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If $ \sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}}=9$, find $x$.

Guest Feb 3, 2015

Best Answer 

 #12
avatar+94083 
+10

Well Bertie I have given it a shot   

 

$$\\\mbox{Prove }\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\\
\\LHS\\
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}\\
f(x+1)=\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}\\\\
f(x)=\sqrt{1+x*f(x+1)}\\\\
f^2(x)=1+x[f(x+1)]$$

 

$$\\RHS\\
g(x)=x+1\\
g(x+1)=x+1+1\\
g(x+1)=1+g(x)\\
g^2(x)=(x+1)^2\\
g^2(x)=x^2+2x+1\\
g^2(x)=1+x(x+2)\\
g^2(x)=1+x[g(x+1)]\\$$

 

$$\\f^2(x)=1+x[f(x+1)]\qquad and\qquad g^2(x)=1+x[g(x+1)]\\\\
$It follows that$\\\\
f^2(x)=g^2(x)\\\\
Now f(x)\ge 0 \qquad $because it is the answer to a square root$\\\\
g(x)=x+1 $ which \;must\; be\; $\ge 0 $ Therefore x is greater than or equal to -1$\\\\$$

 

$$\mbox{The statement cannot be true if x is less than -1 because the}\\ \mbox{RHS would be neg and the Left hand side can never be negative}$$

 

$$\\$so for all real values of x greater than or equal to -1$\\
\\f(x)=g(x) \\\\
\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\qquad Q.E.D.\\$$

 

 

 

As an aside:

$$\\\mbox{A look at two individual cases }\\
If\;\; x=0 \;then\\
LHS=\sqrt{1+0*...}=\sqrt1=1=RHS\\\\
If\;\; x=-1 \;then\\
LHS=\sqrt{1+-1*\sqrt{1+(-1+1)\sqrt...}}=
\sqrt{1+-1*\sqrt{1+0}}=\sqrt{1+-1*1}=\sqrt{0}=0=RHS\\\\$$

 

I am really not completely sure of of this proof.

Is it really true for ALL real values greater of equal to -1?

Is it completely correct Bertie or does it need refining?

Melody  Feb 6, 2015
 #1
avatar+20530 
+10

If $ \sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}}=9$, find $x$.

$$\sqrt{x+9}=9 \quad | \quad ()^2\\
x+9=81 \\
x=81-9\\
x=72$$

heureka  Feb 3, 2015
 #2
avatar+94083 
0

That is really neat Heureka, thank you 

Melody  Feb 3, 2015
 #3
avatar+92367 
+5

Let the expression on the left hand side = N = 9

Square both sides

x + N  = 81  

x + 9 = 81    subtract 9 from both sides

x = 72

 

CPhill  Feb 3, 2015
 #4
avatar+890 
+5

That I think is the second of these nested expansions in just a couple of days.

Here is, probably(?), the first problem of this type to appear in print.

Find the value of

$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots$$

Bertie  Feb 3, 2015
 #5
avatar+92367 
0

Thanks, Bertie......

 

CPhill  Feb 3, 2015
 #6
avatar+94083 
0

So how do you do your one Bertie?  (or anyone else)   

 

$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots$$

Melody  Feb 3, 2015
 #7
avatar+92367 
+5

I believe that the famous Indian mathematician, Ramanujan, proved that this converges to 3

As to how he did it.......that's way above my pay grade....!!!!!

You can read about this mathematical genius, here....

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

 

CPhill  Feb 3, 2015
 #8
avatar+94083 
0

Thanks Chris :))

Melody  Feb 3, 2015
 #9
avatar+890 
+5

Yes Chris, the problem is credited to Srinivasa Ramanujan (1187 - 1920). It appeared in the Journal of the Indian Mathematical Society in 1911. Apparently after about three months no one had come up with a solution, so he published the general result

$$x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\dots}}}}$$

Try proving it as an exercise.

Substituting x = 2 gets you the stated problem with its answer.

Bertie  Feb 3, 2015
 #10
avatar
+5

Wow ! He was more than 700 years old. Was that why he was so good at mathematics ?

Guest Feb 3, 2015
 #11
avatar+94083 
0

Yes I am sure his long gevity did help him.   LOL

Although in reality it appears that Ramanujan was only 33 when he died - he was barely out of nappies.  

Melody  Feb 4, 2015
 #12
avatar+94083 
+10
Best Answer

Well Bertie I have given it a shot   

 

$$\\\mbox{Prove }\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\\
\\LHS\\
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}\\
f(x+1)=\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}\\\\
f(x)=\sqrt{1+x*f(x+1)}\\\\
f^2(x)=1+x[f(x+1)]$$

 

$$\\RHS\\
g(x)=x+1\\
g(x+1)=x+1+1\\
g(x+1)=1+g(x)\\
g^2(x)=(x+1)^2\\
g^2(x)=x^2+2x+1\\
g^2(x)=1+x(x+2)\\
g^2(x)=1+x[g(x+1)]\\$$

 

$$\\f^2(x)=1+x[f(x+1)]\qquad and\qquad g^2(x)=1+x[g(x+1)]\\\\
$It follows that$\\\\
f^2(x)=g^2(x)\\\\
Now f(x)\ge 0 \qquad $because it is the answer to a square root$\\\\
g(x)=x+1 $ which \;must\; be\; $\ge 0 $ Therefore x is greater than or equal to -1$\\\\$$

 

$$\mbox{The statement cannot be true if x is less than -1 because the}\\ \mbox{RHS would be neg and the Left hand side can never be negative}$$

 

$$\\$so for all real values of x greater than or equal to -1$\\
\\f(x)=g(x) \\\\
\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\qquad Q.E.D.\\$$

 

 

 

As an aside:

$$\\\mbox{A look at two individual cases }\\
If\;\; x=0 \;then\\
LHS=\sqrt{1+0*...}=\sqrt1=1=RHS\\\\
If\;\; x=-1 \;then\\
LHS=\sqrt{1+-1*\sqrt{1+(-1+1)\sqrt...}}=
\sqrt{1+-1*\sqrt{1+0}}=\sqrt{1+-1*1}=\sqrt{0}=0=RHS\\\\$$

 

I am really not completely sure of of this proof.

Is it really true for ALL real values greater of equal to -1?

Is it completely correct Bertie or does it need refining?

Melody  Feb 6, 2015
 #13
avatar+92367 
0

Very nice, Melody.....

 

CPhill  Feb 6, 2015
 #14
avatar+3691 
+8

phew! u really thought about this problem! very well done!

BrittanyJ  Feb 6, 2015
 #15
avatar+94083 
+3

Thanks Chris and Brittany  

Melody  Feb 6, 2015
 #16
avatar+890 
+8

Hi Melody

Sorry, but I don't think that your proof is a proof. The problem is at the point where you say ' it follows that '

$$f^{\:2}(x)=g^{\:2}(x)$$.

Just because two functions share the same property, it doesn't follow that they are the same function. For example if we have f(-x) = -f(x) and g(-x) = -g(x), it doesn't mean that f(x) and g(x) are the same function. It is simply saying that both f and g are odd functions, and there will be a whole stack of different possibles for either one of them.

I don't have a proof of this relationship that I really like. That would be a series of steps that took me from

$$f^{\:2}(x)\equiv 1+xf(x+1),$$ 

through to

$$f(x)\equiv1+x.$$

I do have a couple of routines that might just pass.

Suppose (for no particularly good reason that I can think of) that f(x) is a polynomial of degree n. Then, on the LHS we have a polynomial of degree 2n and on the RHS a polynomial of degree n+1. For these to be identically equal we need 2n = n + 1, from which n = 1. Therefore, let f(x) = ax + b. Substitute that in and equate coefficients across the identity and you find that a = b = 1.

If I had to guess how Ramanujan came up with the result I would go for the following. I think that he was doodling, so to speak.

$$(1+x)^{2}=1+2x+x^{2}=1+x(2+x)$$

so

$$1+x=\sqrt{1+x(2+x)}\dots \dots (1) .$$

Similarly,

$$(2+x)^{2}=(1+(1+x))^{2}=1+2(1+x)+(1+x)^{2}=1+(1+x)(3+x)$$

so

$$2+x=\sqrt{1+(1+x)(3+x)}$$

so, substituting that into (1),

$$1+x=\sqrt{1+x\sqrt{1+(1+x)(3+x)}}.$$

Now repeat the routine with 3 + x = 1 + (2 + x) and so on.

Bertie  Feb 7, 2015
 #17
avatar+94083 
+3

BUMMER!      LOL    :D

 

Thanks Bertie :))

Melody  Feb 8, 2015
 #18
avatar+92367 
+3

Well...it SEEMED pretty convincing, Melody....!!!!

I still gave you points........!!!!

 

CPhill  Feb 8, 2015

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