Comparing Two Very Large Numbers
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I was initially surprised, as my first guess was 24^24... but then realised squaring a decimal number approx. doubles the number of digits.
I caught that, and knew, but the demonstration is another matter. :/
Whay do 6iu guys mean ?? Why on earth would you involve decimals at all when these are all integers being multiplied...isn't it kind of random and out of nowhere to involve decimals?
@@leif1075decimal means base-10, not non-integer
@@leif1075 Decimal as in base 10 not decimal points as apposed to say hexadecimal being base 16. If you square a base 10 number it roughly approx. doubles the number of digits. 321 squared is 103041. 3 digits verses 6 digits... just a rough real of thumb. 24! = 620448401733239439360000, 24 digits. Squaring it is 48 digits, being approx. 3.849E47.
Nice and interesting video.
Small error: when starting to look at the general case of comparing (n!)^2 and n^n, you look at (3!)^3 = 216 instead of (3!)^2 = 36, both of which do happen to be larger than 3^3 = 27.
Good point!
Intentional mistake to create more comments ... and it obviously worked ... at least twice 😀(I'm only writing this because I like the video!)
actually that was not intentional. I do not like that approach
@@SyberMath I stopped commenting on errors after I learned from this technique. I'm glad that you don't like it.
No complicated math needed. (24!)^2 has twice as many numbers in the product as 24^24, and most of them are bigger than sqrt(24), so lets try if we can maybe pair them so that each pair is at least 24. And we can. For example
1*24 * 2*23 * 3*22 * 4*21 ...
Since there are 24 pairs in this product and each pair is at least 24, (24!)^2 is larger.
This is essentially what I did. Only difference is I took the sqrt of both sides first so I was only comparing 24! to 24^12, but I still then paired the factors of 24! into 12 pairs each >= 24.
To be even more faster if someone is comparing a factorial no to non factorial then you can just blindly say the factorial one is big
That's... what he did in the video
@@piyushraj8109 Not really. Like the video stated, 24! 2, etc.)
@@pierrecurie Yeah, but in the video it is just stated that you can do this and is not really explained how you could get to that solution from the scratch. The solution is just given without the "why or how".
So I wanted to add my own thought process consisting of simple observations that led me to that solution instead of simply saying "you can solve it like this".
But you are right, I probably should have explained my motives better in the original comment. Looking at it now, it is not obvious what I am trying to achieve there primarily as I thought when writing it.
You could also sqrt both sides, get 24! vs 24^12 and do the same trick.
Yeah that's probably the easiest approach! Mine was (24!)^2=(24*23*22*...*1)^2=((24*24)*(23*23)*(22*22)*...*1), and since all n*n with n >= 5 is bigger than 24 then we have 19 n*n, which means we have 38 ns multiplied meaning (24!)^2>(24)^38>24^24
3!^2 =36.
3^3=27
This is true for all integers over 2
he messed it up in the video he put (3!)^3 instead of (3!)^2
cubed instead of squared
the answer too he put 216 instead of 36 like you said
I actually encountered a question similar to this on a midterm I took recently (and havent got marks back for yet), its a relief to see I was at least mostly on the right track!
Thank You, Sir..
I did it in a more stupid way, but it still worked. I figured that (24!)^2 is the product of the squares from 1 to 24, but since every term from 5 on has to be greater than 24, that gives me something already greater than 24^20. So, I started pairing up (1x24)^2, (2x12)^2, (3x8)^2, and (4x6)^2, which left me with the squares of 5, 7, 9, 10, 11, and 13 through 23. So, 8 terms of 24 and 16 terms that are each larger than 24, which has to be larger than 24^24.
In hindsight, it should have been obvious to just do (1x24) x (2x23) x ... x (24x1), but sometimes you just get stuck in a particular line of thought.
The was easy. I had the exact same idea, to regroup the 48 factors
1*1*...*24*24 to 1*24 * 2*23 * ... * 12*13 * 13*12 * ... * 23*2 * 24*1,
and the two outer products are = 24, all inner products > 24.
Same
This would have been a good video to mention Stirling's approximation to the factorial function.
The last limit was actually cool!
It easily simplifies to (1+1/n)^n /( n+1) = e/(n+1) = 0
A bit of a sketchy equality there, but I agree.
I had a thought to see how many pairings of numbers can give you 24 in (24!)^2 and then see what you habe left. I got 14 pairs and then found 10 other pairs that were larger. The second method I took was to compair the prime factorizations. Definitely took a bit longer but it was a fun exercise
Cool!
use stirling approx. formula to compare with 24^24.
(24!)^2is larger
24 x 23 x 22 x 21. Lets put the first factorial 24! in the correct descending order
1 x 2 x 3 x 4. Let's put the second factorial 24! in ascending order. then multiply both
24 x 26.. 66 this greater than 24^24 or."
24 x 24 x 24 for 24 times
Hence (24!)^2 is greater than 24^24
my approach was to realize that 24^24 involved 24 multiplications, and 24! also had 24, and 24!^2 will involve 24 multiplications of squared numbers. So I compared them one by one, and realized that only 1^2, 2^2, 3^2, and 4^2, the factorial squared grew by a lesser amount than just multiplying by 24. And I just took a pretty mathematically reasonable guess that for 5^2 all the way to 24^2, the amount more that waa being multiplied to the factorial squared more than overturned the lead that 24^24 had at the beginning.
That'x extremely easy. I did the same as in the video, but found the solution in less than 1 second.
😮
compare that factorial expression to 12^48 might be more difficult
Still easy just apply difference of squares
You can prove the general case by mathematical induction as well.
@@callmeshen9754 i meant this are much cliser than original question
I wonder what would be the value to which the infinite series converges?
Given x such that (n!)^x=n^n, x=n*ln(n)/ln(n!). This equation becomes x→1 with n→∞ (which can be easily obtained using Stirling's formula).
Does this mean that n! and n^n are almost the same when n is large?
For an appropriate definition of "almost the same," yes - the actual difference goes to infinity as does the ratio, so you have to be careful about what exactly you're saying. The limit you indicate means that the relative difference in logarithm between the two quantities goes to zero.
I, not feeling like doing any real math, wrote down 1-24 twice and then 24 written down 24 times. I started cancelling from each group. 2x12 3x8 4x6 24 each cancel out a 24 leaving me with 16 24s and two each of 5,7,9,10,11,13,14,15,16,17,18,19,20,21,22,23. (32 numbers) Well every single one of those numbers squared is larger than each 24, so it's obviously (24!)²
At 4:40 I'm missing the proof that each product on the left side is greater than 24. It's not obvious. I think it can be proven using the maximum of a quadratic function.
The proof that X < (X-Y)(Y-X) u mean? The right side expands to -X^2+2XY-Y^2 , so unless Y is a negative number or zero, the right side will always be larger because of the 2XY term
Edit: I didnt consider the case where Y>X, but we should assume Y is always smaller, for this specific case
With our current level of mathematics are we able to figure out what the approximate value "\sum_{n=1}^{\infty}\frac{n^n}{(n!)^2}" converges to when taken to its limit? Btw that formula I put in last sentence is the latex code for the equation written in pink at 8:12 in the video. I was not sure how to translate that formula to plain English for this comment so I put the latex code so anyone curious about it can render it with a latex editor.
Does anyone know what's that sum at the end?
Does anyone know which is larger?
Physics researcher here. Start octave and calculate log10 of each number following the rules as it easily is calculable with double precision. 24*log10(24) < 2*sum(log10(1:24))
My simple approach to the solution:
List out the terms of (24!)^2, or at least understand that there are 48 terms, and 2 of each from 24 doen to 1 (1 doesn't need to be included, but it makes it easy). Then do the same for 24^24. Now, with the first list in order from largest to smallest, take a pair of the first and last number at once, and multiply them. You will notice that these will ALWAYS be at least as large as 24, meaning every one of these pairs of numbers you divide out of the first list can correlate to dividing out a 24 in the other list. And, since there are 24 pairs to coincide with the 24 elements of the other list, it's now only a matter of proving the product of each pair will only be greater than or equal to 24.
Since we have 24 pairs, one element counting from the top, one from the bottom, we can describe the set of pairs as all of the form (n, 24-n+1) where n is from 1 to 12, and every pair is listed twice.
So our first two pairs would be (1,24), the next two (2,23), then (3,22), and so on. Notice that the product of the numbers in each pair takes the form 25n - n^2, which is always greater than or equal to 24 in the domain of n from 1 to 12. In fact, it is far greater than 24 for every n EXCEPT 1, where they are equal, meaning when we divide out correlated terms from both lists, the factorial squared list has substantially more value removed from it than the 24^24 list does, meaning (24!)^2 is larger.
(Of course, you could ignore the division part and simply multiply the first and last terms to make a new list and see that all terms in it are equal to or larger than the 24 24s, but I didn't think of describing it that way until after I wrote the rest out so... yeah)
Here's another approach. Take ln of both values and compare 2*ln(24!) vs 24*ln(24). Now ln(24!) =[ln(2) + ln(3) +ln(4) + ...+ln(24)].
Consider the graph y=ln(x). The rectangle with base from 1 to2 and height ln(2) has area ln(2). The rectangle with base from 2 to 3 and height ln(3) has area ln(3) etc.
The sum of all these rectangles is greater than the area under the curve from 1 to 24. So [ln(2) + ln(3) +ln(4) + ...+ln(24)] > the integral from 1 to 24 of ln(x).
The integral of ln(x) = x*ln(x) - x +C. So the integral from 1 to 24 of ln(x) is 24*ln(24) -24 - [1ln(1) -1] = 24*ln(24) -23.
2*ln(24!) >2*[24*ln(24) -23]. And when compared to 24*ln(24) the difference is 24*ln(24) -46. Now 24>2.8^3 >e^3. So log(24)>3.
24*ln(24) -46 > 24*3 -46 =72-46>0. So LHS>RHS.
i think it's more of a hard way
@@heco. I guess it becomes a matter of taste. I don't find taking the integral of ln(x) terribly hard. Also the formula I developed
extends easily to n*ln(n) -2*(n-1). For n=3 and n=4 direct computation of (n!)^2 and n^n shows (n!)^2>n^n The derivative of
n*ln(n) -2*(n-1) is ln(n)-1 and thus is > 0 for n>e. So the function is increasing for n>e. Also for n=5 n*ln(n) -2*(n-1) =5*ln(5) -8.
Now ln(5)>1.6 (look it up) so 5*ln(5) -8 >0 and so n*ln(n) -2*(n-1) >0 for n>4 thus proving (n!)^2 > n^n for n>2. One final point;
any follower of this channel knows that Syber usually shows several methods for solving a problem, some easier than others
but all having some interesting features in their own right. That's all I intended to do.
Gah! What does the series converge to???
just as a quick look, the (24!)² would be much larger. while 24! would be smaller, the added square makes it way bigger.
It is well known that n^n>(n+1)^(n-1) for n>=3 so we can proceed from there to prove (n!)^2>n^n.
First, for n=3 we have that 36>27 so (using (n!)^2>n^n as IH) we can write ((n+1)!)^2 > (n!)^2 (n+1)^2 > n^n (n+1)^2 > (n+1)^(n-1) * (n+1)^2 = (n+1)^(n+1) QED
The first assumption is true since log(n+1) is monotonic (differentiate and you'll see).
👍
The limit of the sum mentioned at 0:07:23 is approximately 3.548128226 🤠
Thanks! Any closed form?
I would also like to know if there's a closed form, and how to derive it. It wouldn't be a surprise if π and γ were involved. 😜
what's about comparing :
(24!)^2 and 2^(24!)?
With any comparison "a^b" and "b^a" that will be greater where base is closer to the `e` (~2.7) value.
In your case it's 2^(24!)
use stirling approximation of 24! against 24^24
You guys have exclamation points in maths now?
Well factorial growslarger than exponential. If we multiply that bug number by itself, it'll be very big. So my guess it's the factorial
When in doubt plug in smaller numbers that follow the same pattern (4!)^2 = 4 x 3 x2 x 1= 24^2 > 4^4 = 16^2
What does it converge to?
0
It's (3!)² not (3!)³ but the inequality would still hold true.
Before watching video:
Method I used was comparing
24! to 24^12
Front and back of 24! To see if it was greater than 24.
First one 24*1 = 24
Next one 23*2 > 24
Every other one after this will be greater than 24. Since there are an equal number of the above iterations for either function and we know the left is always bigger than the right. We know: 24! > 24^12
Short answer it’s (24!)^2
I think your first comparison was good.
However if you just look at the numbers 1-24…you realize that of all numbers squared only numbers 1 to 4 are less than or equal to 24 so 1-4 squared will all be less than 24.
But all numbers greater than 5 will be much larger than 24. With 24 squared (576) being much larger than 24. So even just from the first set you should be able to see that 24! will be much larger.
I saw this one the other day.. now, take the exponent on the right and put it to the left..........
24 to the 24th power is much larger than 24 to the 2nd power.
there are more ways to arrange a deck of cards on a random deal (top to bottom) than every second that ever existed in the universe....
I sort of brute forced this a little bit, but i broke 24^24 into (24^12)^2, which means we are comparing 24! with 24^12, and sinply by expanding the fracorial and then combining multipes of 24 and then approximating close mutliples we get that 24! is much bigger than 24^12, so 24!^2 must be greater than 24^24
@@TomFoster-en5uc (x^k)^n is equal to x^(k*n)
@@TomFoster-en5uc which means (24^12)^2 is equal to 24^(12*2)
Was looking for this approach, that's how I did it as well. Wouldn't call it brute-forcing, having the same exponent is a classic and nice strategy with those comparing problems.
Elegant.
Thank you!
I think the first one is way larger cause the squaring down results in a solution, that needs to be larger in front of the decimal numbers than a multiplication with itself could provide for.
For example take the easy one (5!)tothepoweroftwo vs. 5tothepowerof5 it is 14400 vs. 3125 so if this paradigm is continuous the first one is way larger, but I do not know if it is... ...I found out in similar problems that it results crucial how large the number with respect to its digits is... ...and I guess some complete induction will show the reasons this being so... ...gonna go for it in restored conditions of my life...
Le p'tit Daniel
At the end is quite teivially the factorial. Using a method similar to gauss we pair 1 and 24, 2 and 23,... There are 12 such pair.
Each of them multiplied is 24 or more, 24, 46,... So the number is bigger than 24^12 and square dis bigger than 24^24
factorial grows faster because 10! is already approx. 3.68 mil
Could this be done more simply by writing the factorial backwards and grouping terms:
24! = 24*23*...*1
24! = 1 * 2 *...*24
(24!)**2 = 24*46*...24
24**24 = 24*24*...*24
Both products have the same number of terms and each term in the product for 24!**2 is at least 24. Therefore the squared factorial is greater.
Thank you so much for getting to the point immediately. The amount of blah, blah,blah on youtube is annoying. liked and subscribed
Pozdrawiam serdecznie i życzę miłego dnia
How I did it (haven't watched the video yet):
24*1=24
23*2>24
22*3>24
...
13*12>24
Therefore 24!>24^12
If a and b are positive numbers such that a>b, then a^2>b^2
24! and 24^12 are positive numbers
(24!)^2>24^24
Edit: literally just what he did in the video
Beh il primo mi sembra mi grande ..anche a occhio
So 24 to the power 24 has (24+24) digits! 😊
An interesting thing would be to find out for which real x (24!)^x = 24^24.
Take the 12th root of the LHS, and then raise it to the 12th power.
[ (24!)^(1/12) ] ^24 >? 24^24
Now take the 24th root to get rid of the exponent of the RHS
(24!)^(1/12) >? 24
The LHS is the Geometric mean of (1 * 2), (3 * 4), (5 * 6) , ... , (23 * 24). The RHS is the arithmetic mean of 24, 24, 24, ... , 24. We have two sequences of 12 elements. All but two of the elements of the GM are larger than 24. This isn't a problem because we can rearrange the RHS to AM(0, 0, 24, 24, ..., 24 + 24 + 24). Now we can easily see that each term in the GM is larger than the corresponding term in the AM. I think this is QED, but I'm kind of rusty on means. :D
This is easy😀
It seems to be simmpel:
Let uss rewrite both terms a bit:
(24!)^2=((24*1)*(23*2)*(22*3)*...*(13*12))^2
24^24=(24^12)^2=(24*24*...*24)^2
Both terms are now squares of a product. Each of the products has 12 factors. Now let us have a look at the factos: the first factor of both products is 24 in both terms, but for all other factors, the factor in the first produt is larger:
(2*23)>24
(3*22)>24
...
(13*12)>24
So the first product is larger, and therefor is the first termm larger than the second:
(24!)^2>24^24
I think 24^(24) needs to learn to stand up for itself.
After you wrote 24! down as 24*...*1 I immediately thought about pairing it with 1*...*24 and getting 24*46*...*46*24 which is obviously more than 24^24, dunno what will you do for the rest of the video
You made a small typo when you started with generalization. (3!) should be squared, not to the third power...
yes
24 24's vs 48 of them.
Before i watch my brain says 24!^2
My intuition was right, but I was very taken aback at how large 24! Is, I mean wow 24^24 isn't exactly small but...
Ezzy..just test it by (3! )^2 and 3^3 , same (4!)^2 and 4^4
We can easily see (n!)^2 is bigger
Rt (24)²⁴~ 5²⁴ and rt (24!)²=24! we know that 5²⁴
6:19 But you just did (3!)³, not (3!)², which would have been part of the general form!!!!!!
6:08 you made an error. You did (3!)^3 rather than ^2
That’s right! Thank you
@@SyberMath 👍🏼
(3!)^2=9*4*1, 3^3=3*3*3 and from this onward it the ratio get's ridiculously in favor of n!....
So, 24^24 is 24*24*24...*24. It'll have 24 24s.
24! Is 24*23*22*...*1. And squared it'll have 2 times as much numbers we need to multiply.
So the question is how much 24*24 does 24! Have in itself, and if there's more then 12 24s, it's bigger.
And since we have 24 numbers that we are multiplying in 24!, lets combine them, to be 12 like that:
24*1*23*2*22*3*21*4...*12*11.
So, as we see, we start at first pair which is 24, and then every of other 11 pairs are bigger then 24.
So 24! Is bigger then 24^12, and so (24!)^2 is bigger then 24^24.
Easy one.
I'd guess 24!² is larger
Am I the only person who really wants to know why the number used is 24?
looks like it. What's your guess? 😁
Solution:
(24!)² is considerably bigger. This can be proven very simply:
(24!)² = 24! * 24! = (24*1) * (23*2) * (22*3) * ... * (1*24)
Now there are 24 bracket terms, each of which is equal to or larger than 24.
Therefore it is bigger than 24²⁴ = 24 * 24 * 24 * ... * 24 (24 times)
If you decompose (24!)² in pairs, it becomes easy to see which one is bigger
(24!)²=(24x23)x(24x23)x(22x21)x(22x21)x(20x19)x(20x19)x(18x17)x(18x17)x...x(2x1)x(2x1)
24^24 can be represented as 24 shown 24 times
(24x23)>24 (1)
(24x23)>24 (2)
(22x21)>24 (3)
(22x21)>24 (4)
...
(2x1)
Prime factorization of (24)²⁴
= 2⁷² × 3²⁴
And,
24!² = 24²×23²×...×2²×1¹
now lets make a number that will be 24!² except, every number between 12² and 23² will be replaced by just 12² and every number between 6² and 11² will be replaced by just 6²
For simplicity, lets call that number "X"
And also lets just ignore the terms after 6² (i.e. 5²,4² etc)
Now X will obviously be smaller than 24!² right?
So we can say that 24!² > X
Now, in X, no. of 12²s will be 12 and no. of 6² will be 6
So our inequality will look like this
=> 24!² > 24² × (12²)^12 × (6²)^6
Lets prime factorize the right hand side...
=> 24!² > 2⁶⁶ × 3³⁸
Now lets take the ratio of 24²⁴ and X
=> (2⁷² × 3²⁴) / (2⁶⁶ × 3³⁸)
=> 2⁸ / 3¹⁴
2⁸ is obviously smaller than 3¹⁴
so that means * X > 24²⁴ *
We now have 2 inequalities,
24!² > X and X > 24²⁴
Which concludes that 24!² > 24²⁴
Big ass proof
Now find x so that (24!)**x = 24**24 ? If so how? ( 24 factorial to the power of x is equal to 24 to the power of 24)
Just take logarithmic and then anti
Answer is here😅
My gut is telling me (24!)² > 24²⁴ because
24² >> 24
23² >> 24
22² >> 24
...
5² > 24
4² < 24
3² < 24
2² < 24
1² < 24
and there's way more greater thans in favor of (24!)².
Easy...
NAH
First
Jesus is larger.
Such a comparison is completely valueless in the big worldview. It's a pure waste of time.
Why?
@@SyberMath Because you can't show me any way in which such a comparison is actually valuable. Nobody can. It provides good drinking water to nobody. It feeds nobody. It clothes or otherwise shelters nobody. It defends nobody. It comforts nobody in times of despair, soothes nobody in times of duress and agony.
Nothing else matters.
24²⁴=(24¹²)²
we're only comparing size to (24!)² so we can drop the square.
so 24¹² vs 24!
They have 12 terms vs 24 terms respectively and typically speaking, having more terms grows faster than higher terms. You can pair up and compare the terms though.
1x24=24
2x23>24
3x22>24
...
12x13>24
conclusion: 24!²>24²⁴
Since the inequality is reversed between 1 and 2, I looked at max( (n^n)/(n!)^2), which occurs at about n=1.4973 and gives about 200453/192818.
We cam just see that when 24! Is squared ..all numbers after 4 are going to bw larger than 24 .. as 4² is 16 and then 5² is 25 ..... And eventually 24².... So basically all individual numbers are larger than 24 so its karger than24²⁴.... Ez
I just solved it by moving my finger in a way that makes UA-cam close, then, using my finger, opening the calculator app, putting in the numbers and then comparing them. Don't know, but this solution was completely different from everyone else's. Hope this helps! 😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊