View Full Version : Infinite surd
8below1994
Jan 1, 2010, 11:48 PM
The sequence:
a1=√(1+√(1 )
a2=√(1+√(1+√(1 )
a3=√(1+√(1+√(1+√(1+... etc.
Calculate the decimal values of the first ten terms of the sequence. Using technology plot the relation between n and an. Describe what you notice. What does this suggest about the value of (a(n))-(a(n+1)) as n gets very large? Use your results to find the exact value for this infinite surd.
Unknown008
Jan 2, 2010, 12:03 AM
If I got it right;
a_1 = \sqrt(1+\sqrt1) \\ a_2 = \sqrt(1+\sqrt(1+\sqrt1)) \\ a_3 = \sqrt(1+\sqrt(1+ \sqrt(1+\sqrt1)))
You'll have to grab a calculator for that, and find the value of each of the 10 terms.
For the second part, I can't help you. Apparently, you have a program which can plot the points of the sequence. Sorry.
For the third part, it's easy if you have plot the graph. You'll see that the value tends to be a constant and the graph makes an asymptote. I'll let you guess the answer to that.
Is it OK?:)
8below1994
Jan 2, 2010, 12:07 AM
Makes sense, but what is the formula for a(n+1) in terms of a(n) ?
Unknown008
Jan 2, 2010, 12:13 AM
Sorry, I don't see the question asking for the formula.
Think a little more now. If the terms approach a certain number, it means that they are nearly the same number, or roughly the same number if n is extremely large.
So, a_n is a term, where n is very large
a_(n+1) is the term just after a_n.
Can you see where I'm getting at?
8below1994
Jan 2, 2010, 12:14 AM
Its making more sense... I asked a broader question... check that?
galactus
Jan 2, 2010, 05:09 AM
This is the nested radical form for the Golden Ratio.
Let L equal the limit. Then, since it is nested and convergent, we can write:
L=\sqrt{1+L}
L^{2}-L-1=0
L=\frac{\sqrt{5}+1}{2}\approx 1.618
If we let L+1=\sqrt{1+(L+1)}
We get \frac{\sqrt{5}-1}{2}\approx 0.618
Unknown008
Jan 2, 2010, 11:19 AM
galactus, can you have a look at the other thread. It's in the same board and it's by the same member.
galactus
Jan 2, 2010, 12:08 PM
Let's do something general.
Start with \sqrt{a+b\sqrt{a+b\sqrt{a+....}}}
and assume it converges when a\geq 0, \;\ b\geq 0.
L=\sqrt{a+b\sqrt{a+b\sqrt{a+....}}}
L=\sqrt{a+bL}
L^{2}-bL-a=0
Using the quadratic formula, we get
L=\frac{b+\sqrt{b^{2}+4a}}{2}
Now, for the given nested radical, a=b=1.
Therefore,
L=\frac{1+\sqrt{1^{2}+4(1)}}{2}=\frac{\sqrt{5}+1}{ 2}
The Golden Ratio, commonly designated as {\phi}