How can I substitute here?

I don't understand how to find the answer to this. Any help will be appreciated!

What is the value of \(x\) if

x=1+21+41+161+2561+x= \sqrt{1+2\sqrt{1+4\sqrt{1+16\sqrt{1+256\sqrt{1+\cdots}}}}}? ​

Note by Vinayak Srivastava
2 weeks, 4 days ago

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I think that the expression does not converge at all. To show that, we will proof n1+n21+n4...>nkn\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n^k for all natural numbers nn and kk. Here is the proof by induction on kk:

START: k = 1 The values of all square roots in the expression are greater than 1, so we have: n1+...>n=n1n\sqrt{1 + ... } > n = n^1

STEP from k to k +1:

We assume n1+n21+n4...>2kn\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > 2^k for all nn and a particular kk.

This also applies to n2n^2, so we have: n21+n4...>(n2)k=n2kn^2\sqrt{1 + n^4\sqrt{...}} > (n^2)^k = n^{2k}

We conclude: n1+n21+n4...>n1+n2k>nn2k=n×nk=nk+1n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n\sqrt{1 + n^{2k}} > n\sqrt{n^{2k}} = n \times n^k = n^{k+1}

The start of induction and the step of induction show the assumption for all kk.

We can apply this statement for n=2n = 2:

21+41+16...>2k2\sqrt{1 + 4\sqrt{1 + 16\sqrt{...}}} > 2^k

This will exceed any real number for large kk and does not converge. Therefore, the whole expression does not converge, the value of xx is undefined.

Finnley Paolella - 2 weeks, 4 days ago

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Thanks a lot @Finnley Paolella! Just asking of curiosity, do you know of any expressions like this(I mean continuing infinitely0), which look like they won't converge, but they do?

Vinayak Srivastava - 2 weeks, 4 days ago

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I think the expression 1+21+41+81+161+\sqrt{1 + 2\sqrt{1 + 4\sqrt{1 + 8\sqrt{1 + 16\sqrt{1 + \dots}}}}} converges.

Finnley Paolella - 2 weeks, 4 days ago

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@Finnley Paolella How can I find the value of this expression?

Vinayak Srivastava - 2 weeks, 3 days ago

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@Vinayak Srivastava According to this code, it is around 4.14031...

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import math
def f(n, depth):
    if depth == 0:
        return 1
    return math.sqrt(1 + n * f(2*n, depth - 1))
print(f(2, 5))
print(f(2, 50))
print(f(2, 200))
print(f(2, 500))

3.6136728719051603

4.140314562141137

4.14031456214126

4.14031456214126

Finnley Paolella - 2 weeks, 3 days ago

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@Finnley Paolella Thank you very much! Any algebraic method?

Vinayak Srivastava - 2 weeks, 2 days ago

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@Vinayak Srivastava I am sure there is... but I haven't found one.

Finnley Paolella - 2 weeks, 2 days ago

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@Finnley Paolella OK, no problem, thanks for spending your time on my question!

Vinayak Srivastava - 2 weeks, 2 days ago

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1+21+41+81+16\sqrt{1+2\sqrt{1+4\sqrt{1+8\sqrt{1+16\sqrt{\cdots}}}}} or 1+21+41+161+256\sqrt{1+2\sqrt{1+4\sqrt{1+16\sqrt{1+256\sqrt{\cdots}}}}}?

Páll Márton - 1 week, 3 days ago

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Vinayak asked for the second expression - which diverges. The first expression is also very interesting, I haven't found the solution to that yet, just estimated the result to be around 4.14031456214126 using python (look at the comments under my solution). If you find the exact result for the first expression, please let me know!

Finnley Paolella - 1 week, 3 days ago

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Ok. So I will start with the first.

Páll Márton - 1 week, 3 days ago

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