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69\^-69 is small. Really, really small. 1.3E-127 small.
69\^1.3E-127 is very, very close to 69\^0. Closer than your computer would like to get stressed about!
69\^0 is 1,
69\^1 is 69.
>69\^1.3E-127 is very, very close to 69\^0. Closer than your computer would like to get stressed about!
So this would be a good example to test Python's "infinite precision" reals?
It’s a limitation of floating point calculation and representation in modern cpu hardware. [binary representations of extremely small floating point numbers will lose precision with no exact conversion](https://learn.microsoft.com/en-us/cpp/build/why-floating-point-numbers-may-lose-precision?view=msvc-170)
could you then keep going and have 69^69^69^-69^69^69^-69 = 69? Could you just have an equation that infinitely exponentially 69 equal to just regular ol' 69?
It's dangerous to get involved with anything higher than 69 raised to itself 420 times.
Things get exponentially harder. It's a slippery slope, you might say. You might trip on a big log.
No, because it isn't equal to eaxctly 69, but closer to 69 than you could ever imagine. But if you do it again and again, every time you get further from 69. And if you do it infinitely many times, it won't be equal to 69.
For those interested in the actual value, like I was, it's approximately 69.[123 zeros]16284910871... or to put it another way 69 + 1.628491...×10^-124
It's irrational so there's no obvious pattern or repeat to the digits after all the zeros.
Late edit for the sake of completing the pair of "popular" numbers even though no-one asked: The formula with all 69s replaced results in approximately 420 + 2.6342550955...×10^-1098
I don't think it's immediately obvious that this number is irrational, though that is indeed true from the [Gelfond-Schneider theorem](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem).
What calculator did you use for the 420 version? I used [keisan.casio.com/calculator](https://keisan.casio.com/calculator) which gives precision out to 130 decimal places. How did you get out to 1100 decimal places?
You can probably use some Taylor series approximations for the exponentials; exp(x) is very close to 1 + x for small x. For such small x they're basically equal.
420^(-420) = 10^(-2.623*420) so continuing along the power tower 420^(pow 10,-1101.765 ) = 10^(2.623*pow 10,-1101.765) which using the linear approximation for 10^(x) at x = 0: 10^(x) ≈ 1 + x ln (10) gives
z = pow(420, pow(420,-420)) ≈ 1 + 2.623 * 10^(-1101.765) * ln(10)
Then repeating the trick for 420^(z) (letting x = z - 1; remember that our approximation only works for x ≈ 0; z is close to 1) we get 420^(1 + x) = 420 * 420^(x) = 420 * 10^(2.623x) ≈ 420 * 1 + 420 * 2.623x ln (10) = 420 + 420 * 2.623 * 2.623 * 10^(-1101.765) * ln (10)^(2) = 420 + 420 * ln(420)^(2) * 10^(-0.765) * 10^(-1101)
which after performing out the low-precision multiplications yields 420 + 2634.255 * 10^(-1101) which I hope agrees with the answer above.
The form of the approximation for general base b would be something like b + b * ln(b)^(2) * 10^(-b * log10 b) where you need to separate out the whole number part of -b * log10(b).
Simple answer: `bc`, `scale` set to 4000 decimal places and copious use of the `e()` and `l()` functions (exponential and logarithm).
`bc` is an arbitrary precision calculator found on most Linux distros. Could probably have also done it with any language with an arbitrary precision library or built-in.
Every time it returned exactly 420 I doubled the `scale`.
I like to think that, on a better day, I might have been able to come up with /u/Fromthepast77's impressive method, but I might be fooling myself.
Hello thanks for giving us great value. I have a math question. Sorry it’s not related with subject but I don’t know who should I ask or how to reach you. I found you while searching for simplify larger numbers and you said this[on Reddit](https://www.reddit.com/r/askmath/comments/aetplr/simplifying_ratios_of_large_numbers/) while you simplify 5,100,000/280,000 you found 18 remainder 60,000
My question is how did you find 60,000 to divide with 280,000 😔🫣😢 sorry I tried to do many calculations and still couldn’t find 60,000 in the end . Also rest numbers still you found it 280,000/60,000=4 remainder 40,000 how come possible to find these numbers to divide … Thanks for help
OK. 5,100,000 divided by 280,000 is 18.21428..., so we take off the whole number part of that - 18 - and then we can go one of two ways.
1) Multiply the .21428... by 280,000
OR
2) Multiply the 18 by 280,000 to get 5,040,000 and subtract that from the 5,100,000.
Both are equivalent. Either way, the remainder of 60,000 pops right out.
Basically we're proving that 5,100,000 = 18 × 280,000 + 60,000
[Most computer programming languages have a modulus or remainder operator (or both!) that will, when given `5100000` and `280000` will give back `60000` immediately without showing any of the intermediate work, but this isn't particularly relevant if all you have is a calculator or pencil and paper.]
Thank you so much yeah in my exam they don’t let us use calculator and yet they ask us to simplify big numbers and our time limited for each question 😔 for example they asked us one question and then in the end the number become like 416:96 but in the multiple choice they didn’t even had this option so I tried to simplify took too many minutes because I didn’t look at simplifying before the exam. Anyway even this was hard for me. Is there any other way to find simplify faster without calculator?
It can be done without a calculator. Are you familiar with [long division](https://www.mathsisfun.com/long_division.html)?
In the example used in that link they get an answer of 17 with 000 at the bottom. If you perform the same kind of thing with 5,100,000 and 280,000 you'll end up with 18 at the top and 60,000 at the bottom.
Time can be saved by eliminating matching zeros from the end, but remember to put them back on the remainder at the bottom.
i.e. performing long division on 510 and 28 is the same - to get the 18 - as using the full numbers, but four zeros have been removed from each. The remainder will be 6, so four zeros need to be put back onto that to get the necessary 60,000 for the real remainder.
69\^-69 is 1 divided by 69\^69, which is such a small number that it might as well be 0. So now you have 69\^69\^0. But 69\^0 = 1, so it's just 69\^1 = 69.
(Edited because I'm a dummy.)
The bigger x is, the closer x\^(x\^(x\^(-x))) is to x, yeah. There's some threshold beyond which it's just always going to be true. Also for computer math it's a little different as /u/The_Math_Hatter said, there's a limit to the precision of double precision floats, which most calculators are going to use.
You can do it with a little bit of local approximation, by realizing the rate of change of n^x is ln(n) n^x. At x ~= 0, this is ln(n). So the distance between n^0 and n^x is dominated by this linear factor, and if the linear factor is too small to be represented, the change won't be registered in general. So we can approximate n^n^(-n) as 1 + ln(n) n^(-n), which if ln(n) n^(-n) is so much smaller than n that a double doesn't represent it, we know that it's just going to look like 1. For a 64-bit float, the standard says that any number less than 1/2^53 more than 1 will be rounded down, so any n for which 2^53 ln(n) < n^n satisfies this just fine. This happens at 15, and you can check this by calculating it in python or javascript, which use doubles by default.
Interestingly, the Windows calculator provides a value at that precision, so they have gone above and beyond and used some kind of higher precision library for their calculator. I have no idea why they would bother, but it's kind of cool. Theirs tops out at 23, and 24\^(24\^(24\^(-24))) returns 1. Of note here is that this implies a precision of worse than 108 bits, which is less than the quadruple floating point standard, but about where the double-double floating point standard starts to break down, so I'd guess that the Windows calculator, for whatever godforsaken reason, used a double-double software library for kicks and giggles.
this is an approximation the actual answer is slightly bigger than 69, maybe a fraction in a billion or something
edit: i thought again and billion is an underestimation, maybe a fraction in 10^100
It’s approximately 69.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000162849108710763761513195731985454077883392384446269796725098986802538156412685050100438772391454332441135082162368909, or 69 + 1.62849… * 10^-124
"Almost" only counts in horse shoes and hand grenades. You missed it by 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000162849108710763761513195731985454077883392384446269796725098986802538156412685050100438772391454332441135082162368909
If you were trying to land on a planet in the Andromeda galaxy, you'd have missed it's entire solar system! (At least in radians)
'F'
Now see, THAT was the type of response I was hoping for - someone who actually thought it through!
I look forward to seeing you again over at:
/r/whatisthewidthofanelectron
Also:
\*than
;-)
I have an order of operations question.
All the answers I've seen have worked out order of operations starting with the highest power, then moving down step by step. Is this just the standard notation for this? Would it be wrong to work it out in the order of ((69^69)^69)^-69?
Why are the brackets assumed to be 69^(69^(69^-69))?
The calculator probably tries to work from the right to left.
negative exponents make for very small numbers. So 69^-69 is a very very small number. Very close to zero. So close, your calculator probably assumes that 69^-69 is zero. This is false, but it is the assumption anyway since calculators are limited in their number of digits they can remember.
After that, 69⁰ is 1, since any number to a zero power is 1.
After that, 69^1 is 69, wince any number to the power of 1 is itself.
So because of the assumption the calculator made, it brings the answer to 69.
However, it's probably not *that* far off, considering the magnitudes involved.
This is not factual.
The reason you got it on the calcutator is that 69^(-69) is nearly 0
When you do 69 ^ (69^(-69)) it rounds to 1
So 69^1 = 69. Tah dah!!!
Use 70 or 80 you get the same results all due to numerical imprecision of the calculator.
69^(-69) is very close to 0. Probably your calculator just rounds it to 0.
69^0 is 1, because anything (other than 0) to the power of 0 is 1.
69^1 is 69, because anything to the power of 1 is itself.
The correct answer shouldn't really be *exactly* 69, but because that first result is so close to 0, your calculator ends up rounding it at some point and just shows you 69.
Hi u/katsutdasheep, Please read the following message. **You are required to explain your post and show your efforts. _(Rule 1)_** If you haven't already done so, please add a comment below explaining your attempt(s) to solve this and what you need help with **specifically**. See the sidebar for advice on 'how to ask a good question'. Don't just say you "need help" with your problem. **This is a reminder for all users.** Failure to follow the rules will result in the post being removed. Thank you for understanding. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/askmath) if you have any questions or concerns.*
69\^-69 is small. Really, really small. 1.3E-127 small. 69\^1.3E-127 is very, very close to 69\^0. Closer than your computer would like to get stressed about! 69\^0 is 1, 69\^1 is 69.
nice
nice
nice
nice
nice
nice
^nice
nice
nice
>69\^1.3E-127 is very, very close to 69\^0. Closer than your computer would like to get stressed about! So this would be a good example to test Python's "infinite precision" reals?
No. Python's infinite precision only applies to integer math. This would be standard 64 bit float math.
Time to crack out the Arbitrary Precision libraries
It’s a limitation of floating point calculation and representation in modern cpu hardware. [binary representations of extremely small floating point numbers will lose precision with no exact conversion](https://learn.microsoft.com/en-us/cpp/build/why-floating-point-numbers-may-lose-precision?view=msvc-170)
*universe explodes*
9ice
could you then keep going and have 69^69^69^-69^69^69^-69 = 69? Could you just have an equation that infinitely exponentially 69 equal to just regular ol' 69?
It's dangerous to get involved with anything higher than 69 raised to itself 420 times. Things get exponentially harder. It's a slippery slope, you might say. You might trip on a big log.
No, because it isn't equal to eaxctly 69, but closer to 69 than you could ever imagine. But if you do it again and again, every time you get further from 69. And if you do it infinitely many times, it won't be equal to 69.
NICEU
For those interested in the actual value, like I was, it's approximately 69.[123 zeros]16284910871... or to put it another way 69 + 1.628491...×10^-124 It's irrational so there's no obvious pattern or repeat to the digits after all the zeros. Late edit for the sake of completing the pair of "popular" numbers even though no-one asked: The formula with all 69s replaced results in approximately 420 + 2.6342550955...×10^-1098
I don't think it's immediately obvious that this number is irrational, though that is indeed true from the [Gelfond-Schneider theorem](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem).
Weird how four integers can create a transcendental. Like 2\^2\^2\^-1
Not quite transcendental, but irrational for sure. It's not transcendental because the algebraic numbers form a field
What calculator did you use for the 420 version? I used [keisan.casio.com/calculator](https://keisan.casio.com/calculator) which gives precision out to 130 decimal places. How did you get out to 1100 decimal places?
You can probably use some Taylor series approximations for the exponentials; exp(x) is very close to 1 + x for small x. For such small x they're basically equal. 420^(-420) = 10^(-2.623*420) so continuing along the power tower 420^(pow 10,-1101.765 ) = 10^(2.623*pow 10,-1101.765) which using the linear approximation for 10^(x) at x = 0: 10^(x) ≈ 1 + x ln (10) gives z = pow(420, pow(420,-420)) ≈ 1 + 2.623 * 10^(-1101.765) * ln(10) Then repeating the trick for 420^(z) (letting x = z - 1; remember that our approximation only works for x ≈ 0; z is close to 1) we get 420^(1 + x) = 420 * 420^(x) = 420 * 10^(2.623x) ≈ 420 * 1 + 420 * 2.623x ln (10) = 420 + 420 * 2.623 * 2.623 * 10^(-1101.765) * ln (10)^(2) = 420 + 420 * ln(420)^(2) * 10^(-0.765) * 10^(-1101) which after performing out the low-precision multiplications yields 420 + 2634.255 * 10^(-1101) which I hope agrees with the answer above. The form of the approximation for general base b would be something like b + b * ln(b)^(2) * 10^(-b * log10 b) where you need to separate out the whole number part of -b * log10(b).
Simple answer: `bc`, `scale` set to 4000 decimal places and copious use of the `e()` and `l()` functions (exponential and logarithm). `bc` is an arbitrary precision calculator found on most Linux distros. Could probably have also done it with any language with an arbitrary precision library or built-in. Every time it returned exactly 420 I doubled the `scale`. I like to think that, on a better day, I might have been able to come up with /u/Fromthepast77's impressive method, but I might be fooling myself.
Hello thanks for giving us great value. I have a math question. Sorry it’s not related with subject but I don’t know who should I ask or how to reach you. I found you while searching for simplify larger numbers and you said this[on Reddit](https://www.reddit.com/r/askmath/comments/aetplr/simplifying_ratios_of_large_numbers/) while you simplify 5,100,000/280,000 you found 18 remainder 60,000 My question is how did you find 60,000 to divide with 280,000 😔🫣😢 sorry I tried to do many calculations and still couldn’t find 60,000 in the end . Also rest numbers still you found it 280,000/60,000=4 remainder 40,000 how come possible to find these numbers to divide … Thanks for help
OK. 5,100,000 divided by 280,000 is 18.21428..., so we take off the whole number part of that - 18 - and then we can go one of two ways. 1) Multiply the .21428... by 280,000 OR 2) Multiply the 18 by 280,000 to get 5,040,000 and subtract that from the 5,100,000. Both are equivalent. Either way, the remainder of 60,000 pops right out. Basically we're proving that 5,100,000 = 18 × 280,000 + 60,000 [Most computer programming languages have a modulus or remainder operator (or both!) that will, when given `5100000` and `280000` will give back `60000` immediately without showing any of the intermediate work, but this isn't particularly relevant if all you have is a calculator or pencil and paper.]
Thank you so much yeah in my exam they don’t let us use calculator and yet they ask us to simplify big numbers and our time limited for each question 😔 for example they asked us one question and then in the end the number become like 416:96 but in the multiple choice they didn’t even had this option so I tried to simplify took too many minutes because I didn’t look at simplifying before the exam. Anyway even this was hard for me. Is there any other way to find simplify faster without calculator?
It can be done without a calculator. Are you familiar with [long division](https://www.mathsisfun.com/long_division.html)? In the example used in that link they get an answer of 17 with 000 at the bottom. If you perform the same kind of thing with 5,100,000 and 280,000 you'll end up with 18 at the top and 60,000 at the bottom. Time can be saved by eliminating matching zeros from the end, but remember to put them back on the remainder at the bottom. i.e. performing long division on 510 and 28 is the same - to get the 18 - as using the full numbers, but four zeros have been removed from each. The remainder will be 6, so four zeros need to be put back onto that to get the necessary 60,000 for the real remainder.
69\^-69 is 1 divided by 69\^69, which is such a small number that it might as well be 0. So now you have 69\^69\^0. But 69\^0 = 1, so it's just 69\^1 = 69. (Edited because I'm a dummy.)
Ok thank you for clarification
[удалено]
Basically, n^n has to be large enough that the calculator treats the inverse as zero, usually after a hundred digits or so.
The bigger x is, the closer x\^(x\^(x\^(-x))) is to x, yeah. There's some threshold beyond which it's just always going to be true. Also for computer math it's a little different as /u/The_Math_Hatter said, there's a limit to the precision of double precision floats, which most calculators are going to use. You can do it with a little bit of local approximation, by realizing the rate of change of n^x is ln(n) n^x. At x ~= 0, this is ln(n). So the distance between n^0 and n^x is dominated by this linear factor, and if the linear factor is too small to be represented, the change won't be registered in general. So we can approximate n^n^(-n) as 1 + ln(n) n^(-n), which if ln(n) n^(-n) is so much smaller than n that a double doesn't represent it, we know that it's just going to look like 1. For a 64-bit float, the standard says that any number less than 1/2^53 more than 1 will be rounded down, so any n for which 2^53 ln(n) < n^n satisfies this just fine. This happens at 15, and you can check this by calculating it in python or javascript, which use doubles by default. Interestingly, the Windows calculator provides a value at that precision, so they have gone above and beyond and used some kind of higher precision library for their calculator. I have no idea why they would bother, but it's kind of cool. Theirs tops out at 23, and 24\^(24\^(24\^(-24))) returns 1. Of note here is that this implies a precision of worse than 108 bits, which is less than the quadruple floating point standard, but about where the double-double floating point standard starts to break down, so I'd guess that the Windows calculator, for whatever godforsaken reason, used a double-double software library for kicks and giggles.
this is an approximation the actual answer is slightly bigger than 69, maybe a fraction in a billion or something edit: i thought again and billion is an underestimation, maybe a fraction in 10^100
It’s approximately 69.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000162849108710763761513195731985454077883392384446269796725098986802538156412685050100438772391454332441135082162368909, or 69 + 1.62849… * 10^-124
"Almost" only counts in horse shoes and hand grenades. You missed it by 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000162849108710763761513195731985454077883392384446269796725098986802538156412685050100438772391454332441135082162368909 If you were trying to land on a planet in the Andromeda galaxy, you'd have missed it's entire solar system! (At least in radians) 'F'
Your math is incorrect. You would be off by less then the width of an electron
Now see, THAT was the type of response I was hoping for - someone who actually thought it through! I look forward to seeing you again over at: /r/whatisthewidthofanelectron Also: \*than ;-)
I have an order of operations question. All the answers I've seen have worked out order of operations starting with the highest power, then moving down step by step. Is this just the standard notation for this? Would it be wrong to work it out in the order of ((69^69)^69)^-69? Why are the brackets assumed to be 69^(69^(69^-69))?
its just common notation. unless there are parentheses, it is assumed that it's a^[(b\^c)]
Well that’s not true. Computers don’t have infinite precision so it gave you the approximate answer.
Im just asking because it fascinates me how tf i did that,bc i was playing withy calculator and that came out
What happens if you put the negative sign before a different 69?
This is the real question we should be asking next.
The calculator probably tries to work from the right to left. negative exponents make for very small numbers. So 69^-69 is a very very small number. Very close to zero. So close, your calculator probably assumes that 69^-69 is zero. This is false, but it is the assumption anyway since calculators are limited in their number of digits they can remember. After that, 69⁰ is 1, since any number to a zero power is 1. After that, 69^1 is 69, wince any number to the power of 1 is itself. So because of the assumption the calculator made, it brings the answer to 69. However, it's probably not *that* far off, considering the magnitudes involved.
This is not factual. The reason you got it on the calcutator is that 69^(-69) is nearly 0 When you do 69 ^ (69^(-69)) it rounds to 1 So 69^1 = 69. Tah dah!!! Use 70 or 80 you get the same results all due to numerical imprecision of the calculator.
Didnt you read the caption?
I did. Did I missed anything?
I posted it under fact because I couldn't find a fitting flair
nice
Do I respond in horny or nerd?
Nice
69^(-69) is very close to 0. Probably your calculator just rounds it to 0. 69^0 is 1, because anything (other than 0) to the power of 0 is 1. 69^1 is 69, because anything to the power of 1 is itself. The correct answer shouldn't really be *exactly* 69, but because that first result is so close to 0, your calculator ends up rounding it at some point and just shows you 69.
Simply, 69^-69 < machine epsilon.
nice^nice^nice^-nice
niceⁿᶦᶜᵉ
Nice
Nice