Thread: Adding of roots....sort of (NOW SOLVED - TOPIC CAN BE CLOSED

Hi all,

A simple question for you all; posed to me by a friend and I'm too dumb to get it.

Basically he asked what was the value of 'a' if:

a = 2^(0.5) +(((2^(0.5) +((2^(0.5)) + .....

I HOPE that I have the notation right. I get the feeling that the answer is two, but some logic would be nice to go with my answer. Thanks in advance,

Michael

if i understand you correctly you want to add 2^0.5 to 2^0.5 an infinite number of times.
i'm pretty sure that makes your answer 2^0.5 times infinity, or a=infinity. i think.

the reasoning:
basically you got 2^0.5(n+1)= F(n+1)

where n is the number of times you add 2^0.5 to 2^0.5.

if F(n+1)=a=2 then (n+1)=2^0.5
which means you only add 2^0.5 to 2^0.5, [2^0.5-1] number of times.

that could be explained better but the holidays have turned my mind into something completely unordered and lazy.

Thanks Wallaby but that is not what I want to do. Sorry that my notation is poor. I will try to explain again:

say that x = 2^0.5.

y = (2+x)^0.5

z = (2+y)^0.5

x is the 1st term in a number series, y is the 2nd term and z is the 3th term and this method will continue infinately up to the infinite term if that makes sense.

So what would the inifinte term be? 2? Why?

Thanks again?

Michael?

2^0.5 as any engineer will tell you off the top of his head is 1.414,

so in your original post it's just 1.414n

in your second post I suggest you could write a simple VB program to compute it, but it's a divergent series so don't wait up for a 'fixed' answer!

Thanks Megabrain. I now know that what I have written in my second post is 100% correct. This was actually a question posted to a friend of mine at a Cambridge interview for his maths degree.

There is some sort of logic which means you know instinctively that the answer is two, and therefore does not require VB programming to compute.

Okay, lets first get the notation out of the way:

You want to know what the limit as n -> infinity of x_n+1 = Sqrt(2+x_n) is. LEts ignore the tedium of proving this is a contraction mapping and that a contraction mapping converges to one of its fixed points and just use the results on faith. So what we have to do is find the fixed points of x_n+1 = Sqrt(2+x_n) i.e.

find t such that t = Sqrt(2+t) => t² - t - 2 = 0 => t = 2 or t = -1.

Now x_n > 0 for all n so we know that x_n must converge to 2.

QED

Yes, that makes perfect sense now. I can hold my head high! Thanks River Rat.

Michael

YOur wish to close the thread - Granted.