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.

@normalone9199

6 ай бұрын

Same here 🎊

@alessiosandro123

5 ай бұрын

The thing is i guess that with higher potencies it gets bigger

@HashiRa248

5 ай бұрын

Sometimes the journey is more important than destination.

@zdenekbina6044

5 ай бұрын

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.

@zdenekbina6044

5 ай бұрын

And that is 2.69 so it supports your output. You got it right. My methid confirmed your result.

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

@OblomSaratov

2 ай бұрын

It usually is, but it's not always the case. For example, 4^4 > 3^5.

@GHOST-RIDER-0

Ай бұрын

2¹ > 1∞

@OblomSaratov

Ай бұрын

@@GHOST-RIDER-0 that's not true because infinity is not a number and 1∞ is undefined.

@hafidmostarhfir2245

Ай бұрын

Only if u are powering numbers greater than 1 ..I think

@CamiloMejiaGeo

27 күн бұрын

@@OblomSaratovthen 1^999999

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.

@jackwilson5542

2 ай бұрын

The easiest solution is comparing 4^5 vs 5^4. There is no reason why the trend won't continue with higher numbers.

@wolf5370

2 ай бұрын

@lson5542 That doesn't match the pattern in the Q. 5^5 vs 4^6 would.

@konglink3359

Ай бұрын

​@@jackwilson5542 but then u have to prove that it's true for all numbers, with concrete evidence

@tontonbeber4555

Ай бұрын

@@konglink3359 only if numbers are > e ... that's the only condition. It doesn't work for 2 and 3 for instance ...

@grapefruitsyrup8185

25 күн бұрын

​@@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 .

@user-qo5jo2qc5q

2 ай бұрын

Now this is a great solution, the video's solution is less elegant as it depends on students having memorized the definition of e. This solution however only relies in algebraic manipulation. It was a bit hard to follow, I originally thought you were wrong there at the end, but upon further analysis, indeed you have proven it! Well done, thank you for sharing this solution, much better than the video's.

@young4783

2 ай бұрын

Good one!

@dragondompyd7171

2 ай бұрын

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.

@santoshkumarvlogs3753

Ай бұрын

Nice solution

@alexbayan8302

Ай бұрын

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.

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.

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

@michaelhartmann1285

5 ай бұрын

That was the first approach that crossed my mind, and a great deal simpler than the algebraic thicket the narrator lays out.

@kanwaljitsingh3248

5 ай бұрын

Good solution

@hrvat7770

5 ай бұрын

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 😉

@justanotherguy469

5 ай бұрын

I'm not that well versed in mathematics. Is it a proof, though?@@michaelhartmann1285

@srinathparimi33

5 ай бұрын

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

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

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.

@epevaldon5421

5 ай бұрын

Oh i got head ache on math. Im so poor on math

@romain1mp

5 ай бұрын

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..

@theupson

5 ай бұрын

@@romain1mp the vid is easily rehabilitated; (1+1/49)^49 is immediately less than e because discrete compound < continuous compound (for positive r)

@lizekamtombe2223

5 ай бұрын

​@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!"

@romain1mp

5 ай бұрын

@@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

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 .

Нужно показать что 50^50

@marlanivanovich1828

2 ай бұрын

Только, пожалуйста, исправьте: квадрат 49 равен 2401.

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 😊

@neiljohnson7914

5 ай бұрын

But you need to use a calculator. This problem asks that you come to a solution according to math principles, not raw calculations

@user-bo5fi7os2p

5 ай бұрын

I knew this without even calculate anything lmao

@chaplainmattsanders4884

2 ай бұрын

i don’t understand that, but I believe you!

@vandemaataram2600

2 ай бұрын

Yes. Your solution is similar to mine.

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.

@manny2092

5 ай бұрын

I like this answer already!

@kaustubhprakash1273

5 ай бұрын

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

@mareshetseleshi2717

3 ай бұрын

Much appreciated

@donmoore7785

2 ай бұрын

How did you calculate 52.966 - a calculator?

@TheSimCaptain

2 ай бұрын

@@donmoore7785 Yep.

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... Or 50/51 Therefore: 50^50 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.

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.

@user-qo5jo2qc5q

2 ай бұрын

That's not a mathematical proof, you can have a gut feel (if you have worked a lot with interest) that your answer is right, but that's not what this question is about.

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.

@thanhquenguyen9462

5 ай бұрын

Agree. Without proving that we can doing the same way as the video to approve 4^4 < 3^5, but it not.

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

@user-ec8ru7je7b

Ай бұрын

why len? int же

@user-hb1vf6lo7p

Ай бұрын

@@user-ec8ru7je7b привет. len посчитает количество символов в каждом из представленных(чтобы не выводить большие числа). Какой из вариантов больше, будет более очевидно.

@madankundu6035

Ай бұрын

that's cheating... ha ha

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

6 ай бұрын

which is actually quite obvious..

@user-gk3on7xp7e

6 ай бұрын

@@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.

@phajgo2

6 ай бұрын

@@user-gk3on7xp7e 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?

@tassiedevil2200

6 ай бұрын

@@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.

@mikaelhakobyan9363

6 ай бұрын

@@phajgo2 No, it isn't. A function can increase at first, then start to decrease, and after that move to it's limit.

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..

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.

You didn't prove that (1 + 1/m)^m infinity. It's true, but you assumed it without proof. You need to show that (1 + 1/n)^n 0.

@tharock220

6 ай бұрын

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.

@thomasdalton1508

6 ай бұрын

​@@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.

@bumbarabun

6 ай бұрын

@@thomasdalton1508 I believe that derivative is n * ( 1 + 1/n ) ^ (n-1) and it cannot be negative for positive n

@thomasdalton1508

6 ай бұрын

@@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.

@bumbarabun

6 ай бұрын

@@thomasdalton1508 you are right, my mistake

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.

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.

@herotb221091

6 ай бұрын

Wow. I love it

@Aut0KAD

5 ай бұрын

nice, using the rule of 72?

@mujtaba21

5 ай бұрын

That's how I thought of it. You wrote it concisely 👏🏽

@mujtaba21

5 ай бұрын

​@@Aut0KADyes

@Btitude

5 ай бұрын

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.

@texasaggiegigsem

4 ай бұрын

I did the same thing, but I like her approach without having to find the natural log, which I reached for my calculator for.

@AbhishekChoudharyB

4 ай бұрын

Shouldn't it be 51 ln 49 How did u get 2450?

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!

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

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.

@larswilms8275

6 ай бұрын

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

@chris8535

5 ай бұрын

Yea this is super obvious with just logic. 50^50 is less than (50-1)^51. Like 2 seconds

@MauuuAlpha

5 ай бұрын

I did something similar

@GarryBoyer

2 ай бұрын

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.

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

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

@RikMaxSpeed

6 ай бұрын

Spot on, this was way over-complicated.

@88kgs

6 ай бұрын

😂😂

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

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

Great explanation

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).

@aniruddhshandilyak3289

5 ай бұрын

A good analysis But this works only when n>=5

@user-zo1bv7hk1l

5 ай бұрын

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

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

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 therefore 50^50 < 49^51

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!

@thomasdalton1508

6 ай бұрын

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.)

Thanks for doing this for "n." So (for future reference) we know that n^n 3^5 but 5^5 < 4^6

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

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.

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.

@sorinturle4599

5 ай бұрын

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.

@atulyaroy8962

5 ай бұрын

@@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

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

@robertoguerra5375

5 ай бұрын

The ratio of the 2 numbers would be approaching n/2.72 Less than 1 for n>2

Thank you for explaining such a terrifying problem so calmly.

Nice solution😊

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

I love these problems, great mental exercise! Thanks.

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

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

Good one, thank you for sharing.

50⁵⁰

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

*General Solution for any N > 1 (50 or not) and 1 (N+a)^(N-a) for all N > 1 and 1 N^N for all N > 4 (a=1) - e.g. 4^6 (4096) > 5^5 (3125) Rule 3 - (N-a)^(N+a) > N^N for all N > 5 (a=2) - e.g. 4^8 (65536) > 6^6 (46656) Rule N-1 - 2^(2N-2) 2 (a = N-2) - e.g. 2^4 (16) And so on ...

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

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

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.

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

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

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 3^5 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

@DesertObserver491

5 ай бұрын

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.

@GolldLining

5 ай бұрын

You proved nothing in the rambling

@yasserahmed-bg7qj

5 ай бұрын

​@@GolldLiningyes but if he continued with what he was doing and learnt mathematical induction he would've proved it

@DesertObserver491

5 ай бұрын

Exactly. He was on the right track and in a multiple choice test, he would have got the correct answer.

@lakshay3745

2 ай бұрын

Bro can you give an example of questions you did without solving which shocked your teacher

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

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

Beautiful explanation! Thanks

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

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

@adamfarmer7665

5 ай бұрын

when x>5 that is, not when x>1

@robertoguerra5375

5 ай бұрын

@@adamfarmer7665that is a good math question: prove that ALL x^x 5

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

@shoodler

2 ай бұрын

The underscore represents a blank for what their relation is

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

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

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

@OblomSaratov

2 ай бұрын

Higher power doesn't always win. 4^4 > 3^5.

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

This is flawed like many such math "solutions" on KZread. 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 go this solved in seconds and with less complexity. The author rather than simplifying the steps instead complicated the steps. 50^50/49^51=50^50/(49^50x49^1)=(50^50/49^50)*(1/49)=(50/49)*(1/49)=50/49^2; answer 50 50^50

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

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

after 6:20 I propose multiply numerators , 150 compare them with DE numerators, 49 squared hence the fraction is less than 1 ,

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

I had this question 35 years ago while living in Soviet Union. Did not use Euler, just natural logarithm. The same exact result.

@thatonenoname

2 ай бұрын

Натуральный логарифм это логарифм по основанию е, так-что почти одно и тоже

I guessed it was the 49^51 because powers mean more than single digits, and found both the answers on a calculator and subtracted them to prove it in about 10 seconds.

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

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

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) Condition: (r sub b) > 1 ; (r sub e) 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) 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. 😱 👀 ⁉️ 🤔

@wolf5370

2 ай бұрын

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).

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

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

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

use logarithms { x=log(50^50) , y=log(49^51) } we can see that y is greater than x. therefore 49^51 is greater than 50^50 .Just remember the value of the first 10 logs, we can solve such questions with much ease if it doesn't involve prime numbers

Divide both side with 49^50, you will get: 50^50/49^50 ...?... 49^51/49^50 simplify will giving: (50/49)^50 ...?... 49 1.0204^50 ...?... 49 And we know that 1.02^50 will lead to a quite small number, (50/49)^192.63855269 only will give us back 49... Hence, 49^51 > 50^50

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.

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

@user-qo5jo2qc5q

2 ай бұрын

That's not a proof!

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?

Thanks for the video. Regards

It's simple. The exponent has a bigger influence than the base, so 49^51 is bigger 🤭

@UltraStarWarsFanatic

6 ай бұрын

Well not necessarily, since 3^2 > 2^3... but in this case yeah, the answer is obvious.

@landpro28

5 ай бұрын

Exactly! Took me 10 seconds to conclude that

@catalinx7301

5 ай бұрын

@@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.

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

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

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

Aa..nice way👍🏻 I can solve it in 2 steps 🙂

50⁵⁰_49⁵¹ ln50⁵⁰_ln49⁵¹ since these numbers are positive or belong to]0, infinite] Equivalent to 50 ln50- 51 ln49 Equivalent to 50 ln50- 51[ ln(50-1)] Equivalent to 50 ln50- 51[ ln50+ln1], we know ln1=0 New equation 50 ln50- 51 ln50 Which equals to (50-51)ln50 Which means 50⁵⁰_49⁵¹

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

@bumpypants3241

2 ай бұрын

That's the correct way to pronounce it 😅

i think this is an easy one in my intermediate i solved many such problem by using binomial theorem , after expansion it can be easily solved.

We need to prove that (1 + 1 / n) ^ n is an increasing function.

Let 50⁵⁰/49⁵¹ > 1, so 5²⁽²⁵⁾/7²⁽⁵¹⁾ > 1, but because the numerator is obvious smaller than the denominator we can state that 50⁵⁰/49⁵¹ < 1, thus 50⁵⁰ < 49⁵¹.

Alternative method is you need to take a function=(x)^1/x and differentiate it after then you can get the answer 👍

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

you could argue that 50^x/49^(x+1) approaches 1 for x going to infinity. if u then calculate for x=1 you can see that thats where the fraction has its max value and is smaller then 1 - therefor for x=50 it will still be!