Tetration of non-integers: is there something wrong with this approach?
$begingroup$
I'm trying to figure out a formula for tetration that will work for
non-integer heights.
I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:
$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$
I also know that $x^y=e^{y ln x}$ for positive $x$.
I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$
Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.
So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?
Thank you.
functions exponentiation tetration
asked Dec 5 '18 at 2:34
user3412516user3412516
183
$endgroup$
add a comment |
$begingroup$
I'm trying to figure out a formula for tetration that will work for
non-integer heights.
I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:
$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$
I also know that $x^y=e^{y ln x}$ for positive $x$.
I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$
Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.
So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?
Thank you.
functions exponentiation tetration
asked Dec 5 '18 at 2:34
user3412516user3412516
183
$endgroup$
$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51
add a comment |
$begingroup$
I'm trying to figure out a formula for tetration that will work for
non-integer heights.
I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:
$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$
I also know that $x^y=e^{y ln x}$ for positive $x$.
I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$
Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.
So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?
Thank you.
functions exponentiation tetration
asked Dec 5 '18 at 2:34
user3412516user3412516
183
$endgroup$
I'm trying to figure out a formula for tetration that will work for
non-integer heights.
I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:
$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$
I also know that $x^y=e^{y ln x}$ for positive $x$.
I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$
Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.
So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?
Thank you.
functions exponentiation tetration
functions exponentiation tetration
asked Dec 5 '18 at 2:34
user3412516user3412516
183
asked Dec 5 '18 at 2:34
user3412516user3412516
183
asked Dec 5 '18 at 2:34
user3412516user3412516
183
asked Dec 5 '18 at 2:34
user3412516user3412516
183
asked Dec 5 '18 at 2:34
user3412516user3412516
183
183
$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51
add a comment |
$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51
$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51
$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.
Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
add a comment |
$begingroup$
Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.
Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
add a comment |
$begingroup$
The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.
Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
add a comment |
$begingroup$
The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.
Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.
Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
answered Dec 5 '18 at 2:58
Milo BrandtMilo Brandt
39.6k475139
39.6k475139
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
add a comment |
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
$endgroup$
– Milo Brandt
Dec 5 '18 at 3:02
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
$begingroup$
I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
$endgroup$
– user3412516
Dec 10 '18 at 19:32
add a comment |
$begingroup$
Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
$endgroup$
add a comment |
$begingroup$
Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
$endgroup$
add a comment |
$begingroup$
Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
$endgroup$
Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
answered Dec 5 '18 at 7:22
Gottfried HelmsGottfried Helms
23.3k24598
23.3k24598
add a comment |
add a comment |
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$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51