I noticed that y^2 is equal to y+x (because it's an infinite serie). You can solve the ecuation y^2 -y-x , taking x as constant by using the cuadratic formula, and you integrate
I tried it without removing the function from the function. Turns out what you really have to do is differentiate while you integrate. Because when u = x + y then du = (1 + dy/dx)dx, and in the u world you end up integrating sqrt(u) - 1/2 du, producing an answer of Y = (x + y)(2y/3 - 1/2) + c.
All you have to do is implicitly differentiate y^2-y=x, which becomes dy/dx= 1/(2y-1), then you substitute the integrand for y and dx for (2y-1)dy and integrate to get 2/3y^3-1/2y^2
Hm. I did this slightly differently. I started with the substitution, got to y=sqrt(x+y), and then to y^2-y=x. But then, I just did u-sub from there, asserting (2y-1)dy = dx. I plugged that into the original integral, and integrated y(2y-1)dy to get (2/3)y^3 - (1/2)y^2 + C, and then I back-substituted the terms of x. So, my answer didn't end up as a closed form. The method in this video is pretty cool. Most parts of the result are similar to my answer. I bet with a little algebra, it could be shown that they're equivalent.
hey what´s your profile picture? it kinda looks lie a p-orbital to me. is that correct? (could be a d-orbital as well but i better ask because my eyes are horrifyingly bad^^)
Very interesting problem and good method to solve it.Ramanujan was greatest mathematician giving a lot of excellent theorems and results regarding infinite series.
The golden ratio is the solution to the equation "x²-x-1 = 0". The equation we have is "y = √(x+y)", which is just the positive solution to "y² - y - x = 0". So for x=1, we have "y²-y-1 = 0", which will give us the golden ratio.
5:14 wait what if x is in the interval range [-0.25, 0)? Then the minus in the plusminus also makes a positive number! No one said an imaginary number can’t appear in an integral!
Regarding your "imaginary number in the integral" it depends on what you're integrating. If the function and it's domain are strictly real, then it's not acceptable to have i's in the integral. (There are some exceptions like sin(x)=(e^ix -e^-ix)/(2i), but those are the exception and not the rule)
The equation "y = √(x+y)" has two solutions for x on (-0.25, 0]. But the equation "y=√(x+√(x+√(x+√(x+...))))" doesn't really make sense for negative x (if you're only dealing with real values).
If you define the function f(x)=√(x+√(x+...)), it seems to me that f(0)=0. But in the expression 1/2 + 1/2√(4x+1), it gives 1. Though in 1/2 - 1/2√(4x+1) it gives 0. Wouldn't the correct sign be - then?
The correct sign is still +. The reason why it doesn't work with 0, but with every other number, is that the limit sqrt(x+sqrt(x+...)) isn't continuous at 0.
f(0) can also be 1. Consider y=sqrt(x+y) => f(0)=sqrt(0+y) which is fullfilled for both 0 and 1 (which is what you would expect, since that's pretty much the equation he solved).
It's because y=√(x+y) has two points where x=0, at y=0 and y=1, whereas y=√(x+√(x+...)) only has the one, at y=0. Therefore both expressions give roots of y=√(x+y), but only the negative one gives the root of f(x)=√(x+√(x+...)).
Ronald K.: But can we freely choose any value for y[0] in the case where x = 0, or is it possible that only y[0] = 0 leads us to the "correct" result? What if sqrt(x + sqrt(x + ...)) is defined this way: lim n->infinity y[n](x), where y[0](x) = sqrt(x) y[1](x) = sqrt(x + sqrt(x)), y[2](x) = sqrt(x + sqrt(x + sqrt(x)) y[n](x) = sqrt(x + sqrt(x + ...)) {containing n+1 "sqrt" terms} Then, when x = 0, we have y[0] = 0, y[1] = 0, y[2] = 0, ... y[n] = 0 for all n. Which, being a sequence of zeroes, trivially converges to zero.
Man. You are soo cool. Continue doing it. I just understand little things, because I'm just in 10 grad but I like watching you explaining this stuff. Can you make a basic video about integration. Why, How... This would be cooool
TH-cam is so incredible! I'm amazed by the amount of excellent resources over TH-cam (especially maths :D). It's such an amazing time to be educated in; all you need is an interest.
at 1:38 can you take a derivative of both sides and get 2ydy-dy=dx (2y-1)dy=dx then plug y for the stuff you're integrating and (2y-1)dy for dx? It doesn't get a closed form but is there anything wrong with that?
In the 5:30sh, if you set x=0, then y as defined under the radical should also be equal to 0. This implies that you have to consider the -ve sign instead of the +ve (here: y= 1/2 +/-1/2 sqrt(4x+1)
The presentation shows that "if" the sequence y_n(x) defined by y_0(x) = sqrt(x); y_(n+1)(x) = sqrt(x + y_n(x)), for n>=0 converges, then it must converge to y(x) = (1 + sqrt(4x +1))/2, for x>0. But it does "not" show that the sequence y_n(x) converges. For example, suppose we define z_0(x) = x; z_(n+1)(x) = x(z_n(x)), for n>=0. If this sequence converges to z(x), then z(x) = x(z(x)), z(x)(1-x) = 0, z(x) = 0, if x isn't 1. But z_n(x) clearly diverges if x > 1 or x = 0 is outlined in my comments to BlackpensRedpen's other video on the derivative of y(x). I again thank him for his very instructive videos.
The problem is the expression is undefined when x < 0, and disconnected at x=0; therefore, you can only define the integral as he did for x > 0. If x is any positive number, the value of the expression is positive, and as x approaches 0 from the positive side, the expression's value approaches 1; however, when x is 0, the expression value is 0. Since the limit as x approaches 0 is different from the actual value when x = 0, you can only define the integral as he did when x > 0. Another way to look at it is this: when the "+/-" is introduced for the square root operation, the "+" is used when x > 0, but the "-" is used when x = 0. So the expression is equal to (1+sqrt(4x+1))/2 when x > 0, or (1-sqrt(4x+1))/2 when x = 0.
I got (2/3)y^3+(1/2)y^2+C by using dx=(1+2y)dy. And this is what I love about math. These answers are so different, but through a little algebra they are equivalent!
5:40 yeah but what if x is so small, like 0.1 for example, then when you minus from a half you will get a positive number. What about then? Plz explain. Thank you and have a nice day.
for the entire domain of the original integrand, it converges to the shown value of y. y has some negative domain, but we don't care about that because integrating the original function will only have positive bounds! so integrating them will be the same for all feasible bounds :)
I wonder wasn't that (-) possibility worth considering? There is a possibility 4x+1 could be smaller than 1 right. In that case (1/2) - (sqrt(4x+1)/2) doesn't become negative.
Reuben Mason I was thinking the same thing. I guess without using software you could calculate the limits for several values of x, just to get an idea of the curve, but I don't know if the infinite function has a limit or diverges. I really don't know. I am super rusty, but it is interesting for sure.
Reuben Mason Or using the simplified form he came up with makes it easier obviously. Some points would be (0,1) (2,2) (6,3) and (12,4). I assume (0,0) would also be a point from the original function, but I really have no idea now. :)
@@RanEncounter that graph is wrong. the function of y=root (x+y) continuous but the nested roots isnt continuous around 0. which is why the equation for the plus root breaks( also when graphing dont square it keep the root as by squaring you change the range)
Could we really substitute a part of this infinite sequence? Do we need to check convergence first? Tho I don’t know how to check this kind of sequence
5:42 However, if you put x=0, using the +ve sign, we get y=1 which is false as y=0. So in that case the -ve sign is correct. Because it is a case where 1 > sqrt(4x+1) so the -ve sign for y still gives a positive result. So what gives? Why is the answer for y sometimes the +ve sign and sometimes the -ve sign? Can you tell me what's the rule here?
@@Anistuffs when you plot the graph it goes from approaching 1 but then when it reaches zero the y coordinate jumps to 0. this is why the function in terms of y doesnt work,( the y function gives 0,0 but with the negative root)
@@JensenPlaysMC ohhhhhhhhhhhhhhhhhh ok yeah, I plotted it on desmos and yeah as I add more and more and more roots, the limit to 0 rises up the y axis. Thanks. Wait, that's bizarre because that would mean sqrt(x+sqrt(x)) should have a discontinuity at 0 but it doesn't So when does the discontinuity start appearing?
@@Anistuffs it does have a discontinuity at 0. its called a jumping discontinuity. but that assumes there are infinitely many defining this in closed form only works for x>0
I think that anti-differentiating an infinitely nested radical is not a valid manipulation in mathematics regardless of what you do in the process, simply because even when you use limiting sequences, they are typically ill-defined.
Profesor Se podría doblar al español estos super interesantes vídeos de redpen-blackpen Seria super por el poco conocimiento de ingles, aunque las demostraciones algoritmicas son barbaras
If anyone is curious here is a graph of the original radical expression and the closed form quadratic. www.desmos.com/calculator/zhacwjrthw We see that the expression y= √x + (√x + (√x + ... gets closer to 1/2+ √(4x + 1) /2 as we nest more radicals
5:37 But you can have: y=0.5(1-sqrt(4x+1)) If: x = [0; -0.25] So your answer should be: 0.5x +/- sqrt((4x+1)^3)/12 + c For: x = [-0.25; 0] And: 0.5x + sqrt((4x+1)^3)/12 + c For: x = [0; +Inf[
For the function to be defined, X has to be positive and a square root is always positive, so the sum X+Y is always positive, no need for absolute value here 😊
It's funny that the "y" you find in 5:46 has a different domain from the original one... Is the imaginary part of the original one thinking of it as a complex function zero in the interval [-1/4,0) as well? can this be proven without the algebraic substitution?
You make solving these fun plus you're so *positive* and *happy* and that's what reaches me ^_^. /THAT'S IT!/
Roarshark12 thank you!!!
I noticed that y^2 is equal to y+x (because it's an infinite serie). You can solve the ecuation y^2 -y-x , taking x as constant by using the cuadratic formula, and you integrate
Are you using a blue pen? I feel cheated lol.
Han Li no he is using a blue expo marker.
Yo dawg, I see you put the function out of the function so that you didn't have to integrate while you integrate.
AndDiracisHisProphet that's correct hahaha
I tried it without removing the function from the function. Turns out what you really have to do is differentiate while you integrate. Because when u = x + y then du = (1 + dy/dx)dx, and in the u world you end up integrating sqrt(u) - 1/2 du, producing an answer of Y = (x + y)(2y/3 - 1/2) + c.
AndDiracisHisProphet give me your WhatsApp number
Shalini Gupta wtf
...What?
"And we're done" *Continues to write down the integration constant* That's how it's done ;-)
The math is easy to understand, but what i don't get is:
Are there a rabbit and a cat, both named Oreo? 🍪🐇🐈
thomas g yes. Owned by different friends
That's because both of them have some black and white fur. Besides, that's the same as the Oreo cookie. (So it's a pun, maybe Cao intended a pun?)
@@blackpenredpen is this not the same,thing as x to the one half plus x to the one fourth plus x to the one eighth, etc.?
@@blackpenredpen sir I will say that the answer is x^2/2
@@ujjubhai6097 no it's wrong
All you have to do is implicitly differentiate y^2-y=x, which becomes dy/dx= 1/(2y-1), then you substitute the integrand for y and dx for (2y-1)dy and integrate to get 2/3y^3-1/2y^2
With the expression y^2-y=x, you have a perfect equation for substitution.
This guy is awesome
You made me understand so easy. Thanks
Dude youre great at this!
Cute rabbit and cute cat! Animals make everything better!
Some good integral is always satisfying. 😊
Hm. I did this slightly differently. I started with the substitution, got to y=sqrt(x+y), and then to y^2-y=x. But then, I just did u-sub from there, asserting (2y-1)dy = dx. I plugged that into the original integral, and integrated y(2y-1)dy to get (2/3)y^3 - (1/2)y^2 + C, and then I back-substituted the terms of x. So, my answer didn't end up as a closed form.
The method in this video is pretty cool. Most parts of the result are similar to my answer. I bet with a little algebra, it could be shown that they're equivalent.
I could do it same way ...easier
y(1) is the golden ratio, as you can see on my channel. :)
Fematika im 9th viewer of it :)):):)
Fematika yes!
Btw, congrats on 200 subs!
hey what´s your profile picture? it kinda looks lie a p-orbital to me. is that correct? (could be a d-orbital as well but i better ask because my eyes are horrifyingly bad^^)
Yeah, that's what it is!
Are you another fellow math and chemistry fan, like myself? :)
Do an integral battle with some Of these crazy functions!
Very interesting problem and good method to solve it.Ramanujan was greatest mathematician giving a lot of excellent theorems and results regarding infinite series.
I noticed that when we substitute 1 in our expression ( 1/2 + sqrt (4x+1) / 2) we get (sqrt(5) + 1)/2 which is the golden ratio. Amazing🔥
The golden ratio is the solution to the equation "x²-x-1 = 0". The equation we have is "y = √(x+y)", which is just the positive solution to "y² - y - x = 0". So for x=1, we have "y²-y-1 = 0", which will give us the golden ratio.
Expressing y in terms x by adding 1/4 and make it perfect square is nice step
Thanks, you just helped us with our math contest
Very good method of solving . Keep it up!
Tomorrow is my physical education exam, idk why I am attracted so badly to ur videos and maths, that I am not even studying for that exam😥
No need to find y, instead you can use the substitution (which you already calculated) x=y^2-y and calculated the integral immediately.
Colour combination is fabulous
holyyy shittt I'm now at the age to finally understand your videos. this is amazing.
Which class plz tell me
5:14 wait what if x is in the interval range [-0.25, 0)? Then the minus in the plusminus also makes a positive number! No one said an imaginary number can’t appear in an integral!
Square root is defined only for positive numbers of x right?
The original domain of the function is x≥0, so the internal you suggested doesn't exist.
Regarding your "imaginary number in the integral" it depends on what you're integrating. If the function and it's domain are strictly real, then it's not acceptable to have i's in the integral. (There are some exceptions like sin(x)=(e^ix -e^-ix)/(2i), but those are the exception and not the rule)
The equation "y = √(x+y)" has two solutions for x on (-0.25, 0]. But the equation "y=√(x+√(x+√(x+√(x+...))))" doesn't really make sense for negative x (if you're only dealing with real values).
If you define the function f(x)=√(x+√(x+...)), it seems to me that f(0)=0. But in the expression 1/2 + 1/2√(4x+1), it gives 1. Though in 1/2 - 1/2√(4x+1) it gives 0. Wouldn't the correct sign be - then?
Oh, it might not be zero because it's an infinite expression. Maybe f goes to 1 as x goes to 0. Would you make a video on that?
The correct sign is still +. The reason why it doesn't work with 0, but with every other number, is that the limit sqrt(x+sqrt(x+...)) isn't continuous at 0.
f(0) can also be 1. Consider y=sqrt(x+y) => f(0)=sqrt(0+y) which is fullfilled for both 0 and 1 (which is what you would expect, since that's pretty much the equation he solved).
It's because y=√(x+y) has two points where x=0, at y=0 and y=1, whereas y=√(x+√(x+...)) only has the one, at y=0. Therefore both expressions give roots of y=√(x+y), but only the negative one gives the root of f(x)=√(x+√(x+...)).
Ronald K.: But can we freely choose any value for y[0] in the case where x = 0, or is it possible that only y[0] = 0 leads us to the "correct" result?
What if sqrt(x + sqrt(x + ...)) is defined this way:
lim n->infinity y[n](x),
where y[0](x) = sqrt(x)
y[1](x) = sqrt(x + sqrt(x)),
y[2](x) = sqrt(x + sqrt(x + sqrt(x))
y[n](x) = sqrt(x + sqrt(x + ...)) {containing n+1 "sqrt" terms}
Then, when x = 0, we have y[0] = 0, y[1] = 0, y[2] = 0, ... y[n] = 0 for all n. Which, being a sequence of zeroes, trivially converges to zero.
Nice Performance
This is fun I'm going to watch this in my free time
Man. You are soo cool. Continue doing it. I just understand little things, because I'm just in 10 grad but I like watching you explaining this stuff. Can you make a basic video about integration. Why, How... This would be cooool
You should watch Khan Academy's already existing guide to integration - its how i learnt when I was 13 yrs old. It's really good!
I have my site. Www.blackpenredpen.com and check out the calc resources
Daniel Shapiro thx
TH-cam is so incredible! I'm amazed by the amount of excellent resources over TH-cam (especially maths :D). It's such an amazing time to be educated in; all you need is an interest.
I don't know what just happened, but it made sense to me.
This suit looks awesome in u
Guilherme Guimarães yep
Guilherme Guimarães thank you!!!
But does it also look good in the x world?
Love from Bangladesh 🇧🇩
Just learning integration, this is insane!
Thx from Syria
That's awesome!
at 1:38 can you take a derivative of both sides and get
2ydy-dy=dx
(2y-1)dy=dx
then plug y for the stuff you're integrating and (2y-1)dy for dx?
It doesn't get a closed form but is there anything wrong with that?
i don't know what a integral is but i did understand most of what you were doing until, 4:04 i started to get lost, but still good video
It can also be solved using "y" substitution and doing implicit diffrentiation to get dx.
amazing proof!
y(0)=1 or 0 depending on your definition... √0+√0+√0+√0+√0+√0...=0 but 1/2+(√0+1)/2=1.
I highly suspect how smoothly you play with the infinity... lol just like Ramanujan in his beginnings
讲的太好啦！
謝謝啦!!!!!!!
In the 5:30sh, if you set x=0, then y as defined under the radical should also be equal to 0. This implies that you have to consider the -ve sign instead of the +ve (here: y= 1/2 +/-1/2 sqrt(4x+1)
@blackpenredpen
Please have a look at my note! Please explain
And you also found a formula for y! Nice!
At 1:36 you are squaring both sides so you should open right hand side with modulus
Can you integrate it in terms of y, right?
The presentation shows that "if" the sequence y_n(x) defined by
y_0(x) = sqrt(x); y_(n+1)(x) = sqrt(x + y_n(x)), for n>=0
converges, then it must converge to
y(x) = (1 + sqrt(4x +1))/2, for x>0. But it does "not" show that the sequence y_n(x) converges.
For example, suppose we define
z_0(x) = x; z_(n+1)(x) = x(z_n(x)), for n>=0.
If this sequence converges to z(x), then z(x) = x(z(x)),
z(x)(1-x) = 0, z(x) = 0, if x isn't 1. But z_n(x) clearly diverges if x > 1 or x = 0 is outlined in my comments to BlackpensRedpen's other video on the derivative of y(x).
I again thank him for his very instructive videos.
I like your new suit it makes you a workman-officer :)
It looks good that such infinite loop reduces a lot...
This is really a awesome and outstanding approach.
This really helps me, and I hope others also.
Thank you very much sir.
The CTS isn't necessary, right? you have y² - y + x = 0 and you apply the quadratic formula, you get the same result.
No ,of course it's not necessary. But so what if it's not necessary? It's easier and less tedious than using the quadratic formula.
Very good brother
What is the practical use of all this?
it´s magic. you´re magic. fantastic.
We should use the negative version as x=0 , y=0 if we use any value of x , y may be divergent
There's something wrong. If x≠0 the equation fails giving 0=1
We've assumed our function Y is greater than 0, so x=0 is a problem
i dont think you can get a zero from an infinitely small expression, when that expression is also always gonna be non-negative
The problem is the expression is undefined when x < 0, and disconnected at x=0; therefore, you can only define the integral as he did for x > 0. If x is any positive number, the value of the expression is positive, and as x approaches 0 from the positive side, the expression's value approaches 1; however, when x is 0, the expression value is 0. Since the limit as x approaches 0 is different from the actual value when x = 0, you can only define the integral as he did when x > 0.
Another way to look at it is this: when the "+/-" is introduced for the square root operation, the "+" is used when x > 0, but the "-" is used when x = 0. So the expression is equal to (1+sqrt(4x+1))/2 when x > 0, or (1-sqrt(4x+1))/2 when x = 0.
@@PsychoMurdoctor The equation "y = √(x+y)" actually has two positive solutions for -0.25
Well done!
handsome, genius, friendly good man
I got (2/3)y^3+(1/2)y^2+C by using dx=(1+2y)dy. And this is what I love about math. These answers are so different, but through a little algebra they are equivalent!
Same answer bro....but how can this be changed to what sir had written?
But there will be a minus sign in the middle
5:40 yeah but what if x is so small, like 0.1 for example, then when you minus from a half you will get a positive number. What about then? Plz explain. Thank you and have a nice day.
How do you know, which way to go?
Something that blows my mind a little is that, when x=0, that infinite nest doesn't equal zero--it equals [1/2][1+ sqrt(1+4*0)] = 1.
Please cut of more explanation
While solving such problems
So that we can comprehend easilly sir.. Thank you for solving such a beautiful problem
It is puzzling to know that some functions are disguised in Halloween. (that is how I imagined this function) Thanks a lot.
That was awesome. A bit advanced for me, but a lot of this will come in handy in the future. My next course is Trigonometry.
if x was between 0 and 1 we could also keep the minus infront of the root;
love your vids
I saw it on Japanese youtuber "予備校のノリで学ぶ数学・物理", called "yobinori".
Wow sir👏👏👏👏
What was that c for ? The integration I don't think needed a constant of integration,
But does sqrt(x+sqrt(x+...)) always converge? What if it doesn't?
for the entire domain of the original integrand, it converges to the shown value of y. y has some negative domain, but we don't care about that because integrating the original function will only have positive bounds! so integrating them will be the same for all feasible bounds :)
That was pretty easy..
Yes I am a jee aspirant...
thanks for uploading
Thank you so much I love you
I wonder wasn't that (-) possibility worth considering? There is a possibility 4x+1 could be smaller than 1 right. In that case (1/2) - (sqrt(4x+1)/2) doesn't become negative.
It seems like the infinite square root for x=0 willl be 1/2 + sqrt(1/2) which is more than 0.
True
can't believe that I was watching this just for fun 😂
What does the graph of this function look like?
Reuben Mason I was thinking the same thing. I guess without using software you could calculate the limits for several values of x, just to get an idea of the curve, but I don't know if the infinite function has a limit or diverges. I really don't know. I am super rusty, but it is interesting for sure.
Reuben Mason Or using the simplified form he came up with makes it easier obviously. Some points would be (0,1) (2,2) (6,3) and (12,4). I assume (0,0) would also be a point from the original function, but I really have no idea now. :)
It is a parabola but 90 degree rotated to the right.
www.wolframalpha.com/input/?i=y%5E2+%3D+x+%2B+y
EDIT: this is not the integral but the y only.
V. G
@@RanEncounter that graph is wrong. the function of y=root (x+y) continuous but the nested roots isnt continuous around 0. which is why the equation for the plus root breaks( also when graphing dont square it keep the root as by squaring you change the range)
practical application of hilberts hotel, nice!
Thanks sir.
now differentiate it to get the original function , what are you going to do with the 1/2 derived from 1/2x?
Lol
Could we really substitute a part of this infinite sequence? Do we need to check convergence first? Tho I don’t know how to check this kind of sequence
What will be the sol for the sqrt of x sqrt of x sqrt of x upto infinity without + symbol
Very nice
Is it possible to write the answer in a similar way as the nested roots?
What is the valid domain of that primitive function?
Fun fact: evaluating that function at x = 1 gives you the golden ratio
Every video incites a blue pen controversy...
5:42 However, if you put x=0, using the +ve sign, we get y=1 which is false as y=0. So in that case the -ve sign is correct.
Because it is a case where 1 > sqrt(4x+1) so the -ve sign for y still gives a positive result.
So what gives? Why is the answer for y sometimes the +ve sign and sometimes the -ve sign? Can you tell me what's the rule here?
pretty sure its to do with discontinuity at 0
@@JensenPlaysMC Wait, uhm, why is there a discontinuity at 0? The function looks pretty continuous for all non-negative values of x.
@@Anistuffs when you plot the graph it goes from approaching 1 but then when it reaches zero the y coordinate jumps to 0. this is why the function in terms of y doesnt work,( the y function gives 0,0 but with the negative root)
@@JensenPlaysMC ohhhhhhhhhhhhhhhhhh
ok yeah, I plotted it on desmos and yeah as I add more and more and more roots, the limit to 0 rises up the y axis.
Thanks.
Wait, that's bizarre because that would mean sqrt(x+sqrt(x)) should have a discontinuity at 0 but it doesn't
So when does the discontinuity start appearing?
@@Anistuffs it does have a discontinuity at 0. its called a jumping discontinuity. but that assumes there are infinitely many defining this in closed form only works for x>0
I think that anti-differentiating an infinitely nested radical is not a valid manipulation in mathematics regardless of what you do in the process, simply because even when you use limiting sequences, they are typically ill-defined.
Profesor
Se podría doblar al español estos super interesantes vídeos de redpen-blackpen
Seria super por el poco conocimiento de ingles, aunque las demostraciones algoritmicas son barbaras
If anyone is curious here is a graph of the original radical expression and the closed form quadratic.
www.desmos.com/calculator/zhacwjrthw
We see that the expression y= √x + (√x + (√x + ... gets closer to 1/2+ √(4x + 1) /2
as we nest more radicals
Why complete the square if the quadratic formula leads to the same result? It seems arbitrary but was there a reason?
5:37
But you can have:
y=0.5(1-sqrt(4x+1))
If:
x = [0; -0.25]
So your answer should be:
0.5x +/- sqrt((4x+1)^3)/12 + c
For: x = [-0.25; 0]
And:
0.5x + sqrt((4x+1)^3)/12 + c
For: x = [0; +Inf[
うひひ
エクセレントですね!
If x = 0, y = root(0+root(0+…) = 0, however if y = 0.5(1+root(4x+1)), then when x = 0, y = 1, how do we approach that?
thank you very beautiful but do you have any idea how to prove that this sequence converge for every positive x if it converges
recursion u_n = sqrt (u_n-1 +x)
1:26 in that case if you square at both sides, shouldn't y^2 be equal to the absolute value of x+y ?
For the function to be defined, X has to be positive and a square root is always positive, so the sum X+Y is always positive, no need for absolute value here 😊
thanks u are really nice : )
It's funny that the "y" you find in 5:46 has a different domain from the original one... Is the imaginary part of the original one thinking of it as a complex function zero in the interval [-1/4,0) as well? can this be proven without the algebraic substitution?
although, if you think about it... for x=0 the function SHOULD be zero. with the substitution you got it's in fact 1.