# Recursive square root inside square root problem

• I
• Gionata
In summary: Then I copied G5 to F6. That gives us a good estimate of f(99).Then I highlight cells F6 and G6. In the lower right hand corner of the highlighted area, there is a little square. I click on that square and drag it down. That copies the formula in the cells G6 and then F6 to cells F7 and G7. And so on, all the way down to cell F85 and G85.But your sequence is recursive, so you don't need Excel. I just used Excel to get an initial estimate for f(100). If you want, you can use Excel to get an initial estimate for f(1000

#### Gionata

I have been debating this issue for days:

I can't find a recursive function of this equation:
##\large{\sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}}}##

Starting value 2 always added with pi

has been trying to find a solution this for days now, is what I have achieved so far:
This sucession result not correct:
##f\left(n\right)=\sqrt{n+1+\pi f\left(n+1\right)}##

http://www.wolframalpha.com/input/?i=f\left(n\right)=\sqrt{n+1+\pi+f\left(n+1\right)}
Unfortunately, I do not know how to move forward, thanks a lot!

Last edited:
I don't know the "correct way" to do this kind of recursive equation, but this was my strategy:
If you start from a large n and work backwards, whatever error you started with is going to get less and less as you move towards n=1. This is because you're doing those square roots over and over again.

So I solved for the f(100) given the recursion formula and given the approximation f(100)=f(101).
That got me f(100) = 11.74269.
Then I worked it backwards to f(1) using an Excel spreadsheet.

If you work out an estimate using this method or some other method, I let you know if it's consistent with what I have.
If it is consistent, I'll give you a Wolfram Alpha link that shows the series.

Thank you very much for the very comprehensive response.
I need the recursive formula because I have to demonstrate with another program (mathlab) that at high sequnze of n (very high) the value tends to a value x.
To do this I cannot be satisfied with an approximation but rather with a recursive formula that calculates exactly the value of x on the basis of (n)

It sounds as though you are saying that you are looking for a version of f(n) that is not recursive.
Or are you saying that you are looking for a recursive formula in the form of ##f(n)=g(f(n-1),n)##? In that form, you would be able to compute a series of f(n) given f(0).

Also, "tends to a value x": I presume x is a different function of n. In rough terms, as n gets very large, f(n) will be approximately ##\sqrt{n} + \pi/2##.
But I was able to get that with the existing f(n) definition - just by presuming that f(n) is approximately equal to f(n+1) for very large n.

I am a looking for a recursive formula in the form of:
##f\left(n\right)=\sqrt{n+1+\pi f\left(n+1\right)}##
in alternative
##f\left(n+1\right)=\frac{1}{\pi}\left(f\left(n\right)^2-n-1\right)##

These examples mentioned above are very close to the solution but unfortunately wrong.

What I want is to be able to get a recursive formula in order to calculate at will, repeating n times the root the exact result

Meanwhile, thank you for your interest:)

Gionata said:
I am a looking for a recursive formula in the form of:
##f\left(n\right)=\sqrt{n+1+\pi f\left(n+1\right)}##
in alternative
##f\left(n+1\right)=\frac{1}{\pi}\left(f\left(n\right)^2-n-1\right)##
OK.
I have that sequence. But I'm not suppose to just give it to you. On this board, I am only suppose to help you calculate it.
So I will give the general form of the answer, but not the exact answer:
http://www.wolframalpha.com/input/?i=f(n)=sqrt(n+1+(pi+f(n+1)));+f(0)=3.5

Warning: The sequence generate by that link is also not what you want.

To get it right, you need to replace that 3.5 value with the correct value.
I told you how to calculate that value earlier in this thread. But if you want more assistance, I will assist.

One more thing: When you are using wolframalpha to work this problem, f(0) to f(19) will be accurate. f(20) to f(24) may be accurate. But they plot values for f(25) to f(30) which are not valid. That's because of limitations on the precision of the floating point values they are using - and the fact that tiny errors in f(0) or f(1) quickly build up after f(19).

That is why you can get much greater precision for f(n) by starting with a good estimate of f(n+20) and working backwards to f(n). This is true no matter what tool you are using: Matlab, Excel, wolfram.

well now I've understood the situation,
I have to find the key value of f0), but I'm currently struggling a lot, if you're patient and want to I'd like if together you help me to calculate the value of f(0). I would be very grateful.

I used Excel.

I reserved column "F" for n. And "G" for f(n).
At cell F4, I put the number 100.
At cell F5, I put =f4-1. (which will be 99).
At cell G3, I put in an estimate for f(101). Almost any value will work, but I put in my estimate of 11.75. Less accurate values (0 or 20) will also work.
At cell G4, I put in your formula: =SQRT(F4+1+PI()*G3)
Then I copied G4 to G5. That gives us a good estimate of f(99).

for f(99)= 11.70016315

for f(0)= 3.609993083
Correct?

Last edited:
Gionata said:
for f(99)= 11.70016315

for f(0)= 3.609993083
Correct?
Yes. BTW: I look for the PF alert. So if you quote something of mine, I will see it a lot quicker.

Now if you use that in Wolfram, you will see the values in the n vs. n+1 look good up to about 20.
You can't see it in the plot because the scale is messed up with f(30).

So that demonstrates:
1) that you formula was basically OK;
2) the value for f(0) to f(20) can be deduced from Wolfram.
3) Values for f(21) and on need another method - but you should have the first 80 of so in your Excel spreadsheet.

So, if you need f(9,000) in MatLab, do you know how to generate it?
For example: f(10000) = 101.5882103

f(1000000) = 1001.572531
Notice how close that is to ##\sqrt{n} + \pi/2##

.Scott said:
So that demonstrates:
1) that you formula was basically OK;
2) the value for f(0) to f(20) can be deduced from Wolfram.
3) Values for f(21) and on need another method - but you should have the first 80 of so in your Excel spreadsheet.

So, if you need f(9,000) in MatLab, do you know how to generate it?
For example: f(10000) = 101.5882103

f(1000000) = 1001.572531
Notice how close that is to ##\sqrt{n} + \pi/2##

I am really happy that it works, after days of work.
Thanks for the explanations, you have a great talent.

You asked me if I can generate f(9,000) in MatLab, the answer is no, do you want to explain it to me? That would be the culmination of the project.

You're going to do the same thing in MatLab that you did with Excel.
Start will 9020 and make a guess. I'd use the formula ##\sqrt{n} + \pi()/2## - but even 0 will work.
Then compute f(9019) from f(9020), f(9018) from f(9019), and so on until you get to f(9000).

.Scott said:
You're going to do the same thing in MatLab that you did with Excel.
Start will 9020 and make a guess. I'd use the formula ##\sqrt{n} + \pi()/2## - but even 0 will work.
Then compute f(9019) from f(9020), f(9018) from f(9019), and so on until you get to f(9000).

well, understood thank you very much!

Last question just to understand:
f(0)= 3.609993083
http://www.wolframalpha.com/input/?...left(n\right)^2-n-1\right);+f(0)=+3.609993083

##\sqrt{2+\pi }## f(0) should not give 2.26751?

from the second link the first value is 3.60999

##\sqrt{n} + \pi()/2## is an approximation to f(n) for large n.
As we said earlier, ##f(0)=3.609993083##.
##f(1) = \sqrt{2+\pi() f(0)} = 3.829920486##.

I'm not sure where you are getting 2.26751.

Oh.
There is a difference between ## \sqrt{2+\pi/2} ## and ## \sqrt{2} + \pi/2 ##.

.Scott said:
Oh.
There is a difference between ## \sqrt{2+\pi/2} ## and ## \sqrt{2} + \pi/2 ##.
oh, my error ignores...

Question, if I want to replace (retaining the formula) only the pi with phi example, and doing all the calculations as we did together should work?
##\large{\sqrt{2+\phi \sqrt{3+\phi\sqrt{4+\phi\sqrt{5+\dotsb}}}}}##

Yup. Given some specific positive value for ##\phi##.

Super
.Scott said:
Yup. Given some specific positive value for ##\phi##.
Super! Thanks a lot for your help !

Gionata said:
I calculated that for phi (Golden section)
f(0)=2.267712876

But it is not correct:
http://www.wolframalpha.com/input/?i=f\left(n+1\right)=\frac{1}{1.6180339887}\left(f\left(n\right)^2-n-1\right);+f(0)=+2.267712876
What's wrong?
I'm not sure anything is wrong. But I got f(0)=2.2677128829.
http://www.wolframalpha.com/input/?i=f(n)=sqrt(n+1+(1.6180339887f(n+1)));+f(0)=2.2677128829

Also, I notice that with 1.618.., the results are only good for about the first 15.
http://www.wolframalpha.com/input/?...1+++n]],+f[0]+==+2.267712882},+f,+{n,+0,+20}]

Last edited:
.Scott said:
Strange, I calculated manually with excel the first 10 series:
I got: 3.128171204
D2= 1.6180339887 (phi)

You're calculating f(1).
But you have some terms backwards.
I set D10 to phi and used this:
=SQRT(2+D10*SQRT(3+D10*SQRT(4+D10*SQRT(5+D10*SQRT(6+D10*SQRT(7+D10*SQRT(8+D10*SQRT(9+D10*SQRT(10+D10*SQRT(11+D10*SQRT(12+D10*3.7)))))))))))
I got 2.560219. If you want greater precision than that, you need to use more terms.

.Scott said:
You're calculating f(1).
But you have some terms backwards.
I set D10 to phi and used this:
=SQRT(2+D10*SQRT(3+D10*SQRT(4+D10*SQRT(5+D10*SQRT(6+D10*SQRT(7+D10*SQRT(8+D10*SQRT(9+D10*SQRT(10+D10*SQRT(11+D10*SQRT(12+D10*3.7)))))))))))
I got 2.560219. If you want greater precision than that, you need to use more terms.

= RADQ (2 + D10 * RADQ (2 + D10)) = 2.2534
different
= RADQ (D10 + 2 * RADQ (D10 + 3)) = 2,4322

the value of pi I wish before the sequence 2,3,(n)

Good morning, Dear Friend Gionata, even if it is evening.

Best Regards,
one of Yours many friends,
Vitalie.

Gionata said:
I have been debating this issue for days:

I can't find a recursive function of this equation:
##\large{\sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}}}##

Starting value 2 always added with pi

has been trying to find a solution this for days now, is what I have achieved so far:
This sucession result not correct:
##f\left(n\right)=\sqrt{n+1+\pi f\left(n+1\right)}##

http://www.wolframalpha.com/input/?i=f\left(n\right)=\sqrt{n+1+\pi+f\left(n+1\right)}
Unfortunately, I do not know how to move forward, thanks a lot!

#### Attachments

• Selection_655.png
54.5 KB · Views: 204
After working on the problem for a day I have found:

Starting with the more general expression, ## \large{\sqrt{1+\pi \sqrt{2+\pi\sqrt{3+\pi\sqrt{4+\dotsb}}}}} ## , we can get the backward recursive relation: $$f\left(n\right)=\sqrt{-1+\pi \sqrt{n+1+f\left(n+1\right)^2}}$$
Putting ## f\left(n+1\right) = i ##, we work our way down until we reach ## f\left(1\right) ##.

For example, let ## n=3 ##.

## f\left(4\right) = i##, ## f\left(3\right) = \sqrt{-1+\pi \sqrt{4+i^2}} = \sqrt{-1+\pi \sqrt{3}}##, until we reach finally: $$f\left(1\right) = \large{\sqrt{-1+\pi \sqrt{1+\pi \sqrt{2+\pi\sqrt{3}}}}}$$.

Squaring, adding one and dividing by ## \pi ##, we are getting ## \sqrt{1+\pi \sqrt{2+\pi\sqrt{3}}} ## as required.

For your particular sequence getting as far down as ## f\left(2\right) ## only, will suffice...

As easy as that!

Quasimodo said:
After working on the problem for a day I have found:

Starting with the more general expression, ## \large{\sqrt{1+\pi \sqrt{2+\pi\sqrt{3+\pi\sqrt{4+\dotsb}}}}} ## , we can get the backward recursive relation: $$f\left(n\right)=\sqrt{-1+\pi \sqrt{n+1+f\left(n+1\right)^2}}$$
Putting ## f\left(n+1\right) = i ##, we work our way down until we reach ## f\left(1\right) ##.
And here is the Mathematica code for ## \sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}} ##, for n=100:

z = I; For[n = 100, n > 1, n = n - 1,
z = N[Sqrt[-1 + \[Pi] Sqrt[n + 1 + z^2]], 100]]; (z^2 + 1)/\[Pi]

## 1. What is the recursive square root inside square root problem?

The recursive square root inside square root problem is a mathematical problem that involves taking the square root of a number, and then taking the square root of that result, and so on for a specified number of times.

## 2. How is this problem different from a regular square root?

This problem is different from a regular square root because it involves taking the square root of a number multiple times, whereas a regular square root only involves taking the square root once.

## 3. What is the purpose of this problem?

The purpose of this problem is to demonstrate the concept of recursion in mathematics. It can also be useful in certain fields of mathematics and physics, such as in the study of fractals.

## 4. How is the recursive square root inside square root problem solved?

This problem is typically solved using a recursive algorithm, where the function calls itself until a specified number of iterations is reached. The algorithm involves taking the square root of the initial number, and then calling the function again with the result as the input, until the desired number of iterations is reached.

## 5. What are some real-world applications of this problem?

One real-world application of this problem is in the calculation of fractal dimensions, which are used to describe complex geometric patterns found in nature. It can also be used in computer graphics to generate intricate and visually appealing images. Additionally, it can be applied in financial modeling and time series analysis.