Everyone knows about the simple operation of addition. And it is also simple that multiplication is merely repeated addition (3*4=3+3+3+3). It is also simple that exponentiation is simply repeated multiplication e.g. 3^4=3*3*3*3. But less people know about the function of repeated exponentiation. This function is called the hyper 4 operator. The hyper 4 operator can be written as x^^y. x^^y equals x^x^x^x... with y x's.

Note: In this post the notation (^^) will represent the hyper 4 operator. Similar notation has be used for other functions. Please do not confuse them.

The hyper 4 operator has two types: lower and upper. The only difference between the lower and higher hyper 4 operators is the placement of the parentheses. For example, 3^^4, with the lower hyper 4 operator equals (3^3)^3)^3=7625597484987 while 3^^4 with the upper hyper 4 operator equals 3^(3^(3^3) which has over a trillion digits! As you can see, the upper hyper 4 operator produces much higher numbers than the lower hyper 4 operator. In this post, assume that all hyper 4 operators are upper, unless otherwise specified.

The hyper 4 operator can also be applied to non-integers. For example 3.5^^5=3.5^3.5^3.5^3.5^3.5. The real difficulty is repeating an exponent fractional times like in 3^^3.5. There is a simple solution to this, however. The solution is 3^(3^(3^(3^.5). Notice that the fourth three isn't whole, but taken to the .5 power. The 3^.5 represents half of a three. Another example is 5^^2.84. The solution is 5^(5^(5^.84) with 5^.84 representing .84 of a three. Even 3^^π can be calculated using an approximation of pi like 3.14 or 3.1415. For more info on pi see here. There are alternative approaches to the generalization of the hyper 4 operator. For more details see the comments below this post.

The hyper 4 operator can be applied to negative numbers as well. As with the previous example, having a negative number for x (in the equation x^^y) poses no issue (-2^^3=-2^(-2^-2), but for y, there is a real dilemma. How can there be a power tower with a negative number of elements in the tower? Just as 3^.5 can represent half a three, 3^-1 can represent a negative three. Using this rule, 3^^-2 would equal (3^-1)^(3^-1) or simply (1/3)^(1/3). Therefore, if x and y are positive numbers, x^^-y equals 1/x^(1/x^...1/x) with y 1/x's.

The rules above can be used to calculate the hyper 4 operator with any two real numbers.

If all functions are named this way, addition would be the hyper 1 operator, multiplication would be the hyper 2 operator, and exponentiation would be the hyper 3 operator. But what about higher operators like the hyper 5 operator? We can create the notation x^^^y for the hyper 5 operator. But what about other operators? How can 4^^^5.6 or even 8^^^^^^7 be solved? Using the rule of hyper operators, the hyper 5 operator must be repeated hyper 4 operators. As with the previous example, there are lower and upper hyper 5 operators. For example, for the lower hyper 5 operator, 3^^^4 would equal (3^^3)^^3)^^3, and the higher hyper 5 operator would be 3^^^4=3^^(3^^(3^^3). Beyond that, the hyper 6 operator would be repeated hyper 5 operators and so on. Higher hyper operators can also be written x^(y)^z with y being the number of karats used (2 karats for the hyper 4 operator, 3 for the hyper 5 operator etc.)

Many of these expressions are much to high for number form and must be expressed in scientific notation. But for things like 3^^65, the power of ten (e.g. x*10^y) will be too high to represent. In this case, a number can be represented as 10^(x*10^y) or even 10^(10^(10^(x*10^y). As you can see, these resemble hyper 4 operators themselves, except in base 10. All numbers can be represented this way.

Although the extension of the hyper 4 operator into all real numbers was my personal theory, the basis of this function and other info about huge numbers can be found here.

This notation leads to a whole new realm of incomprehensibly high numbers which push the limit of infinity.

## Sunday, February 8, 2009

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## 3 comments:

This is a really interesting post. However, I have to disagree with you on the subject of taking the upper hyper-4 operator to non-integers.

Your example was 3^^3.5. Hopefully you don't mind if I reduce that to 3^^2.5 so that the numbers don't spiral out of control. The principle should be pretty much the same anyway.

Say you wanted to find 3^^.5. The easiest way to do this would be to set x^x = 3. The solution to this equation can be found using Newton's method (or a similar algorithm), which gives x ~= 1.825455. (It could also be done more directly by realizing that the solution to the equation x^x = y is ln(y)/(W(ln(y))), but that isn't necessary). This can be confirmed by taking 1.825455^1.825455, which indeed comes out to 3 within the margin of error.

In multiplication, a * b = a * (b-1) + a. Similarly, in exponentiation, a^b = a * a^(b-1). It then bears to reason that in the upper hyper 4 operator, a^^b = a^(a^^(b-1)).

With that, one can find 3^^1.5 because it would have to equal 3^(3^^.5). Doing that puts 3^^1.5 equal to 7.429565. In the same way, 3^^2.5 = 3^(3^^1.5), which makes 3^^2.5 = 3505.9314.

And here we find the difference in methods. Your method would produce 3^(3^(3^.5)), which gives 1581.5907.

I am aware that my method produces discontinuities near every positive integer greater than 2. However, I remain convinced that the hyper 4 operator (and, by extension, all higher hyper operators) produces irremovable discontinuities at these points.

That said, I would love to hear your thoughts on this matter. If I am missing anything obvious (and especially if I am missing something subtle), let me know.

Regards,

David Freiberg

Thanks for your comment.

I must agree that your method has promise as an extension of the hyper-4 operator to non-integers. I encountered such variations shortly after publishing this post, and I have been meaning to do an update for some time.

Among all the variations of hyper 4 extension (and there are a fair few) I believe yours is the most intuitive one which is still reasonably simple. However, there are some aspects to work out.

For example, it is clear that an expression such as 3^^(1/3) would be evaluated under the discussed method as the solution of x^(x^x) = 3. 3^^(2/3) presents a slightly more difficult problem. To the best of my knowledge, the easiest approach is as follows. Take 3^^(2/3) to be equal to 3^^(2*(1/3)) = (3^^2)^(3^^(1/3)) = 27^(3^^(1/3)). This rule seems to mimic the exponential identity a^(b+c) = (a^b)*(a^c) when the +, *, and ^ operators are replaced by *, ^, and ^^, respectively. What are your thoughts.

Additionally, the extension to negative numbers poses a problem. Using the technique of extending exponentiation, x^(-1) is defined as the multiplicative inverse of x, 1/x. Would x^^(-1) following the same pattern not be the exponential inverse of x? This requires the use of the log, but it is unclear how to proceed, as the log requires two inputs (the base and the argument) while finding 1/x only requires x.

Perhaps the attempted application of exponential identities to the hyper 4 operator is an ineffective view, of this I cannot be sure. However, exponents define the hyper 4, so it seems a good place to start. I would be very interested to hear your views.

Thank you,

Professor Quibb

Oops, there was a slight typo in the above argument. 3^^(2/3) should have been the solution to x^x = 27 through the calculations 3^^(2/3) = (3^^2)^^(1/3) = (3^3)^^(1/3) = [the solution of x^x = 27]. The value I provided (27^(3^^(1/3)) I believe is the evaluation of the expression 3^^(2+1/3).

It appears to make more sense if the equation a^(bc) = (a^b)^c is transformed by substituting ^^ for ^, and leaving the multiplication untouched. Otherwise, one obtains a value greater than 3^^2 for 3^^(2/3) and this is clearly incorrect.

I apologize for the inconvenience.

Professor Quibb

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