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avatar+135 

Evaluate 

 Nov 5, 2019
edited by tomsun  Nov 8, 2019
 #1
avatar+2499 
+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.

 

\(x^2+x-2=0\)

 

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

 

So the answer is 1, as the answer can't be negative.

 

 

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
avatar+105411 
+1

Very nice, CU....I saved this one to my "Watchlist"  !!!!!

 

 

 

cool cool cool

CPhill  Nov 5, 2019
 #2
avatar+135 
0

Proof, please.

 Nov 5, 2019
 #4
avatar+105411 
+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

 

 

????

 

 

cool cool cool

 Nov 6, 2019
 #5
avatar+2499 
+1

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

CalculatorUser  Nov 6, 2019

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