+0

# I am stuck on this problem for alcumus contest!

+1
67
5

Evaluate

Nov 5, 2019
edited by tomsun  Nov 8, 2019

#1
+2

Here:

$$\sqrt{2-\sqrt{2-\sqrt{2-...}}}$$

Set it equal to x

$$\sqrt{2-\sqrt{2-\sqrt{2-...}}}=x$$ equation 1

Square both sides

$$2-\sqrt{2-\sqrt{2-...}}=x^2$$equation 2

Because it goes on for infinity, it makes sense for us to susbtitute equation 1 into equation 2,

$$2-x=x^2$$

Now solve.

$$x^2+x-2=0$$

$$x=(-2,1)$$

A trick to remember is this formula.

$$n-\sqrt{n-\sqrt{n-...}}=\frac{n}{2}$$

Does anyone know why it works? A proof?

Nov 5, 2019
#3
+1

Very nice, CU....I saved this one to my "Watchlist"  !!!!!   CPhill  Nov 5, 2019
#2
0

Nov 5, 2019
#4
+3

Here's what I  get  as a "proof"

Let  "a" be the integer under the root....using  CU's quadratic, we have that

a  -  x  = x^2

x^2  + x  -  a  =  0

x^2 + x  =   a              complete the square on  x

x^2 + x  + 1/4   =   a + 1/4

x^2  + x + 1/4  =  (4a + 1)  / 4

(x + 1/2)^2  =   (4a + 1) / 4         take the positive root

x  + 1/2  =    sqrt (4a + 1)  / 2

x =  sqrt  (4a + 1)  - 1

_______________

2

Note   ...

If   a  = 2....then   this evaluates  to   1

If a  =  6.....then  this evaluates to  2

????   Nov 6, 2019
#5
+1

Cool! I can't confirm the answer, but it definitely makes sense.

CalculatorUser  Nov 6, 2019