Which is larger??
ฝัง
- เผยแพร่เมื่อ 20 ก.ย. 2024
- We determine whether 50^99 or 99! factorial is bigger three different ways.
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Fourth way: Overkill - Calculate both numbers by hand
Even more over kill, take the logarithm of both sides and use binets log gamma functions.
@@pneumaniac14 even more overkill, use logarithm and Fermat's approximation
It's called the brute force
I think it is 1st one i.e. simple
As the Michael said
Hyperkill subtract 50^99 FROM 99! In head and write the result in 10 secs
They are equal! I typed them both into my calculator, and they both evaluated to “overflow error”!
Kkkkkkkkkk
@@lucassalomao4882 ok?
@@Fierywell ok o que??
@@lucassalomao4882 oh spanish I see
A vc é gringo kk. Pelo nome "Pedro" achei q fosse BR irmao
The way we know that all the pairings are greater than one is that the denominators (51×49, 52×48, ... ,99×1) are of the form (50+n)(50-n) = 50^2-n^2 < 50^2
This is what I did to know in 5 seconds
Still watching to see whether this is one of his three ways
👍🏻👍🏻
That’s clever
Yeah, was very surprised to not see that by him, feels very intuitive
1:44 "Notice 49 + 49 is 98, plus one is ... 50" I learn something new everyday lol
yeah, that was great :D
This must be that "new" math I've been hearing about.
@@xevira it’s called h (t)= am
I do that sometimes talking not about math
@@xevira alternative math lol
I was expecting wolframalpha for the third method...
"by inspection"
isnt it
Can you help me with number of digits in a factorial without a program...
I really want to prove this by inequality of number of digits.
Edit: Nevermind.. got it.. stumbled upon Stirling’s approximation
BPRP! Nice to see you here :)
😂
At 3:35, I think a great way to show that all of the denominators are smaller than the numerators is by using difference of squares.
You can express every denominator as (50 - n)(50 + n) for 0 < n < 50. This is equivalent to 50^2 - n^2, which is always smaller than 50^2.
I was looking for that.
This is deeply related to 'e'. If you look at the power expansion of 'e', you'll find this form. It turns out the question of when the denominator starts to dominate the numerator is exactly a factor of 'e' away from the number. So, in this case, you'll see that 50 * e will be the place where the denominator starts to dominates the numerator. Now, floor of 50 * e = 135. So, (50^134) / (134!) is greater than 1, but (50^135) / (135!) will be less than 1.
This then ties into the length of the side of the higher dimensional square, given an area of n!. So, 'e' is actually a constant that relates area and parameters between dimensions.
As a consequence, you get the limit n / (n!)^(1/n) as n goes to infinity = e
Try the limit out on wolfram alpha
awesome!
@DukeOfDystopiaeither self taught, or they are a math major at uni.
When I plug this into wolframalpha, I get 50^134/(134!) < 1. The crossover point occurs between 132 and 133.
after having watched the first part: its basically squares vs. rectangles. when you have a set length of all sides combined, the surface area is always biggest when you make it a square. the longer and slimmer it gets, the less area it has (down to a line with no surface)
I was thinking the same thing, he just proved that for a given circumference, a square gives the largest area of any rectangle
Great observation
Thought exactly the same! Also way more elegant than the brute force induction! 👍👍
Yeah
I like to say it in this way
(x)(x) > (x-a)(x+a)
This was exactly my first thought as well! Although I would've explained it much less clearly.
3rd way: super easy, barely an inconvenience
Comments: you could have used amgm
Whoops!
Whoopsie!
Using wolfram alpha is tight
wow wow wow
wow
@@tamarpeer261 whsts that arithmetic versus geometric mean you mean?
I understand your reference
Thanks for another great video!
Though I think it would've been good to note that e.g. 49*51=(50-1)(50+1) , and 48*52=(50-2)(50+2) etc. so all of the denominators are of the form (50-a)(50+a) which is 50^2-a^2 so it's less than 50^2.
I dont think it was necessary to make a rigorous proof of the statement.Though in an exam scenario, it would probably be necessary.
@@divyanshaggarwal6243 Yah, it definitely isn't necessary. It's much easier than calculating though as you don't even have to look at numbers. It's a bit harder to explain I would guess though.
It actually is easier since you dont need to calculate 49*51 etc, just show that a^2
そうそう
Take logs on both sides, we have 99*log(50) > log(1) + ... + log(99) by Jensen's inequality since the logarithm is concave.
Bruh...
I'm weak at inequality...
Jensen's inequality has a less or equal sign in it, not a less than sign.
Or take log base 99 equals !
That's what I first thought! Kudos!
While true, using Jensen here is definitely overkill. The general case follows directly from AM-GM by noting the arithmetic mean of 1,2,...,n is (n+1)/2, so (n+1)/2 is at least the geometric mean, which is the n’th root of n!. Take n’th powers, and we get the general case.
1:45 2:20 ...and that's a good place to check your arithmetic.
10:08 That’s a silent way to stop
Spanish troupe?
@@ivanlazaro7444 confirmamos
@@ivanlazaro7444 confirmo
A 4th way of doing it : the AM-GM inequality yields ((50+k+50-k)/2))^2=50^2>=(50+k)(50-k) so taking the product over the k's between 0 and 49 we get 50^100>=50*99! hence the result
Edit : or (50+k)(50-k)=50^2-k^2
Genius!
This is effectively the first method, only you proved the inequality, while I didn't notice the proof in the video.
I did in less fashionate way, Consider trinomial x²-99x+2500 Since D=99²-10000=99²-100²0 for any x. In particular, k(99-k) n! (second method), from ((n+1)/2+k)((n+1)/2-k) < ((n+1)/2)². I believe this is simpler than by induction.
Better method: th-cam.com/video/KijvVFAu8WI/w-d-xo.html （2 topics included）
I like the ending... for some reasons :p
We could use Stirling's approximation too. n! ~ sqrt(2*pi*n)*(n/e)^n
If we cancel some terms at the end: (1/2)^n > (1/e)^n
we could get a pretty good correlation between two general forms!
I guessed correctly from my experiences in carpentry and ordering materials. I thought it was interesting that perfectly square rooms only had a difference of 1 compared to rooms that had dimensions of the same square room +1 and -1
10 x 10 = 100
9 x 11 = 99
Extrapolating the method further, I learned it was the difference of squares. 10 x 10 example:
From 100
9 x 11 = 99 difference of 1 squared
8 x 12 = 96 difference of 2 squared
7 x 13 = 91 difference of 3 squared
6 x 14 = 84 difference of 4 squared
And so on.
Math can strangely be fun especially showing the kids interesting tricks like this. Thank you.
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Why does this remind me of the “99 bottles of beer” song? 😝
Take the 99th root of both sides and apply the AM-GM inequality.
yeah smh...
wow, nice method!
that was what I thought
There's one more
Method
I'm sure we all figured this out as kids when we wondered which two numbers, that add up to the same sum, would make the biggest rectangle. the longer the rectangle becomes, the smaller the area, and a square is the most efficient rectangle in this way.
The longer the rectangle the smaller the area
I know what you mean but
L
I paired the terms of each series the same way that Michael did (99*1, 98*2, ... 51*49), and noted that each product has the form (50+k)(50-k). That equals 50^2-k^2. However, the pairwise products in 50^99 are always 50*50, and 50^2 is obviously larger than 50^2 minus k^2.
4th way. Apply the “engineer’s function” and make everything equal to 3.
3black1red Reminds me of the old comment about mathematicians want the exact answer, engineers are happy if the numbers are a reasonable match and astronomers are ecstatic if they have the roughly the same order of magnitude
@@ianmoseley9910 brilliant
oh yes, like that pie number
yea pi=3=e
pi^2=g (thats actually pretty damn close due to the old definition of a meter being the length of a pendulum with a period of two seconds. With this definition g would be exactly pi^2 m/s^2 and if you are confused now since g has to be the same even if the def of a meter isnt the same, then you are not completely right. because g is 38622 inch/s^2)
Haha great joke bro. Definitely could not see it coming in this math puzzle video specifically about comparing arithmetic values. Jokes are hilarious when they are as innovative as this!
Forth way: Use log10
(This can be helpful for very large numbers or powers)
Let the symbols be: (n^x, x!)
Your program should be:
double a = 0.0, b = 0.0;
for(int i=0; i
I think we can use also log function.
99!/50^99 = (99/50)(98/50)•••(2/50)(1/50)
Use log function
log(99/50) + log(98/50) + ••• + log(2/50) + log(1/50) < 0 [because, log(50/50) = 0 and (-log(1-(k/n))) > log(1+(k/n)) (n>0, k>0)]
So 50^99 > 99!
(I'm korean so I can't good english speaking. sorry guys)
How do you know -log(1-(k/n)) > log(1+(k/n)) (n>0, k>0)]
Pre-watch guess: 50^99
Reasoning: both have 99 terms. I know that usually the central term multiplied with itself is bigger than outside numbers multiplied with each other. We'll see if it holds up.
That was my logic too.
Another way to think about it is that (x+n)(x-n) is always smaller than x^2 for any integer 'n' that isn't 0. Basically, having 99 of the same number multiplied together must be larger than an equivalent number of different numbers multiplied together. If you did (50^97)×(51)(49), it would also be smaller than 50^99, for the same reason.
I thought the third method was only called "cheating" as sort of an expression to say it's figured out by doing some dirty tricks, like making some sort of a guess we couldn't possibly know and then proving it, but no... It's literally cheating. I didn't see that coming!
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Nice video, as always, but I think that in the simple solution, when you talked about the "denominators increasing but being less than 50^2", you could, instead, just say that those pairs in the denominator are of the form (50-k)(50+k)=50^2-k^2, which is less than 50^2 for every k between 1 and 49 (both included). This way you don't need to explain why the denominators are increasing or even calculate the values of 50^2 and 49x51.
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4th way: inequality of means. Take the 99th root of both terms. The result for 50^99 is just 50. For 99!, you write that the geometric mean is strictly less than the arithmetic mean which is (1+2+3+...+99)/99 = 50. Therefore, 99th root of 99! is less than 99th root of 50^99, so once you raise everything to power 99, you get that 99! < 50^99.
I went straight to thinking about area ... where we know the largest area (multiplication of the two sides) for a given circumference is when the sides are equal. So we know that n^2 > (n-k)*(n+k) for any k {1,n-1}
Excellent! So you basically maximized the volume of a 99-dimensional cube given the sum of the lengths of it's sides is constant. I guess this can also be used to show the general case of
((n+1)/2)^n>n!
For the second method you can just used AM-GM and sub in 1, 2, 3…n and it comes out straight away
4th way : MULTIPLY BOTH SIDE BY ZEROES. And Tadaaaa You Get Equality.
It doesn't work like that tho
I mean it's a good joke, maybe
Fun problem! The first thing I thought of was to use the concavity of log. Which is very simple and also implies the AM-GM inequality and proves a pretty general version of this result.
1:44 "Notice 49 + 49 is 98, plus one is 50" Too many 50s to keep track of, I suspect
XD
I felt like I was so good at maths when I heard this
It is even simpler to show, that (a-1)*a*(a+1) = a^3 - a < a^3, or more generell (a-b)*a*(a+b) = a^3 - b^2*a < a^3 ( for a > 0).
==> 49*50*51 < 50*50*50, 48*49*50*51*52 < 50^5 and so on
I actually solved this question with the simple method as we know that for every positive integer x, x² is greater than (x+y)(x-y) where y is any positive integer.
Except when y = 0 of course.
we can even tell how much greater/more using the formula: x^2 = (x+y) (x-y) + y^2
Don't these rules apply to negative integers for x as well?
(-2)^2 > (-2+1) (-2 - 1)
at least as long as x and y are both elements of Z it seems to work.
x^2 > (x+y)(x-y) because 0 > -y^2 (for y not equal to 0) x^2 > x^2 - y^2 x^2 > (x - y)(x + y).
It applies for every x,y belonging to r, there are no restrictions except from y must not equal to 0, like Blacksmith said
@@petrospatrianakos9166
I take it you meant -y^2>0 (for y not equal to 0), or did you mean 0 > -(y^2) for any y not equal to 0?
(notation methods tend to change over the years and my way may have been outdated by now :)
though if one exchanges the > symbolfor a >= symbol (not sure how to type greater than or equal to symbol on a qwerty keyboard), then even y = 0 works. since if y = 0 then (x+y) (x-y) = x^2 so then it boils down to x^2 >= x^2 which is correct.
any real number not equal to 0 for y gives an y^2 > 0 in the formula x^2 = (x+y) (x-y) + y^2
any real number for y gives an y^2 >= 0
@@BlacksmithTWD -y^2 is smaller or equal to 0, since any number squared is a non-negative number (not sure how it is called in english), and since it has a minus in front of it, it is a non-positive number (negative or 0). So you can either say 0 > -y^2 x^2 > x^2 - y^2 x^2 > (x - y)(x + y) (for y not equal to 0) or say 0 >= -y^2 x^2 >= x^2 - y^2 x^2 >= (x - y)(x + y) for every y, and the equality is true when y=0.
But x and y can be any number, not just a positive integer and y is not necessarily smaller than x like the original comment suggested.
I was thinking about a problem like this the other day, just in two dimensions.
If you’ve taken calc 2 you might know that in order to maximize area of a rectangle of fixed perimeter, you choose a square. This extends beautifully into this problem, which is essentially the same in 99 dimensions. We have 99 factors in both products, and to pick the greatest “volume”, we should choose the “square”, i.e. 50^99.
exactly, its congruent to simple max-min problems
Isn't it easier to use (n+k)*(n-k) =n^2-k^2
Nice observation. This way you don’t need to do the whole induction proof
Isn't this essentially the first way that he showed us?
@@DylanNelsonSA yeah, with proof by example, very hand-wavy
Here's a more useful general theorem that implies 90! < 50⁹⁹:
Let 𝑎 a list of numbers. Construct 𝑏 as a list where 𝑏ᵢ = 𝑎ᵢ for all indices 𝑖 except some 𝑖₀ and 𝑖₁, where 𝑏ᵢ₀ = 𝑎ᵢ₀ + 𝛿 and 𝑏ᵢ₁ = 𝑎ᵢ₁ − 𝛿 with 𝛿>0 such that 𝑏ᵢ₀ ≤ 𝑏ᵢ₁. Then it holds that ∏𝑎 < Π𝑏.
Corollary: for any list 𝑎 where not all entries are equal, ∏𝑎 < 𝑎 ̅ⁿ (i.e. the arithmetic mean to the power of the list length).
Surprised this did not come up
1:44 49+49=98, 98+1=50 🤔😉
You can use different of two square to proof that.
99! = 1x2x3...x49x50x51.....x99
49 = 50-1
51 = 50+1
(49)(51) = 50^2 -1
Do this to every term
99! = 50(51x49)(52x48)(53x47)(54x46)...(1x99)
= 50(50^2 -1^2)(50^2 - 2^2)...
{49 term of 50^2}
But 50^99 = 50(50^2)(50^2)(50^2)...
{49 term of 50^2}
Btw 50^2 > 50^2 - (positive no.)
Therefore 50^99 greater than 99!
The "cheating way" was the best, I laughed a lot
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The first methode can be interpreted geometrically:
Imagine a rectangle with the sides a and b, where a + b = 100.
Calculating a*b means calculating the area of such a rectangle. So we can e.g build following rectangles: (99*1), (98*2), (97*3) and so on (looks familiar, right?)
But as we (should) know: The bigest area of a rectangle with the same circumference (here: 2a + 2b = 200) is the square, in this case the square 50*50
I like the way you think! Clever.
I think you mean with the same perimeter
Me after failing honor precalc test: Im gonna study hard for next test
Also me at mid night: 50^99 or 99! well let's figure it out
Bruh same i have 81 rn
For a fourth method for comparing x=((n+1)/2)^n and y=n! we can calculate ln(x) and ln(y) where the latter is approximated using Stirling's approximation to O(ln(n))
10:15
Don’t be too hard on yourself and don’t forget to stay hydrated. No homework today, sorry folks. If you want a particular topic for the next one, tell me.
Is this the first non spoken "good place to stop"?
@@Guilherme-xp1tv I think it is
Can you do a counting/combinatorics question please?
Guilherme Castro Dela Corte, we need to get his kid to stroll up to the chalkboard and hold up a “that’s a good place to stop” poster.
Something w inveriants
My method to calculate it in the head: log on both sides ad then see, that the mean value of log(50) - log(1)... log (50) - log(49) has to be bigger than the mean value of log(99)-log(50) ... log(51)-log(50). Therfore 50*log(50) has to be bigger than the sum of all log(i)
Oh my ! A concavity inequation (or whatever it's called in English).
Nice and elegant !
I thought the cheating way is to compare 2^3 and 3! 😂
Uh
But you have to prove the method 2 first to ensure that it works in this case.
I wouldn't call that method cheating. Especially if you also compared 4^5 with 5!, 6^7 with 7!, 8^9 with 9! and pointed out the emerging pattern to derrive the conclusion that 50^99 > 99!.
@@BlacksmithTWD
I think is to compare 3^5 with 5! , 4^7 with 7! , 5^9 and 9!
It can be easily understood by the equation (x+y)(x-y)=x^2-y^2 less than x^2.
For example, 99×1
@@user-vv4zy9ss5s
My bad, I was too hasty, the comparisons are when considering in an even number n as follows : n^(2n-1) with (2n-1)!
so that would give us 2^3 with 3!, 4^7 with 7!, 6^11 with 11! etc.
Fairly easy to prove without induction... rewrite the factorial in the denominator as a product of differences of squares:
(2n-1)! =
Product[i,{1,n-1}].n.Product[i,{n+1,2n-1}]=
Product[i,{1,n-1}].n.Product[2n-i,{i,1,n-1}]=
Product[i*(2n-i),{i,1,n-1}]=
Product[(n+i)(n-i),{i,1,n-1}]=
Product[n^2-i^2, {i,1,n/2}] and then examine the two products term by term
Thanks PROF.
I LOVE YOUR TEACHING AND BECAUSE OF YOU I'M LOVING OLYMPIAD MATH
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in the first method, the denominators are products of the kind (a-n)*(b+n), with a=99, b=1, and n any natural from 0 to 48;
if you assume that products are crescent, you get the disequation:(a-n)*(b+n)
This is funny, my wife asked me the same question a few days ago! Fun video, thanks Michael.
I think the simple explanation should be simpler. I answered the question in my head by thinking 9 * 11 < 100, done! That implies that 49*51
Better method: th-cam.com/video/KijvVFAu8WI/w-d-xo.html （2 topics included）
That is what I am saying.
I mean, you can use the Stirling's formula. N! is approximately equal to (N/e)^N * sqrt(2πN) when N is greater than 50 or so. The 2πN term is negligible. You can take the log of both sides for convenience and then plug in N=99. The result will be smaller than log(50^99), which means 99! is smaller than 50^99.
Me at 3 am need to sleep when there is school tomorrow:
Let's watch this cause why not
Another easy to calculate cheat, by comparing:
ln(50^99)=99×ln(50) ≈ 387.29
ln(99!)=ln(99)+ln(98)+...+ln(2)
we can find this sum in many way,
I choose simple python code:
import math
s = 0
for x in range(2,100):
s = s + math.log(x)
print(s)
the result approximately ≈ 359.134
so:
ln(50^99) > ln(99!)
=>
50^99 > 99!
Or even simpler method, by comparing:
exp(50^99), exp(99!)
One of the best ending ever on this channel , I loved it 🤩❤️
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At 2:20, you can explicitly prove that quotients like (50*50 / 51*49) are greater than one, i.e. that the numerator is larger by noting that:
51*49 = (50+1)(50-1) = 50^2 - 1^2 < 50^2 = 50*50
or, more generally, for all real x,a >0,
(x-a)(x+a) = x^2 - a^2 < x^2 = x*x. Thus, expressions of the form (x*x) / [(x+a)*(x-a)] are all greater than 1.
There is another approach for the general case: use AM-GM inequality.
(99!) ^(1/99) < (1+.. +99) /99 = 50
Watch out: the inequality is strict because involved numbers are different!
You can see that the number of zeroes at the end of 99! is 22 so this number is much much less than 50⁹⁹.
For ones who do not know this thing:
To find the number of zeroes at the end on N!, divide N by 5 and do this until the quotient becomes less than 5....
For example, 27! : 27/5 yields quotient 5 . Then 5/5 =1 . We had 5 as first and 1 as second quotient therefore number of zeroes is 5+1=6.
Your subscribers have grown rapidly
When i subscribed you, you were at 36 k
Michael will reach the 100K subs in December. I’d love to see him with the silver button from YT.
Another way is using Stirling approximation and the fact that log is a non decreasing function:
- suppose 50^99 > 99!
- take the log both sides and use Stirling: 99*log(50) > log(99!) ~ 99*log(99) - 99
- divide by 99 and rearrange:
log(99/50) < 1
- exponentiate:
99/50 < e ---> 99 < 50*e
True, because 50*e > 50*2 = 100 > 99
I honestly thought, that for the "cheating" way he'd just take out his calculator
The calculator gives up for factorials bigger than 69!
Mine goes up to 170!
@@SadisticNiles It all depends on what the maximum value your calculator can hold. The difference is large enough that any rounding is irrelevant.
@@ZipplyZane true, but I would guess that for most standard calculators that limit is e100.
@@SadisticNiles Yeah. I was actually thinking of the graphing calculator I used in math classes. I don't remember where it maxed out, but it was over e100. It wouldn't surprise me if it just used 64-bit floats, which max out around e300.
The square is the rectangular shape with the maximal area given a fixed perimeter (it would be a circle if we consider any shape). Meaning that a rectangle with sides 50x50 has a greater area than a rectangle with sides 51x49, 52x48, 53x47, ..., 99x1, all of which are rectangles with the same perimeter as the square of side length 50. This can also been seen in 50^2 > (50+c)(50-c) for non-zero c (and equal for c=0), because (50-c)(50+c) = 50^2 - c^2.
This probably falls under cheating as well, but it got me the answer. The geometric mean of ninety-nine 50's is easy to calculate, it's 50. The geometric mean of the integers 1 through 99 is less than its arithmetic mean and is therefore less than 50. Since both terms can be rewritten as (geometric mean)^99, 50^99 must be bigger since it has the larger geometric mean.
Definitely isn’t cheating. It’s using AM-GM smartly to prove the general case in a more insightful way than the induction.
Better method: th-cam.com/video/KijvVFAu8WI/w-d-xo.html （2 topics included）
I didn't notice the factorial sign at first and thought you were comparing 50^99 to 99
I started panicking when I saw there was hardly a minute for the video to end and he didn’t start to explain the cheating method.
Nice one xp
Another way:
1. Write each in terms of logarithms - eg, 99! = e^\sum{\log i}, etc
2. Apply Jensen's inequality to the exponent of 99!
3. Compare 2. with the logarithmic exponent of 50^99 (the one you calculated in 1.)
4. Voila
This is a quick numerical way someone could do on their calculator: Take ln() of both sides, so we have 99*ln(50) and ln(99*98*...*2*1) = Sum from k=0 to k=99 of ln(k).
This way, one could raise both sides to the power of e after computing numerical values and tell by how much one side is greater than the other! :)
Just use Stirling's approximation log(n!) ~ nlog(n/e). log(99!) ~ 99log(99/e), comparing with 99log(50). 99/e < 50.
我有中文視頻！下面是英文的。
MY SOLUTIONS: th-cam.com/video/UzMyVXS0XMg/w-d-xo.html
th-cam.com/video/Bht9U5j76fY/w-d-xo.html
1:44 "49 + 49 is 98, +1 is 50"
Referring to the 50 in the middle
@@matthewlockard6599 I know, I'm just teasing.
I used my own derrived mathematical formula a x a = (a+b) x (a-b) + b x b (not sure if someone else came up with this before me, it would be nice to know, but it's rather obvious so not that special a formula anyhow). I derrived it for other purposes, but it turns out to be quite usefull to quickly answer this problem as well.
Which makes it easy to see that 50^99 > 99! since 99! equates to the latter without the bxb part when 50^99 equates to the first part. hence 50^99 is about the bxb part more than 99!.
Sorry if I've been a bit sloppy and short on the elaboration of my proof, but I'm quite sure it's sound enough (those interested can still request for more elaboration, and if so I gladly comply) . I consider this approach to be even simpler than the simple approach in the video.
I initially thought this was an overkill video. Missing your overkill vids.
I was expecting the "cheating method" to establish an upper bound on ln(1)+...+ln(99) using an integral from calculus such as 100*ln(100)-99. Subtracting from 99*ln(50) you will get that the ln(lhs/rhs) = 99-101*ln(2)-2*ln(5). Using approximations ln(2)
For the 'cheating' way, I thought you were going to apply the Stirling approximation for n!. Even though it is 'only' an approximation there are bounds on the error term in the approximation that I'd expect to be able to use to turn the argument into a rigorous proof. Haven't actually worked this out on paper.
I believe that MMA has magic to deal with precision
Easy idea. 99! is a sum multiply from 1 to 99. 50^99 is what it's looked like.
However, in these two numbers we could simply PAIR it like this:
the former 99! is a multiply of one 50 and a function of (50-x)(50+x) from x range to integer 1 to 49, (results in a multiple result consists of 99 factors)
the latter 50^99 is a multiply of 99 50s in a row.
Attention: (50-x)(50+x) is 2500-x^2 which is smaller than 50*50 (2500), thus the result of 99! is smaller than 50^99.
(Supplement: x ranges from 1 to 49 which means NO negative values involved in the factor of 2500-x^2, otherwise the previous conclusion would be not valid)
1:48 :"Notice 49+49=98+1=50" ExCuSe Me WtF?!! 😂😂
Referring to the 50 in the middle, the 98th exponent + 1 is 50.
You could also use Stirling's formula; 99! ~ [(99/e)^99 ] * sqrt(198 pi) < 38^99
huuunm
[ (50-n)*(50+n)=50² - n² ]
50² > 50² - n²
Me too
The middle number in 99! is 50. 99! can be recast in the following form: 50*(51*49)*(52*48)*(53*47)*...*(98*2)*(99*1). There are 49 products of the form (50+n)*(50-n). Expanding this product, one gets 2500-n^2. Therefore each product is less than 50^2. Whereas 50^99 can be written in the format: 50*(50*50)*(50*50)*...*(50*50), where there are 49 products of 50*50 in parentheses. Therefore, 50^99 is greater than 99!.
50^99=1.3*10^168 99!=9.33*10^155
I love induction because it’s like answering “Why is this true?” with “Because math says so”
… You’re just showing the previous case implies the current case. Nothing fancy.
Take log.
ln(99!)={sum(ln(x)) from x=1 to 99} is about the area under the curve y=lnx from 1 to 100.
ln(50^99)=99ln(50), is a rectangle with base 99, height ln(50). Log function increases faster in the small number range, so ln(50) looks greater than average height of ln(x) from x=1 to 100.
So the rectangle is larger. 50^99>99!.
The simple method is actually obvious. Why the complicated "general" method?
Basically every term over there is 50*50/(50 - x)(50 + x) which is 50^2/(50^2 - x^2) which is always >= 1. QED
If you take the natural logarithm of both sides you get 99*ln(50) vs ln(99!). Which can be written as ln(99) + ln(98) + ln(97)+.....ln(2)+ ln(1).
99*ln(50) will always be greater than ln(99!) Because of the way the natural logarithm increases very slowly.
Therefore (99*ln(50))/(ln(99!)) >1
So 50^99 > 99!
Thats the kind of question we get in JEE where we don't even have enough time .
we dont get these😂😉😉🙂
I happened to check the ratio 99! / 50^99 and used GM ≤ AM inequality ... and it straight up gave the max value to be 1 , obviously meaning that 50^99 is bigger.
My way was faster, easier and just as reliable! I basically went "idk 50^99 just feels bigger" and, clearly, I was right
The first solution can be described in one sentence: The surface of a rectangle with fixed circumference maximazes when it is a square.
Brilliant
Just posting a comment before it hits again in everyone's recommendation
The solution I came up with uses Karamata's inequality.
Taking the natural logarith on both sides, we have to compere the sum of the logariths up to 99 and the sum of 99 ln(50).
Karamata's inequality states that, for a concave function, if a series majorizes another (in other words, let A be a series (a1, a2, ..., an) and B be a series (b1, b2, ..., bn), then, it is said that A majorizes B if, the elements in each group are ordered in a non-increasing way and the sum of the k first elements in A is bigger than the sum of the k first elements in B for every 0
"98 + 1 is 50" Hmmm, is it?
Fourth way:
Consider the function f(x)=x^(99)-99!
Obviously that's an increasing function Solving f(x)=0 we find that :
exp(1/9 ×(the sum from 1 to 99 of ln(k))
which is equal to 37,62.. is the solution
As 50>37,7
f(50)>0
So 50^(99)>99!
Just a quick note guys: relying blindly on a software without any proof is very dangerous. :D
I'm pretty sure I can trust wolfram alpha
OR you can recognise that the maximum product of numbers that sum to a constant is when those numbers are equal.
eg. x + y + z = c, maximum x*y*z occurs when x=y=z=c/3
The terms of 50^99 have the same sum as 99! (see the pairings as in the first method, pairs sum to 100 with a 50 left over), thus 50^99 will be larger.
Happy diwali
We can use natural logarithm ( ln(x) )
We know, if ln(x) > ln(y) then x > y
ln(50^99) = 99*ln(50) and we can calculate this with normal calculator..
ln(99!) = ln(99*98*.....*2*1) = ln(99) + ln(98) + ..... + ln(1) = summation (k=1 -> 99) ln(k) and we can calculate this too
We find : ln(50^99) > ln(99!) 50^99 > 99!
50^99 > 99!
3:00 love how he says davaide
뭐야뭐야 알고리즘 땜에 여기 온 사람 나밖에 없나
Another "cheat" would be to take logarithms of both numbers. You'd then just compare 99log50 and
log1 + log2 + ...+ log99.
In method 1 there’s no need to calculate . All the denominators (but the center point 50/50) are
(50-k)(50+k) = 50^2 - k^2 < 50^2
You can take the logarithm of both sides and divide by 99, then use Jensen's inequality on the ln(x) function. The left side is ln(mean(S)) and the right side is mean(ln(S)) for S = {1, ..., 99}.