For any 2 numbers greater than e, the one closest to e will be the greatest when raised to the other number's power

@davidseed2939

2 жыл бұрын

yes and the reason is that comparisons of this type can be transformed into comparing x^(1/x) with y^(1/y) in this case 100^(1/100) cf 99^(1/99). 99 is the smaller number and thus 99^(1/99) is the greater

@CAROLUSPRIMA

Жыл бұрын

I have an earned doctorate and you guys are impressing the hell out of me. I got through graduate level statistics and still don’t know how.

@kapsel-yg2sk

Жыл бұрын

Yes, and I'm disappointed that youtube is still flooded with this class of problems

@FlummoxTheMagnificent

Жыл бұрын

So basically the smallest to the power of the largest, if they are greater than e.

@rachityadav5011

Жыл бұрын

@@CAROLUSPRIMA we will get this by the graph of F(x)=lnx/x and we will get that func is increasing upto e and decreasing after e then just putt values then just cross multiply and take numerical part in the power of log and than cancel out log you will get your answer

Take derivatives of ln(x)/x to find that the function is maximised when x=e. ln(99)/99>ln(100)/100 so 99^100 is bigger

@ToiLaChinhMinh

Жыл бұрын

ye 99^100 = 99 multiply by itself 100 times. And with 100^99 that's just 99 times so😅

@kkkkjlkkkkj

Жыл бұрын

bro why tf you just do 100^99(99+1)^99 99^99+1 that is visually smaller than 99^100

@Jinsun202

Жыл бұрын

@@kkkkjlkkkkj "visually" 🤣🤣🤣

@user-mv5lz5qf4q

Жыл бұрын

This solution method is popular in Japanese entrance examination.

You have to learn to write 9 instead of 8

@vjlaxmanan6965

2 жыл бұрын

Yes...agree! :(

@yeshwantphatak4709

2 жыл бұрын

😀😀

@bobbylashley5995

2 жыл бұрын

It's understandable so stop complaining

@tomlee9534

2 жыл бұрын

9 instead of g I reckon.

@chillingsk

Жыл бұрын

Nah that's the gravitational constant

i think my brain is becoming inflexible. i had to keep reminding myself that the characters that looked like 'g' are actually '9' and the characters that look like 'Λ' are actually '1'. when I was younger I would have seen the pattern and just kind of flipped a switch in my brain to be in your local writing style. I'm getting old. i hate it, but it beats the alternative.

@kenhaley4

Жыл бұрын

It is sometimes a problem as her 9's sometimes look like 8's, as the tail of the 9 comes up and touches the upper loop.

It’s very simple. 100^99 = error 99^100 = error 100^99 = 99^100

@shinybagel3407

Жыл бұрын

Lol

@Thorfinnkarlsefni047

Жыл бұрын

Omg it's Albert Einstein

@deleted-something

Жыл бұрын

Genius

@thonydq

Жыл бұрын

This is elon musk

@adhamabdelaty2479

Жыл бұрын

bruh

One can prove (1+1/n)^n

If you look at a still frame around 2:50, you'll see a nice demonstration why you shouldn't use a hook in number 9 because it's really easy to mix it with 8. I've intentionally trained myself to avoid the hook in the bottom of the 9 to make it harder to mix it with 8.

@momonga2850

Жыл бұрын

Man...she said that it was 99..... Please stop complaining and atleast thank her that you came to know something interesting

If e

Thank you for the bringing back the logic! Love this post🎉

I compared 3^4 vs 4^3 and guessed that anything higher would have the same rule

@newthan2

Жыл бұрын

same

@Jinsun202

Жыл бұрын

"guessed" 🤣🤣🤣

@shrikrishnahospital9762

Жыл бұрын

Very good

Reminds me of a blackpenredpen video. Main takeaway is if a

I like the way you explained it, even more than the problem itself. Thanks for sharing, I shall look up more of your videos.

Amazing!!! Whoever you are, I wish I had YOU for a math(s) teacher when I was in my secondary and post-secondary school years!! BTW, I've just now liked this video, subscribed to your channel and requested ALL notifications to it!! Thank you very much!!

Once at a school math olympiad, I got a task to compare 2007^ 2008 and 2008^2007. 15 years have passed, and I still don't know how to solve it

@vigneshwars1331

Жыл бұрын

@Dylan Bradley that would be the right answer because multiplying 2k something extra will change the number largely

@elenabale5755

Жыл бұрын

by the way, now I have known how to solve it 🙂 youtube really helps sometimes

@locomotivetrainstation6053

Жыл бұрын

2007^2008 is biggest

@qtfy

Жыл бұрын

@Dylan Bradley ok but why that didn't work on the video tho

the general solution is if e

I think you need to proof there exists N N

the best approach is to divide each side by 99^99. then R=99 L=(100/99)^99 =(1+1/n)^n for n=99 but.from the definition of e L

@keinKlarname

2 жыл бұрын

You need that (1 + 1/n)^n is increasing when n is increasing.

@davidseed2939

2 жыл бұрын

@@keinKlarname which is true. but i only need that the sequence is bounded above which is also true.

@ericwilliams1832

Жыл бұрын

awesome!

Whenever e100^99. Proof is not hard Let's f(x)=ln x / x... For x >e this function is decreasing since derivative is negative. Therefore ln a /a > ln b/ b. If we multiply this by ab в lna > a ln b. ln (a^b)> ln(b^a) a^b > b^ a

@marcusdecarvalho1354

Жыл бұрын

#thanks

@aca4262

Жыл бұрын

So you are saying 5^2 is bigger than 2^5?

@aca4262

Жыл бұрын

Is that it? cause i'm confused

@marklevin3236

Жыл бұрын

@@aca4262 a and b must be greater than e. 2

@aca4262

Жыл бұрын

@@marklevin3236 Tnx got it.

This is a good solution. I had recently made a video on e^pi vs pi^e with a different approach.

@Mr.kasugai

Жыл бұрын

oh,that's famous problem in Japan because this problem afflicted examinee on entrance examination of elite university in Japan.

@EatThatLogic

Жыл бұрын

@@Mr.kasugai Now you know where the solution is :)

In a math problem where we have to compare the value of x^y with y^x where y=x+1, let's say that x^y=a and y^x=b, only the pair 2^3=8 < 3^2=9 where we have a

@tarikyalcin2819

Жыл бұрын

İ agree with you. Clever way.

I used binomial theorem and arithmetic mean and got the same result. I feel happy. Also I discovered a type of formula for these types of inequalities.

@gddfx8513

Жыл бұрын

Super piotruś

We could just change the 100 ^ 99 into a 99 ^ Log99(100) * 99, since the value inside the log is greater than the base, so the value of Log99(100) must tend to 1 from the right side as it reaches 99, and since 100 is just 1 more than 99, that means that its really close to 1, so we can assume that log99(100) is ~1, so by substituting 1 for log99(100), we get 99 ^ 99 * 1 which is smaller than 99^100

@Shekhuguruji

Жыл бұрын

absolutely amazing mind, you've a great engineered mind, thank you for this

@lunarthicclipse8219

Жыл бұрын

@@Shekhuguruji oh thank u!

Way of teaching is also very excellent

As a general rule, a^(a+1) is always greater than (a+1)^a

@trigeminalneuralgia9889

Жыл бұрын

not really, it's opposite for values of a less than e, 1^2

@shreyjain3197

Жыл бұрын

@@trigeminalneuralgia9889 ok fine but note that I did say "general" and also if it's such small values just calculate it yourself

@jasonrichner2537

Жыл бұрын

You're both right. It works for 3 and above.

@pablojp3498

Жыл бұрын

@@shreyjain3197 rather than say in general say. When a>2.29317, a^(a+1)>(a+1)^a

@Jinsun202

Жыл бұрын

@@shreyjain3197 It's certainly not general since there are infinite values less than e.

Can we do this using binomial theorem?

For any two consecutive numbers, the expression with the smaller base will be greater for bases ≥ 1+√2. For bases less than 1+√2 the greater based expression will be greater.

I'm very poor with math. Why do we divided both in first place if we wanted to see which one is bigger? And at last part how 1+1/99(99) is

Nice solution .

you can use same approach as in the 1.005 vs 2 video -> first take the 100^99 convert it to (99*(100/99))^99 which then equals 99^99 * (1+ 1/99)^99 and then compare the (1+ 1/99)^99 to 99. then you can expand the 99 to: 2/1 * 3/2 * 4/3 ... 98/97 * 99/98 to get the 99th fraction we can split 2/1 into sqrt of 2 which is around 1.41 -> we get sqrt(2) * sqrt(2) * (1 + 1/2) * (1 + 1/3) * (1 + 1/4) ... (1 + 1/97) * (1+ 1/98) and then compare these fractions to the (1+ 1/99)^99 and we can see that every fraction in the first set is larger than any in the second one therefore 99^100 is larger. there is probably better way to get 99 fractions, and splitting the "2/1" seems crude, but it works as 100/99 is much less than sqrt(2)

@elir626

Жыл бұрын

Rather than using complex math, just use logic, if 2^3 is bigger than 3^2 then the same logic must apply to every other scenario accept for 1^2 vs 2^1

@karlisberzins5676

Жыл бұрын

@@elir626 2^3=8 < 3^2=9 it seems that after around 2.29^3.29 vs 3.29^2.29 you are right, but that would have to be shown (i just plotted the y=x^(x+1)-(x+1)^x in wolfram alpha... modern tools are cool)

@elir626

Жыл бұрын

@Kārlis Bērziņš oh shoott I think I was thinking of 3^3 as 27 and not 2^3

Clever, but you’ve left out two things. 1) How do you know that (1+1/n)^n is an increasing function of n? 2) How do you know that your limit = e? Or for that matter, how do you know that (1+1/n)^n is even bounded as n approaches infinity.

@haotiankong6904

Жыл бұрын

I think she thought both of them were given. Most people who know what limits are probably already know about this specific limit.

@joseluisvazquez4221

Жыл бұрын

The limit can be a given. But the increasing function fact is not at all

Very good solving

Thank you Vinamilk

Sorry if there are any mistakes here, english is not my native language. I made a ratio of 99¹⁰⁰/100⁹⁹, got 0.99⁹⁹×99, put 0.99 as an approximate value of 1 and from there I saw that the ratio was greater than 1 and concluded that 99¹⁰⁰ is greater than 100⁹⁹. Edit: I allow you not to write the same type of comments about not rounding 0.99 to 1.

@a_minor

2 жыл бұрын

your english is still better than most native speakers

@cringeconnoisseur6037

Жыл бұрын

@@a_minor he missed a comma 🤬

@shaguna

Жыл бұрын

When you approximated up from 0.99 to 1 you introduced a margin of error, and you cannot use that approximation to just state that the initial number will be higher. For example, consider the function f(x)=x*[(0.9)^x] (x multiplied by 0.9 to the power of x). f(33)=1.018804.... and f(34)=0.945636 .... According to your method, if we round up, both values should be bigger than 1, but they are not.

@cringeconnoisseur6037

Жыл бұрын

@@shaguna if he approximated .9 to 1 he would have been completely wrong as well.

@vadirajpogal4199

Жыл бұрын

since 1*1=1 any number less than 1 multiplied by itself, product should be less than 1. so 0.99 raised to 0.99 has to be less than 1 is a logical conclusion and not an assumption is my humble opinion 🙏

2³ = 8 3² = 9 But the next sequences is 3⁴ = 81 4³ = 64 So i think 100^99 < 99^100

Well, you can apply the logic as 100 = 2 x 2 x 5 x 5 and 99 = 3 x 3 x 11 so addition of their factors would be for 100 ~ 14 and 99 ~ 17 so if the powers were to consider 14² will be always less than 17³. So 100⁹⁹ will always be less than 99¹⁰⁰.

Check the function ln(x)/x.

Me with a calculator: I am 4 parallel universes ahead of you

@shreyjain3197

Жыл бұрын

pretty sure no calculator can compute 99^100 but ok

@LD-dt1sk

Жыл бұрын

@@shreyjain3197 you can just use Google.

Thanks a lot

In my mind, it's: 99ln(100) vs. 100ln(99) And ln() grows slower than linear for larger numbers. So 99^100 is bigger.

Meanwhile me solving it using binomial theorem

You had me at 2.718 😂. - from Jerry Maguire.

It's simple if you make both numbers to the same exponent, the number with the largest base is larger.... Therefore (99x99)^99 is larger than 100^99

@QuartzQuill

Жыл бұрын

99^100 is bigger though... They don't have the same exponent, that's the whole point of the question...

@akatsukiesports-aks-7859

Жыл бұрын

@@QuartzQuill 99^100= 99x99x99.....100 times. Or if you multiply once and make the rest the exponent, you get (99x99)^100-1 Which simplifies to 9801^99

@QuartzQuill

Жыл бұрын

@@akatsukiesports-aks-7859 just do 99^100-100^99, you get a positive number therefore 99^100 is bigger

@akatsukiesports-aks-7859

Жыл бұрын

@@QuartzQuill no

@akatsukiesports-aks-7859

Жыл бұрын

U using a calculator for this?

And I knew without doing any complex math bc if 3^4 is bigger than 4^3 than that should apply all the way up to thos scenario

For the people who dont want to see the whole video, the answer is 99 to the power 100

a^b > b^a For e

Good job

Muito bom!!!

I think there's an easier way to solve it: let's consider b = a - 1. 1) When a=4 and b=3, a^b=64 and b^a=81. 2) When a=5 and b=4, a^b=625 and b^a=1024. Then 1) b^a - a^b = 81 - 64 =17 2) b^a - a^b = 1024 - 625 = 399 This means that b^a grows faster than a^b. So considering a=100 and b=99, 99^100 must be bigger than 100^99

@arion6649

Жыл бұрын

not enough

If base and exponent are simply interchanged, the arrangment which has higher value exponent always larger.

@hhgygy

Жыл бұрын

except when not 1, 2

@isakzadeng3570

Жыл бұрын

@@hhgygy Right, except 0 and 1

When bases are similar, larger power is always bigger especially in large numbers

@nguyentrongnhan6908

Жыл бұрын

Actually, with x larger than 2, x^(x+1) is always larger than (x+1)^x

@adamslajer2203

Жыл бұрын

@@nguyentrongnhan6908 actually not larger than 2 but larger than e(2.718...)

Thanks

x^n > (x+1)^(n-1) for all values x > 2, and n >= 1

Can we generalize this to the case [n^(n-1)]

@RohiNkwama

Жыл бұрын

I followed your solution to this general case and it works for All real numbers n>e+1

This kind of question can be generalized and all have a common answer, (n+1)^n is always smaller than n^(n+1) when n>2 (n is integer)

We could simply divide both nos. by 99⁹⁹ so 100⁹⁹ would become (100/99)⁹⁹ which would be around (1.01)⁹⁹ which will give 2.67 (

Amazing

idk what the video is all about but can't we just solve it by using 100⁹⁹×100/99¹⁰⁰×100=100¹⁰⁰/99¹⁰⁰×100 and using law of exponents (100/99)^100×1/100. 100/99 is 1.01 so 1.01¹⁰⁰/100 and since 1.01¹⁰⁰/100100⁹⁹

100*log99 > 99*log100 Since log is a strictly increasing function in the positive numbers (and 99 and 100 are both positive numbers), then 99^100 > 100^99

@Pao-vo8mf

Жыл бұрын

listen, nerd, your parlance is uninteligible. harness the power of concrete words!

@robertveith6383

Жыл бұрын

You did not justify that top inequality. log(100) > log(99), so you would have to explain how the right-hand side is not large enough to be larger than the left-hand side.

@Lowraith

Жыл бұрын

2^3 vs 3^2 ...

It’s easier to divide the inequation by 99^99. Then you have e to the left and 99 to the right. So left is smaller.

@robertveith6383

Жыл бұрын

You won't have e on the left-hand side, but you will have under approximation to it.

Madam , just take log of both and convert to similar exponent for both terms, it quick to see second term is bigger with anti log . Of both terms, we just need log of 2 and simple log and antilog

@toco1318

2 жыл бұрын

Exactly what I was thinking

@bryanekers3472

Жыл бұрын

@@toco1318 That was my approach, also. 100 = 2^6.6439, therefore 100^99 = 2^(6.6439 x 99) = 2^657.742. 99 = 2^6.6294, therefore 99^100 = 2^(6.6294 x 100) = 2^662.94 Therefore 99^100 is larger, and we can even easily tell by how much, i.e. it's 2^5.198 times larger, or 36.707 times larger.

Doamna Math Window...the proof it liked me very much, but be more careful when you write digit 9, which in a few locations I confuse 9 w/ 8. So, please be a little bit careful. And good luck!

@mathwindow

2 жыл бұрын

Thanks for your remind!

Simply apply log on both sides

even without the approximation it is clear that (1+1/99)^99 will have expansion in the form 1+(99)(1/99) + (99)(98)(1/99)²+.... the rest of the terms will be too small to make difference....and the above number is just around 2-3 so fraction will be 2.../99

sequence a_n=(1+1/n)^n < 3 bdd and finish with n=99

Y to divide

impressive

I looked up the answer on the Internet. If I am correct,it said that 99(100) is greater than 100(99).

@mrtrinity143

2 жыл бұрын

I did also my own computation and the result was 99^100 was the greater..🙄

her way of writing "9" and "lim" concerns me

I thought 100^100 is bigger than (99^100)x100, but it was wrong :-o

Just write it as (1+1/99)^99 x 1/99 = 2/99 (1+x)^n = 1+nx for large values of n and x

If (1+1/n)*n = e, then why is (1+1/99)*99 smaller than e when n can be any number?

what was the point of writing e to 17 significant figures and explaining how easy it is to remember them? all than matters in this problem is that e is less than 99

@oniondesu9633

Жыл бұрын

also, it really should be shown in this solution that (1+1/n)^n is a strictly increasing sequence, even if it does seem trivial

I thought about making the 99^100 as (100-1)^100. Which then becomes 100^100+1^100-(2×100×1). Which makes it 100^100-199 and this is bigger than 100^99

@adityasingh3963

Жыл бұрын

That indentity that you are using to open the brackets only works when you are squaring. For example (a-b)² = a²+b²-2ab, but (a-b)³ = a³ - b³ -3a²b + 3ab². Similarly (a-b)¹⁰⁰ has a different formula. It's not as simple as you thought. Though you got the answer correct the method you used is wrong. And it can give completely wrong answers for comparing numbers like 100⁴⁹⁹ and 99⁵⁰⁰. According to your method 99⁵⁰⁰ should be larger but in actuality 100⁴⁹⁹ is larger.

@kimamaral3015

Жыл бұрын

@@adityasingh3963 if he used the identity of (a-b)^n it would be correct, that is how i've done it. Im just not sure if this is a high school problemn, because if it is, im not sure if in high school this identity is as widely known as in calculus

Which is larger than what?

another way: 100^99 or 99^100 ; divide by 99^99 (100/99)^99 or 99^1 (1+1/99)^99 or 99 from the 1st order binomial approximation (1+x)^n ≈ 1+nx 1+99/99 or 99 2 or 99

Good development. But as a contribution I must comment that, in this case and in other similar cases, the largest is always the one with the smallest base and the greatest exponent.

how can we prove that (1 + 1/99)⁹⁹ < e?

I coded a program to print the total amount of digits in 100^99 and 99^100 And I found that 100^99 has 199 digits But 99^100 has 200 digits so 99^100 isbgreater

Easiest question ever. Exponents are so much bigger than the difference between 99 and 100. So it should be clear that 99 getting an extra ability to double itself would be greater than 100 ^99

@gytisdramblewolfskis8521

Жыл бұрын

Good point.

math got confusing when she said -the number 'e'-

(9, g, 8, 3) (1, 7, λ)

Спасибо за урок

It's simple 100^99 = 10^100 10^100 < 99^100

I do math too, this question is easy one.

By which percentage X^100 is greater than X^99?

99^100 is just (100*0.99)^100. Take out 100^99 and cancel out the 100^99 on the top. Leaving us with 1 / 100*0.99^100. 100*99^100 is probably bigger than 1.

Now, my lad, how do you prove that the above limit in infinity is indeed e?

I understand everything except for why the 9s look like either an 8 or a g

My brain after this video 🥇

Thumbs up for ur nice accent 🤗🤩

Well maybe i am wrong, but if you open a calculator and try something like 3^4 and 4^3,. 3^4 is greater again try 5^4&4^5, 4^5 is greater and how many ever you try for numbers above 3 you will always get the smaller number with greater value , so 99^100 should be greater Ofc numbers 1&2 and 2&3 are exceptions

I am still finding which is greater.

유퀴즈 보고 왔어요~~

There's a more useful fact you did use but didn't correctly prove. That is, this sequence is monotone increasing. The monotone increasing part follows from a clever application of AM-GM inequality. Thus e is not only the limit but also a valid upper bound. In some books, it is easier to show the sequence is upper bounded by 3. That also works.

Why you write g's for 9's ?

It can be solved by logarithmic numbering Of course 100 pow 99 greater than 99 pow 100 100 pow 99 is 99 log 100 99 pow 100 is 99 log 99 + 1 The ratio is 99 log 100/99 + log 1/99 more than 1

Limit dışındaki çözümleride merak ediyorum.

@furkanpolat1843

Жыл бұрын

Limit olmadan çözüm baya zorlaşır . Ama bende merak ettim

I tried it with smaller number like 9^8 vs 8^9 and calculating it manually, to arrive at a similar answer.

@Lucifez6160

Жыл бұрын

Yes, when a number is positive and higher than e (2.71828), the number with the higher exponent will always be higher. But if it is below the e, the higher base will always be higher.

You need to prove that (1+1/x)^x is always increasing for n>=1

Why do 9s look like g and 8? How do we KNOW that (1 + 1/99)^99 is less than e?