I am nowhere near fluent enough in math to calculate this problem. I went with 49^51 being larger based on one simple thought.... If you multiply 50 by itself 50 times you get a number that is astronomically huge. If you then multiply 49 with itself 50 times you get a smaller number that is also astronomically huge. But that smaller number will then get multiplied with 49 one more time, which will be a much bigger number. Then, to test my idea I tried smalle numbers. 10^10 is 10 billion. 9^11 is more than 3 times higher.

Logically, the power is always the most powerful part of a number.

The simplest solution would be to let 50=n, 49=n-1, and 51=n+1. Taking logarithms of both numbers: Log n^n = nlogn and log (n-1)^(n+1)=(n+1)log(n-1). For a large number n, log n is approx. =log (n-1). Therefore, we are left with 2 values n and n+1 and obviously. n+1 is greater than n, which is why 49^51 is bigger than 50^50.

We need to compare f = (n+1)^(n+1)/n^(n+2) with 1 for a large n. This expression can be rewritten as f = (1+1/n)^n * (1+1/n)/n. Then using the standard binomial expansion, we have (1+1/n)^n = 1 + {n}*1/n + {n*(n+1)/2}*1/n^2 + {n*(n+1)/3!}*1/n^3 + ... < 1 + 1 + 1/2! + 1/3! + 1/4! + ... < 1 + 1 + 1/2 + 1/4 + 1/8 = 3. Therefore, for a large n such as 49, f should be smaller than 1, i.e., 50^50 < 49^51.

A simpler solution: 49^51 / 50^50 = 49 x (49/50)^50 If take the square root which needs to be over 1 if 49^51 is larger we get: 7 x (49/50)^25 . And (49/50)^25 > 49/50 - (25x1/50) as each power of 49/50 will reduce the number less than 1/50. So (49/50)^25 > 24/50. Then, 7 x (49/50)^25 > 7 x 24/50 > 1. As a result 49^51 > 50^50 .

Thanks for the fun mental challenge, and sharing the lateral thinking and inference thinking that goes with it. I enjoyed watching your video ;)

You need to set up an axillary function to analyse the rate of the fuction at a point. 50^50 ? 49^51 ln(49)/(49+1) ? ln(50)/(50+1). Then axillary f(x)=ln(x)/(x+1). Find the derivative and approximate extremum point. Maximum of f(x) is at x on (3, 4). f(x) at x=49 is monotonic and decreasing. Hence, ln(49)/50 > ln(50)/51 49^51 > 50^50 .

You have to analyze the function lnx/(x+1). This has a max value when x is somewhere between 3 and 4 (where its derivative is equal to 0), after that decreasing but staying > 0. So for any x>4 we have lnx/(x+1) > ln(x+1)/(x+2). Making x = 49 we have (ln49)/50) > (ln50)/51, thus 49^51 > 50^50.

Нужно показать что 50^50

Take the logarithm of both numbers. For numbers above zero, a > b if log a >log b. Taking log of both sides reduces the problem to 50 * log 50 which is between 84 and 85 51 * log 49 is between 86 and 87

Use compound interest logic, you’ll do this much earlier. 50/49 is 1.02 approx raised to 50, will be much below 49, think of it as getting a 2% interest on your bank deposit. It will take maybe 25-30 years to double, and would max be 4 times in 50 years - hence 50/49 raised to 50 is def below 4, and when you divide by 1/49, it’s clearly below 1.

No, you still have to prove (1+1/49)^49 < 3. This is because the limit may or may not be monotonically increasing or decreasing. Therefore, you have to prove the limit is monotonically increasing to justify your claim.

My intuitive solution is: Let x = 50, and apply natural logarithm on both sides, we have x*ln(x), and (x+1)*ln(x-1). Now look at two terms: (x+1) / x and lnx / ln(x-1), they are just gradient of functions y=x, and y=ln(x) respectively, with dx = 1. Obviously y = x has constant gradient of 1, while y=ln(x) has decreasing gradient (always < 1) approaching to zero. Thus we have (x+1) / x > ln(x) / ln(x-1), therefore, x*ln(x) < (x+1) / ln(x-1). so we have proved 50^50 < 49^51 in a very simple way.

Chứng minh được mệnh đề tổng quát, (bằng phương pháp quy nạp toán học): n^n > (n+1)^(n-1). Với n = 50 là bài toán mà bạn nêu ở trên..

By using Log you come to the answer in a few Seconds; Log 50^50 = 50 Log 50 = almost 85 Log 49^51 = 2 x 51 x Log 7 = almost 86 So 49^51 is almost 10 times bigger than 50^50 😊

Its much simpler than the video. You find the 50th root of both numbers. That means you divide each exponent by 50 such that you get 50/50 and 51/50. Then you just calculate 50 to the power of 50/50. And then 49 to the power of 51/50. 50 to the power of50/50 is 50 to the power of 1 or just 50. 49 to the power of 51/50 is 49 to the power of 1.02 which is 52.966. Because the fiftieth root of 49 to the power of 51/50 is bigger, then that number must be bigger.

Your voice is very sweet to listen... Loving and enjoying your voice

Since you appeal to recognising Euler's number as the limit as n-> infinity, you need to show that it approaches that limit from below to use it as a bound for finite n=49 case.

Spasibo. If you can't wait, quickly in python: print(len(str(50**50)), len(str(49**51)))

Should have applied natural logarithm rule. Very easy... ln( 50^50 ) ...?... ln( 49^51 ). Take any log of both sides. 50x (ln 50 ) ...?... 51x ( ln 49 ) ---> 50/51 ...?... ln 49 / ln 50 ---> 0.98... < 0.99... (Reduced to numeric 2-decimal places on both sides). Or 50/51 < ln 49 / ln 51. Therefore: 50^50 < 49^51 No need for advanced formulas or calculus. Just logarithm rule and basic arithmetic. Someone was close to this but complicated the simplicity of basic power rule of logarithms with functions, extrema, etc. All exponential power comparisons are pre-calculus algebra or arithmetic. Variations, infinitesimals or their limits are unnecessary.

I ended up here on an insomniac night (very counterintuitive) and this is just the perfect voice I needed to calm down and have a good night. Thx!

maybe simplify this greatly- which is smaller- 3 to the power of 3, or 2 to the power of 4? (or 4 to the power of 2). 3 to the power of 3 is the smallest so must be a minimum point if you graphed these

Good one, thank you for sharing.

Great explanation

Finding an unnecessarily complicated, inelegant and difficult solution is not a sign that one should be a mathematician.

Your voice was soothing and gave me peace while my mind was screaming inside

I took the natural log of both sides and saw that it was 50 ln 50 vs 2450 ln 49. This showed a clear difference. The other method suggested is a more general approach that I didn't think of. Works better in the long term.

I love these problems, great mental exercise! Thanks.

Beautiful explanation! Thanks

I don't care about what your teaching. I just came here cuz your voice is soothing and your handwriting is crystal clear.

You didn't prove that (1 + 1/m)^m < lim (1 + 1/n)^n as n -> infinity. It's true, but you assumed it without proof. You need to show that (1 + 1/n)^n < (1 + 1/(n+1))^(n+1) for all integers n > 0.

Thank you for explaining such a terrifying problem so calmly.

Since I've lost access to my logarithm skills I took the approach of graphing the first terms in the series to observe convergence. At first it's unclear with 1^3=1 vs 2^2=2, but 2^4=16 narrows with 3^3=27, and again 3^5=81*3=243 vs 4^4=16*16=256. Intuitively, as n grows, the relative effect of the larger exponent will overtake the base value - 2^64 is much more than 64^2. Since the problem doesn't call for a more specific answer, it can end after plotting an estimate of the crossover. I plugged it in a graphing caculator to check and it's very obvious that before n=5 you've already crossed over.

This is an excellent problem and great way to resolve , did learn a lot

Write 49^51 as (50-1)^51 and write out the first few (5 should do it) terms of the expansion. Use combinatorics to find the coefficients. Rewriting it in terms of 50^50 gets you to see its about 18*50^50.

6:53 How did one < times one > times one < ended up as one < symbol? Does it only bother me?

Well done. It's pretty clever how you've managed to prove it

Take 'log', then the problem wil become too easy. 👍👍👍 But I think, the problem is of arithmatic. That's why we are having these complicated solutions.

Its so simple left side become (49+1)^50 then on simplification it becomes (49^50)+1^50 direct compare to righ side which is greater than left side.

You can see by binomial approximation that 49^51 is greater. Though I believe the gap is big enough that the error wouldn't matter. Since LHS will have factor of around 2 and RHS will have factor of 49 which is quite large.

My simple approach to guessing, just simply the problem as (50)^1 = 50 and (49)^2 = 2401, so RHS will be bigger

If you needed to apply Euler’s, why did you need to extract (1+ 1/49)^49* (1+1/49) Wouldn’t that hold true directly for (1+1/49)^50?

Nice solution😊

Divide both sides with 49^50. 50/49 is about 1.02 therefore we can compare 1.02 ^50 with 49. Rule of thumb: to double 1.02 we have to multiply it 35 times so 1.02^50 is less than 4. Since 4 is less than 49 we get the same answer as the video.

Thanks for doing this for "n." So (for future reference) we know that n^n < (n-1)^(n+1) for relatively large n. What is the lower cutoff (using integers), where the inequality sign switches? 4^4>3^5 but 5^5 < 4^6

Move everything to one side and you're asking "Which is bigger: (50^50)/(49^51), or 1?" Then that simplifies to [(50/49)^50] versus 1... which is "number very close to e, divided by 49" versus 1. Pretty obvious then than 49^51 is larger.

I think there's no need for such a laborious approach... We can get this done in two steps Step one divide both terms and multiply and divide by 50 Step two (50÷49)^51 /50 is final expression which is clearly< 1

Dividing 50^50/49^51, if numerator is larger then it will be >1, otherwise it will be

Your pen is awesome. Which one brand?

I wrote the equation as: 50^50 / ( 50-1)^50 = 50-1, then, replace 50 by X. But still cant solve, it any advice?

That looked like a very long complicated approach: I took a log on both sides and approximated ln(50) = ln(49)+1/49 (ie: first order derivative and taylor series). Way simpler!

I just started with small numbers and gradually incremented them. By the time you get to 5^5 vs. 4^6, the 2nd evaluation is outpacing the first, so you can safely assume the trend will continue for numbers up to 50.

By Modular Arithmetic for divisibility of 50 50^50 ___ 49^51 0 ____ (-1)^51 0 ___ (-1) 0 ___ 49 (since the remainder must be positive) 0 < 49 therefore 50^50 < 49^51

Eu assisti em outro idioma e entendi, por isso eu amo a matemática ❤

I solved the problem in 2 seconds with my intuition ...and the answer was correct, so it's true that imagination is more important than the knowledge - Einstein

Same answer result. But it glaringly shows how Mathematician and an Economics and Finance pips answered this numerical logic query step by step relative to their learned principles.

You generalize (x-1)^(x+1) / x^x and take limit at 1 and infinity. You took see it's a diverging function and hence 49^51 > 50^50. Or in general any (x-1)^(x+1) > x^x for x >>1.

And if there’s anyone who knows a harder way to do this, the ball is in your court now

good explanation proffessor

Just compare logarithmic values. 51 ln 49 vs 50 ln 50 51 ln 49 vs 50 (ln 49 + ln 50/49) ln 49 vs 50 ln (50/49) ln 49 vs 50 ln (1+1/49) Right hand side is smaller than it's first degree approximation since ln (1+x) is concave. So right hand side is smaller than 50/49, which is way smaller than ln 49 = 2 ln 7 > 2 since 7 > e. Left hand side is way bigger. So 49^51 is bigger than 50^50.

Since 1/49 will diminish rapidly in the binomial expansion of (1+1/49)^50 you can approximate it to 1+50/49= 2+1/49 to the first order so (50^50)/(49^51) will be (2*49+1)/(49^2)=99/2399

This is much simpler to analyze. Make it easily envisioned by reducing the bewilderingly large powers. Get them down to human scale. 50^1 compared to 49^2. 50^1=50. 49^2 is bigger than 50 by inspection. Therefore, by induction, n^n < (n-1)^(n+1).

It’s easy, man. For all x,y,a,b > 1: x^a > y^b if a>b AT ALMOST CASE

freaking mic drop at the end there. sheesh. Super cool video, thanks for making it!

Thanks for the video. Regards

The log derivative of (a-x)^(a+x) wrt x is -(a+x)/(a-x)+log(a-x). This is greater than -51/49+log(49) for a=50 and x=0,…,1, which is much larger than 0 as can be easily estimated (log(49)>log_3(49)>log_3(27)=3, for example). Hence monotonously increasing, hence 49^51 is greater.

I tried making a graph of this relation in the form of x^y_(x-1)^(y+1). For values of x from 1 to 4 (integer) LHS will always be greater (regardless of value of y) and for any nunber greater than 4 RHS will always be greater regardless of the value of y

I knew the answer but i couldn’t stop the video in between because of that voice. That damn voice ❤

beautifully explained.

Higher power wins for all numbers > 5^5 49^51 is actually almost 20 times larger than 50^50. (bruteforced it by excel)

wa oh! your voice is so good, It feels like I am hearing asmr. And it helps me to sleep.

I went with binomial expansion by splitting 49^51 into (50-1)^51. Then in its expansion we have 50^51 with the additions of many more digits.

To compare which number is larger between 50^50 and 49^51, we can calculate their values. 50^50 ≈ 8.881784 × 10^69 49^51 ≈ 1.319507 × 10^77 Comparing the two values, 49^51 is larger than 50^50.

Interesting. If you use 4^4, that's > 3^5, but starting with n=5, we have n^n < (n-1)^(n+1). I think the left hand is Lambert's w-function?

Cool trick. Important to remember defn of e.

I expanded 50^50 as (49+1)^50 using the binomial theorem and compared each term with 49^51 which is 49^50 + 49^50 + ... + 49^50 (49 times). It ends up like each term in the second expression is larger than the ones in the first expression.

I’m not sure why you back away from the numbers at the end. Sure it’s less than 1, intuitively one might have guessed that a small difference in powers will have more of an impact than a small difference in bases. But without that simplification you’d have an idea of how large the difference actually is. 3x50 (150) over 49 squared isn’t that bad; square 50 (2500) then subtract 50 to get 50x49 then subtract 49 to get 49 squared, or 2500-99 or 2401. It’s still not exact by any means, since the 3 is an approximation, but you can easily tell at a glance then that 49^51 is between 15 and 20x larger than 50^50

I used simple intuition. It makes more sense to me that the (49x49 ... x49) falls behind (50x50... x50), however the extra instance of multiplying x49 accounts for all of the previous distance between those equations. If you are repeating something 50x, and 1x49 is just 1 less than 50. We haven't gotten far enough exponentially to create more than 1 digit of a gap. We know from multiplication rules it will go around 2x10^20 for either equation, but it just intuitively makes sense that 49^51 > 50^50.

49 could be expressed as 50 ,-1 and then apply binomial expansion. Take the ratio of given two terms and can be solved easily. Just see whethr ratio is less than or greater than 1

This is true for any x^y and (x-1)^ (y+1). Well, for positive numbers greater than 1. Didn't look at others, but probably reversed for others.

I think it's simpler using binomial expansion of 49^51 = (50 -1)^51 = 50^51 + other stuff and therefore > 50^50

Awesome mam

I did a really simple equation, whats bigger 50x50 or 49x49x49, I deducted that it would deviate acordingly, and apparently it works for any positive number greater than 1

Take (a - b) and divide both sides by 50^50. That is (50^50)- ((50-1)^(50+1)) = x. Divide x by 50^50.. You will get x as negative meaning the latter is bigger

How can we confirm that (1+1/n)^n is monotone increasing function?

This is flawed like many such math "solutions" on UKposts. The fact that the limit of a sequence is < 3 does not guarantee that the terms of the sequence are equally so. Like other viewers pointed out.

basically doing log at the end. you could have done the log in the beginning and solve it early. 50*log(50) vs 51*log(49)

Pretty easy. It’s easy to calculate in your head that 4^6 > 5^5. 4096 > 3125. So for any number greater or equal to n = 5, n-1^ n+1 > n^n.

(5×10)^50 on one side and (10×4,9)^51 on other side. Now we silplify by 10^50 and we get 5^50 and 10x4,9^51. Next 4,9/5 is 0.98. Then we have 10x0.98^51x5x5^50. 50x0.98^51 is greater than 1 so 49^51 is more than 50^50.

49^51. There is one additional zero there. Triumphs the other easily. No need to complicate the problem.

could that be extended to all numbers ? this example : 50 over 50 versus 49 over 51 general example : n over n versus (n-1) over (n+1)

As a calculus fan, to solve this I analyzed the function y=(50-x)^(50+x), where y(0)=50^50, y(1)=49^51. Its derivative is y'=(50-x)^(50+x) (ln(50-x)-(50+x)/(50-x)). We only need to consider x on the interval [0; 1]. Now, to find the sign of y', let's do following estimations: 1) 0≤x≤1 => 50≤50+x≤51 and 49≤50-x≤50; 2) consequently, ln49≤ln(50-x)≤ln50; 3) e log(3; 49) ln50

Too easy let p=50^50 and q=49^51 Then take log(p) and log(q). Since logx function will always evaluate bigger positive numbers for x>1 you can easily find out which one is bigger. In this case log(q)>log(p)

glad to hear someone pronounce Euler as I did in the past lol

I feel... Between n^n and (n-1)^(n+2), It looks when n5, (n-1)^(n+2) is bigger. (n=natural number) But I don't know if it's true, or how to porve it.

You can just see it by looking at the numbers, that's how exponentials work Just like you can intuitively see that 1045 is larger than 983.

If we take binomials expansion, then easy confirm (1+1/49) n of firs two order is bigger than 1. 50/49 is also bigger than 1 , then any 1.xxxxxxxxx multiple

ok, but why for all n there is (1 + 1/n) ^ n < 3 ? maybe for a given n an so but not for all. is the string ascended starting from 1 ? maybe yes but please confirm.

Applying binomial theorem, ( 50-1) power 51, then expand then common 50 to power 50 is so simple

50^50 os 3.125 *10^12 e 49^51 is 1.6 * 10*16. De second one is 4 times greater than the first one.