A Canadian Problem

[@pojuantsalo3475]

The solution in this video is genius! I am astonished! I would have never figured this out.

[@RyanLewis-Johnson-wq6xs]

Never stop learning because those who stop learning have stopped living is good advice.

[@butch2kow549]

I liked your comment, " Never stop learning .... living."

[@sciphyskyguy4337]

It works in the other direction, too. “Those that stop living stop learning.” 😀 It’s commutative!

[@leonardobarrera2816]

I don’t know how many years you have been studying But in my case is until at one point, not all the life A true story that I lived

[@michaelbujaki2462]

That explains why dumb people seem immortal; they never started living in the first place.

[@Grecks75]

@Prime Newtons: In the description you asked us to verify whether your method is sufficient for a proof. Yes, it absolutely is, I find it very conclusive, clear and easy to follow. It doesn't raise any questions for me. Thanks for showing! ❤

[@PrimeNewtons]

Thank you!

[@LnlyCloud]

I found what I think is an easier way to prove the LHS. Take the number A, and multiply it by 1999. We now have (1/2 * 3/4 * ... * 1997/1998) * 1999. However, we can rearrange these fractions by shifting each numerator to the left (and dividing by 1 to get an extra denominator). So we have 1/1 * (3/2 * 5/4 * 7/6 * ... * 1999/1998). We can remove the factor of 1/1 to get just 3/2 * 5/4 * ... * 1999/1998. This number, 1999A, is made up of factors which are all greater than 1. Therefore 1999A > 1, and A > 1/1999

[@timirbiswas3834]

Yes..very nicely done.

[@benkahtan6802]

Clever!

[@Grecks75]

Nice. I think it's pretty much the same as what Newtons did. Whether you multiply A by a number 0 < B < 1 to get exactly AB = 1/1999, or you multiply A by 1999 to get some C=1999A > 1 is essentially the same. We have the correspondence C=1/B that links both arguments.

[@lorrainefigueroa4624]

I love when he looks at the camera, knowing you see it too! I love these videos!!!

[@butch2kow549]

Excellent video. I liked your comment, the number is 44.something. Students should know the number sense short-cut for numbers ending on 5. Also, I liked your approach to the solution of the problem.

[@ThePhotonMan110]

Massive congratulations for reaching 200k subs!

[@benkahtan6802]

Really elegant solution. Thanks so much for sharing.

[@oscarcastaneda5310]

Very Creative, Thanks for sharing this way of looking at things : )

[@waterlord6969]

Thanks! Very informative ❤

[@martinmonath9541]

Beautiful concise solution.

[@apnakaamkrelala]

6:22 the look man 👽

[@MeQt]

Sus ass look

[@rakeshkalshan2302]

Sir your video's are great ❤

[@baidonchandipo2804]

This guy is awesome and enjoyable to watch

[@CasiMediocre]

Never knew that there were other right-handed people who start drawing the digit "8" going right and down instead of left and down

[@deyesed]

I go right and up

[@Grecks75]

I do too, go right and down. Very peculiar I find the way he draws the "9". 😃

[@joseluishablutzelaceijas928]

Thanks for the video and the nice solution. My solution was a bit different: As I like proving inequalities and my eye is therefore trained for recognizing expression where e.g. AM-GM can be applied, I obtained the RHS through 1*3*5*...*1995*1997 = sqrt(1*3)*sqrt(3*5)*sqrt(5*7)*...*sqrt(1995*1997)*sqrt(1997) < 2*4*6*...*1996*sqrt(1997) = 2*4*6*...*1996*1998*sqrt(1997)/1998, and sqrt(1997)/1998 < 1/44 as 44^2=1936 and therefore 1997*1936 < 1998^2. For the LHS I observed that 2 < 3, 4 < 5, 6 < 7, ... and 1998 < 1999, if one thus multiplies all these inequalities, one obtains 2*4*6*...*1998 < 3*5*7*...*1999, which is equivalent to the LHS.

[@guidichris]

Great video. Update for 2024, and you end up with 1/2025 and 1/45 on the bounds.

[@tessfra7695]

Elegant.......thank you.

[@cret859]

Nice explanations as always. I like this chanel. At the end, instead of the "44.something" trick. We may have notice that 44²=1936. Knowing that 1/1999 < 1/1936, from the inéquation A² < 1/1999 we get A² < 1/1999 < 1/1936 that lead to A < 1/44. No need to extract any square root.

[@robertveith6383]

If you "may have noticed that 44^2 = 1936," *then you are indirectly estimating a square root!*

[@cret859]

@@robertveith6383 You right, I correct my comment. Computing the square of 44 is much easier than extracting the square root from 1999. Sorry my for my English

[@michaelbujaki2462]

The engineer's solution: Open spreadsheet. A1 = 1. B1 = 2. C1 = A1/B2. A2 = A1+2. B2 = B1+2. C2 = A1/B2*C1. Copy A2, B2 and C2 and paste them from A3 to C999. Observe that 1/1999 < C999 < 1/44. Your way is much more interesting.

[@miniong360]

Bravo! Keep these videos coming, bro.

[@camilotiznado5325]

excellent channel!

[@RyanLewis-Johnson-wq6xs]

A^2

[@makehimobsessedwithyou6412]

very interesting problem

[@michaelz2270]

For the first part you can use 3/4 > 2/3, 5/6 > 3/4, 7/8 > 4/5, and so on, to get that A > 1/2 *2/3 * 3/4 *... * 998/999 = 1/999 which is larger than 1/1999.

[@downrightcyw]

44.something (Can we write that in examination ?)

[@mybarca8083]

You can also make direct comparison: 1/1999 < 1/2 * 2/4 * 4/6 *...... 1994/1996 *1996/1998 = 1/1998 < 1/2 * 3/4 *....... 1997/1998 because each Multiple is bigger -> by transitivity it is correct Similarly for other part of the given exercise

[@Xebtria]

12:00 I know it ends up being correct, but the reasoning does not 100% satisfy me. from that reasoning alone we can argue that sqrt 1999 is less than 45, that is clear, but I do not know what it is. it could be 43 or 42 or whatever. as a result, 1/43.something or 1/42.something would definitely be bigger than 1/44, so that would make the proof false. we need the additional square of 44 (which is 1936) to conclude that sqrt 1999 MUST actually be 44.something (because 1936 < 1999 < 2025, therefore sqrt 1999 is between 44 and 45, and therefore 44.something. and with that, the rest of the argument works just fine. TL;DR: I think the square of 44 (1936) needs to be mentioned in the proof to make it complete and unambiguous

[@Grecks75]

Technically you are correct, but it seemed too obvious to me to say something about it.

[@Creativemathlearning]

Có cách khác không

[@maburwanemokoena7117]

I don’t know how I’ll solve this but I think i’ll just log everything and turn it into a summation. Let’s see how he solves it!

[@myparasan]

"Those who stop learning, stop living" black screen, video over, no more learning, no more living

[@kianushmaleki]

❤️

[@RyanLewis-Johnson-wq6xs]

I did it in my head.

[@davidbrisbane7206]

I did it in your head 😂.

[@robertveith6383]

Produce your work in a thread. "I did it in my head" does not count.

[@timirbiswas3834]

Soon you will get into headache and then you will become the cause of other people's headache. Then you will say I have solved it in my tail which is my ancestors' property.

[@GrifGrey]

me too!

[@RyanLewis-Johnson-wq6xs]

2 101= 27318619677157413541998666579156061420147177666088128046591030596082725294498066722338505744902120368830900788923839991099564447458450075226030128555294655577015766113909738825769262480452415909200510101 =101^101

[@aaryavbhardwaj6967]

Hi can anyone pls twll me that why Integral of tanx= ln|sec x|+C But shouldn't integral of tanx = -ln|cos x|

[@Jeremy-i1d]

Yes but the two answers are the same because of the log power law: - ln cos = ln cos^-1 = ln sec

[@satrajitghosh8162]

Bit of work gives the term at the middle (between two inequality sign) is (1998)! /( (2 ^999) (999)!) ^2 =( 1998_C_999) /(2 ^1998) Again 2n_C_n / 2 ^ (2 n) is nearly equal to ( 2n/e) ^ (2n) /( n^(2n)) * √ ( 4 π n) /(2 π n) /( 2 ^ (2 n)) = 1 /√ (π n) Now 1/ √ ( π * 1998) = 1/80 Since 1/9999 < 1/(80)

[@daboffey]

Expression = 1999! / (999! × 2 ^ 999)^2, which is approx. 0.0178.

[@Grecks75]

LOL 🤣You didn't consider this a solution, or did you? I mean, you could have just written 1/2 * 3/4 * 5/6 * ... * 1997/1998 is approximately 0.0178. By the way, note that the factorial expression given by you is _not_ correct, it must be 1998! / (2^999 * 999!)^2. Which in turn means that your factorial expression can never be 0.0178..., because it is actually larger by a factor of 1999. So how did you calculate it? This still leaves two questions: 1) What exactly do you mean by "approximately"? (Since -1,000 is also in the proximity if you ask me.) and 2) Can you prove the stated proximity to 0.0178? (This means to give a conclusive logical reasoning for it instead of referring to some unknown, dubious electronic device as an authority.) Oh, wait, this was actually the problem to be solved.

[@Creativemathlearning]

Hay quá

[@ionelpatriche6866]

Felicitari,pentru solutie!

[@Grecks75]

Before watching: I found a much stronger inequality for the given product P that goes like this: 1/56.04 < 1/sqrt(999.5*pi) < P < 1/sqrt(999*pi) < 1/56.02. It is based on (an interesting relation with) the Wallis product.

[@souverain1er]

Nice

[@carpediem3420]

재밌다... 이해하기 쉬워서 그런가 ㅋㅋ

[@davidmexicotte9862]

nice

[@baconboyxy]

Im assuming this is no calculator, but my immediate instinct is just to do this in python nevertheless lol

[@ronaldjensen2948]

numA = range(1, 1998, 2) num, den = 1, 1 for i in numA: num *= i den *= i+1 print('{} < {} < {}'.format(1/1999, num/den, 1/44)) # 0.0005002501250625312 < 0.017847935113411026 < 0.022727272727272728

[@baconboyxy]

@@ronaldjensen2948 pretty much the same thing I did. Using .format instead of an f string hurts me tho lol

[@RyanLewis-Johnson-wq6xs]

A

[@SanePerson1]

I used a similar trick of inserting numbers between the factors given, except I elected to use 2/2, 4/4, 6/6, ...1998/1998. If N is the numerator and D is the denominator of the original 999 term mystery number, the mystery number can than be expressed as N/D = N•D/D², but also it is 1998!/2¹⁹⁹⁸•(999!)². So I was looking at comparing the binomial coefficient of 1998 choose 999, 1998!/(999!)², multiplied by 1/2¹⁹⁹⁸. Unfortunately, I was then tempted by Stirling's approximation: N! ≈ √(2πN)(N/e)^N. It should work, but the method shown here is so simple and more elementary.

[@christodoulostsilopoulos7932]

A

[@RyanLewis-Johnson-wq6xs]

A^2

[@RyanLewis-Johnson-wq6xs]

AB=1/1999

[@Sleepystranger]

f(x) = x / f'(x)

[@michelebrun613]

First inequality 1/1999

[@hazalouldi7130]

we must prove that A

[@robertveith6383]

Your post is wrong! I don't know why you think it is correct without the needed grouping symbols: n ÷ (n + 1) < (n + 1) ÷ (n + 2). Or, write it this way: n/(n + 1) < (n + 1)/(n + 2).

[@RyanLewis-Johnson-wq6xs]

1/Sqrt[1999]=Sqrt[1999]/1999