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

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cakey
cakey 2014-9-24
评论： cakey 2014-9-25
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.
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cakey
cakey 2014-9-24
I am not sure what that entrails. I keep trying to figure out a formula but nothing works.
Alberto
Alberto 2014-9-24
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.

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采纳的回答

Image Analyst
Image Analyst 2014-9-24
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
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Image Analyst
Image Analyst 2014-9-25
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
cakey 2014-9-25
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?

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更多回答（3 个）

Stephen23
Stephen23 2014-9-24
编辑：Stephen23 2014-9-24
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.
Roger Stafford
Roger Stafford 2014-9-24
编辑：Roger Stafford 2014-9-24
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.
Roger Stafford
Roger Stafford 2014-9-25
编辑：Roger Stafford 2014-9-25
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.

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