I don't know if you know but this video is being blatantly copied, here's the link: ua-cam.com/video/Um53h6yq_6o/v-deo.html If you look at the comments here there are various ways in which you can solve the problem but in the specific video they have the same script/format with how you systematically solved the problem, even down to the explanation of the constant "e" and how they used 1/6 instead of 1/49. Even the title is copied.
Thank you for your remind! Not only is this copied, but also my other videos! Not only are my videos copied, but many videos from other channels as well! Even the duplicates have more views than the original ones! SHAMEFUL!!!
@@mathwindow Glad to be of help, what's worse is that youtube recommended that specific video to me. It might be the case that the algorithm pushed it to me because I watch your videos. Shameful behavior indeed!
I divided both sides by base 49 to the power of 50. That yielded base 50/49 to the power of 50 compared with 49. Algebra shows that the base 50/49 requires a power greater than 193 to exceed 49, so base 50 to the power of 50 is less than base 49 to the power of 51.
@@BalderOdinson We want to find k such that (50/49)^k >= 49. Log is monotone, so we can take logs of both sides, giving k log(50/49) >= log 49, and solve for k. Pretty lame for an Olympiad problem to be so easily solvable with a calculator. Edit: actually, with a calculator, you can just instantly see that 50^50 and 49^51 differ by two orders of magnitude, so the whole thing is pointless. I'd assumed they'd at least look equal to the precision that a standard calculator would give.
@@beeble2003 proving it without a calculator is the whole point. Any proof that involves picking up a calculator could be substituted by just using a calculator on the original question.
I saw a lot of people doing complicated solutions but i did it in a simpler way, don't know if it would work in all cases. I first tested this same case with smaller numbers that you can actually calculate: 4^4 and 3^5, in this case 4^4 > 3^5 and the diference between them is 13. Then i tested it again with 5^5 and 4^6, and in this case 5^5 < 4^6, and the differencre between is -971. With that we can see the difference will just keep getting bigger with higher values, so from that point on the number with the highest exponent will be greater, so 49^51 > 50^50.
@@kaustubhnadiger3387 in induction proofs, a base case means all numbers below it are disregarded. x=4,3,2,1... and on are all considered irrelevant where the base case is x=5. after that, you just need to prove that the statement holds for all cases where (x+1) replaces (x) and use the inductive hypothesis to assume the statement involving x is already true, which holds if the base case is true, which is true for x=5.
Divide both by 50⁵⁰: 1 vs. (49/50)⁵⁰ × 49 = (0.98)⁵⁰ × 49 Take natural logs of both and rearrange slightly: 0 vs. 50•ln(1 - 0.02) + ln49 To a very good approximation, ln(1 - 0.02) ≈ -0.02, so this is to a good approximation, 0 vs. 50(-0.02) + 2ln7 or 0 vs. -1.0 + 2ln7 The rhs is easily positive, so the circle encloses a "
Or just note that if a is larger than c and 2a = b + c is larger than 2 times Euler´s number, then a^a is larger than b^c by basic calculus.
Рік тому
Well, here c is larger than a in my last post. The largest exponent gives the largest value if the conditions in my last post are met. (51^49 is less than 50^50 which is less than 49^51).
Рік тому
Nice calculation
Рік тому
Of course there is the requirement that the numbers are "close". For example 12^188 is larger than 100^100, But 11^189 is less than 100^100. Here logarithms come in handy. 188 ln(12) is greater than 100 ln(100), which is greater than 189 ln(11). (51 ln(49) is greater than (50 ln(50).
If you go in a calculus approach, you can consider the function f(a) = (50-a)^(50+a) = e^{(50+a)ln(50-a)} You can take the derivative, which is f’(a) = (50-a)^(50+a) [(50+a)/(50-a) + ln(50-a)] One can see that f(a) monotonously increases at least in the range such that (50+a)/(50-a)>0 and (50-a)>1 => f monotonously increases for -50 < a f(0) < f(1) => 50^50 < 49^51 Which is what we wanted. And it is nice to see one can immediately get 51^49 < 49^51 And the like exercises.
(1-1/50)^50 is not 1/e. Don't be confused with the sign. We could argue that (1 - 1/50) is less than 1, which implies (1-1/50)^50 < 1. The conclusion is same. Great approach.
The math is interesting to be shown but I really feel this more of a logic test as with the base numbers being so close just being multiplied that 1 extra time is a massive jump over the other. So it had to be the bigger of the two.
True, but if it wasn't proven are you 100% sure tho? Maybe 70% or 80% but never 100%. Math is all about 100% right solutions(at least 99% of math cuz theoretical maths)
I did that with derivatives, that was tough! You should just take derivative from x^(100-x) then solve the transcendental equation, the solution is easily guessed: it is somewhere near 24 or 25. On infinity the derivative is negative, so the function is lesser for 50^50 as 50 > 49 and both 50 and 49 are greater than 25.
Take the log of each side, know the log power rule making 50log(50) vs 51log49 and we all know that log grows sublinearly, but monotonically increasing, therefore 49^51 is easily bigger.
Because it has a decay function. But with the smallest number that makes sense... (1+1/2)^2 to 1.5×1.5= 2.25 okay small but maybe a BIG number works (1+1/9001)^9001≈2.71813 No matter what, this function will not pass its limit for positive numbers, which is eulur's number, that is < 3
See, I just used a calculator. But 50^50 and 49^51 are so large they would probably just give an error, so instead i took _the log of both,_ and 51log49 is larger than 50log50.
Only with access to a calculator. Otherwise it is only kicking the can down the road, so to speak - the problem of evaluating the logarithm by hand is not any easier, and will probably use a similar trick.
@@la.zanmal. Right but it's pretty lame to have an Olympiad problem that's so easily solved with a calculator. Edit: actually, don't even bother with logs. Your calculator can do 50^50 and 49^51 and they differ by more than two orders of magnitude. I was expecting that they'd be close enough together that they'd look equal to the precision that a standard calculator gives.
@@gatoordinario94 Your argument is incomplete. You need to show that log 50/log 49 < 51/50. You've correctly argued that the left hand side is close to 1. That's not enough, as the right hand side (1.02) is also close to 1, and we need to know which of these numbers is bigger.
Take log on both sides, take the exponent in the front as a constant, divide by (51*50). This yields a comparison between 2 values of function ln(x)/(x+1). The left side is x=50 and the right side is x=49. This function is clearly decreasing. So we have a solution.
Extract root 50 both sides which yields 50 ? 49x49^(1/50), and knowing that numbers < 1 rised to any positive power online be 1 at the infinite, so 49x0,... will be < 50.
I just used binomial theroem at (1+(1/49))^50 *(1/49)≈ (1+(50/49))*(1/49)=99/49^2 so yeah maybe not the best but it did the job basically it states that (1+x)^n ≈ 1 + nx
Once you rewrote the problem as a relationship to one, the only question is whether or not it’s less than or greater than. 50 to the 50th power divided by 49 to the 50th power is less than one with that quotient being multiplied by 1 over 49. So you are asking do two numbers less than 1 multiply to a number larger or smaller than one. The answer is they will always be less than one so 49 to 51 power is greater than 50 to the 50th power.
Nobody here has an intuitive solution. The clear proof is to write 49^51 as 7^102 and 50^50 as 5sqrt(2)^100. We can the approximate 5sqrt(2) as 5*1.414 = 7.07: => 7.07^100 < 7^102 => 7.07^100 < 7^100 * 7^2 /7^100 => 1.01^100 < 1 * 49 => e < 49 Which is true. No frills, no fancy theorem, and no extra computation, just the knowledge of the approximate value of sqrt(2) and the value of e.
Well IIRC, if the number of digits on the base numbers were in the same range (like 10 with 11 being 2-digited numbers), even if one of the base numbers were bigger, the bigger exponent will always have bigger value no matter which one has it. So by that logic, because the difference between 50 and 49 was small enough, I can assume that the value of 49^51 would be at least 10 times larger than 50^50.
Since the question was which side is greater, without requiring proof, I intuitively knew that the side with the higher power is greater for anything greater than 3^2 vs 2^3
I think i found an easier way: you subsitite x with 50, you get on one side x^x for 50^50 and for 49^51 you get (x-1)^(x+1) and then using binomial formulas and etc you get: 50^50 < 50^51 - 2498 (as 2489 is much smaller than 50^50 x 50) Tell me what you think!
You could also have done it using logarithms Taking log ( base 49) of both sides and the approximating value of log(base 49 ) (50) and getting the answer
generic solution x^x vs (x-1)^(x+1) transforming it to 1/(x-1) vs (1-1/x)^x then left part is 0 < 1/(x-1) < 1/4 for x>5 right part: 1/4 < (1-1/x)^x < e for x>2, thus for any x> 5 right part is always larger. Done
Every time you are increasing the power like 10^6 and 10^7, it becomes a huge difference because it would have more zeros at the end, so the number with the larger power would likely to be larger than smaller one. Unless it more complicated, I could not say that 2^3 is larger than 4^2, so it varies on cases, but since the digits are close, I think you can tell.
I solved it with the properties of logs. Taking the log of both sides simplifies to 50 log 50 and 51 log 49. Log 50 and log 49 are almost the same due to log based anything big, squishes changes hard. Therefore, 50 and 51 are compared alone.
if you formulate the question x*x and (x-1)*x+1 and start to put small numbers in these formulas you would see except for one (4*4 and 3*5), (x-1) * (x+1) is always greater than x*x
The suggested solution is incorrect. The fact that (1+1/n)^n converges to e as n→∞ says nothing about how large (1+1/49)^49 might be. It only says something about the asymptotic behavior of the function that maps n to (1+1/n)^n. An easy and correct way is to use the inequality ln(1+1/n) < 1/n, derived from the Taylor expansion of the natural logarithm, to deduce ln( (1+1/49)^49 ) = 49·ln(1+1/49) < 49·1/49 = 1, which implies (1+1/49)^49 < e. To participants in a math olympiad this should be a trivial exercise.
A simpler step could be dividing the (50/49)^50 inside term, that will be 1. smth and overall approx to 1 now this 1/49 is obviously small term hence the final answer is that 50^50 is smaller than 49^51
I just used a calculator to observe that 5^5 is less than 4^6. I therefore generalized that for any positive number n greater than 2, n^n < (n-1)^(n+1).
No need. A slightly bigger number multiplied certain times is way smaller than a slightly smaller multiplied one more time. Because the last added multiplacation is done to the total number reached before it while the increase in the multiplied number is an increment.
@mathwindow I know 1/49 is less than 1/6. I just don't see where the 1/6 come in to play to solve the problem. You wrote it down, but I don't see where that value came from. I watched that part 5 times and still did not see it. Maybe I overlooked. Just curious about that. Thanks.
Just an easy way, even no pen or paper needed: 3^3 > 2^4; and 4^4 > 3^5, but already next group 5^5 < 4^6 with a gap increasing on 6^6 < 5^7 and so on. Safe to interpolate 50^50 < 49^51.
Anyone who's interested in solving IMO problems doesn't need more than two minutes (from 0:54 to 2:59) to get that (50^50)/(49^51) = (50/49)^50 x 1/49.
When x is large, (x-1)^(x+1) > x^x. Let x be a positive integer. We want to compare x^x and (x-1)^(x+1). We address here the case when x is large (>>1). To do so, we take the log: log ((x-1)^(x+1)) = (x+1) log(x-1). x being large, log(x-1) = log(x(1-1/x)) = log(x) + log(1-1/x) \approx log(x) - 1/x because log(1+u) \approx u when u is small. So, (x+1) log(x-1) \approx (x+1) (log(x) - 1/x) = x log(x) + log(x) - (x+1)/x We conclude that when x is large, (x+1) log(x-1) - x log(x) \approx log(x) - (x+1)/x And obviously, log(x) - (x+1)/x is also dominated by log(x) which not only is positive, but diverges! So, when x is large, (x-1)^(x+1) > x^x. Remains the question of 'large'. Well, instead of developing the log(1-1/x) around zero, you can keep its exact value and study the difference, (x+1) log(x-1) - x log(x) = log(x) + (x+1) log(1-1/x). Starting from x=5, you can show that this difference is monotonic, increasing, so large is not so large, it's x=5.😉
49^51 is larger. I solved it with a simple arithmetic calculator and a logarithm table. 50^50 49^51 log 50^50 log 49^51 50 log 50 51 log 49 50 x 1.6989 51 x 1.6901 84.948 86.200 Inv log Inv log 8.8716^84 1.5849^86
We can compare the two numbers by taking their ratio: (50^50) / (49^51) Simplifying this expression, we can rewrite it as: (50/49) ^ 50 Since 50/49 is greater than 1, raising it to the 50th power will give us a number that is greater than 1. Therefore, (50^50) / (49^51) is greater than 1, which means that 50^50 is greater than 49^51. Therefore, 50 to the power of 50 is larger.
one of the simplest way to solve this question by taking logarithm both side and solve it it easily showed 49^51 is greater than 50^50 its very simple problem if we didn't use your method my friend
7:33 when did 1/6 came from? Is there any reason to be specifically 1/6 or it's just to simplify the multiplication... (but then we could just use 1/3 and it would be 50/49 and LARGER than 1)
We could use any number bigger than 1/49,for example 1/48,1/47,1/46....etc. 1/6 is one of these numbers chosen by the lady to be accurate and convient enough to solve the problem. if you choose 1/3,It's not accurate if we know the result is smaller than 50/49,what we need to know is that whether the result is smaller than 1.
Or you could just know that bigger exponent than base is bigger than bigger base than exponent for the same numbers, as long as both numbers are greater than e. And this problem is just using a midpoint which would obviously be between them. Not a hard proof but it works. Yeah 2^3
Just consider (50/49)^50 vs 49. (50/49)^50 has 51 terms with only one term 1 and one term (50/49) larger than 1. You can take the excess 1/49 and any one of the remaining terms and merge them with any one term and still have 49 terms with only two equal to one nad rest less than one. This would always be less than 49.
i knew the correct answer immediately just based on the principle that in large numbers, the exponent is always more influential than the base; for example although in the case of 2^3 < 3^2 the base matters more . . . this is an exception to the rule because the numbers are so small . . . the rule that "exponents have more influence than the base" applies for even slightly bigger numbers; for example 5^6 > 6^5. I have no proof to offer, but there is a general rule that holds when comparing numbers x^y vs (x-1)^(y+1) as long as both x and y are greater than 2 then (x-1)^(y+1) will always be greater than x^y
I started looking at (1+1/n)^n and soon figured out that it tends to e (the base of the natural logarithm) which implies that the second number is (very) approximately 50/e times larger.
I did it in a very simple manner using basics of binomial... I wrote 50^50 as (49+1)^50 and expanded it a bit as.. (50c0.49^50 + 50c1.49^49 + 50c2.49^48+ - - - - + 50c50.1).... EQN. 1 Now i evaluated the numerically greatest term in the expansion using (n+1) /(1+|a/b|) - 1
Honestly these are waaay more simple than they seem, just graph x^x=y and (x-1)^(x+1), then compare ‘em, they intersect at around 4.136 and never look back
My estimate would be 49^51 greater. Because for two numbers being only 1 unit appart, also being powered at least 50 times, wich is a lot, 51>50. But i do not know the verification yet.
Another way that I approached it employed the binomial theorem and root approximation: 50^50 ? 49^51 (49+1)^50 ? 49*49^50 (1+1/49)^50 ? 49 1+1/49 ? 49^(1/50) Now x^(1/a) ~ t+ (x-t)/a*t^(a-1) where t is a close estimate to the root letting t = 1, then 49^(1/50)= 1+ (49-1)/50*1^(50-1) = 1+48/50*1^49 = 1+48/50 Comparing now 1 + 1/49 ? 1 + 48/50, leaves 1/49 ? 48/50 so 50 ? 48*49 where ? must be "
What you're doing is reducing both exponents by 48. This is not doing the same to both terms. One term is devided by 50^48 and the other by 49^48. If your change really would be the same on both, then you would be right. To better wrap your head around consider this: You have a lake with swimming lake flowers. On the first day, 1m² of the lake is covered and from then on each day the covered size doubles. If you went back from day 14 to day 12 or from day 8 to day 6, it would both times be 2 days but the amount of lake flowers would be much more in the jump from 14 to 12, as "doubeling" the size from day 12 is much more than doubeling the size from day 6.
I'd like to know: if you take two integers x^x and (x-1)^(x+1), is the second one always larger? 4^4 = 256 and 3^5 = 243. Then 5^5 = 3125 but 4^6 = 4096. Now the second is larger. Does that relationship continue always now, as x gets even larger?
We first note that comparing x^x vs (x-1)^(x+1) is equivalent to comparing x.ln(x) vs (x+1).ln(x-1), which itself is equivalent to comparing ln(x)/(x+1) vs ln(x-1)/x. So we want to study the variations of the function x -> ln(x)/(x+1). The derivative is (1+1/x - ln(x))/(x+1)^2 which is positive for small x's but becomes negative afterwards (I believe the sign changes between 3 and 4?). Coming back to our studied function, it is increasing for x small but quickly becomes decreasing. Therefore, for x big enough, ln(x)/(x+1) < ln(x-1)/x and so we get that x^x < (x-1)^(x+1)
If you're going to use a calculator, just ask it 50^50 and 49^51, and it'll tell you that they differ by two orders of magnitude (about 10^84 and 10^86).
I don't know if you know but this video is being blatantly copied, here's the link: ua-cam.com/video/Um53h6yq_6o/v-deo.html
If you look at the comments here there are various ways in which you can solve the problem but in the specific video they have the same script/format with how you systematically solved the problem, even down to the explanation of the constant "e" and how they used 1/6 instead of 1/49. Even the title is copied.
Thank you for your remind!
Not only is this copied, but also my other videos! Not only are my videos copied, but many videos from other channels as well!
Even the duplicates have more views than the original ones! SHAMEFUL!!!
@@mathwindow Glad to be of help, what's worse is that youtube recommended that specific video to me. It might be the case that the algorithm pushed it to me because I watch your videos. Shameful behavior indeed!
Blud doesn't even get copyright 💀
@@spthepero2282 it aint even copied
But the video in the link is newer than this bruv
I divided both sides by base 49 to the power of 50. That yielded base 50/49 to the power of 50 compared with 49. Algebra shows that the base 50/49 requires a power greater than 193 to exceed 49, so base 50 to the power of 50 is less than base 49 to the power of 51.
That's how I did it too! It took just a minute. The method in this video is such a waste of ink.
Please define "Algebra Shows"
@@BalderOdinson Yeah, that sounds more like arithmetic than algebra.
@@BalderOdinson We want to find k such that (50/49)^k >= 49. Log is monotone, so we can take logs of both sides, giving k log(50/49) >= log 49, and solve for k. Pretty lame for an Olympiad problem to be so easily solvable with a calculator.
Edit: actually, with a calculator, you can just instantly see that 50^50 and 49^51 differ by two orders of magnitude, so the whole thing is pointless. I'd assumed they'd at least look equal to the precision that a standard calculator would give.
@@beeble2003 proving it without a calculator is the whole point.
Any proof that involves picking up a calculator could be substituted by just using a calculator on the original question.
I saw a lot of people doing complicated solutions but i did it in a simpler way, don't know if it would work in all cases. I first tested this same case with smaller numbers that you can actually calculate: 4^4 and 3^5, in this case 4^4 > 3^5 and the diference between them is 13. Then i tested it again with 5^5 and 4^6, and in this case 5^5 < 4^6, and the differencre between is -971. With that we can see the difference will just keep getting bigger with higher values, so from that point on the number with the highest exponent will be greater, so 49^51 > 50^50.
Can it be proven to hold for all cases?
I have done it the same way
@@1mol831 you can probably use induction to prove x^x < (x-1)^(x+1) for all integers starting with some base case like x=5
@@Vapor817 nope, easily disproved for x=2
@@kaustubhnadiger3387 in induction proofs, a base case means all numbers below it are disregarded. x=4,3,2,1... and on are all considered irrelevant where the base case is x=5. after that, you just need to prove that the statement holds for all cases where (x+1) replaces (x) and use the inductive hypothesis to assume the statement involving x is already true, which holds if the base case is true, which is true for x=5.
Divide both by 50⁵⁰: 1 vs. (49/50)⁵⁰ × 49 = (0.98)⁵⁰ × 49
Take natural logs of both and rearrange slightly: 0 vs. 50•ln(1 - 0.02) + ln49
To a very good approximation, ln(1 - 0.02) ≈ -0.02, so this is to a good approximation, 0 vs. 50(-0.02) + 2ln7 or 0 vs. -1.0 + 2ln7
The rhs is easily positive, so the circle encloses a "
👍🏻👍🏻👍🏻💟💟💟
Or just note that if a is larger than c and 2a = b + c is larger than 2 times Euler´s number, then a^a is larger than b^c by basic calculus.
Well, here c is larger than a in my last post. The largest exponent gives the largest value if the conditions in my last post are met. (51^49 is less than 50^50 which is less than 49^51).
Nice calculation
Of course there is the requirement that the numbers are "close". For example 12^188 is larger than 100^100, But 11^189 is less than 100^100. Here logarithms come in handy. 188 ln(12) is greater than 100 ln(100), which is greater than 189 ln(11). (51 ln(49) is greater than (50 ln(50).
If you go in a calculus approach, you can consider the function
f(a) = (50-a)^(50+a) = e^{(50+a)ln(50-a)}
You can take the derivative, which is
f’(a) = (50-a)^(50+a) [(50+a)/(50-a) + ln(50-a)]
One can see that f(a) monotonously increases at least in the range such that
(50+a)/(50-a)>0 and (50-a)>1
=> f monotonously increases for
-50 < a f(0) < f(1) => 50^50 < 49^51
Which is what we wanted. And it is nice to see one can immediately get
51^49 < 49^51
And the like exercises.
great idea
Divide both by 49*50^50. Then LHS=1/49, the RHS =(49/50)^50=(1-1/50)^50 which is very much 1/e. e=2.718
(1-1/50)^50 is not 1/e. Don't be confused with the sign. We could argue that (1 - 1/50) is less than 1, which implies (1-1/50)^50 < 1. The conclusion is same. Great approach.
The math is interesting to be shown but I really feel this more of a logic test as with the base numbers being so close just being multiplied that 1 extra time is a massive jump over the other. So it had to be the bigger of the two.
exactly what I was thinking
Yup, in a timed exercise, this proof is unnecessary and as a teaching aid it is too convoluted.
Or we can just use logs to figure out the number of digits.
True
True, but if it wasn't proven are you 100% sure tho? Maybe 70% or 80% but never 100%. Math is all about 100% right solutions(at least 99% of math cuz theoretical maths)
I did that with derivatives, that was tough! You should just take derivative from x^(100-x) then solve the transcendental equation, the solution is easily guessed: it is somewhere near 24 or 25. On infinity the derivative is negative, so the function is lesser for 50^50 as 50 > 49 and both 50 and 49 are greater than 25.
I multiply every thing by zero. No more problem. Back to dog videos.
make this more popular
Lol
Work out value of x when x^x = (x-1)^(x+1). For any +ve n>x: the result of n^n < (n-1)^(n+1).
I think this is just for knowing how numbers work.
You can just do 10^3 and 11^2.
It is 1000 for 10^3 and 121 for 11^2
I choose 49^(51) because it is usually the bigger exponent who would give the bigger value.
Yh 😊
And also it’s only 50&49 as base difference of 1
Take the log of each side, know the log power rule making 50log(50) vs 51log49 and we all know that log grows sublinearly, but monotonically increasing, therefore 49^51 is easily bigger.
6:57 Why should the limit of (1 + 1/n)^n, as n goes to infinity, is smaller than 3 imply that (1 + 1/n)^n is smaller than 3, for ALL n?
Because it has a decay function. But with the smallest number that makes sense... (1+1/2)^2 to 1.5×1.5= 2.25 okay small but maybe a BIG number works
(1+1/9001)^9001≈2.71813
No matter what, this function will not pass its limit for positive numbers, which is eulur's number, that is < 3
It doesn't by itself
50 to the fiftieth power is approximately 8.8817E84.
49 to the fifty-first power is approximately 1.5848E86.
See, I just used a calculator. But 50^50 and 49^51 are so large they would probably just give an error, so instead i took _the log of both,_ and 51log49 is larger than 50log50.
I don't think this is right. 50^50 in log would be something like log base 50 x = 50 ( i use x because we don't know the answer
Logarithm is the most viable/ easiest method
Nice approach, but you could easily take “log” from each side and the answer would appear much sooner
Only with access to a calculator. Otherwise it is only kicking the can down the road, so to speak - the problem of evaluating the logarithm by hand is not any easier, and will probably use a similar trick.
@@la.zanmal. Right but it's pretty lame to have an Olympiad problem that's so easily solved with a calculator. Edit: actually, don't even bother with logs. Your calculator can do 50^50 and 49^51 and they differ by more than two orders of magnitude. I was expecting that they'd be close enough together that they'd look equal to the precision that a standard calculator gives.
No need calculator because log 50/log 49 ~= 1. The exponents drop which gives the factor constants 50 and 51. How 50 < 51 hence 50^50 < 49^51
@@gatoordinario94 Your argument is incomplete. You need to show that log 50/log 49 < 51/50. You've correctly argued that the left hand side is close to 1. That's not enough, as the right hand side (1.02) is also close to 1, and we need to know which of these numbers is bigger.
Take log on both sides, take the exponent in the front as a constant, divide by (51*50). This yields a comparison between 2 values of function ln(x)/(x+1). The left side is x=50 and the right side is x=49. This function is clearly decreasing. So we have a solution.
Extract root 50 both sides which yields 50 ? 49x49^(1/50), and knowing that numbers < 1 rised to any positive power online be 1 at the infinite, so 49x0,... will be < 50.
x^x>(x-1)^(x+1) as long as x>4.141041525..., the value being easily obtained from Newton's iterative formula for non-linear equations.
I just used binomial theroem at (1+(1/49))^50 *(1/49)≈ (1+(50/49))*(1/49)=99/49^2 so yeah maybe not the best but it did the job basically it states that (1+x)^n ≈ 1 + nx
Once you rewrote the problem as a relationship to one, the only question is whether or not it’s less than or greater than. 50 to the 50th power divided by 49 to the 50th power is less than one with that quotient being multiplied by 1 over 49. So you are asking do two numbers less than 1 multiply to a number larger or smaller than one. The answer is they will always be less than one so 49 to 51 power is greater than 50 to the 50th power.
Nobody here has an intuitive solution.
The clear proof is to write 49^51 as 7^102 and 50^50 as 5sqrt(2)^100. We can the approximate 5sqrt(2) as 5*1.414 = 7.07:
=> 7.07^100 < 7^102
=> 7.07^100 < 7^100 * 7^2
/7^100 => 1.01^100 < 1 * 49
=> e < 49
Which is true. No frills, no fancy theorem, and no extra computation, just the knowledge of the approximate value of sqrt(2) and the value of e.
Well IIRC, if the number of digits on the base numbers were in the same range (like 10 with 11 being 2-digited numbers), even if one of the base numbers were bigger, the bigger exponent will always have bigger value no matter which one has it.
So by that logic, because the difference between 50 and 49 was small enough, I can assume that the value of 49^51 would be at least 10 times larger than 50^50.
Exactly how I thought, didn’t need to do any math.
Since the question was which side is greater, without requiring proof, I intuitively knew that the side with the higher power is greater for anything greater than 3^2 vs 2^3
But this question is about comparing x^x vs (x-1)^(x+1), not about comparing x^(x-1) with (x-1)^x.
I think i found an easier way:
you subsitite x with 50, you get on one side x^x for 50^50 and for 49^51 you get (x-1)^(x+1) and then using binomial formulas and etc you get:
50^50 < 50^51 - 2498 (as 2489 is much smaller than 50^50 x 50)
Tell me what you think!
You could also have done it using logarithms
Taking log ( base 49) of both sides and the approximating value of log(base 49 ) (50) and getting the answer
generic solution
x^x vs (x-1)^(x+1) transforming it to 1/(x-1) vs (1-1/x)^x then left part is 0 < 1/(x-1) < 1/4 for x>5 right part: 1/4 < (1-1/x)^x < e for x>2, thus for any x> 5 right part is always larger. Done
Every time you are increasing the power like 10^6 and 10^7, it becomes a huge difference because it would have more zeros at the end, so the number with the larger power would likely to be larger than smaller one. Unless it more complicated, I could not say that 2^3 is larger than 4^2, so it varies on cases, but since the digits are close, I think you can tell.
I took the ln on both sides and solved it really easily. Admittedly a calculator is required. ln50^50 = ln49^51
50 ln 50 = 195
51 ln 49 = 198
I solved it with the properties of logs. Taking the log of both sides simplifies to 50 log 50 and 51 log 49. Log 50 and log 49 are almost the same due to log based anything big, squishes changes hard. Therefore, 50 and 51 are compared alone.
lol yes the easiest way
"The best solution is always the easiest." This is it..
it is a god aproximation in most exponecials cases this will solve
I really knew this answer in a second with just pure logical thinking man
make them to the nth power,(let n -> infinity ) take log , you will find right side is bigger than left side
consider n^n < (n-1)^(n+1) for n >= 5, then prove it by induction is what I did. Your solution is definitely more elegant!
if you formulate the question x*x and (x-1)*x+1 and start to put small numbers in these formulas you would see except for one (4*4 and 3*5), (x-1) * (x+1) is always greater than x*x
In our jee preparation we can solve these type of problems orally .
You can use binomial to quickly get answer
The suggested solution is incorrect. The fact that (1+1/n)^n converges to e as n→∞ says nothing about how large (1+1/49)^49 might be. It only says something about the asymptotic behavior of the function that maps n to (1+1/n)^n. An easy and correct way is to use the inequality ln(1+1/n) < 1/n, derived from the Taylor expansion of the natural logarithm, to deduce ln( (1+1/49)^49 ) = 49·ln(1+1/49) < 49·1/49 = 1, which implies (1+1/49)^49 < e. To participants in a math olympiad this should be a trivial exercise.
A simpler step could be dividing the (50/49)^50 inside term, that will be 1. smth and overall approx to 1 now this 1/49 is obviously small term hence the final answer is that 50^50 is smaller than 49^51
I just used a calculator to observe that 5^5 is less than 4^6. I therefore generalized that for any positive number n greater than 2, n^n < (n-1)^(n+1).
Great video and great question and great solution to problem
No need.
A slightly bigger number multiplied certain times is way smaller than a slightly smaller multiplied one more time. Because the last added multiplacation is done to the total number reached before it while the increase in the multiplied number is an increment.
e ＜ 3, it is OK, however, how do you proof (1 + 1/n)^n is less than 3 when n equals to 3?
I do not think it is obvious.
Please see my comment for a proof that (1+1/n)ⁿ
Maybe I should make a video on it
@mathwindow I know 1/49 is less than 1/6. I just don't see where the 1/6 come in to play to solve the problem. You wrote it down, but I don't see where that value came from. I watched that part 5 times and still did not see it. Maybe I overlooked. Just curious about that. Thanks.
Just an easy way, even no pen or paper needed: 3^3 > 2^4; and 4^4 > 3^5, but already next group 5^5 < 4^6 with a gap increasing on 6^6 < 5^7 and so on.
Safe to interpolate 50^50 < 49^51.
Anyone who's interested in solving IMO problems doesn't need more than two minutes (from 0:54 to 2:59) to get that (50^50)/(49^51) = (50/49)^50 x 1/49.
I think the proof using binomial expansion of 50^50 (== (49+1)^50) is simpler. Just my opinion
That's right!
¿How would this proof go?
When x is large, (x-1)^(x+1) > x^x.
Let x be a positive integer. We want to compare x^x and (x-1)^(x+1).
We address here the case when x is large (>>1).
To do so, we take the log: log ((x-1)^(x+1)) = (x+1) log(x-1).
x being large, log(x-1) = log(x(1-1/x)) = log(x) + log(1-1/x) \approx log(x) - 1/x because log(1+u) \approx u when u is small.
So, (x+1) log(x-1) \approx (x+1) (log(x) - 1/x) = x log(x) + log(x) - (x+1)/x
We conclude that when x is large, (x+1) log(x-1) - x log(x) \approx log(x) - (x+1)/x
And obviously, log(x) - (x+1)/x is also dominated by log(x) which not only is positive, but diverges!
So, when x is large, (x-1)^(x+1) > x^x.
Remains the question of 'large'.
Well, instead of developing the log(1-1/x) around zero, you can keep its exact value and study the difference,
(x+1) log(x-1) - x log(x) = log(x) + (x+1) log(1-1/x).
Starting from x=5, you can show that this difference is monotonic, increasing, so large is not so large, it's x=5.😉
49^51 is larger. I solved it with a simple arithmetic calculator and a logarithm table.
50^50 49^51
log 50^50 log 49^51
50 log 50 51 log 49
50 x 1.6989 51 x 1.6901
84.948 86.200
Inv log Inv log
8.8716^84 1.5849^86
This is a great video!
We can compare the two numbers by taking their ratio:
(50^50) / (49^51)
Simplifying this expression, we can rewrite it as:
(50/49) ^ 50
Since 50/49 is greater than 1, raising it to the 50th power will give us a number that is greater than 1. Therefore, (50^50) / (49^51) is greater than 1, which means that 50^50 is greater than 49^51. Therefore, 50 to the power of 50 is larger.
The exponents are 50 and 51 so by binomial theorem there will be 51 and 52 terms so 49^51>50^50
5 second question
In fact I became interested in the question for which integers m>1 the expression m^m < (m-1)^(m+1) is valid, which is m>4. A nice one.
3^3 < 2^4, just wonder how come it’s m>4? I mean m = 3 is already valid
I thought of other method that you write 49 as (50-1) and then apply binomial expansion on it and neglecting some terms. Then compare it
one of the simplest way to solve this question by taking logarithm both side and solve it it easily showed 49^51 is greater than 50^50 its very simple problem if we didn't use your method my friend
You can just do 1+nlog(x) and find the number of digits
7:33 when did 1/6 came from? Is there any reason to be specifically 1/6 or it's just to simplify the multiplication... (but then we could just use 1/3 and it would be 50/49 and LARGER than 1)
We could use any number bigger than 1/49,for example 1/48,1/47,1/46....etc. 1/6 is one of these numbers chosen by the lady to be accurate and convient enough to solve the problem. if you choose 1/3,It's not accurate if we know the result is smaller than 50/49,what we need to know is that whether the result is smaller than 1.
Or you could just know that bigger exponent than base is bigger than bigger base than exponent for the same numbers, as long as both numbers are greater than e. And this problem is just using a midpoint which would obviously be between them. Not a hard proof but it works. Yeah 2^3
WOW, You are really clever with this one. Hats off to you!
💟❣️💟❣️💟
Use smaller numbera and verify...
5⁵ and 4⁶
5⁵ = 5x5x5x5x5 = 3125
4⁶ = 4x4x4x4x4x4 = 4096
If you have any interesting & splendid questions to provide, just comment! I will choose some of them to make Shorts ❤
Find the sum from 1 to infinity. of 3/n(n+3)
Just consider (50/49)^50 vs 49. (50/49)^50 has 51 terms with only one term 1 and one term (50/49) larger than 1. You can take the excess 1/49 and any one of the remaining terms and merge them with any one term and still have 49 terms with only two equal to one nad rest less than one. This would always be less than 49.
i knew the correct answer immediately just based on the principle that in large numbers, the exponent is always more influential than the base; for example although in the case of 2^3 < 3^2 the base matters more . . . this is an exception to the rule because the numbers are so small . . . the rule that "exponents have more influence than the base" applies for even slightly bigger numbers; for example 5^6 > 6^5. I have no proof to offer, but there is a general rule that holds
when comparing numbers x^y vs (x-1)^(y+1)
as long as both x and y are greater than 2
then (x-1)^(y+1) will always be greater than x^y
same, but i also dont know how to apply any of these formulaes as i am in grade 8
If x = 3 and y = 4, then (x - 1)^(y + 1) = 2^5 = 32; and x^y = 3^4 = 81. So, unfortunately, the hypothesis is false.
@@SkinSlayer26 It works if the numbers are relatively close to each other and none of the 4 numbers drops below 3
Never-ending, I was not thinking straight. I got it now.
Write 49 as 50-1 and use binomial expansion
In physics all things work the same but they can have different scales
4⁶>5⁵ which means 49⁵¹>50⁵⁰
I started looking at (1+1/n)^n and soon figured out that it tends to e (the base of the natural logarithm) which implies that the second number is (very) approximately 50/e times larger.
I did it in a very simple manner using basics of binomial... I wrote 50^50 as (49+1)^50 and expanded it a bit as..
(50c0.49^50 + 50c1.49^49 + 50c2.49^48+ - - - - + 50c50.1).... EQN. 1
Now i evaluated the numerically greatest term in the expansion using (n+1) /(1+|a/b|) - 1
You can already see which is bigger below (not a definite mathematical proof though...)
50^50= (49+1)^50=49^50+2*49*1+1^50
49^51=
49^50*49
I know more about my weak spot in Mathematics in this video, I shall train more about this skill.
Bro simple apply log on both sides and check by basic values
my quastion is why or from where he decided to make e < 3 like why 3?
Excellent math question and more excellent explanation thereon😊
Appreciate ur work... WELLLLL DONEEE
In x^y if x+y = 100 always as constant then 24^76 will be greatest among all
Honestly these are waaay more simple than they seem, just graph x^x=y and (x-1)^(x+1), then compare ‘em, they intersect at around 4.136 and never look back
This video is basically a tutorial of how to turn a simple problem to a larger one to look cool
Can t we just take example like 50 ²
Can't we do comparative analysis? For eg. 50^3 = 125,000 and 49^4 = 5,764,801, hence 50^3 < 49^4 so 50^50 should be lesser than 49^51
My estimate would be 49^51 greater. Because for two numbers being only 1 unit appart, also being powered at least 50 times, wich is a lot, 51>50. But i do not know the verification yet.
you can otherwise just use the log, its done in less than 3 lines
I thought of it in a simplar way. We have that 50^2 is less than 49^3, then 50^n is less than 49^(n+1), so 50^50 is less than 49^51.
Pretty simple expnentional increases makes this easy to figure out without even doing any math
Would it be possible to simply time bith sides by 50? This makes lefthand side equal to 50^51 and righthand (49*50)^51 which is obviously greater?
0:16 what is the name of pen?
oh it is just the simple matter of solving N^N = (N-1)^(N+1). For X>N this result is true .
Another way that I approached it employed the binomial theorem and root approximation:
50^50 ? 49^51
(49+1)^50 ? 49*49^50
(1+1/49)^50 ? 49
1+1/49 ? 49^(1/50) Now x^(1/a) ~ t+ (x-t)/a*t^(a-1) where t is a close estimate to the root
letting t = 1, then 49^(1/50)= 1+ (49-1)/50*1^(50-1) = 1+48/50*1^49 = 1+48/50
Comparing now
1 + 1/49 ? 1 + 48/50, leaves 1/49 ? 48/50 so 50 ? 48*49 where ? must be "
take log of both the numbers and compare them
I'm all praise for this Prof. I myself am prof of mathematics
really nice the way you do it.
And can something be done with the root of ((50/49)^50)/49 = ((50/49)^25)/7 ?
Please ma'am solve this question...what is the remainder of 128^2023 divided by 126.... It's a Olympiad question. Please sir solve this
Ok, i will give it a shot!
If you are just checking to see which is larger do you absolutely need to keep the exponents at 50 and 51 or can you "simplify" this to 50^2 and 49^3?
What you're doing is reducing both exponents by 48. This is not doing the same to both terms. One term is devided by 50^48 and the other by 49^48. If your change really would be the same on both, then you would be right.
To better wrap your head around consider this: You have a lake with swimming lake flowers. On the first day, 1m² of the lake is covered and from then on each day the covered size doubles. If you went back from day 14 to day 12 or from day 8 to day 6, it would both times be 2 days but the amount of lake flowers would be much more in the jump from 14 to 12, as "doubeling" the size from day 12 is much more than doubeling the size from day 6.
Just take log on both side we can easily bring power down
50^50/(49^50)
I'd like to know: if you take two integers x^x and (x-1)^(x+1), is the second one always larger? 4^4 = 256 and 3^5 = 243. Then 5^5 = 3125 but 4^6 = 4096. Now the second is larger. Does that relationship continue always now, as x gets even larger?
We first note that comparing x^x vs (x-1)^(x+1) is equivalent to comparing x.ln(x) vs (x+1).ln(x-1), which itself is equivalent to comparing ln(x)/(x+1) vs ln(x-1)/x.
So we want to study the variations of the function x -> ln(x)/(x+1). The derivative is (1+1/x - ln(x))/(x+1)^2 which is positive for small x's but becomes negative afterwards (I believe the sign changes between 3 and 4?).
Coming back to our studied function, it is increasing for x small but quickly becomes decreasing. Therefore, for x big enough,
ln(x)/(x+1) < ln(x-1)/x and so we get that x^x < (x-1)^(x+1)
Here is a way simple way of saying it
49⁵¹ = 2401⁵⁰
50⁵⁰ < 2401⁵⁰
Best method
Check by comparing 51log49 and 50log50; shows 50exp50 is GREATER than 49exp51.
If you're going to use a calculator, just ask it 50^50 and 49^51, and it'll tell you that they differ by two orders of magnitude (about 10^84 and 10^86).
0.03>1/49>0.02😅😅
e(50/49)
49 almost = 100/2. So roughly estimating whereas 49^51 will be close to a 102 digit number but 50^50 will be close to a 100 digit number.
I thought this was rather obvious. The higher exponent is going to result in a much bigger number with the bases being so close in value.
все куда проще - находим производную функции (a-x)^(a+x) a=50 убеждаемся, что на больше нуля
I made 50 to the power of 2, then, I made 49 to the power of 3. It worked fine but idk if this was a matter of luck or this really works