What is the maximum value of this nested radical?












19














I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:




What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?




Some observations follow.




  • Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time - we can no longer write $F_infty$ as a function of itself to be solved.


  • Here is a plot of $F_{15}$:



enter image description here




  • What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.




  • Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.




I'd appreciate proofs to my thoughts, especially the last one!










share|cite|improve this question




















  • 1




    How exactly does the expression for $F_k$ look? I can't actually work out easily how it terminates at the k^th step.
    – T_M
    yesterday
















19














I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:




What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?




Some observations follow.




  • Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time - we can no longer write $F_infty$ as a function of itself to be solved.


  • Here is a plot of $F_{15}$:



enter image description here




  • What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.




  • Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.




I'd appreciate proofs to my thoughts, especially the last one!










share|cite|improve this question




















  • 1




    How exactly does the expression for $F_k$ look? I can't actually work out easily how it terminates at the k^th step.
    – T_M
    yesterday














19












19








19


6





I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:




What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?




Some observations follow.




  • Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time - we can no longer write $F_infty$ as a function of itself to be solved.


  • Here is a plot of $F_{15}$:



enter image description here




  • What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.




  • Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.




I'd appreciate proofs to my thoughts, especially the last one!










share|cite|improve this question















I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:




What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?




Some observations follow.




  • Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time - we can no longer write $F_infty$ as a function of itself to be solved.


  • Here is a plot of $F_{15}$:



enter image description here




  • What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.




  • Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.




I'd appreciate proofs to my thoughts, especially the last one!







functions recursion maxima-minima nested-radicals golden-ratio






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edited yesterday

























asked yesterday









TheSimpliFire

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  • 1




    How exactly does the expression for $F_k$ look? I can't actually work out easily how it terminates at the k^th step.
    – T_M
    yesterday














  • 1




    How exactly does the expression for $F_k$ look? I can't actually work out easily how it terminates at the k^th step.
    – T_M
    yesterday








1




1




How exactly does the expression for $F_k$ look? I can't actually work out easily how it terminates at the k^th step.
– T_M
yesterday




How exactly does the expression for $F_k$ look? I can't actually work out easily how it terminates at the k^th step.
– T_M
yesterday










2 Answers
2






active

oldest

votes


















8














If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:



$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.



Proof attempt for the domain:



While I'm not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.



Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.



Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.



So, in $F_{2k}(x)$, consider:



$$ frac{cdots}{x - sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} }}$$
$$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$



Now, we will show that the "bottom" of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:



$$ sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
$$ frac{x^{2k-1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
$$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$



The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that's messier in my opinion.



Using this result, it follows that:



$$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^(k+1)} } implies F_{2(k+1)}(x) notin mathbb{R} $$



Now, to prove the "bottom" will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$.



So, for any $x$ there's an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I'm not certain this is rigorously extends to the infinite case.



Extra notes: I've rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.



['0: 8.0', '1: 8.0', '2: 10.0', '3: 14.0', '4: 22.0', '5: 38.0', '6: 68.0', '7: 130.0', '8: 252.0', '9: 500.0', '10: 992.0', '11: 1978.0', '12: 3948.0', '13: 7890.0', '14: 15774.0']



As you can see, as $x$ gets twice as close to 1, it takes almost twice as many terms to go non-real. I've tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.






share|cite|improve this answer



















  • 1




    The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
    – Zachary Hunter
    yesterday






  • 1




    I have added the steps towards solving the problem :)
    – TheSimpliFire
    yesterday










  • If you can prove the domain as well please do add it to your post!
    – TheSimpliFire
    yesterday










  • Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
    – Zachary Hunter
    2 hours ago



















1














Self partial answer: (next improvement is to prove that $H'<1$)



Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F'=frac1{2F}cdotfrac{1(x+G)-x(1+G')}{(x+G)^2}>0impliedby G-xG'>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.



Now this is implied by $$G'=frac1{2G}cdotleft(1+frac{H'x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H'x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H'x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H'x^2-H^2\&impliedby x^2>H'x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H'$. Of course, the latter two are only close approximations to the real distribution of $H$.



enter image description here






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    2 Answers
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    8














    If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:



    $$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.



    Proof attempt for the domain:



    While I'm not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.



    Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.



    Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.



    So, in $F_{2k}(x)$, consider:



    $$ frac{cdots}{x - sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} }}$$
    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$



    Now, we will show that the "bottom" of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:



    $$ sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
    $$ frac{x^{2k-1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
    $$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$



    The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that's messier in my opinion.



    Using this result, it follows that:



    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^(k+1)} } implies F_{2(k+1)}(x) notin mathbb{R} $$



    Now, to prove the "bottom" will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$.



    So, for any $x$ there's an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I'm not certain this is rigorously extends to the infinite case.



    Extra notes: I've rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.



    ['0: 8.0', '1: 8.0', '2: 10.0', '3: 14.0', '4: 22.0', '5: 38.0', '6: 68.0', '7: 130.0', '8: 252.0', '9: 500.0', '10: 992.0', '11: 1978.0', '12: 3948.0', '13: 7890.0', '14: 15774.0']



    As you can see, as $x$ gets twice as close to 1, it takes almost twice as many terms to go non-real. I've tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.






    share|cite|improve this answer



















    • 1




      The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
      – Zachary Hunter
      yesterday






    • 1




      I have added the steps towards solving the problem :)
      – TheSimpliFire
      yesterday










    • If you can prove the domain as well please do add it to your post!
      – TheSimpliFire
      yesterday










    • Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
      – Zachary Hunter
      2 hours ago
















    8














    If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:



    $$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.



    Proof attempt for the domain:



    While I'm not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.



    Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.



    Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.



    So, in $F_{2k}(x)$, consider:



    $$ frac{cdots}{x - sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} }}$$
    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$



    Now, we will show that the "bottom" of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:



    $$ sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
    $$ frac{x^{2k-1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
    $$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$



    The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that's messier in my opinion.



    Using this result, it follows that:



    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^(k+1)} } implies F_{2(k+1)}(x) notin mathbb{R} $$



    Now, to prove the "bottom" will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$.



    So, for any $x$ there's an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I'm not certain this is rigorously extends to the infinite case.



    Extra notes: I've rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.



    ['0: 8.0', '1: 8.0', '2: 10.0', '3: 14.0', '4: 22.0', '5: 38.0', '6: 68.0', '7: 130.0', '8: 252.0', '9: 500.0', '10: 992.0', '11: 1978.0', '12: 3948.0', '13: 7890.0', '14: 15774.0']



    As you can see, as $x$ gets twice as close to 1, it takes almost twice as many terms to go non-real. I've tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.






    share|cite|improve this answer



















    • 1




      The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
      – Zachary Hunter
      yesterday






    • 1




      I have added the steps towards solving the problem :)
      – TheSimpliFire
      yesterday










    • If you can prove the domain as well please do add it to your post!
      – TheSimpliFire
      yesterday










    • Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
      – Zachary Hunter
      2 hours ago














    8












    8








    8






    If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:



    $$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.



    Proof attempt for the domain:



    While I'm not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.



    Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.



    Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.



    So, in $F_{2k}(x)$, consider:



    $$ frac{cdots}{x - sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} }}$$
    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$



    Now, we will show that the "bottom" of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:



    $$ sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
    $$ frac{x^{2k-1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
    $$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$



    The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that's messier in my opinion.



    Using this result, it follows that:



    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^(k+1)} } implies F_{2(k+1)}(x) notin mathbb{R} $$



    Now, to prove the "bottom" will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$.



    So, for any $x$ there's an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I'm not certain this is rigorously extends to the infinite case.



    Extra notes: I've rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.



    ['0: 8.0', '1: 8.0', '2: 10.0', '3: 14.0', '4: 22.0', '5: 38.0', '6: 68.0', '7: 130.0', '8: 252.0', '9: 500.0', '10: 992.0', '11: 1978.0', '12: 3948.0', '13: 7890.0', '14: 15774.0']



    As you can see, as $x$ gets twice as close to 1, it takes almost twice as many terms to go non-real. I've tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.






    share|cite|improve this answer














    If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:



    $$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.



    Proof attempt for the domain:



    While I'm not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.



    Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.



    Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.



    So, in $F_{2k}(x)$, consider:



    $$ frac{cdots}{x - sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} }}$$
    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$



    Now, we will show that the "bottom" of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:



    $$ sqrt{ frac{x^{2k-1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
    $$ frac{x^{2k-1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
    $$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$



    The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that's messier in my opinion.



    Using this result, it follows that:



    $$ x < sqrt{ frac{x^{2k-1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^(k+1)} } implies F_{2(k+1)}(x) notin mathbb{R} $$



    Now, to prove the "bottom" will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$.



    So, for any $x$ there's an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I'm not certain this is rigorously extends to the infinite case.



    Extra notes: I've rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.



    ['0: 8.0', '1: 8.0', '2: 10.0', '3: 14.0', '4: 22.0', '5: 38.0', '6: 68.0', '7: 130.0', '8: 252.0', '9: 500.0', '10: 992.0', '11: 1978.0', '12: 3948.0', '13: 7890.0', '14: 15774.0']



    As you can see, as $x$ gets twice as close to 1, it takes almost twice as many terms to go non-real. I've tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 8 hours ago

























    answered yesterday









    Zachary Hunter

    53110




    53110








    • 1




      The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
      – Zachary Hunter
      yesterday






    • 1




      I have added the steps towards solving the problem :)
      – TheSimpliFire
      yesterday










    • If you can prove the domain as well please do add it to your post!
      – TheSimpliFire
      yesterday










    • Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
      – Zachary Hunter
      2 hours ago














    • 1




      The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
      – Zachary Hunter
      yesterday






    • 1




      I have added the steps towards solving the problem :)
      – TheSimpliFire
      yesterday










    • If you can prove the domain as well please do add it to your post!
      – TheSimpliFire
      yesterday










    • Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
      – Zachary Hunter
      2 hours ago








    1




    1




    The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
    – Zachary Hunter
    yesterday




    The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
    – Zachary Hunter
    yesterday




    1




    1




    I have added the steps towards solving the problem :)
    – TheSimpliFire
    yesterday




    I have added the steps towards solving the problem :)
    – TheSimpliFire
    yesterday












    If you can prove the domain as well please do add it to your post!
    – TheSimpliFire
    yesterday




    If you can prove the domain as well please do add it to your post!
    – TheSimpliFire
    yesterday












    Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
    – Zachary Hunter
    2 hours ago




    Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
    – Zachary Hunter
    2 hours ago











    1














    Self partial answer: (next improvement is to prove that $H'<1$)



    Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F'=frac1{2F}cdotfrac{1(x+G)-x(1+G')}{(x+G)^2}>0impliedby G-xG'>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.



    Now this is implied by $$G'=frac1{2G}cdotleft(1+frac{H'x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H'x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H'x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H'x^2-H^2\&impliedby x^2>H'x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H'$. Of course, the latter two are only close approximations to the real distribution of $H$.



    enter image description here






    share|cite|improve this answer


























      1














      Self partial answer: (next improvement is to prove that $H'<1$)



      Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F'=frac1{2F}cdotfrac{1(x+G)-x(1+G')}{(x+G)^2}>0impliedby G-xG'>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.



      Now this is implied by $$G'=frac1{2G}cdotleft(1+frac{H'x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H'x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H'x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H'x^2-H^2\&impliedby x^2>H'x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H'$. Of course, the latter two are only close approximations to the real distribution of $H$.



      enter image description here






      share|cite|improve this answer
























        1












        1








        1






        Self partial answer: (next improvement is to prove that $H'<1$)



        Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F'=frac1{2F}cdotfrac{1(x+G)-x(1+G')}{(x+G)^2}>0impliedby G-xG'>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.



        Now this is implied by $$G'=frac1{2G}cdotleft(1+frac{H'x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H'x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H'x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H'x^2-H^2\&impliedby x^2>H'x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H'$. Of course, the latter two are only close approximations to the real distribution of $H$.



        enter image description here






        share|cite|improve this answer












        Self partial answer: (next improvement is to prove that $H'<1$)



        Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F'=frac1{2F}cdotfrac{1(x+G)-x(1+G')}{(x+G)^2}>0impliedby G-xG'>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.



        Now this is implied by $$G'=frac1{2G}cdotleft(1+frac{H'x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H'x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H'x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H'x^2-H^2\&impliedby x^2>H'x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H'$. Of course, the latter two are only close approximations to the real distribution of $H$.



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 16 hours ago









        TheSimpliFire

        12.6k62260




        12.6k62260






























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