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
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.
@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! ❤
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.
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.
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 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
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.
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.
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
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
Sir I have a request if you could take up questions from B.Stat and B.Math of ISI (Indian Statistical Institute) entrance exam and CMI(Chennai Mathematical Institute) entrance exam it would be much helpful.These colleges are one of most premium research colleges for math and statistics.And I assure you and give guarantee that you will love the questions they are quite tricky but when you will solve them you will get amazed.Please Sir it's a request.🙏🙏 Love from India 🇮🇳🇮🇳 ❤️❤️
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.
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.
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.
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
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.
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).
The solution in this video is genius! I am astonished! I would have never figured this out.
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
Yes..very nicely done.
Clever!
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.
Never stop learning because those who stop learning have stopped living is good advice.
I liked your comment, " Never stop learning .... living."
It works in the other direction, too. “Those that stop living stop learning.” 😀 It’s commutative!
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
That explains why dumb people seem immortal; they never started living in the first place.
@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! ❤
Thank you!
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.
I love when he looks at the camera, knowing you see it too! I love these videos!!!
Very Creative,
Thanks for sharing this way of looking at things : )
Thanks! Very informative ❤
6:22 the look man 👽
Noticed that😂
Sus ass look
Massive congratulations for reaching 200k subs!
Really elegant solution. Thanks so much for sharing.
Beautiful concise solution.
Sir your video's are great ❤
Elegant.......thank you.
Great video. Update for 2024, and you end up with 1/2025 and 1/45 on the bounds.
This guy is awesome and enjoyable to watch
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.
Never knew that there were other right-handed people who start drawing the digit "8" going right and down instead of left and down
I go right and up
I do too, go right and down. Very peculiar I find the way he draws the "9". 😃
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.
If you "may have noticed that 44^2 = 1936," *then you are indirectly estimating a square root!*
@@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
Felicitari,pentru solutie!
excellent channel!
Bravo! Keep these videos coming, bro.
A^2
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.
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.
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
44.something (Can we write that in examination ?)
Magnificent solution
very interesting problem
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
Technically you are correct, but it seemed too obvious to me to say something about it.
Sir I have a request if you could take up questions from B.Stat and B.Math of ISI (Indian Statistical Institute) entrance exam and CMI(Chennai Mathematical Institute) entrance exam it would be much helpful.These colleges are one of most premium research colleges for math and statistics.And I assure you and give guarantee that you will love the questions they are quite tricky but when you will solve them you will get amazed.Please Sir it's a request.🙏🙏
Love from India 🇮🇳🇮🇳 ❤️❤️
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)
❤️
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!
I did it in my head.
I did it in your head 😂.
Produce your work in a thread. "I did it in my head" does not count.
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.
me too!
2
101=
27318619677157413541998666579156061420147177666088128046591030596082725294498066722338505744902120368830900788923839991099564447458450075226030128555294655577015766113909738825769262480452415909200510101
=101^101
Expression = 1999! / (999! × 2 ^ 999)^2, which is approx. 0.0178.
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.
Hi can anyone pls twll me that why
Integral of tanx= ln|sec x|+C
But shouldn't integral of tanx = -ln|cos x|
Yes but the two answers are the same because of the log power law: - ln cos = ln cos^-1 = ln sec
Có cách khác không
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.
"Those who stop learning, stop living" black screen, video over, no more learning, no more living
Hay quá
Im assuming this is no calculator, but my immediate instinct is just to do this in python nevertheless lol
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
@@ronaldjensen2948 pretty much the same thing I did. Using .format instead of an f string hurts me tho lol
Nice
재밌다... 이해하기 쉬워서 그런가 ㅋㅋ
nice
A
AB=1/1999
f(x) = x / f'(x)
A^2
First inequality 1/1999
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.
we must prove that A
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).
1/Sqrt[1999]=Sqrt[1999]/1999