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.
I thought the opposite. 50 is larger base, so you can calaculte the 1.02 to the power of 50. If its larger than 49, its bigger. If not 49^51 is bigger.
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.
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 .
Maybe I don't understand it fully but as you said, we can only use the squareroot while holding inequality if the value (49^51/50^50) is greater than one. And since that is exactly the thing we are trying to prove, we can't assume that it is greater than 1.
This is a very nice "high school" solution. That is to say you will get a high mark if you wrote this solution in a Math Olympiads. But conceptually there are simpler solutions;just binomial expansion.
I really like this Number Theoretic Solution, don't like logarithmic or the solution presented in the video. Can you explain the step why (49/50)^25> 49/50 - (25x1/50)? If each power reduces 49/50 by less than 1/50 for each power, how does it make sure it's greater than 49/50 - (25x1/50)? Edit: Nevermind, I understand it now
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
But how do you calculate log50 and log49 without a calculator, to come to the conclusion the right side us larger then the left one? I'm sure the point is to solve the problem without a calculator as otherwise you can just calculate both initial terms and see which number is larger 😉
By using properties of log, we can write log50 as log 5 + log 10 and log49 as 2log7. Now, log5 and log7 values can be approximately substituted. log5~0.698 and log7~0.845
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.
@@TheSoteriologisti think they said something like 'pleasw understand that they have a habit of making simple things complex' The translation is not entirely correct, sorry
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.
Thanks for that! The demonstration on the video is not acceptable (unless if the goal is only to get the most probable answer without caring about how you get there). Using a limit to justify an inequality is not sufficient at all! Your method (even though I didn’t verify it) is more rigourous..
@romain1mp That would make the squeeze theorem wrong and also disprove Archimedes, strict inequalities are very useful proofs, but they have to be strict. "Equalities are for children, real men deal with inequalities!"
@@lizekamtombe2223who talked about equalities here? I am just saying that if Lim(f(n)) is smaller than L … when n is close to infinity…. You cannot conclude that f(n=49) is smaller than L without more inputs… For example you need to demonstrate quickly that f is increasing function from a certain level p (i.e. p
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.
The solution in the video is actually pretty clean if you just divide the other way. 49^51/50^50 = (49/50)^50 * 49 = (1-1/50)^50 * 49. The first term is close to 1/e, which is about 0.36. Multiply by 49 and it's much bigger than 1.
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.
@@phajgo2 No, it isn't obvious! Unless you have bounds of the accuracy. On the other hand, the result is correct because it is well known that the sequence (1+1/n)^n is strictly increasing.
@@Γιώργος-κ2ζ exactly this is why I think it is obvious. For n=1, (1+1/n)^n = 2, for n=2, (1+1/n)^n=(3/2)^2=9/4=2,25. and at infinity we know it is e so it's approaching from the left. Isn't that enough?
@@phajgo2 - I guess that provided one adds @user-gk3on7xp7e insight and says "the limit as n-> Infinity is Euler's number and it is well known that the sequence is monotonically increasing" you'd deserve the marks? This rather makes it a general knowledge test in my opinion. One can use calculus to prove that (1+1/x)^x is monotonically increasing for x>0. I'm not convinced that listing the first few terms of a sequence is a proof, although you seem actually in furious agreement with @user-gk3on7xp7e and the "it's well-known" proof by assertion. I looked at raios of successive terms and didn't see a quick proof of monotonicity. Just my 2c.
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.
This is a good answer. However, you would need to show how 49^(1.02) = 52.966. For this you can do it by binomial expansion and it should be easy to do
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.
Also, a convenient rule of thumb is that you need to acrue about 70% of interest "in total" to double the amount. 70 years with 1% or 10 years with 7% interest
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 .
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.
I found my think-a-like buddy. I thought exactly the same way applying the rule of 72. I thought no one else would be smart enough. You are really smart, mate.
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.
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.
@@bumbarabun Why do you believe that? You are differentiating with respect to n, so you can't use the rule for differentiating x^n with respect to x. The variable we are differentiating with respect to appears in both the base and the exponent, so it is a complicated differentiation. It's like differentiating x^x, but worse.
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.
I rewrote 50 as 49+1, since this gives only positive terms. But same reasoning. end up with it being less than 2*49^50 so the fraction would be less than 2/49 which is less than 1 so 49^51 is greater
The binomial expansion would be a good solution if it were more of a close call. But if you have done enough problems like this you can quickly recognize the limit to e and come up with an answer more quickly without doing much math.
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.
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
You can do it logarithmically without a calculator by first expressing the logarithms ln(50^50) = 50 * ln(50) and ln(49^51) = 51* ln(49). You can then use the Taylor expansion around ln(50) to linearlly approximate the values or each. For example, using ln(49) ≈ L - 1 / 50 since ln(49) ≈ ln(50) - 1 / 50. You then solve by substitution, ln(50^50) = 50L, where ln(49^51) ≈ 51 * (L - 1 / 50) = 51L - 51/50. Now, just compare the two values given 51L - 51/50 = 50L + L - 51/50. Here you can clearly see the difference is the term 'L - 51/50' and you need to check if this term is positive, which it is as it equals 2.892. You can then conclude that 51L - 51/50 is > 50L.
To use this method more rigourously, you would have to check if the approximation is valid, by bounding the error in the expansion to ensure it is small. You might also like to look at monotonicity, simply looking at the differences in the observed approximation and actual value.
This is what I did and I got 50^50 < 49^51 By using the rules of Indices, I split 49^51 into 49^50 x 49^1 i.e ( a^m x a^n = a^m+n). Now, 50^50= (50^5)^10 & (49^5)10 i.e [ (a^m)^n = a^mxn ]. Ignoring the power of power i.e 10 50^5= 312,500,000 49^6= 13,841,287,201 ......[ 49^5 x 49^1 = 49^6] P.s. I used calculator for multiplying 😬
In school I was too lazy to do math, but I had a good sense of logic So I usually cheated the calculations with small numbers to have a guess what was going on with the math Something like this: 2^4 < 3^3 (16 < 27) 3^5 < 4^4 (243 < 256) 4^6 > 5^5 (4096 > 3125) inverted gap 5^7 > 6^6 (you could stop here because you already have a proof what is going on) 49^51 > 50^50 "teacher, I don't know how to do the math, but 49^51 > 50^50 for the logic reason listed above, the gap inverted and keeps increasing". "A" 😂. I did so much of this when I was a kid. I still remember there was a 5 question exam once and I did all 5 questions without a single math, only writing sentences explaining why the answer would be X or Y / True False/ how many how much. Teacher told me I "cheated" but still gave me an "A" because he had never seen a math test done correctly without any calculations, only with pure logic. Good memories
Nice going. I wouldn't call it cheating. Rather it's unconventional solving, like MacGyver. Did you follow the syllabus? No. Did you nail the concept and solve the problem? Yes. I'd love to know what you ended up doing as work or hobby using these skills.
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.
but there is need of calculator for ln49/ln50 :) which destroys thee point of even taking log when one could just evaluate the initial values and compare using any calculator or program
I'm honoured to be thousandth commenter. As the video progresses the enthusiasm or the energy in your voice decreases gradually. I knew the answer just by looking which number had more exponent to its power as both are almost near to each other. I just watched fully to discover any other type of solving it quickly. If I do these kind of calculations for solving 1 problem in my aptitude exam then I will run out of time for other questions so I will rather trust my intuition and choose the answers without using up more time😅
@@ЦЕАканал привет. len посчитает количество символов в каждом из представленных(чтобы не выводить большие числа). Какой из вариантов больше, будет более очевидно.
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.
Instead of bringing e into this, you could have found the sum of the first 2 terms in the binomial expansion of (1+1/49)^50 (=1+50/49). We know that the full expansion is greater than this. It's trivial from here on...
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).
right hand side is increasing at a slower rate than the left hand side, so at some point when n increases, left hand side should overtake the right hand side
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.
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!
If you are being rigorous, you would need to put bounds on the error in the Taylor approximation and show that they can't change the conclusion. You can certainly do that, but it gets a little messy. (Doing it the way in the video, you need to prove that (1+1/n)^n is monotonically increasing, which also isn't straightforward.)
Just have to study the function f(x) = ln(x)/(x+1). For x > 0, the derivative is the same sign of g(x) = 1+1/x - ln(x). g is decreasing and g(10) < 0, then g(x) < 0 between 49 and 50. So f(50) < f(49) or 50*ln(50) < 51*ln(49).
Taking a log we compare 50 ln 50 to 51 ln 49. Take a function y(x)=(50+x) ln (50-x), taking a derivative we have dy/dx=ln(50-x) + (50+x)/(50-x). For any -50
In general y(x)=(a+x) ln (a-x) this proof will work for -a < x < a-1. The exact interval boundaries are not so simple to calculate analytically, maybe with Lambert W or impossible?
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
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.
When the numbers are close, the exponent (power) beats the base (the number). In fact, the bigger these numbers are, the smaller number with the bigger exponent (only by one unit) can go lower and lower from 50% of the higher number and the exponentiasion will be higher. Of course, this is only the answer, not the demonstration. More rigurous, but still simple is using the logarithms.
@@sorinturle4599 well even at near x=0 there is crazy separation for exponential graph with higher bases so it makes sense differing by 1 in base doesn't matter as much differing by power by 1
Let y1=49^(x+1) , y2=50^x We know that both logarithm 49^(x+1) & 50^x, increasing exponentially. Y1>Y2 from starting point (0,0) Y1 5^4, then 49^51>50^50 And also clarify to those who said we can just imagine the base is expanding, it not works like this as different numbers have extremely different point to intersect.
This is flawed like many such math "solutions" on TH-cam. 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.
I checked the sequence n^n - (n-1)^(n+1). The limit is -inf. Interestingly after the first few terms it becomes sharply negative and dives down pretty fast. Here's the first few terms: 1 3 11 13 -971 Notice that it seems pretty "tame" for the first 4 terms and suddenly dives down steeply. It is strictly decreating from that point until at least 51st term. I didn't check after that, but the 51st term is -2.098525e+88 Or on order 10^88 and negative. So I can say with confidence that any term after n =5 will be negative.
Exactly… I did something similar. 50^50 is n^n and 49^51 is (n-1)^(n+1) and then, as you did, changed the n for a much smaller number, so I could easily do the calculations, I chose n=2 so I had 2^2 in one side and 1^3 on the other and it appears that the second part is not bigger, but as n grows over 2, you go getting bigger numbers each time on the (n-1)^(n+1) side
Ok this could be an intuitive answer, but it is so obvius to me that 49^51 is higher simply because there is one more multiplication in it. I’m pretty sure that 49^50*10 would be also higher.
@@thomasdalton1508 Intuition is not a lucky guess. I didn’t say that any multiplication is enough here. That could be *0.1 too, which is also a multiplication and obviously wrong. You can say that this is not an acceptable proof, but the luck has nothing to do with.
It does seem intuitive cuz of that one additional multiplication. But it's not true for 3^3, and 4^4. From 5^5, the latter expression is larger So Is 50^50 > 49^51. Answer: No. Cuz m^m < m-1^m+1, where m=/>5 👽
@@Selendeki But, as Ashton pointed out, increasing the exponent by one doesn't always do the job - it only works for 5 and greater. I don't think it is at all intuitive that the threshold is 5. That requires doing the calculations. Just because your intuition gives the right answer doesn't mean it is good intuition. The exact same intuition would have led you astray for different numbers.
@@Selendeki Intuition certainly has its uses and developing your intuition is a very important part of learning mathematics. There are two problems with the OP's comment. First, it is bad intuition - there is no intuitive reason that multiplying more numbers should get a larger result than multiplying larger numbers. It depends on how many more and how much larger. And second, it is completely wrong to say something is "obvious" based on intuition. That just isn't how you do maths. You don't guess and then say your guess is obviously correct.
I don’t know the proofing method I’d use but.. if I was asked what the comparison between the two number is, I would’ve gotten the same answer by a simpler means. The base number and exponent is off by one but one is exponentially more valuable. Therefore it will weigh more heavily. 2^3 < 3^2
Disclaimer: not a math person. Issue I see is that the left hand side is 50^50 (same same), so 2^3 vs 3^2 isn't a relevant pattern. Should be 3^3 vs 2^4, or 4^4 vs 3^5, which would give the wrong answer here. I think only starting from 5^5 vs 4^6 is left hand side lesser.
In such cases, I always try with random numbers For example ; 3^4 = 81 > 4^3 =64 or 4^5 = 1024 > 5^4 = 625 It can go on like that to forever Only exception from 0 to 2 That means n^n+1 > n+1^n always I just wanted to explain, I mean it can be solved within 10 sec if you can see that
But without proof, it's crystal ball gazing. Try 3^3 vs 2^4 ... 🤓 By your logic, the higher number should be the one with the higher exponent, but it's not... 😱
@@Stepan_HIt's better to have proof instead of assumptions, but it's also true that generally higher exponents mean high numbers overall. Yes, 3^3 > 2^4, but even just raising each base by 1 will already make the no. w/ higher exponent larger with 4^3 < 3^4. Keep increasing the bases by 1 after that and the number raised to 4 will always be larger. Even if you had a difference of 2 for the bases, with 4^3 > 2^4, increase the bases by 1 you'll still get 5^3 > 3^4, yes, but increase it one more time and everything starting with and after it'll always be 6^3 < 4^4 or n^3 < (n-2)^4 as long as n > 5. Proof is good, but we're in the comments, not a math comp. Most people here just want the answer, so might as well let them know a "trick" even if it isn't always true. These kinda problems aren't really encountered by the general public often anyways, and anytime they do it'll involve huge numbers with little differences in bases and exponents, at which point the number with the higher exponent is larger 99.9% of the time. By the very small chance it isn't, well it's just youtube so does it really matter? Haha
That's how I looked at it too. I didn't know how to prove it. But using common sense I figured if decrease the base by 1 and raise the exponent by 1. Then the 1 with the raised exponent is larger
unless asked for the difference, they should be consistent, so simple (university degree) trick is to use 0 or 1. 1^1 > 0^2 because 1 > 0^2 2^2 > 1^3 because 4 > 1 3^3 > 2^4 because 27 > 16 4^4 > 3^5 because 256 > 9x9x3 = 243 5^5 > 4^6 because 25x125 = 3125 > 4^6 = 16^3 = 1024 6^6 > 5^7 because 36^3 = 46k and 25^3*5 = 78k 7^7 < 6^8 because 49^3*7 -= 823k and 36^4 = 1.7M So the turning point is around 7, which is weird, because up to 6 the difference were larger and larger.
I think it was totally unnecessary though it was just to make the inequality simpler by canceling using a multiple of 3. She should've just calculated (3 x 50)/(49 x 49) which isn't difficult and very evidently much smaller than 1.
I am an indian and i did binomial expansion and did it in like 10 seconds.Just equate (50/49)^50 to 49 first and 50/49 is 1.02 ,which can be written as (1+0.02).After that just multiply 50 to 0.02 so tha answer will come 1+(50×0.02)=2 which is less than 49. So 50^50 is way way smaller than 49^51.
Why simply not divide on with the other and see what happens? (50/49)^50 * 1/49 and see if it's greater or less than 1 ? If it's less than one, then 49^51 is greater than 50^50. Now, eyeballing this, 49 is by 1 lesseer than 50, which is 2%, so 50/49 is probably around 1,002 -1,003. Do I believe 1.003^50 is greater than 49? I don't believe it's even greater than 2, so if I was a betting man, I'd say 49^51 is definitely greater than 50^50. (Since 1,003^3 / 49 is certainly lesser than 1) Now I can go and see what you did here in 8 minutes...
Even if your answer ls right, You are wrong in some some aspects, so be careful in the future. 50/49 it is a difference of 2% SO it would be around 1,02-1,03. Then 1,03^50 is not greater than 49 but it is greater than 2
Simpler solution Just take it as [(49)³]¹⁷ & [50².⁹⁵]¹⁷ (2.95 is approx but higher than the value of 50/17) We get 117,649¹⁷ & 102,792.52¹⁷ We can easily deduce that the number with higher base would yeild a higher solution when raised to the same exponent. So 49⁵¹ > 50⁵⁰.¹⁵ so it has to be greater than 50⁵⁰ This only works for simple bases and simple powers.
I looked at both numbers and could immediately see that that 49^51st is larger. Visualize 50^50th as 50x50... all the way to the 50th one. Visualize 49^51th as 49x50th ... x the 51st 49. Clearly that will result in 49^51st being larger. You can even use a simpler model to prove it. 50 ^ 3rd vs 49 ^ 4th; 50x50x50 = 125k; 49x49x49x49 = 5,764,801. Notice that the difference between 50 and 49 is only 1. That seems to be true for any two consecutive numbers starting with the number 3.
GSR Theorem (Logical Approach): Given: 50^50 and 49^51. Question: Which is bigger? Solution: Let (assumption) 50^50 =49^51. Left Hand (LH) = Right Hand (RH) Let A = base 50 B = base 49 A > B 50 > 49 Let r sub b = ratio of bases, (50/49) > 1 Let r sub e = ratio of exponents, (50/51) < 1. Condition: (r sub b) > 1 ; (r sub e) < 1. Analyses: (r sub b, greater than 1)/ (r sub e , less than 1), then the result is > 1. OR (r sub b /r sub e) > 1. Therefore: r sub b > r sub e. It follows that (Theorem) if LH: r sub b > 1 AND r sub e r sub e . Meaning, (A/B) > r sub e, OR A > (r sub e)*(B). IF A = (r sub e )*B, THEN B or RH must be BIGGER to COMPENSATE for A (LH) since (r sub e) < 1. THEREFORE: RH > LH --> OR 49^51 > 50^50 --> OR 50^50 < 49^51. ANSWER.🥰
I’m glad she’s not my introductory algebra teacher or I’d go insane. I may be ignorant on this convoluted mathematical solution but I just assumed the following which gave me the correct “guess” to the problem given. I worked out a simpler similar equation in my addled mind: of 4 squared vs 3 cubed, answer: 16 vs 27, therefore 49 to the 51st power is larger. 😱 👀 ⁉️ 🤔
But its an Olympiad Q - you need to show it, not just be content you know the answer. Indeed, 4^4 > 3^5 which does not follow the seeming general rule n^n < (n-1)^(n+1) - and in you sample 3^3 vs 4^2 does not match the pattern in the Q: you have n^n (n+1)^(n-1).
x^x - (x-1)^(x+1) is only positive for x=5. There's a very similar problem, but I forgot the details, where the limit is e. Can anyone figure out what that problem is?
It comes for a problem like 50^49 < 49^50. Then the function to study is ln(x)/x . This function his maximum at e. In this problem, the function is ln(x)/(x+1). See my comment just below.
She chose 6 to make it simple, as long as it was smaller than 49 it would have worked. In a sense, she gave it a try, to see if the whole equation would be smaller than 1 this way, if not, she could have tried bigger denominator, if still smaller than 49, because anyway, the whole reasoning with e is based on not having to be precise if the equation is smaller than 1. If it was bigger, equal to 1, or really close to 1, thing that we are not supposed to know at that moment, then the reasoning would be probably no conclusive, but it was worth giving it an easy try.
I solved it with a logaithmic table and a basic calculator: 50^50 vs 49^51 log 50^50 log 49^51 50 log 50 51 log 49 50 x 1.69897 51 x 1.69020 84.9485 86.2 49^51 is larger
@@phajgo2 Maybe you are right. However most Olympiad competitors start early. In our school, we started early coaching by 12. Most competitive exams prep starts by 10 to 12. I am talking about India and how most successful candidate crack exams.
@@banjo4us1 I'm from Poland so there may be difference in programs :) we also start olympiads early (around 10-12 as you say) but they usually do not go beyond the program of math classes for the given age as the intention for children is not to incentivize learning extensive material before you're supposed to but it is rather to find smart solutions within the knowledge you have. Still I'm curious to know what age was that question intended for because I found it quite difficult :)
@@phajgo2 I agree with you. Most problem in these Olympiad could be solved either using higher theory or rudiment maths. I remember once there was a problem of a bird catching a fish and then perching on a tree. The problem could have been solved using Snell's law, but it could also be solved using Similar Triangles, albeit it is a longer solution. I always studied upto 3 years ahead so that I could solve these kind of exams. Even in IIT papers here in India.... most questions has higher maths solutions.
Taking a logarithm is way easy and short one idea. 50Log50, 51log49 50log 50= 50log(7^2 +1) or we can find sqaure root of 50 as 7.07 aprox. ( You can calculate any square root of a non perfect number by looking difference of two consecutive perfect square numbers eg. 7^2= 49,8^2=64, difference=64-49=15 now 50 is 49+1 so squareroot of 50= 7+1/15= 7.07 aprox. ) So 50log50=50log(7.07)^2 =50*2log7.07=100log7.07 Also 51log49=102log7 Clearly 102log7 is bigger than the other
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
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
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
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
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.
I think 49 is being multiplied by itself one more time than the L.H.S. so I thought RHS is greater than LHS 😅. You proved it mathematically though. I thought it was a trick question so you'd have a trick solution but guess not.
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.
Same here 🎊
The thing is i guess that with higher potencies it gets bigger
Sometimes the journey is more important than destination.
I thought the opposite. 50 is larger base, so you can calaculte the 1.02 to the power of 50. If its larger than 49, its bigger. If not 49^51 is bigger.
And that is 2.69 so it supports your output. You got it right. My methid confirmed your result.
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.
The easiest solution is comparing 4^5 vs 5^4. There is no reason why the trend won't continue with higher numbers.
@lson5542 That doesn't match the pattern in the Q. 5^5 vs 4^6 would.
@@jackwilson5542 but then u have to prove that it's true for all numbers, with concrete evidence
@@konglink3359 only if numbers are > e ... that's the only condition. It doesn't work for 2 and 3 for instance ...
@@tontonbeber4555 but why e? That's so specific how do you figure the explanation?
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 .
Good one!
Maybe I don't understand it fully but as you said, we can only use the squareroot while holding inequality if the value (49^51/50^50) is greater than one. And since that is exactly the thing we are trying to prove, we can't assume that it is greater than 1.
Nice solution
This is a very nice "high school" solution. That is to say you will get a high mark if you wrote this solution in a Math Olympiads. But conceptually there are simpler solutions;just binomial expansion.
I really like this Number Theoretic Solution, don't like logarithmic or the solution presented in the video.
Can you explain the step why (49/50)^25> 49/50 - (25x1/50)? If each power reduces 49/50 by less than 1/50 for each power, how does it make sure it's greater than 49/50 - (25x1/50)?
Edit: Nevermind, I understand it now
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
That was the first approach that crossed my mind, and a great deal simpler than the algebraic thicket the narrator lays out.
Good solution
But how do you calculate log50 and log49 without a calculator, to come to the conclusion the right side us larger then the left one? I'm sure the point is to solve the problem without a calculator as otherwise you can just calculate both initial terms and see which number is larger 😉
I'm not that well versed in mathematics. Is it a proof, though?@@michaelhartmann1285
By using properties of log, we can write log50 as log 5 + log 10 and log49 as 2log7. Now, log5 and log7 values can be approximately substituted. log5~0.698 and log7~0.845
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.
Finding an unnecessarily complicated, inelegant and difficult solution is not a sign that one should be a mathematician.
Please ye samajhiye k inhe simplicity ko complex bana k ghuma fira k baat krne k aadat hogi😂😂😂😂😂
@@neelkamal5274 What ?
@@TheSoteriologisti think they said something like 'pleasw understand that they have a habit of making simple things complex'
The translation is not entirely correct, sorry
And if there’s anyone who knows a harder way to do this, the ball is in your court now
Spot on, this was way over-complicated.
😂😂
I used binomial theorem
😂😂😂😂😂
Calculate both terms by normal multiplication
Logically, the power is always the most powerful part of a number.
It usually is, but it's not always the case. For example, 4^4 > 3^5.
2¹ > 1∞
@@GHOST-RIDER-0 that's not true because infinity is not a number and 1∞ is undefined.
Only if u are powering numbers greater than 1 ..I think
@@OblomSaratovthen 1^999999
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.
Oh i got head ache on math. Im so poor on math
Thanks for that!
The demonstration on the video is not acceptable (unless if the goal is only to get the most probable answer without caring about how you get there). Using a limit to justify an inequality is not sufficient at all! Your method (even though I didn’t verify it) is more rigourous..
@@romain1mp the vid is easily rehabilitated; (1+1/49)^49 is immediately less than e because discrete compound < continuous compound (for positive r)
@romain1mp That would make the squeeze theorem wrong and also disprove Archimedes, strict inequalities are very useful proofs, but they have to be strict.
"Equalities are for children, real men deal with inequalities!"
@@lizekamtombe2223who talked about equalities here?
I am just saying that if Lim(f(n)) is smaller than L … when n is close to infinity…. You cannot conclude that f(n=49) is smaller than L without more inputs…
For example you need to demonstrate quickly that f is increasing function from a certain level p (i.e. p
Нужно показать что 50^50
Только, пожалуйста, исправьте: квадрат 49 равен 2401.
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.
Agree. Without proving that we can doing the same way as the video to approve 4^4 < 3^5, but it not.
th-cam.com/video/hqWkYV4d_PQ/w-d-xo.htmlsi=773t5VKHSSulzbiH
i agree. i used a calculator, and it's true, but she made an assumption here
euler already proved this. youre a clown.
(1 + 1/n)^n < 3 for any natural n. I didn't listen to the whole thing to if they were proving it or not.
The solution in the video is actually pretty clean if you just divide the other way. 49^51/50^50 = (49/50)^50 * 49 = (1-1/50)^50 * 49. The first term is close to 1/e, which is about 0.36. Multiply by 49 and it's much bigger than 1.
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.
which is actually quite obvious..
@@phajgo2 No, it isn't obvious! Unless you have bounds of the accuracy. On the other hand, the result is correct because it is well known that the sequence (1+1/n)^n is strictly increasing.
@@Γιώργος-κ2ζ exactly this is why I think it is obvious. For n=1, (1+1/n)^n = 2, for n=2, (1+1/n)^n=(3/2)^2=9/4=2,25. and at infinity we know it is e so it's approaching from the left. Isn't that enough?
@@phajgo2 - I guess that provided one adds @user-gk3on7xp7e insight and says "the limit as n-> Infinity is Euler's number and it is well known that the sequence is monotonically increasing" you'd deserve the marks? This rather makes it a general knowledge test in my opinion. One can use calculus to prove that (1+1/x)^x is monotonically increasing for x>0. I'm not convinced that listing the first few terms of a sequence is a proof, although you seem actually in furious agreement with @user-gk3on7xp7e and the "it's well-known" proof by assertion. I looked at raios of successive terms and didn't see a quick proof of monotonicity. Just my 2c.
@@phajgo2 No, it isn't. A function can increase at first, then start to decrease, and after that move to it's limit.
My simple approach to guessing, just simply the problem as (50)^1 = 50 and (49)^2 = 2401, so RHS will be bigger
Wont happen with 2^4 and 4^2
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 😊
But you need to use a calculator. This problem asks that you come to a solution according to math principles, not raw calculations
I knew this without even calculate anything lmao
i don’t understand that, but I believe you!
Yes. Your solution is similar to mine.
th-cam.com/video/hqWkYV4d_PQ/w-d-xo.htmlsi=773t5VKHSSulzbiH
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.
I like this answer already!
This is a good answer. However, you would need to show how 49^(1.02) = 52.966.
For this you can do it by binomial expansion and it should be easy to do
Much appreciated
How did you calculate 52.966 - a calculator?
@@donmoore7785 Yep.
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.
Also, a convenient rule of thumb is that you need to acrue about 70% of interest "in total" to double the amount. 70 years with 1% or 10 years with 7% interest
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 .
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.
Wow. I love it
nice, using the rule of 72?
That's how I thought of it. You wrote it concisely 👏🏽
@@Aut0KADyes
I found my think-a-like buddy. I thought exactly the same way applying the rule of 72. I thought no one else would be smart enough. You are really smart, mate.
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 did the same thing, but I like her approach without having to find the natural log, which I reached for my calculator for.
Shouldn't it be 51 ln 49
How did u get 2450?
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.
That's a good point. Maybe take the derivative and show it's always positive? Either way it's a good explanation and doesn't require a calculator.
@@tharock220The derivative is rather messy, so I don't think it will be easy to prove it is greater than zero, but I expect it is possible.
@@thomasdalton1508 I believe that derivative is n * ( 1 + 1/n ) ^ (n-1) and it cannot be negative for positive n
@@bumbarabun Why do you believe that? You are differentiating with respect to n, so you can't use the rule for differentiating x^n with respect to x. The variable we are differentiating with respect to appears in both the base and the exponent, so it is a complicated differentiation. It's like differentiating x^x, but worse.
@@thomasdalton1508 you are right, my mistake
Try 3^3 & 2^4
4^4 & 3^5
5^5 & 4&6
At 5^5 and above, the second expression becomes larger and larger.
You can solve it graphically too
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!
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.
I rewrote 50 as 49+1, since this gives only positive terms. But same reasoning. end up with it being less than 2*49^50
so the fraction would be less than 2/49 which is less than 1 so 49^51 is greater
Yea this is super obvious with just logic. 50^50 is less than (50-1)^51. Like 2 seconds
I did something similar
The binomial expansion would be a good solution if it were more of a close call. But if you have done enough problems like this you can quickly recognize the limit to e and come up with an answer more quickly without doing much math.
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.
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
Lol ..😂
Solution provide karaao
Divide both side (49)^50
L.h.s=(1.0204)^50 which is approx 2.74
R.h.s =49
R.h.s is much larger than L.h.s
(49)^51 Is larger
You can do it logarithmically without a calculator by first expressing the logarithms ln(50^50) = 50 * ln(50) and ln(49^51) = 51* ln(49). You can then use the Taylor expansion around ln(50) to linearlly approximate the values or each. For example, using ln(49) ≈ L - 1 / 50 since ln(49) ≈ ln(50) - 1 / 50. You then solve by substitution, ln(50^50) = 50L, where ln(49^51) ≈ 51 * (L - 1 / 50) = 51L - 51/50. Now, just compare the two values given 51L - 51/50 = 50L + L - 51/50. Here you can clearly see the difference is the term 'L - 51/50' and you need to check if this term is positive, which it is as it equals 2.892. You can then conclude that 51L - 51/50 is > 50L.
To use this method more rigourously, you would have to check if the approximation is valid, by bounding the error in the expansion to ensure it is small. You might also like to look at monotonicity, simply looking at the differences in the observed approximation and actual value.
Your voice is very sweet to listen... Loving and enjoying your voice
Very AMSR
This is what I did and I got
50^50 < 49^51
By using the rules of Indices, I split 49^51 into 49^50 x 49^1 i.e ( a^m x a^n = a^m+n). Now, 50^50= (50^5)^10 & (49^5)10 i.e [ (a^m)^n = a^mxn ].
Ignoring the power of power i.e 10
50^5= 312,500,000
49^6= 13,841,287,201 ......[ 49^5 x 49^1 = 49^6]
P.s. I used calculator for multiplying 😬
In school I was too lazy to do math, but I had a good sense of logic
So I usually cheated the calculations with small numbers to have a guess what was going on with the math
Something like this:
2^4 < 3^3 (16 < 27)
3^5 < 4^4 (243 < 256)
4^6 > 5^5 (4096 > 3125) inverted gap
5^7 > 6^6 (you could stop here because you already have a proof what is going on)
49^51 > 50^50
"teacher, I don't know how to do the math, but 49^51 > 50^50 for the logic reason listed above, the gap inverted and keeps increasing". "A" 😂.
I did so much of this when I was a kid. I still remember there was a 5 question exam once and I did all 5 questions without a single math, only writing sentences explaining why the answer would be X or Y / True False/ how many how much. Teacher told me I "cheated" but still gave me an "A" because he had never seen a math test done correctly without any calculations, only with pure logic. Good memories
Nice going. I wouldn't call it cheating. Rather it's unconventional solving, like MacGyver. Did you follow the syllabus? No. Did you nail the concept and solve the problem? Yes.
I'd love to know what you ended up doing as work or hobby using these skills.
You proved nothing in the rambling
@@GolldLiningyes but if he continued with what he was doing and learnt mathematical induction he would've proved it
Exactly. He was on the right track and in a multiple choice test, he would have got the correct answer.
Bro can you give an example of questions you did without solving which shocked your teacher
Your solution is not suitable for Olympiad. You need to attempt it without using Limits.
It depends of year of olympiad. In USSR limits was part of school program.
@@PARPROX777 Nice ... Which year calculus is introduced. Russians make some great mathematicians
we studied limits in 9th grade, it was 2015, but it is mathphysics school
@@IGAgamesIn Russia every non math school teach limits in 10 grade
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.
but there is need of calculator for ln49/ln50 :)
which destroys thee point of even taking log when one could just evaluate the initial values and compare using any calculator or program
Mathematically, this is the best answer.
I'm honoured to be thousandth commenter. As the video progresses the enthusiasm or the energy in your voice decreases gradually. I knew the answer just by looking which number had more exponent to its power as both are almost near to each other. I just watched fully to discover any other type of solving it quickly. If I do these kind of calculations for solving 1 problem in my aptitude exam then I will run out of time for other questions so I will rather trust my intuition and choose the answers without using up more time😅
Spasibo.
If you can't wait, quickly in python:
print(len(str(50**50)), len(str(49**51)))
why len? int же
@@ЦЕАканал привет. len посчитает количество символов в каждом из представленных(чтобы не выводить большие числа). Какой из вариантов больше, будет более очевидно.
that's cheating... ha ha
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.
Instead of bringing e into this, you could have found the sum of the first 2 terms in the binomial expansion of (1+1/49)^50 (=1+50/49). We know that the full expansion is greater than this. It's trivial from here on...
It sucks how quickly you forget math that you don't use all the time 😢
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).
A good analysis
But this works only when n>=5
right hand side is increasing at a slower rate than the left hand side, so at some point when n increases, left hand side should overtake the right hand side
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.
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!
If you are being rigorous, you would need to put bounds on the error in the Taylor approximation and show that they can't change the conclusion. You can certainly do that, but it gets a little messy. (Doing it the way in the video, you need to prove that (1+1/n)^n is monotonically increasing, which also isn't straightforward.)
Just have to study the function f(x) = ln(x)/(x+1). For x > 0, the derivative is the same sign of g(x) = 1+1/x - ln(x). g is decreasing and g(10) < 0, then g(x) < 0 between 49 and 50. So f(50) < f(49) or 50*ln(50) < 51*ln(49).
Aa..nice way👍🏻
I can solve it in 2 steps 🙂
I love these problems, great mental exercise! Thanks.
Thank you for explaining such a terrifying problem so calmly.
Taking a log we compare 50 ln 50 to 51 ln 49. Take a function y(x)=(50+x) ln (50-x), taking a derivative we have dy/dx=ln(50-x) + (50+x)/(50-x). For any -50
In general y(x)=(a+x) ln (a-x) this proof will work for -a < x < a-1. The exact interval boundaries are not so simple to calculate analytically, maybe with Lambert W or impossible?
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
Your voice was soothing and gave me peace while my mind was screaming inside
Can't we take log base 10 both sides then by properties of log and knowing the value of log2 log5 log7 we can easily calculate
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.
When the numbers are close, the exponent (power) beats the base (the number). In fact, the bigger these numbers are, the smaller number with the bigger exponent (only by one unit) can go lower and lower from 50% of the higher number and the exponentiasion will be higher. Of course, this is only the answer, not the demonstration. More rigurous, but still simple is using the logarithms.
@@sorinturle4599 well even at near x=0 there is crazy separation for exponential graph with higher bases so it makes sense differing by 1 in base doesn't matter as much differing by power by 1
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.
It's simple. The exponent has a bigger influence than the base, so 49^51 is bigger 🤭
Well not necessarily, since 3^2 > 2^3... but in this case yeah, the answer is obvious.
Exactly! Took me 10 seconds to conclude that
@@UltraStarWarsFanatic 3 and 2 are small numbers. Exponential starts to grow after a while, so my logic is for numbers a little bigger than 1.
Let y1=49^(x+1) , y2=50^x
We know that both logarithm 49^(x+1) & 50^x, increasing exponentially.
Y1>Y2 from starting point (0,0)
Y1 5^4, then 49^51>50^50
And also clarify to those who said we can just imagine the base is expanding, it not works like this as different numbers have extremely different point to intersect.
This is flawed like many such math "solutions" on TH-cam. 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.
As a coder we just write a program and find answer single click 😁😁
Python code---
num_1=50**50
num_2=49**51
if num_1
you can use a calculator too, in that logic.
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
Thanks!
Thank you gabirican😊
I had this question 35 years ago while living in Soviet Union. Did not use Euler, just natural logarithm. The same exact result.
Натуральный логарифм это логарифм по основанию е, так-что почти одно и тоже
I checked the sequence n^n - (n-1)^(n+1). The limit is -inf. Interestingly after the first few terms it becomes sharply negative and dives down pretty fast. Here's the first few terms: 1 3 11 13 -971
Notice that it seems pretty "tame" for the first 4 terms and suddenly dives down steeply. It is strictly decreating from that point until at least 51st term. I didn't check after that, but the 51st term is
-2.098525e+88
Or on order 10^88 and negative. So I can say with confidence that any term after n =5 will be negative.
One more trick is you can see what 5^5 is less than 4^6 by a good margin. With that logic a bigger number such as 50^50 would be less than 49^51
Just don't make the mistake and use even lower numbers to make your point: 2^4 and 3^3
Math is exact science(c) 😎
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..
i just compared 10^3 and 9^4. since 9^4 is bigger i thought 49^51 would be bigger
Exactly… I did something similar. 50^50 is n^n and 49^51 is (n-1)^(n+1) and then, as you did, changed the n for a much smaller number, so I could easily do the calculations, I chose n=2 so I had 2^2 in one side and 1^3 on the other and it appears that the second part is not bigger, but as n grows over 2, you go getting bigger numbers each time on the (n-1)^(n+1) side
Higher power wins for all numbers > 5^5
49^51 is actually almost 20 times larger than 50^50.
(bruteforced it by excel)
Higher power doesn't always win. 4^4 > 3^5.
Ok this could be an intuitive answer, but it is so obvius to me that 49^51 is higher simply because there is one more multiplication in it. I’m pretty sure that 49^50*10 would be also higher.
You are correct, but that's just a lucky guess. 49^50*2 also has one more multiplication, but isn't higher.
@@thomasdalton1508 Intuition is not a lucky guess. I didn’t say that any multiplication is enough here. That could be *0.1 too, which is also a multiplication and obviously wrong.
You can say that this is not an acceptable proof, but the luck has nothing to do with.
It does seem intuitive cuz of that one additional multiplication. But it's not true for 3^3, and 4^4. From 5^5, the latter expression is larger
So
Is 50^50 > 49^51. Answer: No. Cuz m^m < m-1^m+1, where m=/>5 👽
@@Selendeki But, as Ashton pointed out, increasing the exponent by one doesn't always do the job - it only works for 5 and greater. I don't think it is at all intuitive that the threshold is 5. That requires doing the calculations. Just because your intuition gives the right answer doesn't mean it is good intuition. The exact same intuition would have led you astray for different numbers.
@@Selendeki Intuition certainly has its uses and developing your intuition is a very important part of learning mathematics. There are two problems with the OP's comment. First, it is bad intuition - there is no intuitive reason that multiplying more numbers should get a larger result than multiplying larger numbers. It depends on how many more and how much larger. And second, it is completely wrong to say something is "obvious" based on intuition. That just isn't how you do maths. You don't guess and then say your guess is obviously correct.
I don’t know the proofing method I’d use but.. if I was asked what the comparison between the two number is, I would’ve gotten the same answer by a simpler means.
The base number and exponent is off by one but one is exponentially more valuable. Therefore it will weigh more heavily.
2^3 < 3^2
Disclaimer: not a math person. Issue I see is that the left hand side is 50^50 (same same), so 2^3 vs 3^2 isn't a relevant pattern. Should be 3^3 vs 2^4, or 4^4 vs 3^5, which would give the wrong answer here. I think only starting from 5^5 vs 4^6 is left hand side lesser.
In such cases, I always try with random numbers
For example ; 3^4 = 81 > 4^3 =64 or 4^5 = 1024 > 5^4 = 625
It can go on like that to forever
Only exception from 0 to 2
That means n^n+1 > n+1^n always
I just wanted to explain, I mean it can be solved within 10 sec if you can see that
That’s weird. I had a hunch 49^51 is bigger than 50^50 because exponent is always much bigger than a base, without doing all the arithmetic 😂
But without proof, it's crystal ball gazing. Try 3^3 vs 2^4 ... 🤓
By your logic, the higher number should be the one with the higher exponent, but it's not... 😱
@@Stepan_HIt's better to have proof instead of assumptions, but it's also true that generally higher exponents mean high numbers overall. Yes, 3^3 > 2^4, but even just raising each base by 1 will already make the no. w/ higher exponent larger with 4^3 < 3^4. Keep increasing the bases by 1 after that and the number raised to 4 will always be larger.
Even if you had a difference of 2 for the bases, with 4^3 > 2^4, increase the bases by 1 you'll still get 5^3 > 3^4, yes, but increase it one more time and everything starting with and after it'll always be 6^3 < 4^4 or n^3 < (n-2)^4 as long as n > 5.
Proof is good, but we're in the comments, not a math comp. Most people here just want the answer, so might as well let them know a "trick" even if it isn't always true. These kinda problems aren't really encountered by the general public often anyways, and anytime they do it'll involve huge numbers with little differences in bases and exponents, at which point the number with the higher exponent is larger 99.9% of the time. By the very small chance it isn't, well it's just youtube so does it really matter? Haha
That's how I looked at it too. I didn't know how to prove it. But using common sense I figured if decrease the base by 1 and raise the exponent by 1. Then the 1 with the raised exponent is larger
unless asked for the difference, they should be consistent, so simple (university degree) trick is to use 0 or 1.
1^1 > 0^2 because 1 > 0^2
2^2 > 1^3 because 4 > 1
3^3 > 2^4 because 27 > 16
4^4 > 3^5 because 256 > 9x9x3 = 243
5^5 > 4^6 because 25x125 = 3125 > 4^6 = 16^3 = 1024
6^6 > 5^7 because 36^3 = 46k and 25^3*5 = 78k
7^7 < 6^8 because 49^3*7 -= 823k and 36^4 = 1.7M
So the turning point is around 7, which is weird, because up to 6 the difference were larger and larger.
I didn't understood step at 6:28 where you take 1/6 instead of 1/49
I think it was totally unnecessary though it was just to make the inequality simpler by canceling using a multiple of 3. She should've just calculated (3 x 50)/(49 x 49) which isn't difficult and very evidently much smaller than 1.
Because (1 + 1/49)^50.... Is smaller than 3x 50/49x1/49. So,if that thing Is smaller,It has also to be smaller than 3x50/49x1/6.
Lost me. Need a different explanation. Anyone??
How is (1+1/49)^49 lesser than 3 this lim n~infinite (1+1/n)^n = e used here n is 49 which is finite
after 6:20
I propose
multiply numerators , 150
compare them with DE numerators, 49 squared
hence the fraction is less than 1 ,
You are good, but it becomes so convoluted I give up before the answer. Is there an short cut to do the problem?
That’s math for ya
There's no route only for nobles in math
WOW, it is great but a bit lengthy..!🥰
I am an indian and i did binomial expansion and did it in like 10 seconds.Just equate (50/49)^50 to 49 first and 50/49 is 1.02 ,which can be written as (1+0.02).After that just multiply 50 to 0.02 so tha answer will come 1+(50×0.02)=2 which is less than 49. So 50^50 is way way smaller than 49^51.
6:53 How did one < times one > times one < ended up as one < symbol? Does it only bother me?
Why simply not divide on with the other and see what happens?
(50/49)^50 * 1/49 and see if it's greater or less than 1 ? If it's less than one, then 49^51 is greater than 50^50.
Now, eyeballing this, 49 is by 1 lesseer than 50, which is 2%, so 50/49 is probably around 1,002 -1,003. Do I believe 1.003^50 is greater than 49? I don't believe it's even greater than 2, so if I was a betting man, I'd say 49^51 is definitely greater than 50^50.
(Since 1,003^3 / 49 is certainly lesser than 1)
Now I can go and see what you did here in 8 minutes...
@@English_shahriar1 OMG! I think you've just helped me realize I'm in an early stage of Alzheimer's :(
But thanks, I guess....
Even if your answer ls right, You are wrong in some some aspects, so be careful in the future. 50/49 it is a difference of 2% SO it would be around 1,02-1,03. Then 1,03^50 is not greater than 49 but it is greater than 2
Simpler solution
Just take it as
[(49)³]¹⁷ & [50².⁹⁵]¹⁷ (2.95 is approx but higher than the value of 50/17)
We get 117,649¹⁷ & 102,792.52¹⁷
We can easily deduce that the number with higher base would yeild a higher solution when raised to the same exponent.
So 49⁵¹ > 50⁵⁰.¹⁵ so it has to be greater than 50⁵⁰
This only works for simple bases and simple powers.
I looked at both numbers and could immediately see that that 49^51st is larger. Visualize 50^50th as 50x50... all the way to the 50th one. Visualize 49^51th as 49x50th ... x the 51st 49. Clearly that will result in 49^51st being larger. You can even use a simpler model to prove it. 50 ^ 3rd vs 49 ^ 4th; 50x50x50 = 125k; 49x49x49x49 = 5,764,801. Notice that the difference between 50 and 49 is only 1. That seems to be true for any two consecutive numbers starting with the number 3.
I think you mean 5 and higher. (n^2 > (n-1)^(n+1))
GSR Theorem (Logical Approach): Given: 50^50 and 49^51. Question: Which is bigger?
Solution: Let (assumption) 50^50 =49^51.
Left Hand (LH) = Right Hand (RH)
Let A = base 50
B = base 49
A > B
50 > 49
Let r sub b = ratio of bases, (50/49) > 1
Let r sub e = ratio of exponents, (50/51) < 1.
Condition: (r sub b) > 1 ; (r sub e) < 1.
Analyses: (r sub b, greater than 1)/ (r sub e , less than 1), then the result is > 1. OR (r sub b /r sub e) > 1.
Therefore: r sub b > r sub e.
It follows that (Theorem) if LH: r sub b > 1 AND r sub e r sub e .
Meaning, (A/B) > r sub e, OR A > (r sub e)*(B).
IF A = (r sub e )*B, THEN B or RH must be BIGGER to COMPENSATE for A (LH) since (r sub e) < 1.
THEREFORE: RH > LH --> OR 49^51 > 50^50 --> OR 50^50 < 49^51. ANSWER.🥰
I’m glad she’s not my introductory algebra teacher or I’d go insane. I may be ignorant on this convoluted mathematical solution but I just assumed the following which gave me the correct “guess” to the problem given.
I worked out a simpler similar equation in my addled mind: of 4 squared vs 3 cubed, answer: 16 vs 27, therefore 49 to the 51st power is larger. 😱 👀 ⁉️ 🤔
But its an Olympiad Q - you need to show it, not just be content you know the answer. Indeed, 4^4 > 3^5 which does not follow the seeming general rule n^n < (n-1)^(n+1) - and in you sample 3^3 vs 4^2 does not match the pattern in the Q: you have n^n (n+1)^(n-1).
x^x - (x-1)^(x+1) is only positive for x=5.
There's a very similar problem, but I forgot the details, where the limit is e.
Can anyone figure out what that problem is?
It comes for a problem like 50^49 < 49^50. Then the function to study is ln(x)/x . This function his maximum at e. In this problem, the function is ln(x)/(x+1). See my comment just below.
@@HenriLaporte-kv6qq thank you!
How did you reduced 49 as to 6?
She chose 6 to make it simple, as long as it was smaller than 49 it would have worked. In a sense, she gave it a try, to see if the whole equation would be smaller than 1 this way, if not, she could have tried bigger denominator, if still smaller than 49, because anyway, the whole reasoning with e is based on not having to be precise if the equation is smaller than 1. If it was bigger, equal to 1, or really close to 1, thing that we are not supposed to know at that moment, then the reasoning would be probably no conclusive, but it was worth giving it an easy try.
She should have explained that substitution maybe a little better. The rest of the video was quite easy to follow
I solved it with a logaithmic table and a basic calculator:
50^50 vs 49^51
log 50^50 log 49^51
50 log 50 51 log 49
50 x 1.69897 51 x 1.69020
84.9485 86.2
49^51 is larger
Excellent!! Your writing system of "9" is totally exceptional 😮
This is an excellent problem and great way to resolve , did learn a lot
What age group is targeted?
12 years
@@banjo4us1 no way! at least 14-15. There is no way you can have that limes part at 12
@@phajgo2 Maybe you are right. However most Olympiad competitors start early. In our school, we started early coaching by 12. Most competitive exams prep starts by 10 to 12. I am talking about India and how most successful candidate crack exams.
@@banjo4us1 I'm from Poland so there may be difference in programs :) we also start olympiads early (around 10-12 as you say) but they usually do not go beyond the program of math classes for the given age as the intention for children is not to incentivize learning extensive material before you're supposed to but it is rather to find smart solutions within the knowledge you have. Still I'm curious to know what age was that question intended for because I found it quite difficult :)
@@phajgo2 I agree with you. Most problem in these Olympiad could be solved either using higher theory or rudiment maths. I remember once there was a problem of a bird catching a fish and then perching on a tree. The problem could have been solved using Snell's law, but it could also be solved using Similar Triangles, albeit it is a longer solution. I always studied upto 3 years ahead so that I could solve these kind of exams. Even in IIT papers here in India.... most questions has higher maths solutions.
The answer is correct but the proof is not complete. The limit is e but it is still lack of proof that it is smaller than 3.
Eu assisti em outro idioma e entendi, por isso eu amo a matemática ❤
Abstract solution:
For 50^50 v.s. (50-1)^51,
(50-1)^51=
50^51 (“-“ a*50^50“+” b*50^49“-“ c*50^48 …+-+-…)-1
Where, (“-“ a*50^50“+” b*50^49“-“ c*50^48 …+-+-…) are merely negligible, compared with 50^51.
Hence, 49^51 is larger.
More Abstract solution, write 49^51 = (49×49)^50 which means
50^50 ___ (49×49)^50. And this clearly states that right hand side is bigger.
@ your equation could be false. 49^51 is not equal to (49*49)^50. It’s 49*49^50
@@aaaaaaaaaaaaa. Yeah you are correct, I miss typed it. Thanks for the correction.
Time management is also part of the test. The answer is C.
Taking a logarithm is way easy and short one idea.
50Log50, 51log49
50log 50= 50log(7^2 +1) or we can find sqaure root of 50 as 7.07 aprox. ( You can calculate any square root of a non perfect number by looking difference of two consecutive perfect square numbers eg. 7^2= 49,8^2=64, difference=64-49=15 now 50 is 49+1 so squareroot of 50= 7+1/15= 7.07 aprox. )
So 50log50=50log(7.07)^2 =50*2log7.07=100log7.07
Also 51log49=102log7
Clearly 102log7 is bigger than the other
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
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?
I don't care about what your teaching. I just came here cuz your voice is soothing and your handwriting is crystal clear.
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
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
The underscore represents a blank for what their relation is
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
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.
Thank you very much my dear friend ❤
I think 49 is being multiplied by itself one more time than the L.H.S. so I thought RHS is greater than LHS 😅.
You proved it mathematically though. I thought it was a trick question so you'd have a trick solution but guess not.