is going to infinity :

y = sqrt(x+sqrt(x+…+y)

So What is the real correct answer ? Nice video btw, glad to see more and more videos of maths on your channel these times!

Thanks for this, I am in calc BC and we have been doing logarithmic differentiation. I thought it would be fun to play with some infinite functions, like x^x^x…. What a coincidence!

Isn't 1/sqrt(1+4x) another (or even better) way to write the same thing though?

Awesome

Hi!, your videos are awesome!.

I would really appreciate if you help me with this interesting integral: x/(e^x -1) from 0 to infin. , the result is π^2/6. I don't know how to get that result.

Regards!

Now I know how to impress girls. Wish me luck!

Don't forget to rationalize the denominator! :-p

g(x) = sqrt(x+sqrt(x+sqrt(x+…)))

g(x) = sqrt(x + g(x) )

g(x)^2 – g(x) – x = 0

g(x) = (1+sqrt(1+4x))/2

easy from there

4:48 I like to use y' in these cases, so isn't so messy

Excellent Intuition!

Great example!!

2:04 hahahaha, thanks, man 😀

It would be great if you could do a video to integrate this differential equation 🙂

Nice shirt

Is the last step even allowed ? Wouldn't you have to prove that y can't be 1/2 so that you are not diving by 0 ?

in these problems you always want to find the infinite sequence again, but somewhere else, then substitute.

y = sqrt(x+sqrt(x+…

if you take the square you get

y²=x+ sqrt(x+sqrt(x+…

then you can move the x to the left and you get

y² – x = sqrt(x+sqrt(x+…

From what we started with we know that y = sqrt(x+sqrt(x+…

therefore:

y² – x = y

Then it's much simpler and you can use implicit differentiation 🙂

I have a question. Is it not better to just express it in terms of x by solving the quadratic in the beginning? I had the same idea you had and solved it by Implicit Differentiation but then realized I could also do this:

y^2-y-x=0

D=1+4x

Given that y is properly defined as a real number – albeit an irrational one maybe – it stands to reason that it is defined for all x:(1+4x>0). Moreover, since y=sqrt(x+y), if x<0, then y becomes the sqrt of something negative+the square root of something negative + …, which cannot be even a real number, let alone a positive one. Therefore, the domain for this function is (all x>0).

Consequently, we solve the quadratic and obtain the following formula.

y=1/2*(1+sqrt(1+4x)) , the negative-sqrt solution is considered invalid because:

sqrt(1+4x)>1, for x>0

1/2(1-sqrt(1+4x))<0, for x>0

Therefore, y would be negative, which cannot be true given that y equals a square root of a positive real number.

Now we can simply perform Explicit Differentiation onto the above formula and arrive at the derivative expressed in finitely many terms of x:

dy/dx=2/sqrt(1+4x)

What do you think?

Edit: After taking a second look, the negative sqrt solution is also valid for 0>=x>=(-1/4)

So, actually y is not a well defined function unless we specify the domain first.

If x belongs between [-1/4,+inf), then y=1/2*(1+sqrt(1+4x))

If x belongs between [-1/4,0], then y=1/2*(1-sqrt(1+4x))

Either case, we can also use E.D.

can we do a derivation/integral to a recursive function?

Can't you just differenciate the square root using product rule and chain rule?

Would not be different if you (at step 3) wrote two x's then you write y ..so it will be

Y=sqrt(x+sqrt(x+y)) then differentiate..?

let Y =1+1+1+1+… : if you can write Y = 1 + Y then 0=1

Mr Chow, how could I even impress a girl without rationalizing the denominator?

Now integrate

Once you have y^2 – y = x, can't you just complete the square to solve for y? Then you'll have a closed form expression for y. Differentiate that, and get a much simpler answer for dy/dx.

Why not keep going and solve the differential equation? http://www.wolframalpha.com/input/?i=y%27+%3D+1%2F(2y-1)

https://www.wolframalpha.com/input/?i=dy%2Fdx%3D1%2F(2y-1)

An equivalent form is y = (1 ± sqrt(4 x + 1))/2, https://www.wolframalpha.com/input/?i=Plot%5B%7B(1+%2B+Sqrt%5B1+%2B+4+x%5D)%2F2,+(1+-+Sqrt%5B1+%2B+4+x%5D)%2F2%7D%5D

Wow I actually found it!!!!

Am I crazy or that's the same than y = 1 + sqrt(1+4x)/2 and dy/dx = 1/sqrt(1+4x)???

Here is how:

y=sqrt(x+sqrt(x+sqrt(x+…. Squaring both sides…

y^2=x+sqrt(x+(sqrt(x+…=x+y

y^2-y-x=0 (quadratic equation with a=1, b=-1 and c=-x)

y = (-b +/- sqrt(b^2-4ac))/2a

y= (1 +/- sqrt(1+4x))/2

Since x needs to non-negative (for how the function was defined first, wither you would have a sqrt of a negative number), the smallest result of the sqrt(1-4x) can be sqrt(1), and hence the subtraction in the numerator (if I use the the – option) would be negative for any x except zero, what would in turn would give a negative y, which is not possible with the original function if I take the sqrt sign as the principal root (the positive one). Hence I will discard the – option in the +/- and keep:

y= (1+sqrt(1+4x))/2. which is easily differentiable:

y=1/2+1/2*(1+4x)^1/2

dy/dx= 1/2*1/2*(1+4x)^(-1/2)*4

dy/dx= 1/sqrt(1+4x)