Rule for evaluating a power tower & tetration. Reddit rule of exponents r/askmath
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- čas přidán 1. 07. 2024
- We will discuss the order of exponentiation. Does 2^3^4^5 mean ((2^3)^4)^5 or 2^(3^(4^5)). We will also look at power towers and tetration. This question is from Reddit r/askmath / tgnjcpkcer
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#math #algebra #mathbasics
0:00 Order of exponentiation, i.e. power tower 2^3^4^5
2:42 More examples
6:33 Quick introduction to tetration
Come on, Steve, admit it.
This entire video was made for the sole purpose so you can show off your ninja powers of writing numbers on a blackboard without looking!!! 😅
😆
The number 65536 was a number I always dreaded seeing coming up in error messages when programing in the early 90's. That meant that I had an unhandled uint issue that I had to track down.
A note here: there is a lot of confusion when using calculators. If you don’t know what you’re doing. Different calculators have different capabilities. Some may not handle the order of operations for you, especially older calculators. This isn’t a bug per se. It’s just that you’re supposed to know that. What I mean is you have to know your calculator.
I'm but a simplied man.
Same
A simp man?
@@CutleryChipsHe implied the mistake at 5:35
@@sohailansari07289 I saw too, I jus simp the expression further
Didn't know that exp rule. I remember getting that question somewhere before and I just evaluated it from left to right. Didn't know I was wrong until now.
@bprp to calculate 2^3^4^5, take the logarithm from base 2 to base 10 so first 4^5 is 1,024 and to calculate 3^1,024 do (10^log10(3))^1,024 which will make it 10^(lg3*1,024) that is about 10^488.57 so we can minus the integers left with the decimals and take the log 10 again so that 3^4^5 will be 10^488*10^0.57 that is 3.715*10^488 and to do 2^10^488.57 do the same way make it 10^(lg2*3.715*10^488) making it 10^10^(1.118*10^488) also we can see that in 2^(3^4^5) the change of base 2 to base 10 barely even decreases just from 3 times 10 to the 488 to 10 to the 488 and if it is 2 to the (2^3^4^5) then it basically doesn’t decreases if we make the base 2 to base 3 that’s also the reason why the base of tetration is useless if the height is high and as long as the base is bigger than the e(th) root of e
Actually calculators do 2^3^4 from left to right giving 4096. Some will force parenthesis so you'll have to write 2^(3^(4)) giving 2.4*10^24. Many modern calculators actually allow wiring it in mathematically proper way.
Ok so basically
(3^3)^3 is just 3^9 and doesn’t require a new operation.
So tetration is obviously trying to evaluate the other case which means 3^(3^3) which blows up quickly
Edit:
If parenthesis is left group first,
2^3^4^5 = (2^3^4)^5 = ((2^3)^4)^5 = (2^(3*4))^5 = (2^12)^5 = 2^(12*5) = 2^60 which is a simplification.
If parenthesis is right grouping,
2^3^4^5 = 2^(3^4^5) = 2^(3^(4^5)) = 2^(3^1024)
Blows upo indeed. 2^3^4^5 has I have no idea how many digits but the number of digits is a number with 489 digits in it. A googolplex has a measly 10^100+1 digits in it which has only 101 digits in it. Tetration is scary stuff.
You have your parens backwards. (3^3)^3 is evaluated with the parens first so 9^3
@@MightyBiffer
I think you need to have another go at it =)
Yes, inside the parenthesis first, but (3³)³ does not equal (9)³ •••
You're saying (3³), but calculating it as (3²) [(3²)=9].
(3³) = 27
It should be like these:
(3³)³ = (27)³ = 19,683
(3³)³ = (3³ˣ³) = 19,683
(3³)³ = (3⁹) = 19,683
@@MightyBiffer Omg I confused myself again for a moment. Let me edit my comment to have 1 more number in the chain
@@GFlCh oops so using all 3 was a bad example because specifically true that (3^3)^3 = 3^(3^3)
Got to say, huge respect for the continuity in the cut at 6:33 .
What cut? Are you doubting his powers? 🙃
It wasn't immediately obvious how ³2 is different since 2⁴ and 4² both equate 16. However, the difference in order becomes apparent thereafter: ⁴2=2^2^2^2=2¹⁶=65536, yet ((2^2)^2)^2=(4²)²=16²=256. And ⁵2=2^2^2^2^2=2⁶⁵⁵³⁶=ALOT, yet (((2^2)^2)^2)^2=(16²)²= only 65536
But you can write the number in binary form like "1" and 65535 zeroes: 1000....000
3:20 the teach in me is screaming "scientific notation, there's no reasonable why you need all those digits" ... but then again there's no reasonable excuse for using a "tower of power" or whatever it is.
FYI 2^3^4 in the Windows calculator give 4096, so order of operations does matter.
Just tried it, the Windows calculator calculated 2 ^ 3 the moment I entered the second ^, so ended up with 8 ^ 4 = 4096. A programming bug, clearly, but then the Windows calculator really isn't set up for complicated math.
Windows calculator also gives 16 for 2+2×4, it has no order of operations and the moment you click an operation button (+,-,×,÷,^...) it evaluates what it had so far and only then continues.
Thats because windows calculator works the same as an old fashioned calculator: it evaluates expressions immediately as they are entered, not at the end, so order of operations is lost.
@@krzychxyz8358 That much I know is not true. Are you not using the calculator in scientific mode?
It does get 2^3^4 wrong, though.
@@ZipplyZane How do you turn scientific mode in Windows calc? I'm faily sure mine does not have that...
I've avoided exponentiation like the plague, but now I'm fairly confident on it. Nice.
The question implies Wolfram calculates left-to-right. Is the questioner just confused?
I just tried it with 2^3^4 and it evaluated it right-to-left.
It seems so. I just checked with WA. Plugging in 2^3^4^5 gives the right answer. Then I checked by changing to "Math Input" and used the exponent button to build the power tower and it gave the same answer. Both times 10 to the power of 10 to the 488.05... (what you see for a short moment at 2:44).
Not sure how this dude on Reddit managed to get the wrong result.
He may have used Windows calculator, which evaluates everything left to right.
@@MisteribelPossible but the dude specified that "Wolfram calculated" it that way in the Reddit post that is shown at the beginning of the video.
@@jensraab2902 ah, gotcha!
At 5:21, you say:
4
3
2 •••
Can't be “simplified”.
I'm sure if you think about a bit, you'll realize it can...
Look at that point, to the top line. You'll see:
4)
(3 81
2 = 2
What you wrote is just calculation, not simplification. Simplification is where terms are able to be removed and/or operations are able to be changed without knowing a concrete underlying value. He shows this using the exponent multiplication rule when the problem is not a tower but rather has parenthesis.
e.g. (a^b)^c = a^(b*c) [simplification from exponent b^c to multiplication of b*c]
a^b^c = no simplification
I dunno, the JavaScript console is saying 2^3^4 is 5
that's because ^ is the XOR operator, not exponentiation
@@flotspe sir, do you know what a joke is?
@@Kokice5to be fair, the two group of math enthusiasts and web devs almost don't have intersection. Such joke will be lost here if not explained.
@@flotspethanks for getting me on the inside of this joke. It's pretty cozy in here.
For those other math people, you would actually do the math like this in JavaScript:
2**3**4
Though, if you want all the digits, you'd better use Big Integer mode, like this:
2n**3n**4n
WOW... an amazing sensei you are.
Where do we actually apply exponents of exponents?
2^3^4 gives 4096 on my old TI-36X.
Reddit is really screwing up its notation, isn't it? Old Reddit has no problem with putting a superscript inside a superscript.
2^81 is 2.4 septillion XD
What's the meaning of inventing a " (^4)2 " when you can simply write 2^16?
simplied put
I love this. Since discovering Mosers number first and then Grahams number (the concept, not the actual number) on the internet, I am obsessed with this.
You could even take that further.
If you want to know which number is bigger of 2 similar big power towers, you could do base conversion.
For example: 3^^3 would translate to 3^3^3 would translate to 3^27 and give you 7.625.597.484.987. Taking this on step further (not one operation, just one step), how would you describe 3^^4? Or taking one step closer to 3^^^3? Just one step!
It would be 3 ^ that number from before, right?
So how would you possibly describe 3^7.625.597.484.987 ?
Well it's not so easy, but you use a similar concept in the video.
By doing a base conversion and turning the 3 into a 10, you could actually count the numbers of digits.
So 3 would be 10^0.4771212547196624372950279032551153092001288641906958648298656403052291527836611230429683556476163015
By using this concept and replacing the 3's with 10's and some random number, you are able to get the exact number of digits at least and also the exact number itself.
3 is just the example, that if the exponent number is smaller than one, it will be automatic the number we are looking for, our FIRST DIGIT of the number.
Taking that concept, you can calculate the first digit of fairly large exponents.
So take n^x , work out how n can be replaced by 10^y, use the logarithms base 10 to do that, and then calculate it.
Numbers equal or greater than 1 can be represented easily, giving you the size of the number (1000 would be 3 in 10^3), numbers smaller than 1 represent your digits in base 10. (0.123456). So in that short example you get 1.23456 * 10^3 . This is the representation that's important.
With this base conversion you can represent numbers like x^n for example 3^7625 in a better understandable y*10^z number.
I think that this is the only way for getting more digits for large constants like G64 or Mosers number.
It might even serve as one key aspect in getting the first numbers.
But as I only used it in a concept with 10^something to have it replacing 3, it has yet to be proven, if this concept holds true in general (or have I made a mistake), and if I'm correct, could it be used to convert "very difficult to calculate" operations like pi^pi^pi^pi in a different 10^x form for being able to getting closer and closer to that numbers actual value in base 10. There would always be an error with transcendental numbers however, but are we able to get close enough, that we can calculate a fairly large ammount of it and be positive that those first digits will be acurate to a certain point we know of?
Therefore logarithms are powerful tools when it comes to base conversion.
In your video 2^81 could be converted into:
2=10^0.301029995663981195213738894724493026768189881462108541310427461127108189274424509486927252118186172
and take that 0.3010299... number, use base conversion, logarithms.
That way you get your first digits.
You can calculate is on
arndt-bruenner.de/mathe/java/rechnergz.htm
Anyway,
If I'm wrong, just tell me. If not, use this information to make a video guide on how to convert large numbers like n^m into readable x*10^y format. It's a game changer.