Hi all, I was hoping someone would respond to this thread, as the original puzzle seems natural and elementary enough, yet I don't understand the constant that apparently comes out of it. [And my result differs from what CL suggests. See below.] So maybe someone can explain... For large n, the recurrence f(n+1) = (1 + 1/f(n))^n will typically alternate between large and small values. (By small, I mean values close to 1.) The large values will either occur on the even indices or the odd indices depending on the initial choice of f(1). If you begin large, e.g., choose f(1)=100, then the alternation clearly continues with large values on the odd indices, but if you begin small, e.g., f(1)=1/100, then the alternation continues with large values on the even indices. The definition is a continuous function of f(1), so one wonders what happens in the middle where the two behaviors meet. Can it approach infinity “smoothly”? (scare quotes because it is discrete in n) I analyze it as follows: If the series is not alternating between large and small values, then the smooth transition between the two behaviors approximately satisfies f(n+1)=f(n). Letting x=f(n), we need to solve x=(1+1/x)^n. Multiply both sides by x^n and take logs to get: x^(n+1) = (x+1)^n (n+1) log(x) = n log(x+1) (n+1)/n = log(x+1)/log(x) Approximating log(x+1) with one term based on its derivative: (n+1)/n = (log(x) + 1/x) / log(x) which is easily manipulated to get: n = x log(x) This can't be solved for x with elementary functions, but the Lambert W function (a.k.a. Omega function or Product Logarithm) is defined as the inverse of x*e^x, so the solution is: x = n/W(n) In terms of the original problem, the asymptotic solution that doesn't alternate between large and small is then: f(n) = n/W(n) for large n What I don't understand is where you have to start when choosing f(1) in order to get there. A quick numerical search with floating point calculations gives me the critical value approximately: f(1) = 1.1874523511 Below is my Mathematica code. (I just ran it repeatedly, increasing the initial value if f[40] was large and decreasing it if f[40] was small.) Examining the plot shows this initial value is accurate enough to postpone the alternating behavior for about 40 terms and that the curve matches the asymptotic solution very well. (Note: ProductLog[] is the Mathematica syntax for the W function.) F[1] = 1.18745235112; f[n_] := (1+1/f[n-1])^(n-1) //N; ListPlot[{Table[{i,f[i]},{i,1,40}], Table[{i+1,i/ProductLog[i]},{i,1,40}] }] But what more can be said about f(1)? I don't know how one might back out from the asymptotic solution an exact starting value. I don't find the value 1.187452351 in the online inverse symbolic calculator as having a simple expression. I don't understand why my value is different from the 1.84 value that CL reported. Can anyone shed light on all this for me? George http://georgehart.com On 9/1/2020 7:25 PM, christopher landauer wrote:

hihi, all -

since i didn't know what to do with this question, i wrote a program to generate some values;

i assumed that if f(n) is really large, then the term in parentheses is only a little more than 1,

so that f(n+1) may not be very large (depending on n, so a kind of slow divergence might be possible)

it turned out much weirder than that -

there appears to be a threshold t ~ 1.84 (i have more digits) with the property that

if f(1) > t, then the sequence eventually alternates

    f(n) = 1 for odd n, f(n) = 2 ^ n for even n,

and if f(1) < t, then the sequence eventually alternates

    f(n) = 1 for even n, f(n) = 2 ^ n for odd n

the reason this is only partly a spoiler is that this depends on finite precision computation,

so that when n is large enough,

    1 + 1/2^n = 1,

and the rest follows easily,

and of course that means that this might be completely bogus for the actual sequence

but this might stimulate some thought (and the threshold seems strange)

more later,

chris

On 2020-08-30 19:12, Dan Asimov wrote:

...

I found this interesting question in an undisclosed location:

Let f : Z+  —>  R be a function such that

a)   f(1) > 0

and

b)   f(n+1)  =   (1 + 1/f(n))^n

for all n >= 1.

Is it possible that f(n) —> oo as n —> oo ???

—Dan

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

If you allow for f(1) < 0 and if I did my analysis correctly, then I see a point similar to George's 1.187452351 at f(1) ~ -0.71774114214285. Taking this problem as a fractal iteration: z(n+1) = (1+1/z(n))^n with z(1) = pixel, then there is a fractal contour connecting the points 1.187452351+0i and -0.71774114214285+0i. Kerry On Fri, Sep 4, 2020 at 8:14 AM George Hart <george@georgehart.com> wrote:

Hi all,

I was hoping someone would respond to this thread, as the original puzzle seems natural and elementary enough, yet I don't understand the constant that apparently comes out of it. [And my result differs from what CL suggests. See below.] So maybe someone can explain...

For large n, the recurrence f(n+1) = (1 + 1/f(n))^n will typically alternate between large and small values. (By small, I mean values close to 1.) The large values will either occur on the even indices or the odd indices depending on the initial choice of f(1). If you begin large, e.g., choose f(1)=100, then the alternation clearly continues with large values on the odd indices, but if you begin small, e.g., f(1)=1/100, then the alternation continues with large values on the even indices. The definition is a continuous function of f(1), so one wonders what happens in the middle where the two behaviors meet. Can it approach infinity “smoothly”? (scare quotes because it is discrete in n)

I analyze it as follows: If the series is not alternating between large and small values, then the smooth transition between the two behaviors approximately satisfies f(n+1)=f(n). Letting x=f(n), we need to solve x=(1+1/x)^n.

Multiply both sides by x^n and take logs to get:

x^(n+1) = (x+1)^n

(n+1) log(x) = n log(x+1)

(n+1)/n = log(x+1)/log(x)

Approximating log(x+1) with one term based on its derivative:

(n+1)/n = (log(x) + 1/x) / log(x)

which is easily manipulated to get:

n = x log(x)

This can't be solved for x with elementary functions, but the Lambert W function (a.k.a. Omega function or Product Logarithm) is defined as the inverse of x*e^x, so the solution is:

x = n/W(n)

In terms of the original problem, the asymptotic solution that doesn't alternate between large and small is then:

f(n) = n/W(n) for large n

What I don't understand is where you have to start when choosing f(1) in order to get there. A quick numerical search with floating point calculations gives me the critical value approximately:

f(1) = 1.1874523511

Below is my Mathematica code. (I just ran it repeatedly, increasing the initial value if f[40] was large and decreasing it if f[40] was small.) Examining the plot shows this initial value is accurate enough to postpone the alternating behavior for about 40 terms and that the curve matches the asymptotic solution very well. (Note: ProductLog[] is the Mathematica syntax for the W function.)

F[1] = 1.18745235112; f[n_] := (1+1/f[n-1])^(n-1) //N; ListPlot[{Table[{i,f[i]},{i,1,40}], Table[{i+1,i/ProductLog[i]},{i,1,40}] }]

But what more can be said about f(1)? I don't know how one might back out from the asymptotic solution an exact starting value.

I don't find the value 1.187452351 in the online inverse symbolic calculator as having a simple expression.

I don't understand why my value is different from the 1.84 value that CL reported.

Can anyone shed light on all this for me?

George http://georgehart.com

On 9/1/2020 7:25 PM, christopher landauer wrote:

...

hihi, all -

since i didn't know what to do with this question, i wrote a program to generate some values;

i assumed that if f(n) is really large, then the term in parentheses is only a little more than 1,

so that f(n+1) may not be very large (depending on n, so a kind of slow divergence might be possible)

it turned out much weirder than that -

there appears to be a threshold t ~ 1.84 (i have more digits) with the property that

if f(1) > t, then the sequence eventually alternates

f(n) = 1 for odd n, f(n) = 2 ^ n for even n,

and if f(1) < t, then the sequence eventually alternates

f(n) = 1 for even n, f(n) = 2 ^ n for odd n

the reason this is only partly a spoiler is that this depends on finite precision computation,

so that when n is large enough,

1 + 1/2^n = 1,

and the rest follows easily,

and of course that means that this might be completely bogus for the actual sequence

but this might stimulate some thought (and the threshold seems strange)

more later,

chris

On 2020-08-30 19:12, Dan Asimov wrote:

...

I found this interesting question in an undisclosed location:

Let f : Z+ —> R be a function such that

a) f(1) > 0

and

b) f(n+1) = (1 + 1/f(n))^n

for all n >= 1.

Is it possible that f(n) —> oo as n —> oo ???

—Dan

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

If anyone is interested, here is the fractal boundary that I found looking at the Foias' constant iteration using complex numbers. http://www.kerrymitchellart.com/temp/foias-constant.htm Kerry On Fri, Sep 4, 2020 at 11:17 AM Kerry Mitchell <lkmitch@gmail.com> wrote:

If you allow for f(1) < 0 and if I did my analysis correctly, then I see a point similar to George's 1.187452351 at f(1) ~ -0.71774114214285.

Taking this problem as a fractal iteration:

z(n+1) = (1+1/z(n))^n

with z(1) = pixel, then there is a fractal contour connecting the points 1.187452351+0i and -0.71774114214285+0i.

Kerry

On Fri, Sep 4, 2020 at 8:14 AM George Hart <george@georgehart.com> wrote:

...

Hi all,

I was hoping someone would respond to this thread, as the original puzzle seems natural and elementary enough, yet I don't understand the constant that apparently comes out of it. [And my result differs from what CL suggests. See below.] So maybe someone can explain...

For large n, the recurrence f(n+1) = (1 + 1/f(n))^n will typically alternate between large and small values. (By small, I mean values close to 1.) The large values will either occur on the even indices or the odd indices depending on the initial choice of f(1). If you begin large, e.g., choose f(1)=100, then the alternation clearly continues with large values on the odd indices, but if you begin small, e.g., f(1)=1/100, then the alternation continues with large values on the even indices. The definition is a continuous function of f(1), so one wonders what happens in the middle where the two behaviors meet. Can it approach infinity “smoothly”? (scare quotes because it is discrete in n)

I analyze it as follows: If the series is not alternating between large and small values, then the smooth transition between the two behaviors approximately satisfies f(n+1)=f(n). Letting x=f(n), we need to solve x=(1+1/x)^n.

Multiply both sides by x^n and take logs to get:

x^(n+1) = (x+1)^n

(n+1) log(x) = n log(x+1)

(n+1)/n = log(x+1)/log(x)

Approximating log(x+1) with one term based on its derivative:

(n+1)/n = (log(x) + 1/x) / log(x)

which is easily manipulated to get:

n = x log(x)

This can't be solved for x with elementary functions, but the Lambert W function (a.k.a. Omega function or Product Logarithm) is defined as the inverse of x*e^x, so the solution is:

x = n/W(n)

In terms of the original problem, the asymptotic solution that doesn't alternate between large and small is then:

f(n) = n/W(n) for large n

What I don't understand is where you have to start when choosing f(1) in order to get there. A quick numerical search with floating point calculations gives me the critical value approximately:

f(1) = 1.1874523511

Below is my Mathematica code. (I just ran it repeatedly, increasing the initial value if f[40] was large and decreasing it if f[40] was small.) Examining the plot shows this initial value is accurate enough to postpone the alternating behavior for about 40 terms and that the curve matches the asymptotic solution very well. (Note: ProductLog[] is the Mathematica syntax for the W function.)

F[1] = 1.18745235112; f[n_] := (1+1/f[n-1])^(n-1) //N; ListPlot[{Table[{i,f[i]},{i,1,40}], Table[{i+1,i/ProductLog[i]},{i,1,40}] }]

But what more can be said about f(1)? I don't know how one might back out from the asymptotic solution an exact starting value.

I don't find the value 1.187452351 in the online inverse symbolic calculator as having a simple expression.

I don't understand why my value is different from the 1.84 value that CL reported.

Can anyone shed light on all this for me?

George http://georgehart.com

On 9/1/2020 7:25 PM, christopher landauer wrote:

...

hihi, all -

since i didn't know what to do with this question, i wrote a program to generate some values;

i assumed that if f(n) is really large, then the term in parentheses is only a little more than 1,

so that f(n+1) may not be very large (depending on n, so a kind of slow divergence might be possible)

it turned out much weirder than that -

there appears to be a threshold t ~ 1.84 (i have more digits) with the property that

if f(1) > t, then the sequence eventually alternates

f(n) = 1 for odd n, f(n) = 2 ^ n for even n,

and if f(1) < t, then the sequence eventually alternates

f(n) = 1 for even n, f(n) = 2 ^ n for odd n

the reason this is only partly a spoiler is that this depends on finite precision computation,

so that when n is large enough,

1 + 1/2^n = 1,

and the rest follows easily,

and of course that means that this might be completely bogus for the actual sequence

but this might stimulate some thought (and the threshold seems strange)

more later,

chris

On 2020-08-30 19:12, Dan Asimov wrote:

...

I found this interesting question in an undisclosed location:

Let f : Z+ —> R be a function such that

a) f(1) > 0

and

b) f(n+1) = (1 + 1/f(n))^n

for all n >= 1.

Is it possible that f(n) —> oo as n —> oo ???

—Dan

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun