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Hi!One good way to tackle this problem is to ask the following question:My name is Krishna. Before when I was in the math club, they posed a question. The question is:

what is the 'sqrt[1+sqrt 1+sqrt 1 ........ ?

The square root is abbreviated with 'sqrt'. Also note that the sqrt sign is within the sqrt sign and so on.

Thank you.

Krishna.

Assuming this number exists, is there an equation that it must satisfy, an equation which is simple enough that it can be easily solved?If we let x denote this number, notice that we have

x = sqrt[ 1 + sqrt[ 1 + sqrt [ 1 + ... ] ] ] *****************************

and the part underlined with asterisks is the same thing as x itself. This means that x must satisfy the equation x = sqrt[ 1 + x] which can be solved by squaring both sides to get x^2 = 1 + x, using the quadratic formula to find

x = (1 plus or minus sqrt[5])/2.

Since (1-sqrt[5])/2 is negative, but x is positive, x has to be the other root, namely (1+sqrt[5])/2.

What this method does is it tells you that, *if* such a number
x exists, then you can figure out what it has to be.

That's probably all you were asking for, but strictly speaking it isn't a complete answer. It leaves open the question: Does this number exist at all?

To prove that it does, you need some ideas from calculus: *every
bounded, increasing sequence has a limit*. Here we are looking for
the limit of the sequence

sqrt[1], sqrt[1 + sqrt[1]], sqrt[1 + sqrt[1 + sqrt[1]]], ...

Can you show that this sequence is bounded and increasing? (Hint: you will need to use mathematical induction. Boundedness is the trickiest one to prove; try proving by induction that all of the terms are less than 2. Use the fact that the nth term a(n) and (n-1)st term a(n-1) are related by

a(n) = sqrt[1 + a(n-1)]

to show that, if a(n-1) is less then 2, then a(n) must be also).

Post another question if you want more of an answer on this part!

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