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@MathQueenSusanne ngl I just used hypercalc by javascript, but I love you method much better, actually helps you understand why the answer is the correct one, thank you
Efficiency principle gives the answer quickly and intuitively. 7*7 is larger than 6*8 (by 1). 10*10 is larger than 11*9 (by 1). The square of two numbers is always larger than (n+1)*(n-1) by 1. The further you go (n+2)(n-2) etc, the less efficient, the lower it is compared to n squared. This is just an axiom people can memorize.
Basic calculus teaches that for a rectangle with a fixed perimeter, the greatest possible area is a square. That means that if two numbers add to x, they have the greatest product when each of them is x/2. Both numbers are the product of 99 numbers. 50^99 makes all the numbers equal to have the greatest product. To give an example with small numbers, 2^3 = 8, and 3! = 6, so 2^3 > 3!
we could also just do for last pair of 51*49 and say that it's smaller than 50*50 and since 51*49 will be the largest of all pairs, hence all pairs will be smaller
They are actually "relatively" close. There's only a factor of 1690545151688.5560277... between the two numbers :) That may look like a lot, but both numbers are huge. 50^99 == 1.57*10^168 while 99! == 9.33*10^155
50*50 = 2500 and 51*49=2499, Now take the two values you will multiply example: (51*49) one easy way to analyze is averaging both numbers into a same number in this case 51+49 = 100 and its average would be 100/2 = 50, now express the multiplication as (51*49)=(50*50) - the squared positive amount your initial values deviate from your average value in this case 51 and 49 deviate by 1, so (51*49) = (50*50) - 1^2 = 2500 - 1 = 2499. Now (52*48)=(50*50) - 2^2 = 2500 - 4 = 2496 and so on so on notice as we deviate from the center value (50) we always decrease the amount by (deviation)^2.
I also thought about pairing the numbers immediately. After that it is easy because the area of a rectangle is greatest when it is a square, meaning X * X > Y * Z if 2*X = Y + Z and Y != Z. In this case: 50 * 50 (area of a square) > 99 * 1 or 51 * 49 - both rectangles with smaller area, but identical circumference.
99! equals the geometric mean of {1, 2, 3, ..., 99}, raised to the 99th power. AM-GM inequality says that the geometric mean of {1, 2, 3, ... 99} is less than the arithmetic mean of the same set, which happens to be 50.
That was my approach too. I see other people in the comments saying "it's trivial with the Stirling approximation" but I don't think it gets any more trivialized than this.
In case anyone was wondering how much smaller/larger: 99! ~= 9.33 x 10^155 50^99 ~= 1.58 x 10^168 So 50^99 is not only larger, it's larger by about 12 orders of magnitude.
I cheated and used Excel to get these answers (any spreadsheet works). It could calculate 99!; for 50^99 = 9.3326E +155 with the FACT function. I used 99 x log (50) which is 168.19803 or 10^0.19803 x 10^168 or 1.5777 E+168. Either of these could have been calculated manually with a table of logarithms though adding the logarithms of 99 numbers for the factorial would have been a chore!
I combined the pairs just as you, but as soon as I saw pairs like (99*1) which is of course smaller than (50*50), I jumped to verify that (51*49) is also smaller than (50*50). Since it's all linear, I knew that all terms in between were also smaller than (50*50) and had my answer.
I'd never considered factorial as being linear. Coming from a computer science background, we consider factorial as being the fastest growing and therefore the worst time complexity i.e. factorial > exponential.
@@toby9999 I agree with you, the computation time for factorial isn't linear. But what I meant was the products of each pair of numbers is... well 'monotonic', each product pair getting smaller and smaller with no 'bumps' or other rises. The highest value is the pair 50*50, and the product of all other pairs get smaller and smaller, continually shrinking until you get to the last pair, 1*99. So 50*50 is the largest product possible among all the pairs.
I paused the video and solved it before watching. I used the idea that , with the perimeters being equal, the area of a rectangle is always less than the area of a square. But i really like how you showed it as 50^2 - n^2 . That’s super clear!
the story of Carl Friedrich Gauss-who, as an elementary student in the late 1700s, amazed his teacher with how quickly he found the sum of the integers from 1 to 100 to be 5,050. Gauss recognized he had fifty pairs of numbers when he added the first and last number in the series, the second and second-last number in the series, and so on. For example: (1 + 100), (2 + 99), (3 + 98), . . . , and each pair has a sum of 101. This problem reminded me a lot of this old one.
Figured it out in my head in about 20 seconds. You line up 99 50's and you line up 1 to 99 underneath. Take the first and last of both lines: 50 times 50 is more than 1 times 99. Then the next pair: 50 times 50 is more than 2 times 98. Keep pairing them off and the 50 times 50 wins every time. So 50^99 is MUCH bigger than 99! No algebra or equations needed. Just common sense.
I actually did it a completely different way. If you take logs of both numbers (you get the same answer because log is an increasing function), you get the sum from n=1 to 99 of log 50 vs the sum from n=1 to 99 of log n. Then, because log is concave down, the box through the midpoint is larger than the area under the curve (the values of log that are larger than log 50 increase by less than the values that are smaller than log 50 decrease). The techniques are related, in that mine can be seen as pairing up contributions from each side of the midpoint, but I think it's neat to do it based only on the general shape of a curve and no specific calculations at all.
Very good, but that approach feels even less intuative. I got it wrong, sadly. And even after watching the proof, I still "feel" that factorial should beat exponential, all else being equal.
You are the best math explainer on the planet. The only thing I would have added is showing how the product of the binomial pair is the first digit squared minus the second digit squared. A quick multiplication would show how the two middle numbers cancel out.
I have been watching your german channel for quite a while. It is great. Funny that you changed the shape of the number "1" compared to the german style of writing. Grüße und viel Erfolg.
Hey Mark, nice to see you here as well! 😍 Yes, an American friend of mine told me, that no one writes the 1 like I do 😅 So I adjusted it for the English videos. It’s not easy though, because I was used to write it in a different way my whole life. But I like challenges!
@@MathQueenSusanne Ich habe mir die amerikanische Schreibweise auch in Deutschland angewöhnt, damit ich nicht immer wechseln muss. Meine Tochter (Grundschule) beäugt aber meine Zahlen argwöhnisch....
By intuition, a square will always have more area than a rectangle that has the same perimeter. However, the actual demonstration here I would never thought of, pretty cool.
The solution that i thought of was take log of both. We know that if a>b and a,b>0 and base is greater than 1 then log(a)>log(b) Now we calculate 99log(50) which is 99+99log(5) which is roughly 165. Now log(99!) According to stirling approximation it is approximately 99log(99) - 99log(e) which comes out to be 154 approx. So it means 50^99 is greater
I did not do the calculation but instead looked at a similar concept, 10x10 vs 5x15 to see which is higher. This gave me an idea of which would likely be greater in a more complicated situation. That's how I guessed 50^99 > 99!.
I did it pretty much as you did; rearranged the factors in the factorial expansion and then noticed that they were all pairing up into sub-expressions of the form (50 - x) * (50 + x), which is going to be equal to 50 ** 2 - x ** 2. This is clearly smaller than the 50 ** 2 from the corresponding positions in the expansion of 50 ** 99.
I loved the explanation❤❤❤ Made med understand how and why I immediately said 50⁹⁹ is larger. Your explanation made me remember the methodes I probably forgot two decades ago, but still remembered the results, without being able to prove it so nicely. ❤❤❤
I figured all those smaller than 50 numbers would drag down the total and the pure 50s would prevail. Especially since the last term of the factorial was 1. Whereas the 50s were just cranking along each time doing their 50 thing. Goes to show that steady, even power is better than a flash in the pan. Many thanks, great vid!
The power had a significant head start, already half the distance with its base. I wonder what the base has to be to make it even or have the factorial outgrow the power. Interesting follow up question
Nice method to get a rigorously proof! As a quick way to get a feel for the answer, I was looking at the ratios of the factors from 99! to those of 50^99. The one in the middle, 50/50, does not favor either side. The ratios going towards 99 all give 99! a small boost between 1 and 2. The equal number of ratios towards 1, on the other hand, get smaller much faster ending up at 1/50. So they have a much bigger effect in favor of 50^99.
Well done Math Queen ,The prince of Math Friedrich Gaus" The PRINCE of MATH" used the principle of pairing also when he easily found the sum of all numbers from 1 to 100 and multiplied the sum by 50 .
Stirling approximation is allowed?? ln(x!) = x ln(x) - x approximately, which for x=99 is about 356. And ln(50^99) = 99 ln(50) is about 387. Since exp(387) > exp(356) we have 50^99 > 99!
Impressive, but I didn't take it nearly to that length. However, I know the power of compound interest and my mind applied it immediately so I got the correct answer!
8×10^67 is the probability of shuffling a deck of cards the same way twice. It's also the estimated amount of atoms in the universe. So when I saw 50^50 i knew that must've been the one! 😅 Great video today, I would have struggled to show my work!
For me, I care more about where to find the answer, than how to find the answer. It saves a lot of time. Wolfram Alpha gave me the answer in less than 30 seconds.
Good explanation. Just I saw it I saw the answer is 50^99 because the multiplication of a.b where a + b = c get bigger when a and b get closer to each other. For example among positive integers whose sum is 10, 5x5 will give the highest value of multiplication (25) and you are farther away from the other number, the smaller the multiplication (4x6 = 24, 3x7 = 21, 2x8 = 16, 1x9 = 9, 0x10 = 0). Conveniently the given numbers in the question have exactly the same pattern so I immediately found the answer.
Just an intuitive guess: 99log(50) has to be greater than log1 + log2 + ….+ log99 as log x steeply rises and then flattens with increasing x. log 99 is not way too far from log 50 when compared with difference between log 1 log 50. log x < log 50 for all x < 50. So, on one hand log 1 … log 50 have a huge deficit with compared with 50 log 50. And then the right side of 50 does not add too much to match the remaining 49 log 50 as the logarithm curve flattens rather fast.
My intuitive thinking was that the numbers bigger than 50 are only slightly bigger but the numbers less than 50 are a lot less (in terms of ratios, which is what matters when we are multiplying them). Therefore, the numbers that are less will dominate, so 99! is less. Your method is much easier to make rigorous, though.
Thinking more about it, I can make my method rigorous: Replace the 99 to 51 with 50*2 Replace the 49 to 26 with 50 Replace the 25 to 13 with 50/2 Replace the 12 to 6 with 50/4 Replace the 5 to 3 with 50/8 Replace the 2 with 50/16 Replace the 1 with 50/32 All of those make the numbers bigger. So we have 99!
Місяць тому
@@thomasdalton1508 Yes. Good thinking. One could just since multiplication is commutative rearange th factors (for example 1•2•98•99 is the same as 1•99•2•98) and 1•99, 2•98 etcetera are all clearly less than 2500. Your proportionality argument is great. (Of course it is dependent on the actual numbers here. Can be more tricky in general). Best Greetings from Sweden.
It's such an easy question to answer, no need to even start watching the video. 99*1 and 98*2 are both much smaller than 50*50, and so is every other combination.
I feel like at the difference of squares part, instead of all that you can just actually multiply a few of the pairs of numbers and see what they are. You can quickly tell that 50 x 50 is the largest product of numbers that sum to 100.
"Simply"? And why start with n=4? It's already true for n greater than 1; see below: n=2: 3!=6 < 2^3=8 n=3: 5!=120 < 3^5=27*9=180+67= And instead of total induction, (i.e. if a claim is true for lower indeces, you can derive for a next higher index it must also be true and you also have enough examples of the claim being true for low indeces to apply this at once, then claim must be true for al the indeces starting from the examples and higher up), I'd rather just directly derive it from the claim itself via the method showed in the video. i.e. (2n-1)! has factors of the shape (n-k)(n+k)=n^2-k^2 for each k=1...n-1, so n-1 such factors in total, and because each of these factors is smaller than n^2 , which n^(2n-1) also has n-1 times as factor and the single not yet compared factor in (2n-1)! and n^(2n-1) is the one where they're equal: n, (and so, following my 'because') (2n-1)!1 and n is an integer.
What I'm interesting in, is how much 99! is, without computating it. At least, how much zero's does the result have. So, 10log1+10log2+10log3...10log99. And then use some kind of algorithm to calculate those logs and add them up, without computating them 1 by 1.
Fun but good exercises Thanks. I just thought about the simpler problem going from 1 to 10 and comparing pairs like 4x6=24 3x7=21 always going to be less than the pairs 5x5 so its clear the factorial loses for once! Did it in my head in about 1 min 30 sec. :)
Nice. I did a rough and ready estimate to guess the answer quickly: 99! is no more than about 2^50 (50!^2) = (2^{25} 50!)^2, and 2^{25} 50! is no more than about (50!/25!)^2. But 50^99 is about (50^25)^4, so we are left to compare 50^{25} with 50!/25!, and the first is clearly much larger. I've been a bit sloppy, but the difference is so large that it swamps any errors in the estimates. Although I'm a mathematician, I find it helpful to think like an engineer sometimes, and with a bit more care these estimates could be turned into a proof, now that we know which way to do the inequalities in the estimates.
I just calculated 9! and 5^9 and compared the results. The difference was largely in favor for 5^9 and so I assumed, that what happens with the litle it is true for the big. And in that case it is true, because the structure of this problem has the same proportion of my minature approach.
Since I was a kid I noticed that factors with an equal total sum are most "effective" as the factors become closer to each other. So 50^99 is a no-brainer. Never really thought about a n-dimensional area/circumference metric.
It is interesting to solve the equation a^n=n!. Using stirling's approximation for factorials and after simplifying it turns out the relation between a and n is linear for larger n. The relationship is a=n/e or a=0.38n
From the thumbnail, I did it essentially the same but without writing anything. All of the pairs from 99! are less than 50*50, so 99! is smaller, and by quite a lot. It's good that you did a rigorous proof, but that's really not necessary. You've got a lovely, soothing voice. I could listen to you talk all day.
Usually I come up with a different method from the video, but this time, I was surprised, we used the same method! Rewrite 99 factorial as differences of squares. Out of curiosity, I asked ChatGPT. ChatGPT suggested comparing the log of 50^99 with Stirling's approximation for the log of 99!. ln(50^99) = 99 ln 50, versus ln(99!) ≈ (99 ln 99) - 99 + a small error term. Interesting start, but ChatGPT started making numbers up after that step.
Yeah, ChatGPT is incapable of doing math - especially higher math - correctly. It is a TEXT-based program meaning it just generates text, that is likely to be the result of your query. So it doesn't calculate or understand your question, it just generates a semi-correct string of words.
My gut lazy math made me seek a near minimum using rounding and scientific notation. I floored all 2 digit numbers to 10 and all 1 digit numbers to 1. 50^99 led to 1e100, mentally. 99! led to 1e90, mentally. I don't trust my mental math accuracy because 1 off errors are all too common, so my approximations felt like they have an error range of +-1e2. 1e100+-1e2 > 1e90+-1e2, hence my near minimum led me to believe 50^99 is larger. A 1e6 difference is quite sizable. Doing this mentally was quite quick. I surprised myself that the true answer reflected my hunch. I would like to thank the game Balatro for getting me reintroduced to scientific notation math, which most likely led to this thought process. My goal wasn't accuracy, but quick and rough comparison.
I like her work. We must ,however, be careful if we have x by x by x for example to say we use it as a factor three times not that they are multiplied three times. Math is fun.
Hi Susanna. Awesome video. Thank you. A mind expanding exercise. Your English is fantastic. One small suggestion. You use the word “structure” in the video. In my opinion the word “pattern” is a better English word for this use case.
I remember that for a given circumference a square is the rectangle with the largest area, so after the pairing of numbers (49--1, 48--2...) it quickly became obvious 50^99 is larger than 99!.
When you highlight things in green for example; the bigger number number is 50 to the power of 99. But yeah I get this is about math like in another way and so on. Thanks Queen.
i put 49x50x51 in a calculator and saw it was slightly smaller than 50x50x50, like many in the comments point out, I could have skipped the middle 50's. I assume that constitutes an official proof because you can argue that this effect will get ever larger when you go further out from the middle.
You can think about scaling a number with a coefficient equal to the ratios of factors such as 50/99 and 50/1. 50/99 in logarithmic sense is very ineffective in scaling compared to 50/1. So 50’s take over the scaling, and scale the term (that is 50 to some power between 1 and 99) much more in positive direction.
I experimented with 5^2 and 10*9, then 5^3 and 10*9*8, until I had 5^10 and 10!. Each had a different slope, and different final number, both of which indicated to me by analogy that 50^99 would be larger than 99!.
If you think about it the average multiplier of the factorial is 50 and there are 99 factors but as their disparity is more like 4x4 is larger than 3x5 so 50^99 will be larger
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Hey math friends! If you’re enjoying this video, could you double-check that you’ve liked it and subscribed to the channel? It’s a simple equation: your support + my passion = more great content! Thanks for helping me keep this going - you’re the best!
Harmonic is always superior
Dr PK also presented this same question using three different methods🎉
Well, the more difficult question would be: How much smaller is 99! than 50^99 ?
nicee video, may allah bless you.
@MathQueenSusanne ngl I just used hypercalc by javascript, but I love you method much better, actually helps you understand why the answer is the correct one, thank you
Efficiency principle gives the answer quickly and intuitively.
7*7 is larger than 6*8 (by 1).
10*10 is larger than 11*9 (by 1).
The square of two numbers is always larger than (n+1)*(n-1) by 1. The further you go (n+2)(n-2) etc, the less efficient, the lower it is compared to n squared.
This is just an axiom people can memorize.
Basic calculus teaches that for a rectangle with a fixed perimeter, the greatest possible area is a square. That means that if two numbers add to x, they have the greatest product when each of them is x/2. Both numbers are the product of 99 numbers. 50^99 makes all the numbers equal to have the greatest product. To give an example with small numbers, 2^3 = 8, and 3! = 6, so 2^3 > 3!
Simply explained with great patience and clarity, wonderful!
Thank you so much for your kind words!
excuse me, simply????
we could also just do for last pair of 51*49 and say that it's smaller than 50*50 and since 51*49 will be the largest of all pairs, hence all pairs will be smaller
This is how I got the answer.
Once you know this, you realize 50^99 is the larger of the two expressions.
(50 + n)(50 - n) = 50² - n² < 50². I am actually surprised however, intuitively I thought the factorial would win.
Perhaps it would be more intuitive to think that the largest area rectangles for any given perimeter are squares.
106! > 50^99 and 133! > 50^133
@@timbombadil4046 My thought, too (before watching the video). But I generalised to (x-a)(x+a)=x²-a²
They are actually "relatively" close. There's only a factor of 1690545151688.5560277... between the two numbers :) That may look like a lot, but both numbers are huge. 50^99 == 1.57*10^168 while 99! == 9.33*10^155
@@Bunny99s I guess the balance of my bank account is close to elon musk's then
So nice to listen to English spoken with a natural voice! (not AI) Vilen Dank, Susanne!
I think the world would be a far better place if there were more maths teachers like Susanne
Aaaaw, how cute is that! Thank you so much ❤️
Crystal clear, delivered cheerfully. What more can you ask for?
It's interesting. We can determine immediately which one is larger by comparing (50*50) and (51*49).
How
50*50 = 2500 and 51*49=2499, Now take the two values you will multiply example: (51*49) one easy way to analyze is averaging both numbers into a same number in this case 51+49 = 100 and its average would be 100/2 = 50, now express the multiplication as (51*49)=(50*50) - the squared positive amount your initial values deviate from your average value in this case 51 and 49 deviate by 1, so (51*49) = (50*50) - 1^2 = 2500 - 1 = 2499. Now (52*48)=(50*50) - 2^2 = 2500 - 4 = 2496 and so on so on notice as we deviate from the center value (50) we always decrease the amount by (deviation)^2.
I also thought about pairing the numbers immediately. After that it is easy because the area of a rectangle is greatest when it is a square, meaning X * X > Y * Z if 2*X = Y + Z and Y != Z.
In this case: 50 * 50 (area of a square) > 99 * 1 or 51 * 49 - both rectangles with smaller area, but identical circumference.
This is what I thought as well intuitively.
99! equals the geometric mean of {1, 2, 3, ..., 99}, raised to the 99th power. AM-GM inequality says that the geometric mean of {1, 2, 3, ... 99} is less than the arithmetic mean of the same set, which happens to be 50.
That was my approach too. I see other people in the comments saying "it's trivial with the Stirling approximation" but I don't think it gets any more trivialized than this.
wow, your method is amazing, thanks for expanding my knowledge ❤
Literally the first thing that popped into my head was that the sum of numbers from 1 to 99 is 99 times 50. So the AM GM inequality is useful here.
at what point would the second number be bigger? if we substitute the 50 with 40 or 30?
heh i tested Leo AI on this and it approximated 4.595
Why is that relevent? I don't understand the significsnce of those means
In case anyone was wondering how much smaller/larger:
99! ~= 9.33 x 10^155
50^99 ~= 1.58 x 10^168
So 50^99 is not only larger, it's larger by about 12 orders of magnitude.
I cheated and used Excel to get these answers (any spreadsheet works). It could calculate 99!; for 50^99 = 9.3326E +155 with the FACT function. I used 99 x log (50) which is 168.19803 or 10^0.19803 x 10^168 or 1.5777 E+168. Either of these could have been calculated manually with a table of logarithms though adding the logarithms of 99 numbers for the factorial would have been a chore!
I combined the pairs just as you, but as soon as I saw pairs like (99*1) which is of course smaller than (50*50), I jumped to verify that (51*49) is also smaller than (50*50). Since it's all linear, I knew that all terms in between were also smaller than (50*50) and had my answer.
That's what my thoughts were also. I thought she made is needlessly complicated ;).
I'd never considered factorial as being linear. Coming from a computer science background, we consider factorial as being the fastest growing and therefore the worst time complexity i.e. factorial > exponential.
@@toby9999 I agree with you, the computation time for factorial isn't linear. But what I meant was the products of each pair of numbers is... well 'monotonic', each product pair getting smaller and smaller with no 'bumps' or other rises. The highest value is the pair 50*50, and the product of all other pairs get smaller and smaller, continually shrinking until you get to the last pair, 1*99. So 50*50 is the largest product possible among all the pairs.
I paused the video and solved it before watching. I used the idea that , with the perimeters being equal, the area of a rectangle is always less than the area of a square. But i really like how you showed it as 50^2 - n^2 . That’s super clear!
your method is pretty interesting, could please explain that in detail to me,please.
the story of Carl Friedrich Gauss-who, as an elementary student in the late 1700s, amazed his teacher with how quickly he found the sum of the integers from 1 to 100 to be 5,050. Gauss recognized he had fifty pairs of numbers when he added the first and last number in the series, the second and second-last number in the series, and so on. For example: (1 + 100), (2 + 99), (3 + 98), . . . , and each pair has a sum of 101.
This problem reminded me a lot of this old one.
Randomly opened the video, just wanted to note that you are very pretty, and explaining math problems makes you so much more pretty 😊 have a great day
Knowing the optimal rectangle (area wise) is a square makes it easy to see this right away.
Figured it out in my head in about 20 seconds. You line up 99 50's and you line up 1 to 99 underneath. Take the first and last of both lines: 50 times 50 is more than 1 times 99. Then the next pair: 50 times 50 is more than 2 times 98. Keep pairing them off and the 50 times 50 wins every time. So 50^99 is MUCH bigger than 99! No algebra or equations needed. Just common sense.
Agreed. This video would be the most detailed illustration and proof of why that's the case but overall it's far simpler and more intuitive.
I actually did it a completely different way. If you take logs of both numbers (you get the same answer because log is an increasing function), you get the sum from n=1 to 99 of log 50 vs the sum from n=1 to 99 of log n. Then, because log is concave down, the box through the midpoint is larger than the area under the curve (the values of log that are larger than log 50 increase by less than the values that are smaller than log 50 decrease).
The techniques are related, in that mine can be seen as pairing up contributions from each side of the midpoint, but I think it's neat to do it based only on the general shape of a curve and no specific calculations at all.
Very good, but that approach feels even less intuative. I got it wrong, sadly. And even after watching the proof, I still "feel" that factorial should beat exponential, all else being equal.
Wait a minute ! You have a German channel of mathematics 😮😲
Her mind is strong, intuitively I knew the answer but I wouldn’t be able to prove it as elegantly as her
Didn't view the video. Only viewed the 10:54 and realized: right you are! Saved me 9 minutes!
You are the best math explainer on the planet. The only thing I would have added is showing how the product of the binomial pair is the first digit squared minus the second digit squared. A quick multiplication would show how the two middle numbers cancel out.
I have been watching your german channel for quite a while. It is great. Funny that you changed the shape of the number "1" compared to the german style of writing. Grüße und viel Erfolg.
Hey Mark, nice to see you here as well! 😍 Yes, an American friend of mine told me, that no one writes the 1 like I do 😅 So I adjusted it for the English videos. It’s not easy though, because I was used to write it in a different way my whole life. But I like challenges!
@@MathQueenSusanne Ich habe mir die amerikanische Schreibweise auch in Deutschland angewöhnt, damit ich nicht immer wechseln muss. Meine Tochter (Grundschule) beäugt aber meine Zahlen argwöhnisch....
Great puzzle! I'm not a mathematician, but your videos make me love math!
By intuition, a square will always have more area than a rectangle that has the same perimeter. However, the actual demonstration here I would never thought of, pretty cool.
am-gm inequality gives this result!
And the square has more area because it is a better approximation to the optimal - a circle
The solution that i thought of was take log of both.
We know that if a>b and a,b>0 and base is greater than 1 then log(a)>log(b)
Now we calculate 99log(50) which is 99+99log(5) which is roughly 165.
Now log(99!) According to stirling approximation it is approximately 99log(99) - 99log(e) which comes out to be 154 approx.
So it means 50^99 is greater
They are both the nintynineth element of different sequences. Just check which sequence grows faster.
Very nicely explained...
11-12 year English speaking students can also understand easily.
I did not do the calculation but instead looked at a similar concept, 10x10 vs 5x15 to see which is higher. This gave me an idea of which would likely be greater in a more complicated situation.
That's how I guessed 50^99 > 99!.
I did it pretty much as you did; rearranged the factors in the factorial expansion and then noticed that they were all pairing up into sub-expressions of the form (50 - x) * (50 + x), which is going to be equal to 50 ** 2 - x ** 2. This is clearly smaller than the 50 ** 2 from the corresponding positions in the expansion of 50 ** 99.
not yt recommending me this channel after me knowing the german channel for a few years
It's great to see you here, too! 😊
It was so simple after you explained it😭
Wow, that was a lotta steps but it worked quite swimmingly. Thanks' for the tutorial. I really love this channel!
Largest factorial pair 51.49 < 50.50
All factorial pairs < 50.50
Therefore 50^99 > 99!
I loved the explanation❤❤❤
Made med understand how and why I immediately said 50⁹⁹ is larger.
Your explanation made me remember the methodes I probably forgot two decades ago, but still remembered the results, without being able to prove it so nicely.
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Enjoyed the process!
I figured all those smaller than 50 numbers would drag down the total and the pure 50s would prevail. Especially since the last term of the factorial was 1. Whereas the 50s were just cranking along each time doing their 50 thing. Goes to show that steady, even power is better than a flash in the pan. Many thanks, great vid!
That's such a wild way of saying it, I love it
The power had a significant head start, already half the distance with its base.
I wonder what the base has to be to make it even or have the
factorial outgrow the power.
Interesting follow up question
Simple comparison 50^2=2500 99X1=99. The largest number on the right is 49X51=2499. The terms on the right will always be smaller than on the left.
Nice method to get a rigorously proof!
As a quick way to get a feel for the answer, I was looking at the ratios of the factors from 99! to those of 50^99.
The one in the middle, 50/50, does not favor either side. The ratios going towards 99 all give 99! a small boost between 1 and 2. The equal number of ratios towards 1, on the other hand, get smaller much faster ending up at 1/50. So they have a much bigger effect in favor of 50^99.
Well done Math Queen ,The prince of Math Friedrich Gaus" The PRINCE of MATH" used the principle of pairing also when he easily found the sum of all numbers from 1 to 100 and multiplied the sum by 50 .
Enjoy.....we love your teaching style....keep up the good work Tiger 🐅......we all love and respect you...👍👍💯💯
This is what internet supposed to be. Thank for such cool content.
Stirling approximation is allowed?? ln(x!) = x ln(x) - x approximately, which for x=99 is about 356. And ln(50^99) = 99 ln(50) is about 387. Since exp(387) > exp(356) we have 50^99 > 99!
Impressive, but I didn't take it nearly to that length. However, I know the power of compound interest and my mind applied it immediately so I got the correct answer!
Thank you. Really beautiful!
8×10^67 is the probability of shuffling a deck of cards the same way twice. It's also the estimated amount of atoms in the universe. So when I saw 50^50 i knew that must've been the one! 😅 Great video today, I would have struggled to show my work!
Enjoy your video from Indonesia🇮🇩
For me, I care more about where to find the answer, than how to find the answer. It saves a lot of time. Wolfram Alpha gave me the answer in less than 30 seconds.
Good explanation. Just I saw it I saw the answer is 50^99 because the multiplication of a.b where a + b = c get bigger when a and b get closer to each other. For example among positive integers whose sum is 10, 5x5 will give the highest value of multiplication (25) and you are farther away from the other number, the smaller the multiplication (4x6 = 24, 3x7 = 21, 2x8 = 16, 1x9 = 9, 0x10 = 0). Conveniently the given numbers in the question have exactly the same pattern so I immediately found the answer.
You have such a beautiful brain! 🧠
Just an intuitive guess: 99log(50) has to be greater than log1 + log2 + ….+ log99 as log x steeply rises and then flattens with increasing x. log 99 is not way too far from log 50 when compared with difference between log 1 log 50. log x < log 50 for all x < 50.
So, on one hand log 1 … log 50 have a huge deficit with compared with 50 log 50. And then the right side of 50 does not add too much to match the remaining 49 log 50 as the logarithm curve flattens rather fast.
My intuitive thinking was that the numbers bigger than 50 are only slightly bigger but the numbers less than 50 are a lot less (in terms of ratios, which is what matters when we are multiplying them). Therefore, the numbers that are less will dominate, so 99! is less. Your method is much easier to make rigorous, though.
Thinking more about it, I can make my method rigorous:
Replace the 99 to 51 with 50*2
Replace the 49 to 26 with 50
Replace the 25 to 13 with 50/2
Replace the 12 to 6 with 50/4
Replace the 5 to 3 with 50/8
Replace the 2 with 50/16
Replace the 1 with 50/32
All of those make the numbers bigger. So we have 99!
@@thomasdalton1508 Yes. Good thinking. One could just since multiplication is commutative rearange th factors (for example 1•2•98•99 is the same as 1•99•2•98) and 1•99, 2•98 etcetera are all clearly less than 2500. Your proportionality argument is great. (Of course it is dependent on the actual numbers here. Can be more tricky in general). Best Greetings from Sweden.
The method you describe is exactly the method in the video.
@@thomasdalton1508 Yes of course
This was my first thought too. The small numbers are many times less than 50, but the big ones are only less that 2 times bigger.
great to see you on your english channel now, good luck!
Thank you so much! Just started a few weeks ago, but I’m so happy about all the feedback!
Very nice approach
Thank you for this video. Using the same grouping by 2, we can simply show x.(100-x) =0, which is true for all x.
It's such an easy question to answer, no need to even start watching the video. 99*1 and 98*2 are both much smaller than 50*50, and so is every other combination.
Yes, but in these types of questions they usually ask you to provide a proof
This comment made me lose neurons
Don't be a jackass. You wouldn't make this comment on a video made by a man.
@@adityakhanna113he is right though
Well said
Very confusing for moderators! It shouldn't be this long video
So much confusing... But very well explained! Very nice
I picked 50 to the power of 99 and jumped to the end...... I can not handle equations like this. 😵💫😭😱
After pairing the numbers, it was easy to observe a few facts:
99.1 = 99
Not only was my first instinct wrong, it was really wrong. This became crystal clear around the 4:14 mark.
I feel like at the difference of squares part, instead of all that you can just actually multiply a few of the pairs of numbers and see what they are. You can quickly tell that 50 x 50 is the largest product of numbers that sum to 100.
Wow!! Nice proof.
Nice! it's also known that the area of a rectangle is largest when it's a square. so in the 99! you have the square only at the middle.
Simply prove that for any n greater than 3, (2n-1)! < n^(2n-1) by induction
"Simply"?
And why start with n=4? It's already true for n greater than 1; see below:
n=2: 3!=6 < 2^3=8
n=3: 5!=120 < 3^5=27*9=180+67=
And instead of total induction, (i.e. if a claim is true for lower indeces, you can derive for a next higher index it must also be true and you also have enough examples of the claim being true for low indeces to apply this at once, then claim must be true for al the indeces starting from the examples and higher up), I'd rather just directly derive it from the claim itself via the method showed in the video. i.e. (2n-1)! has factors of the shape (n-k)(n+k)=n^2-k^2 for each k=1...n-1, so n-1 such factors in total, and because each of these factors is smaller than n^2 , which n^(2n-1) also has n-1 times as factor and the single not yet compared factor in (2n-1)! and n^(2n-1) is the one where they're equal: n, (and so, following my 'because') (2n-1)!1 and n is an integer.
What I'm interesting in, is how much 99! is, without computating it. At least, how much zero's does the result have. So, 10log1+10log2+10log3...10log99. And then use some kind of algorithm to calculate those logs and add them up, without computating them 1 by 1.
Fun but good exercises Thanks. I just thought about the simpler problem going from 1 to 10 and comparing pairs like 4x6=24 3x7=21 always going to be less than the pairs 5x5 so its clear the factorial loses for once! Did it in my head in about 1 min 30 sec. :)
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Super lieb von dir, Dankeschön! Ich bin echt überrascht wie gut der Kanal angelaufen ist!
Very nice^ thank you!
Nice. I did a rough and ready estimate to guess the answer quickly: 99! is no more than about 2^50 (50!^2) = (2^{25} 50!)^2, and 2^{25} 50! is no more than about (50!/25!)^2. But 50^99 is about (50^25)^4, so we are left to compare 50^{25} with 50!/25!, and the first is clearly much larger. I've been a bit sloppy, but the difference is so large that it swamps any errors in the estimates. Although I'm a mathematician, I find it helpful to think like an engineer sometimes, and with a bit more care these estimates could be turned into a proof, now that we know which way to do the inequalities in the estimates.
Really astonishing ❤🎉 u gained a New subscriber (nee)
I just calculated 9! and 5^9 and compared the results. The difference was largely in favor for 5^9 and so I assumed, that what happens with the litle it is true for the big. And in that case it is true, because the structure of this problem has the same proportion of my minature approach.
This reminds me of the famous time when the little boy Gauss added all the numbers from 1 to 100 and amazed his teacher.
The general problem would as below.
Given equation: x^y
Since I was a kid I noticed that factors with an equal total sum are most "effective" as the factors become closer to each other. So 50^99 is a no-brainer. Never really thought about a n-dimensional area/circumference metric.
It is interesting to solve the equation a^n=n!. Using stirling's approximation for factorials and after simplifying it turns out the relation between a and n is linear for larger n. The relationship is a=n/e or a=0.38n
Very cool problem! Each step i was like - ok i get that, but where to next? I'd never figure this out, not in 99! years.
Finally, someone worthy of my love
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Math is worthy of all of our love! 😂
I love this VDO clip.
Thank you so much!
thanks for this video ❤❤
Thank you so much for watching! ❤️
Thank you.. You are very smart
From the thumbnail, I did it essentially the same but without writing anything. All of the pairs from 99! are less than 50*50, so 99! is smaller, and by quite a lot. It's good that you did a rigorous proof, but that's really not necessary.
You've got a lovely, soothing voice. I could listen to you talk all day.
Usually I come up with a different method from the video, but this time, I was surprised, we used the same method! Rewrite 99 factorial as differences of squares.
Out of curiosity, I asked ChatGPT. ChatGPT suggested comparing the log of 50^99 with Stirling's approximation for the log of 99!. ln(50^99) = 99 ln 50, versus ln(99!) ≈ (99 ln 99) - 99 + a small error term. Interesting start, but ChatGPT started making numbers up after that step.
Yeah, ChatGPT is incapable of doing math - especially higher math - correctly. It is a TEXT-based program meaning it just generates text, that is likely to be the result of your query. So it doesn't calculate or understand your question, it just generates a semi-correct string of words.
My gut lazy math made me seek a near minimum using rounding and scientific notation. I floored all 2 digit numbers to 10 and all 1 digit numbers to 1. 50^99 led to 1e100, mentally. 99! led to 1e90, mentally. I don't trust my mental math accuracy because 1 off errors are all too common, so my approximations felt like they have an error range of +-1e2. 1e100+-1e2 > 1e90+-1e2, hence my near minimum led me to believe 50^99 is larger. A 1e6 difference is quite sizable. Doing this mentally was quite quick. I surprised myself that the true answer reflected my hunch. I would like to thank the game Balatro for getting me reintroduced to scientific notation math, which most likely led to this thought process. My goal wasn't accuracy, but quick and rough comparison.
I like her work. We must ,however, be careful if we have x by x by x for example to say we use it as a factor three times not that they are multiplied three times. Math is fun.
Nice work
Hi Susanna. Awesome video. Thank you. A mind expanding exercise. Your English is fantastic. One small suggestion. You use the word “structure” in the video. In my opinion the word “pattern” is a better English word for this use case.
Thank you so much for your kind feedback! I try to use this word instead then. Awesome, that you help me getting better!
I remember that for a given circumference a square is the rectangle with the largest area, so after the pairing of numbers (49--1, 48--2...) it quickly became obvious 50^99 is larger than 99!.
If you know Stirling's approximation, then it is trivial. This way is more intuitive though.
When you highlight things in green for example; the bigger number number is 50 to the power of 99. But yeah I get this is about math like in another way and so on. Thanks Queen.
I am in love ( with math )
I just learned of young Carl Gauss and his quickly solving by realizing 1+100,=101 2+99=101, 3+98=101... Then 50 pairs of 101 were 5050.
jac7895 said it better
i put 49x50x51 in a calculator and saw it was slightly smaller than 50x50x50, like many in the comments point out, I could have skipped the middle 50's. I assume that constitutes an official proof because you can argue that this effect will get ever larger when you go further out from the middle.
Excellent mam❤❤
Thank you so much! ❤️
You can think about scaling a number with a coefficient equal to the ratios of factors such as 50/99 and 50/1. 50/99 in logarithmic sense is very ineffective in scaling compared to 50/1. So 50’s take over the scaling, and scale the term (that is 50 to some power between 1 and 99) much more in positive direction.
How would you go about proving that 99! < 40^99? By the Stirling formula we have N! ≈ √(2πN)•(N/e)^N. On closer look one finds 37^99 < 99! < 38^99.
I experimented with 5^2 and 10*9, then 5^3 and 10*9*8, until I had 5^10 and 10!. Each had a different slope, and different final number, both of which indicated to me by analogy that 50^99 would be larger than 99!.
If you think about it the average multiplier of the factorial is 50 and there are 99 factors but as their disparity is more like 4x4 is larger than 3x5 so 50^99 will be larger
Difference of squares...nice
Well, thats the first time ive seen "!" used in mathematics notation other than as an exclamation.
Thanks
I've been seeing ! used for 30 years but still got it wrong :)
n! read aloud as "n factorial" shows up a lot in combinatorics. You will see it used in probability and statistics.