How do I get formula for the nth term of this on matlab?

 sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...))))

I know this limit is 3...but I need to get matlab to give me the first 40 terms. I am confused on how to code it.

[Adam]

Doesn't it need some kind of initialisation condition?

[cakey]

I am not sure what that entrails. I keep trying to figure out a formula but nothing works.

[Alberto]

I think the sucession should be like this:

a_1=sqrt(1)
a_2=sqrt(1 + 2*sqrt(1))
a_3=sqrt(1 + 2*sqrt(1 + 3*sqrt(1)))
...

Has the same limit and doesn't need initial value for recursion.

Answer 1

[Image Analyst]

What would be inside the parentheses of the 40th sqrt()? Just a 1?

Try a for loop and see what happens

s(40) = 1;
for k = 39 : -1 : 1
  s(k) = k * sqrt(s(k+1)+1)
end

[cakey]

Hello, I don't understand why s(40)=1...s(40) is too large for me to put in my calculator.

[cakey]

I believe s(1)=1 the sequence starts with 1, so I am a bit confused

[Image Analyst]

Calculator?! We're using MATLAB programs here. Did you actually even try the code?

[cakey]

Yes, I tried it. it gave me a lot of numbers. I'm sorry I am a noob to all this. :s

[cakey]

How come you got s(40)=1?

[cakey]

Okay, upon examination, I see you started a(1)=1 and you are going backwards?

[Image Analyst]

There is no "a". If you want, put a semicolon at the end of the s(k) line and just put s on its own line after the loop to have it print out the whole array.

[cakey]

I keep, but it won't give me whole array appears to only give value of s(n) only. After I have array, how can I plot it to see graph?

Answer 2

[Stephen23]

You could try writing a for loop. The loop would just need to increment down itr = 40:-1:1, and calculates itr*sqrt(1+last_val) , with last_value defined before the loop (what value?).

Check it first with a small number of iterations first (1, 2, 3), to confirm that it calculates the expected values. Then try it with more iterations.

Answer 3

[Roger Stafford]

It doesn't matter what you initialize it at, the limit as n approaches infinity is always 2, not 3.

Correction: You were right. I was in error. The limit is always 3 no matter what your initial value is.

Answer 4

[Roger Stafford]

Since I made an error in my first answer, here is a bit more information. The problem can be expressed this way:

   x(1) = sqrt(1+2*x(2))
   x(2) = sqrt(1+3*x(3))
   x(3) = sqrt(1+4*x(4))
   ...
   x(n-1) = sqrt(1+n*x(n))

Now suppose x(n) were equal to n+2. Then

   x(n-1) = sqrt(1+n*(n+2)) = sqrt((n+1)^2) = n+1
   x(n-2) = sqrt(1+(n-1)*(n+1)) = sqrt(n^2) = n
   ...
   x(1)   = 3

However, if x(n) is not equal to n+2, express its ratio to n+2 as x(n)/(n+2) = 1+e(n). Then we have

   1+e(n-1) = x(n-1)/(n+1)
            = sqrt((1+n*(n+2)*(1+e(n)))/(n+1)^2)
            = sqrt(1+n*(n+2)/(n+1)^2*e(n))

   e(n-1) = sqrt(1+n*(n+2)/(n+1)^2*e(n)) - 1

This will always approach zero for sufficiently large n to start with and hence the limit for x(1) must be 3 no matter what the initial value is.