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How can this be an integer??

One of my friends told me that \(\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}\) comes out to be an integer. Can anyone help me out with guessing that single digit integer?

Note by Pranjal Jain
2 years, 1 month ago

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The method to solve this kind of nested radical is to define a function \[f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}\] Clearly \[f(x)^2-1=xf(x+1)\] Now note that \[f(0)=1\] We will use induction to show that \[f(x)=x+1\] for all positive integers \(x\) (which will in turn give us the value of \(f(2)\)) First note the base case \[f(0)=1=0+1\] Now suppose \[f(k)=k+1\] for some positive integer \(k\). Then \[f(k)^2-1=kf(k+1)\implies (k+1)^2-1=kf(k+1)\implies k^2+2k=kf(k+1)\implies f(k+1)=k+2\] and the induction is done.

We conclude that \[f(x)=x+1\] for all positive integers \(x\).

Now we compute the value of \(f(2)\) simply as \(2+1=3\). :) Nathan Ramesh · 2 years, 1 month ago

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@Nathan Ramesh Thanks dude! Wikipedia had a more general one which was more difficult to be understood! Keep it up! Pranjal Jain · 2 years, 1 month ago

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@Pranjal Jain Thanks! :)

There is still a slight issue with the induction though. (I think). See if you can find it. Nathan Ramesh · 2 years, 1 month ago

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@Nathan Ramesh Yeah sure! I just checked till \(f(x)^{2}-1=xf(x+1)\) Pranjal Jain · 2 years, 1 month ago

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@Pranjal Jain Here is the issue (which I'm not sure if it's really an issue, maybe @Calvin Lin can clear it up)

When going from \(f(0)\) to \(f(1)\) using the recurrence relation and solving for \(f(1)\), what do we get? Nathan Ramesh · 2 years, 1 month ago

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@Nathan Ramesh Oh yeah! \(\frac{0}{0}\) indeterminate form!!! Pranjal Jain · 2 years, 1 month ago

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@Nathan Ramesh Not quite. In essence, you have not proven anything as yet. You need the fact that \( f(1) = 2 \) to conclude that \( f(k) = k+1 \).

The induction step is only valid if you can divide by \(k \) (right at the end). Hence, this need not hold true for \(k = 0 \). As such, your base case needs to be \(k=1 \). Calvin Lin Staff · 2 years, 1 month ago

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@Calvin Lin Can you help me in proving this? I guess @Nathan Ramesh 's method is not bad and "something" can be done with induction. Pranjal Jain · 2 years, 1 month ago

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@Calvin Lin @Calvin Lin does this prove the base case?

\[2=\sqrt{1+1\cdot 3}\] \[\sqrt{1+1\cdot 3}=\sqrt{1+1\sqrt{1+2\cdot 4}}\] \[\sqrt{1+1\sqrt{1+2\cdot 4}}=\sqrt{1+1\sqrt{1+2\sqrt{1+3\cdot 5}}}\]

We can continue this indefinitely Nathan Ramesh · 2 years, 1 month ago

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@Nathan Ramesh This way you can directly prove mine one to be 3!!!!

\(3=\sqrt{1+2×4}=\sqrt{1+2\sqrt{1+3×5}}\) and so on as you did in your solution!! Pranjal Jain · 2 years, 1 month ago

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@Pranjal Jain Yes, of course. Nathan Ramesh · 2 years, 1 month ago

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It is 3! And it is reported in http://en.wikipedia.org/wiki/Nested_radical Chew-Seong Cheong · 2 years, 1 month ago

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@Chew-Seong Cheong Is there any other possible approach? Pranjal Jain · 2 years, 1 month ago

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@Pranjal Jain Not one that I know. Chew-Seong Cheong · 2 years, 1 month ago

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Its value is 3

This is very famous nested radical given by ramanujan. Krishna Sharma · 2 years, 1 month ago

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@Krishna Sharma yes i also approached same question on this site, the key is ramanujan Mardokay Mosazghi · 2 years, 1 month ago

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From: \[f(x)^2 -1 = xf(x+1)\] \[ ( f(x)+1 ) ( f(x)-1 ) = xf(x+1)\]

Suppose \(f(x) = x+1\), then: \[(x+1+1)(x+1-1)=x(x+2)\] \[x(x+2) \equiv x(x+2)\]

It appears that \(f(x)=x+1\) satisfies the first equation for all real values of \(x\). The special case of \(x=0\) is also shown to be \(f(0) = 0 +1\).

I hope this is adequate. Chew-Seong Cheong · 2 years, 1 month ago

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