9 ms·

Alternative notation for exponents, logs and roots? (2011)

177 points by aleyan 6y ago

yunruse 6y ago
While the typesetting might not be very portable or compact, this is beautifully useful for teaching. The two extra identities found appear difficult to prove, but in this notation are blindingly obvious. I wonder if this can be used to apply to other sets of functions, or if the geometry of chaining functions so can be extended to other such geometrically-obvious proofs.
- adamjb 6y ago
  I don't know how hard they would be to come up with ex nihilo, but they're fairly straightforward to prove if you know that logb(x) = log(x)/log(b) and log(x^y)=y * log(x). For logy√z(z) = y logy√z(z) = log(z)/log(z^1/y) = log(z)/(1/y × log(z)) = y * log(z)/log(z) = y For logx(z)√z = x logx(z)√z = z^(1/(logx(z))) log(LHS) = log(z^(1/logx(z)) = log(z)/logx(z) = log(z)/(log(z)/log(x)) = log(x) = log(RHS) Therefore LHS = RHS
- contravariant 6y ago
  The two other identities are only tricky because they're using logarithms with a different base and they're writing their roots with the radical symbol. If you write them using log_b(x) = log(x)/log(b) they become: log(z)/log(z^(1/y)) = y z^(log(x)/log(z)) = x the first is obvious from the fact that log(x^y) = y log(x) and the latter is why log(x)/log(z) is also considered the base z logarithm. The only reason it looks nonobvious is because the notation they chose makes it non-obvious that log_b(y) = 1/log_y(b) (and that the yth root of x is x^(1/y)).
Zhyl 6y ago
I think it's worth linking to the 'Notation as a tool of thought' thread from the other day, as it gave me quite a lot to read, watch and think about [0]. It also reminds me of Graphical Linear Algebra [1] which I occasionally see mentioned here. And, as included in my comment in that thread, the notion of using Tau as the circle constant in equations [2]. Notation is a weird topic to tackle. Like with new technologies or languages on HN, there seem to be those who get [a new notation when it is proposed] and evangelise it, and those who see it as pointless and vocally dismiss it. Posts like the article where you're weighing up and exploring benefits and limitations of notation seem rare - and even those that do exist seem to be pitching for their new notation to be a global replacement rather than as a pedagogical or epistemological tool. [0] https://news.ycombinator.com/item?id=25249563 https://news.ycombinator.com/item?id=25249563 [1] https://graphicallinearalgebra.net/about/ https://graphicallinearalgebra.net/about/ [2] https://tauday.com https://tauday.com
- sriku 6y ago
  While a notation for this case makes for an interesting discussion (which the post is), the "blast radius" of this situation is so small that (to me) it wouldn't be worth adopting any of the proposals (not saying anyone there pushed them). For a notation to be worth adoption,its impact must be far reaching. The "graphical linear algebra" makes a good case with a larger blast radius, for example. However even that has nowhere near the impact of, say, Feynman diagrams. The x^y notation is repurposed elsewhere - ex: power set of a set S is written 2^S .. with no implication that the "logarithm" of the power set to the base 2 is S. Same for matrices raised to a power, where matrices are usually not thought of as a base for doing logarithms. Same for operators in calculus (ex: laplacian .. now that would be confusing to club with a triangle!)
  - alanbernstein 6y ago
    I'm not sure what you mean by blast radius, but if you were to think about the amount of people learning these concepts - as explained in the SE post here - the order would obviously be the reverse of what you suggest (exponents most important).
    - sriku 6y ago
      By "blast radius" I meant the impact of the notation beyond the originating case into other areas. I wasn't referring to the number of people it would impact. The notation in the OP doesn't play well with "nearby" areas like powerset and matrices .. which are well served by the conventional exponentiation notation.
  - umvi 6y ago
    Speaking of Feynman, I remember when he was in high school he invented his own notation[0] for sin/cos/tan to use little angular lines, but then the issue he ran into was that nobody could decipher his math notation and he had trouble helping others learn, so he conformed to the sin/cos/tan notation. [0] https://tex.stackexchange.com/questions/274463/feynman-trig-notation-creating-custom-characters https://tex.stackexchange.com/questions/274463/feynman-trig-...
contravariant 6y ago
Although really the first two already have a unified notation, namely x^y and x^(1/y). There are some minor differences in usage between x^(1/y) and the yth root, but those come down to the fact that the yth root isn't uniquely determined. And there's just 1 logarithm, which has the property log(x^y) = y log(x). You don't need the ones with a different base.
- Jenz 6y ago
  > You don't need the ones with a different base. To add on that, we need only one logarithm in the sense that all other then follows. For any base b, we have log(x; b) = log(x)/log(b)
  - 6gvONxR4sf7o 6y ago
    I think to complete the point, it’s also worth pointing out that for any bases b, c and d: log(x; b) = log(x; c)/log(b; c) = log(x; d)/log(b; d) The second equality is why you can ignore c and d, and just pretend there's one logarithm when you do log(x; b) = log(x)/log(b)
- qsort 6y ago
  Yeah, their assumption that there's a ternary relation between those immediately breaks down when you consider sets other than the real numbers. And you don't even need to go that far, roots as used on the real numbers don't make sense in C. The way I see it is that there's only one fundamental function, which is the exponential function, and log is its inverse. Everything else, including a^b, is syntactic sugar. (If you define exp on C, even sin and cos...) I guess a different notation could have some meaning pedagogically, math notation is incredibly inconsistent at times, but there really is no "deeper truth" here.
  - contravariant 6y ago
    Exponentiation does have some scenarios where it can be defined without an (obvious) exponential function, and roots may not be uniquely defined (or rather they are almost never unique), which means you need to be a bit careful, and which means that in theory the 3rd root could differ from the definition of x^(1/3). However in the cases where you need to be careful most of stuff you'd use the more general notation for wouldn't be applicable anyway, you'd have a high chance of writing down an expression that has no unique value, or can't even be evaluated.
hclimente 6y ago
Related video by 3Blue1Brown: https://youtu.be/EOtduunD9hA https://youtu.be/EOtduunD9hA EDIT: correct link below.
- skovorodkin 6y ago
  "This is the corrected version of the one I put out a month or so ago, in which my animation for all the inverse operations was incorrect": https://www.youtube.com/watch?v=sULa9Lc4pck https://www.youtube.com/watch?v=sULa9Lc4pck.
- kortex 6y ago
  This really hammers home the advantage of this notation. It leads naturally to the question, "What is the operation when we leave the bottom right constant?" which Grant calls "O-plus" (tex call it \oplus), which is in fact related to the harmonic mean (which is n times the o-plus of the terms). I don't know if there's a better term for "o-plus" other than "reciprocal sum of reciprocals". Maybe "optical sum" which kind of makes o-plus make even more sense? https://en.wikipedia.org/wiki/Optic_equation https://en.wikipedia.org/wiki/Optic_equation https://en.wikipedia.org/wiki/List_of_sums_of_reciprocals https://en.wikipedia.org/wiki/List_of_sums_of_reciprocals
  - multidim 6y ago
    > which is in fact related to the harmonic mean (which is n times the o-plus of the terms). I don't know if there's a better term for "o-plus" other than "reciprocal sum of reciprocals" I would call it the "harmonic norm", which is consistent with is being the "norm version" of the harmonic mean. It might also be called the "(p=-1) norm" since it would be a p-norm with p=-1. Also "L-1 norm" to put it in the "L norm" family. https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm
deleted 6y ago
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jonsen 6y ago
So that’s why Johnny can’t subtract. There’s a corresponding problem with + and - . For a + b = c we should of course write the equivalent a = b - c . If 2 + 3 = 5 then 2 = 3 - 5 .
- deleted 6y ago
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zests 6y ago
log_a(b)*log_b(c) = log_a(c) My favorite identity.
jerf 6y ago
I think the triangle notation is actually terrible; it breaks the = symbol. A fully-filled-in triangle with all three components is essentially an equation, or a set of equations, and I can't think of any other (common) situation in which we hide the equality symbol like that. Having only two of them filled out makes me feel like it's a math problem where we're being asked to fill out the rest of the equation, not an operator. I also don't like that this is far from the only set of operations that might fit into a triangle of some sort. In fact I've seen math problems from school using it for + and - already. I haven't seen it for * and / but it's easy to imagine. It's possible this notation is already ruined for teaching students by the common core stuff already in use. And the mere fact that the operators can be arranged in a triangle is not sufficiently unique to give the triangle to this particular set of them. One could argue that the "=" symbol could use a rethink, but I would consider this not a terribly good place to begin that argument just because one set of operators happens to have this particular relationship. Putting up and down arrows under exponents/roots is also not that great; it looks fine when you have one letter above the arrow but it's not going to scale well. I'd happily argue that standard exponentiation doesn't scale particularly well either once the exponents start getting complicated, but putting another symbol below it doesn't help. Putting them as inline operators flows better, but may hide the lede too much, so to speak; while the exponentiation operator we use today may have some issues, at least it's clearly visible. Really, the problem isn't the three of exponents, roots, and log, the problem is just log. The whole "three letter operator" thing seems to have a lot of problems; see also the trig functions and their bizarre standards for sticking powers on them (where -1 is supermagical). That said, there probably isn't a problem large enough to be solvable here because the solution isn't going to be better enough to overcome inertia.
- alisonkisk 6y ago
  There's not much inernia to overcome. It can just be something to show to learners as a visual aid while teaching the standard notation, similar to how kids learn 10 different visual ways to add and multiply.
  - madhadron 6y ago
    > It can just be something to show to learners as a visual aid while teaching the standard notation My son's teacher used number pyramids like that for addition and subtraction a few weeks ago.
amelius 6y ago
I guess you'd want to do the same for the other operations then (plus/minus, mulitply/divide), and perhaps even generalize: https://en.wikipedia.org/wiki/Hyperoperation https://en.wikipedia.org/wiki/Hyperoperation
- macromaniac 6y ago
  Exponentiation isn't commutative, so each position on the triangle matters. For multiplication there would be two equivalent triangle states for instance since you can swap xy=z and yx=z. Not saying its a bad idea though.
  - Someone 6y ago
    It doesn’t make me a fan of this notation, but multiplication isn’t necessarily commutative, either. Matrix multiplication is a simple example (https://en.wikipedia.org/wiki/Commutative_property#Matrix_multiplication https://en.wikipedia.org/wiki/Commutative_property#Matrix_mu...)
mncharity 6y ago
Perhaps adaptive edtech will enable greater use of variant notations? Two decades back, I did toy context where you could pull-down select the notation of a page, and mouseover to see alternate forms. Personalized notation can compromise communication, but like some students benefiting from Feynmanesce drawing of equations in multiple colors, variant notations might be used tactically, for targeted disruption of misconceptions and such.
- alisonkisk 6y ago
  Latex/mathjax can prety much do that. Mathjax already supports toggling between multiple renderers (used mainly for image formats, but could be used for more drastic variants too)
potiuper 6y ago
"Though the ancient Egyptians used heap as a general term for an unknown quantity. Diophantus, a Greek mathematician in Alexandria about 300 AD, was probably the original inventor of an algebra using letters for unknown quantities. Diophantus used the Greek capital letter delta (not for his own name!) for the word power (dynamis; compare dynamo, dynamic, and dynamite), which is therefore one of the oldest terms in mathematics. A conjunction has been used to raise a function to a power. This syntax brings out the parallelism between raising a number to a power and applying a function an equal number of times. The algorithm fails when the number of doublings is further increased." A proposed non-commutative infix binary operator inverse to similar to the non-commutative infix binary exponentiation operator is "[x's] 'log base' [base]": Operator symbols: [] = implicit; vertical orientation / higher potential = implicit increasing position on number "line" : | = addition, - = subtraction; two vertices. triangle "ratio" : ▽ = multiplication, △ = division; three vertices. square "The power of a line is the square of the same line" [x^2] : ◇ = exponentiation, □ = log base; four vertices... [0|]y=y : | = next() grouping operator [0]-y : - = inverse operator [0|]y [|]-y=0 : 0 = identity operand [1▽]y=y : ▽ = | grouping operator [1]△y : △ = inverse operator [1▽]y [▽](1△y)=y△y=1 : 1 = identity operand [y◇(1△y)◇]y=y : ◇ = ▽ grouping operator [y◇(1△y)] □ y = 1△y : □ = inverse operator [y◇(1△y)◇]y [◇]((y◇(1△y)) □ y) =y◇(1△y) : y◇(1△y) = identity operand <in the infinite limit = e>
- deleted 6y ago
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OJFord 6y ago
I'm in slight disbelief that I can't find anyone pointing out that the y'th root of x can also be written x^(1/y) - it must be in there somewhere...
ajkjk 6y ago
Don't particularly care for any of these. "fill-in-the-blank" notations don't really work for operators. You can drop the exponents and write x^y = z as y ln x = ln z, and then everything commutes and is solved nicely. Same idea (until you get to complex numbers, maybe). With multiplication, xy = z is solved by y = z/x or x = z/y, which works becuase of the commutativity. If it wasn't commutative, though, you would need to use left- and right- division: x = z/y but y = x\z (I guess), implying y = (x^-1) z. In the same vein, x^y = z has the radical symbol as a specialized notation to invert it on one side: x = √^y z, which we can parse as a non-commutative operator that acts like f(z) = z^(1/y). But it helps that the raising to a power has an inverse operation that is _also_ raising to a power (x^y)^(1/y) = x. Whereas 'being raised to a power' doesn't have an inverse operation that is also 'being raised to a power'. The other problem is that when you apply a logarithm operator to a term, powers switch to being multiplied. They 'change domains' in a sense. So it's not possible to do anything to the 'x' in x^y on its own, because that would result in f(x)^y which is still exponentiating by y. You need the 'y' to 'move' into the main line of the equation, out of the exponent. I think a good way to model this would be to imagine allowing x^y = z shifting so that the 'y' is the main line of the equation, becoming something like 1_x y = 1_z. 1_x and 1_z would ideally have the subscript on the left side, to avoid confusion with other uses of subscripts, and to look like a shifted version of x^1 and z^1. These are literally log x and log z in some base, but they're just numbers, so you can solve the equation as y = 1_z/1_x. Then you have identities like x^1_x = e, so x^(1_z/1_x) = e^(1_z) = z. I think you just do away with the notation log_x z entirely; it's too odd compared to everything else. So basically I propose y = 1_z/1_x, but I don't think you can reconcile this with the square root notation at all, as they're too different. But it does, at least, keep things consistent with using a division operation for the inverse, akin to x = z^(1/y).
diffeomorphism 6y ago
We have a unified notation already? x^y = z x = z^{1/y} \log(x) y = \log(z) Some of these generalize well to complex numbers/matrices/groups/flows/etc. some don't.
- cuspycode 6y ago
  I would like a more unified notation for the concept of "inverse". But I have no good suggestions for how to implement that. However, I have some bad suggestions just to illustrate what I mean: INV(*) x, instead of 1/x INV(+) x, instead of -x INV(f) x, instead of f^{-1}(x) The last one should really be (INV(°) f) x = f^{-1}(x), where ° denotes function composition as the group operator. But involving this operator in the notation would probably be overkill in most circumstances.
syrrim 6y ago
Ideally, we would treat e^x and ln(x) = log~e(x) as the default. We know that a^x = e^(x ln a), and log~a(x) = ln x / ln a. So if we introduced an operator ^(x) = e^x, and another v(x) = ln(x), we could write a^x = ^xva, a√x = ^(vx/a), log~a(x) = vx/va. Common identities would be written thus: ^xva * ^yva = ^(xva + yva) = ^(x + y)va ^yv(^xva) = ^yxva (note that this is just the identity v(^x) = x) ln(a^x) = x ln a is just v(^xva) = xva ^yvx = z <=> yvx = vz <=> y = vz/vx ^yvx = z <=> yvx = vz <=> vx = vz/y <=> x = ^(vz/y) (alternate derivation:) ^yvx = z <=> ^(v^yvx/y) = ^(vz/y) <=> x = ^(vz/y) Differentials are thus: D(^f) = Df^f D(vf) = Df/f d/dx(^xva) = va^xva d/dx(^nvx) = d/dx(nvx)^nvx = n/x * ^nvx = n ^-1vx ^nvx = n^(n-1)vx d/dx(vx/va) = 1/xva
bandie91 6y ago
i vote for this notation: b^p -> b^p \root p \of x -> x^(1/p) {\log_b} x -> b^? x

177 points by aleyan 6y ago

While the typesetting might not be very portable or compact, this is beautifully useful for teaching. The two extra identities found appear difficult to prove, but in this notation are blindingly obvious. I wonder if this can be used to apply to other sets of functions, or if the geometry of chaining functions so can be extended to other such geometrically-obvious proofs.

adamjb 6y ago
I don't know how hard they would be to come up with ex nihilo, but they're fairly straightforward to prove if you know that logb(x) = log(x)/log(b) and log(x^y)=y * log(x). For logy√z(z) = y logy√z(z) = log(z)/log(z^1/y) = log(z)/(1/y × log(z)) = y * log(z)/log(z) = y For logx(z)√z = x logx(z)√z = z^(1/(logx(z))) log(LHS) = log(z^(1/logx(z)) = log(z)/logx(z) = log(z)/(log(z)/log(x)) = log(x) = log(RHS) Therefore LHS = RHS
contravariant 6y ago
The two other identities are only tricky because they're using logarithms with a different base and they're writing their roots with the radical symbol. If you write them using log_b(x) = log(x)/log(b) they become: log(z)/log(z^(1/y)) = y z^(log(x)/log(z)) = x the first is obvious from the fact that log(x^y) = y log(x) and the latter is why log(x)/log(z) is also considered the base z logarithm. The only reason it looks nonobvious is because the notation they chose makes it non-obvious that log_b(y) = 1/log_y(b) (and that the yth root of x is x^(1/y)).

Zhyl 6y ago

I think it's worth linking to the 'Notation as a tool of thought' thread from the other day, as it gave me quite a lot to read, watch and think about [0]. It also reminds me of Graphical Linear Algebra [1] which I occasionally see mentioned here. And, as included in my comment in that thread, the notion of using Tau as the circle constant in equations [2]. Notation is a weird topic to tackle. Like with new technologies or languages on HN, there seem to be those who get [a new notation when it is proposed] and evangelise it, and those who see it as pointless and vocally dismiss it. Posts like the article where you're weighing up and exploring benefits and limitations of notation seem rare - and even those that do exist seem to be pitching for their new notation to be a global replacement rather than as a pedagogical or epistemological tool. [0] https://news.ycombinator.com/item?id=25249563 https://news.ycombinator.com/item?id=25249563 [1] https://graphicallinearalgebra.net/about/ https://graphicallinearalgebra.net/about/ [2] https://tauday.com https://tauday.com

sriku 6y ago
While a notation for this case makes for an interesting discussion (which the post is), the "blast radius" of this situation is so small that (to me) it wouldn't be worth adopting any of the proposals (not saying anyone there pushed them). For a notation to be worth adoption,its impact must be far reaching. The "graphical linear algebra" makes a good case with a larger blast radius, for example. However even that has nowhere near the impact of, say, Feynman diagrams. The x^y notation is repurposed elsewhere - ex: power set of a set S is written 2^S .. with no implication that the "logarithm" of the power set to the base 2 is S. Same for matrices raised to a power, where matrices are usually not thought of as a base for doing logarithms. Same for operators in calculus (ex: laplacian .. now that would be confusing to club with a triangle!)
- alanbernstein 6y ago
  I'm not sure what you mean by blast radius, but if you were to think about the amount of people learning these concepts - as explained in the SE post here - the order would obviously be the reverse of what you suggest (exponents most important).
  - sriku 6y ago
    By "blast radius" I meant the impact of the notation beyond the originating case into other areas. I wasn't referring to the number of people it would impact. The notation in the OP doesn't play well with "nearby" areas like powerset and matrices .. which are well served by the conventional exponentiation notation.
- umvi 6y ago
  Speaking of Feynman, I remember when he was in high school he invented his own notation[0] for sin/cos/tan to use little angular lines, but then the issue he ran into was that nobody could decipher his math notation and he had trouble helping others learn, so he conformed to the sin/cos/tan notation. [0] https://tex.stackexchange.com/questions/274463/feynman-trig-notation-creating-custom-characters https://tex.stackexchange.com/questions/274463/feynman-trig-...

contravariant 6y ago

Although really the first two already have a unified notation, namely x^y and x^(1/y). There are some minor differences in usage between x^(1/y) and the yth root, but those come down to the fact that the yth root isn't uniquely determined. And there's just 1 logarithm, which has the property log(x^y) = y log(x). You don't need the ones with a different base.

Jenz 6y ago
> You don't need the ones with a different base. To add on that, we need only one logarithm in the sense that all other then follows. For any base b, we have log(x; b) = log(x)/log(b)
- 6gvONxR4sf7o 6y ago
  I think to complete the point, it’s also worth pointing out that for any bases b, c and d: log(x; b) = log(x; c)/log(b; c) = log(x; d)/log(b; d) The second equality is why you can ignore c and d, and just pretend there's one logarithm when you do log(x; b) = log(x)/log(b)
qsort 6y ago
Yeah, their assumption that there's a ternary relation between those immediately breaks down when you consider sets other than the real numbers. And you don't even need to go that far, roots as used on the real numbers don't make sense in C. The way I see it is that there's only one fundamental function, which is the exponential function, and log is its inverse. Everything else, including a^b, is syntactic sugar. (If you define exp on C, even sin and cos...) I guess a different notation could have some meaning pedagogically, math notation is incredibly inconsistent at times, but there really is no "deeper truth" here.
- contravariant 6y ago
  Exponentiation does have some scenarios where it can be defined without an (obvious) exponential function, and roots may not be uniquely defined (or rather they are almost never unique), which means you need to be a bit careful, and which means that in theory the 3rd root could differ from the definition of x^(1/3). However in the cases where you need to be careful most of stuff you'd use the more general notation for wouldn't be applicable anyway, you'd have a high chance of writing down an expression that has no unique value, or can't even be evaluated.

hclimente 6y ago

Related video by 3Blue1Brown: https://youtu.be/EOtduunD9hA https://youtu.be/EOtduunD9hA EDIT: correct link below.

skovorodkin 6y ago
"This is the corrected version of the one I put out a month or so ago, in which my animation for all the inverse operations was incorrect": https://www.youtube.com/watch?v=sULa9Lc4pck https://www.youtube.com/watch?v=sULa9Lc4pck.
kortex 6y ago
This really hammers home the advantage of this notation. It leads naturally to the question, "What is the operation when we leave the bottom right constant?" which Grant calls "O-plus" (tex call it \oplus), which is in fact related to the harmonic mean (which is n times the o-plus of the terms). I don't know if there's a better term for "o-plus" other than "reciprocal sum of reciprocals". Maybe "optical sum" which kind of makes o-plus make even more sense? https://en.wikipedia.org/wiki/Optic_equation https://en.wikipedia.org/wiki/Optic_equation https://en.wikipedia.org/wiki/List_of_sums_of_reciprocals https://en.wikipedia.org/wiki/List_of_sums_of_reciprocals
- multidim 6y ago
  > which is in fact related to the harmonic mean (which is n times the o-plus of the terms). I don't know if there's a better term for "o-plus" other than "reciprocal sum of reciprocals" I would call it the "harmonic norm", which is consistent with is being the "norm version" of the harmonic mean. It might also be called the "(p=-1) norm" since it would be a p-norm with p=-1. Also "L-1 norm" to put it in the "L norm" family. https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm

deleted 6y ago

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jonsen 6y ago

So that’s why Johnny can’t subtract. There’s a corresponding problem with + and - . For a + b = c we should of course write the equivalent a = b - c . If 2 + 3 = 5 then 2 = 3 - 5 .

deleted 6y ago
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zests 6y ago

log_a(b)*log_b(c) = log_a(c) My favorite identity.

jerf 6y ago

I think the triangle notation is actually terrible; it breaks the = symbol. A fully-filled-in triangle with all three components is essentially an equation, or a set of equations, and I can't think of any other (common) situation in which we hide the equality symbol like that. Having only two of them filled out makes me feel like it's a math problem where we're being asked to fill out the rest of the equation, not an operator. I also don't like that this is far from the only set of operations that might fit into a triangle of some sort. In fact I've seen math problems from school using it for + and - already. I haven't seen it for * and / but it's easy to imagine. It's possible this notation is already ruined for teaching students by the common core stuff already in use. And the mere fact that the operators can be arranged in a triangle is not sufficiently unique to give the triangle to this particular set of them. One could argue that the "=" symbol could use a rethink, but I would consider this not a terribly good place to begin that argument just because one set of operators happens to have this particular relationship. Putting up and down arrows under exponents/roots is also not that great; it looks fine when you have one letter above the arrow but it's not going to scale well. I'd happily argue that standard exponentiation doesn't scale particularly well either once the exponents start getting complicated, but putting another symbol below it doesn't help. Putting them as inline operators flows better, but may hide the lede too much, so to speak; while the exponentiation operator we use today may have some issues, at least it's clearly visible. Really, the problem isn't the three of exponents, roots, and log, the problem is just log. The whole "three letter operator" thing seems to have a lot of problems; see also the trig functions and their bizarre standards for sticking powers on them (where -1 is supermagical). That said, there probably isn't a problem large enough to be solvable here because the solution isn't going to be better enough to overcome inertia.

alisonkisk 6y ago
There's not much inernia to overcome. It can just be something to show to learners as a visual aid while teaching the standard notation, similar to how kids learn 10 different visual ways to add and multiply.
- madhadron 6y ago
  > It can just be something to show to learners as a visual aid while teaching the standard notation My son's teacher used number pyramids like that for addition and subtraction a few weeks ago.

amelius 6y ago

I guess you'd want to do the same for the other operations then (plus/minus, mulitply/divide), and perhaps even generalize: https://en.wikipedia.org/wiki/Hyperoperation https://en.wikipedia.org/wiki/Hyperoperation

macromaniac 6y ago
Exponentiation isn't commutative, so each position on the triangle matters. For multiplication there would be two equivalent triangle states for instance since you can swap xy=z and yx=z. Not saying its a bad idea though.
- Someone 6y ago
  It doesn’t make me a fan of this notation, but multiplication isn’t necessarily commutative, either. Matrix multiplication is a simple example (https://en.wikipedia.org/wiki/Commutative_property#Matrix_multiplication https://en.wikipedia.org/wiki/Commutative_property#Matrix_mu...)

mncharity 6y ago

Perhaps adaptive edtech will enable greater use of variant notations? Two decades back, I did toy context where you could pull-down select the notation of a page, and mouseover to see alternate forms. Personalized notation can compromise communication, but like some students benefiting from Feynmanesce drawing of equations in multiple colors, variant notations might be used tactically, for targeted disruption of misconceptions and such.

alisonkisk 6y ago
Latex/mathjax can prety much do that. Mathjax already supports toggling between multiple renderers (used mainly for image formats, but could be used for more drastic variants too)

potiuper 6y ago

"Though the ancient Egyptians used heap as a general term for an unknown quantity. Diophantus, a Greek mathematician in Alexandria about 300 AD, was probably the original inventor of an algebra using letters for unknown quantities. Diophantus used the Greek capital letter delta (not for his own name!) for the word power (dynamis; compare dynamo, dynamic, and dynamite), which is therefore one of the oldest terms in mathematics. A conjunction has been used to raise a function to a power. This syntax brings out the parallelism between raising a number to a power and applying a function an equal number of times. The algorithm fails when the number of doublings is further increased." A proposed non-commutative infix binary operator inverse to similar to the non-commutative infix binary exponentiation operator is "[x's] 'log base' [base]": Operator symbols: [] = implicit; vertical orientation / higher potential = implicit increasing position on number "line" : | = addition, - = subtraction; two vertices. triangle "ratio" : ▽ = multiplication, △ = division; three vertices. square "The power of a line is the square of the same line" [x^2] : ◇ = exponentiation, □ = log base; four vertices... [0|]y=y : | = next() grouping operator [0]-y : - = inverse operator [0|]y [|]-y=0 : 0 = identity operand [1▽]y=y : ▽ = | grouping operator [1]△y : △ = inverse operator [1▽]y [▽](1△y)=y△y=1 : 1 = identity operand [y◇(1△y)◇]y=y : ◇ = ▽ grouping operator [y◇(1△y)] □ y = 1△y : □ = inverse operator [y◇(1△y)◇]y [◇]((y◇(1△y)) □ y) =y◇(1△y) : y◇(1△y) = identity operand <in the infinite limit = e>

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OJFord 6y ago

I'm in slight disbelief that I can't find anyone pointing out that the y'th root of x can also be written x^(1/y) - it must be in there somewhere...

ajkjk 6y ago

Don't particularly care for any of these. "fill-in-the-blank" notations don't really work for operators. You can drop the exponents and write x^y = z as y ln x = ln z, and then everything commutes and is solved nicely. Same idea (until you get to complex numbers, maybe). With multiplication, xy = z is solved by y = z/x or x = z/y, which works becuase of the commutativity. If it wasn't commutative, though, you would need to use left- and right- division: x = z/y but y = x\z (I guess), implying y = (x^-1) z. In the same vein, x^y = z has the radical symbol as a specialized notation to invert it on one side: x = √^y z, which we can parse as a non-commutative operator that acts like f(z) = z^(1/y). But it helps that the raising to a power has an inverse operation that is _also_ raising to a power (x^y)^(1/y) = x. Whereas 'being raised to a power' doesn't have an inverse operation that is also 'being raised to a power'. The other problem is that when you apply a logarithm operator to a term, powers switch to being multiplied. They 'change domains' in a sense. So it's not possible to do anything to the 'x' in x^y on its own, because that would result in f(x)^y which is still exponentiating by y. You need the 'y' to 'move' into the main line of the equation, out of the exponent. I think a good way to model this would be to imagine allowing x^y = z shifting so that the 'y' is the main line of the equation, becoming something like 1_x y = 1_z. 1_x and 1_z would ideally have the subscript on the left side, to avoid confusion with other uses of subscripts, and to look like a shifted version of x^1 and z^1. These are literally log x and log z in some base, but they're just numbers, so you can solve the equation as y = 1_z/1_x. Then you have identities like x^1_x = e, so x^(1_z/1_x) = e^(1_z) = z. I think you just do away with the notation log_x z entirely; it's too odd compared to everything else. So basically I propose y = 1_z/1_x, but I don't think you can reconcile this with the square root notation at all, as they're too different. But it does, at least, keep things consistent with using a division operation for the inverse, akin to x = z^(1/y).

diffeomorphism 6y ago

We have a unified notation already? x^y = z x = z^{1/y} \log(x) y = \log(z) Some of these generalize well to complex numbers/matrices/groups/flows/etc. some don't.

cuspycode 6y ago
I would like a more unified notation for the concept of "inverse". But I have no good suggestions for how to implement that. However, I have some bad suggestions just to illustrate what I mean: INV(*) x, instead of 1/x INV(+) x, instead of -x INV(f) x, instead of f^{-1}(x) The last one should really be (INV(°) f) x = f^{-1}(x), where ° denotes function composition as the group operator. But involving this operator in the notation would probably be overkill in most circumstances.

syrrim 6y ago

Ideally, we would treat e^x and ln(x) = log~e(x) as the default. We know that a^x = e^(x ln a), and log~a(x) = ln x / ln a. So if we introduced an operator ^(x) = e^x, and another v(x) = ln(x), we could write a^x = ^xva, a√x = ^(vx/a), log~a(x) = vx/va. Common identities would be written thus: ^xva * ^yva = ^(xva + yva) = ^(x + y)va ^yv(^xva) = ^yxva (note that this is just the identity v(^x) = x) ln(a^x) = x ln a is just v(^xva) = xva ^yvx = z <=> yvx = vz <=> y = vz/vx ^yvx = z <=> yvx = vz <=> vx = vz/y <=> x = ^(vz/y) (alternate derivation:) ^yvx = z <=> ^(v^yvx/y) = ^(vz/y) <=> x = ^(vz/y) Differentials are thus: D(^f) = Df^f D(vf) = Df/f d/dx(^xva) = va^xva d/dx(^nvx) = d/dx(nvx)^nvx = n/x * ^nvx = n ^-1vx ^nvx = n^(n-1)vx d/dx(vx/va) = 1/xva

bandie91 6y ago

i vote for this notation: b^p -> b^p \root p \of x -> x^(1/p) {\log_b} x -> b^? x