Find the closest integer

[@anoopyadav5617]

3:50 1=A(n+2)+B(n-2)

[@platypi_otbs]

I love watching you solve. I think it's your attitude while you're doing it. You seem excited but patient.

[@savaesmek]

Tbh i think this is just it for me i used to like the videos but now hes just doing easier and easier questions and stretching them out for 15 minutes

[@albajasadur2694]

By shifting the index, the sum of the partial fraction can be written as this form: 1000/4* [sum(1/k) - sum(1/j)], where k=1=9998 and j=5=10002. We can observe the cancellation and the first 4 terms left from k expression and last 4 terms left from j expression. The numerical answer is 520.73 (to 2 d.p.), and so the nearest 521

[@itsphoenixingtime]

I used partial fractions to expand and then evaluated that: S = 250 (1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002) Assuming this is a no calculator problem, it would be silly for me to try and find a common denominator and then divide. That would be too hard! So I first took that S ~ 250 (1 + 1/2 + 1/3 + 1/4 - 1/10000 - 1/10000 - 1/10000 - 1/10000) Because subsequent harmonic numbers are very close to each other for large N. I then combined the sum. S ~ 250 (12/12 + 6/12 + 4/12 + 3/12 - 1/10000 - 1/10000 - 1/10000 - 1/10000) S ~ 250 (25/12 - 4/10000) S ~ 250 (24/12 + 1/12 - 4/10000) S ~ 250 (2 + 1/12 - 1/2500) S ~ 500 + 250/12 - 250/2500 S ~ 500 + 240/12 + 10/12 - 1/10 S ~ 500 + 20 + 5/6 - 1/10 S ~ 500 + 20 + 25/30 - 3/30 S ~ 500 + 20 + 25/30 - 3/30 S ~ 500 + 20 + 22/30 S ~ 521 So the closest integer is 521. The approximation of the sum here is 520.73333333...., while the actual sum according to Wolfram Alpha is 520.73333833183353328834133168365326884613076884523090380923765236... or 520 4889899947767/6667999933320. Clearly it is too difficult to try and find LCM!

[@Grecks75]

Great job. This time your solution development was more straight-forward than mine. My thought process went something like this: It was immediately clear to me that the corresponding infinite series converges, and I thought I could use its limit as an approximation. I didn't see from the start that I could actually calculate it, let alone the given partial sum. So I thought I could bound it with Euler's Basel series (zeta(2)) and work with its limit (pi^2/6) instead. It took me a while to realize that it's actually a telescoping sum that we can directly calculate, while you seemed to know this right from the start. I then rewrote the sum as the limit of the infinite series (25/48) minus some four small terms. I then used an inequality for the harmonic mean to bound those four remaining terms and find out how good the approximation with the limit was. It turned out be good enough to justify the answer, even after multiplication with 1000.

[@itsallaboutcomputersandit7925]

wow nice solution

[@keshavchauhan6290]

I am a highschool student and I love watching your videos in my free time. I love Mathematics and your videos present it in a very calming and peaceful way which is exceptional.

[@graf_paper]

Blessed to have you as a math teacher! Thanks all the videos.

[@TheMarkSpurs]

Good video but I struggled with: 1. @ 3:51 PrimeNewtons (PN) writes B(n-1) in place of the actually correct B(n-2). I didn't catch this error in the moment only afterwards by reading the comments. And by choosing the number 2 the corrected B side of the addition becomes 0, leaving just the A term. Lovely!! 2. @10:40 it is not obvious, IMO, that you would even look for all future fractions to be cancelled except the last 4. That was quite the jump, possibly made easier if you've studied higher mathematics. I do love PN's style, clarity on the board, and the calmness during his explanations. Thanks a lot. :)

[@jakehobrath7721]

I love your embrace of learning rooted in faith. You make me want to be a better person watching the energy you give the world.

[@張謙-n3l]

for the final step, I think we can re-index the two parts get the answer more easily sum from 3 to 10000 of (1/(n-2) - 1/(n+2)), for the first term we can see the denominator is from 1 to 9998, and the second one will be from 5 to 10002, the part from 5 to 9998 will be cancelled. Then the sum will actually be 1+1/2+1/3+1/4 - 1/9999-1/10000-1/10001-1/10002

[@uwepleban3784]

I asked my favorite LLM o1-mini, and it solved this without a hitch. Here is its closing summary: 1. Factored the denominator using partial fractions to simplify the summation. 2. Telescoped the series to reduce the summation to a manageable form. 3. Calculated the remaining terms accurately to evaluate the sum. 4. Scaled the sum by 1000 and identified the closest integer to the resulting value

[@Grecks75]

That's a bit scary. 👻 I often think this is the nail in the coffin for education. Add laziness on top. Who's gonna do original thinking and provide original ideas in the future?

[@ChristopherBitti]

1/(n^2 - 4) = 1/4 * [1/(n - 2) - 1/(n + 2)] so if the sum is S we have 4S = 1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002 because most of the sum telescopes. You can see this because n - 2 where 3

[@jamesatkinson12]

nicely explained

[@RomanOrekhov]

To make the result more rigorous you should have proved that 250*(1/9999+1/10000+1/10001+1/10002) is < 1/3 so that when it's subtracted from 520+5/6 you get smth greater than 520.5

[@陈瀚凌-t5b]

I send you many thanks, as a Chinese I am going to apply to Cambridge, and the interview part is really harsh for me. The biggest problem is that I can’t really talk about a math question in English but you really helped me get used to that and get familiar with Olympiad style questions!

[@srisaishravan5512]

Yayy! My favourite math youtuber iploads again! Thank you for uploading regularly newton sir.

[@low-litlight3438]

There is a way of approaching this without a creative technique. If you approximate a lower bound on the the sum’s infinite version as being ζ(2) - 1 - 1/4 = π^2/6 - 5/4, you can improve the accuracy of the bound by adding 1/(n^2 - 4) - 1/n^2 from n=3 to n=16. Multiplying by 1000, this will give you roughly 520.53, which rounds to 521.

[@chriscalderon1337]

I worked this out and accidentally noticed that you can approximate the sums as sums to infinity and you get the same result a bit more cleanly.

[@random_Person347]

If I was doing this in an exam situation, after the fraction decomposition, I would have split the summation into two separate series with indices adjusted to start at 1 and 5 respectively. It is then easier to show that all the overlapping terms cancel and you are left with the first 4 terms of the one series and the last 4 terms of the other one.

[@red2go56]

2:29 the denominator is getting bigger by 7, 9, 11. so the dinominator are getting bigger by the next unevan number, at least for the first 4.

[@arimermelstein9167]

I had a go at this before watching your answer and got the same answer and was so excited. The only thing I did differently is I argued that the numbers at the end were going to be so small as to not throw off the answer that much. I’m not sure that is justified in general

[@Grecks75]

Correct. In general you need to find lower and upper bounds for the remaining terms to make the reasoning rigorous.

[@eve_the_eevee_rh]

Partial fractions: 1/(n-2)(n+2) ≡ A/(n-2) + B/(n+2) 1 = A(n+2) + B(n-2) n = 2: 4A = 1, A = ¼ n = -2: -4B = 1, B = -¼ 1/(n²-4) ≡ 1/4(n-2) - 1/4(n+2) sum from 3 to 10000: = { 1/4(3-2) - 1/4(3+2) + 1/4(4-2) - 1/4(4+2) + 1/4(5-2) - 1/4(5+2) + 1/4(6-2) - 1/4(6+2) + 1/4(7-2) - 1/4(7+2) + 1/4(8-2) - 1/4(8+2) + 1/4(9-2) - 1/4(9+2) ... + 1/4(9994-2) - 1/4(9994+2) + 1/4(9995-2) - 1/4(9995+2) + 1/4(9996-2) - 1/4(9996+2) + 1/4(9997-2) - 1/4(9997+2) + 1/4(9998-2) - 1/4(9998+2) + 1/4(9999-2) - 1/4(9999+2) + 1/4(10000-2) - 1/4(10000+2)} = ¼ {1/1 - 1/5 + 1/2 - 1/6 + 1/3 - 1/7 + 1/4 - 1/8 + 1/5 - 1/9 + 1/6 - 1/10 + 1/7 - 1/11 ... + 1/9992 - 1/9996 + 1/9993 - 1/9997 + 1/9994 - 1/9998 + 1/9995 - 1/9999 + 1/9996 - 1/10000 + 1/9997 - 1/10001 + 1/9998 - 1/10002} = ¼(1 + 1/2 + 1/3 + 1/4 + (1/5 - 1/5) + (1/6 - 1/6) + (1/7 - 1/7) ... + (1/9996 - 1/9996) + (1/9997 - 1/9997) + (1/9998 - 1/9998) - 1/9999 - 1/10000 - 1/10001 - 1/10002) = ¼(1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002 Multiply all this by 1000 250(1+1/2+1/3+1/4-1/9999-1/10000-1/10001-1/10002) ≈250(1+1/2+1/3+1/4) (1/9999+1/10000+1/10001+10002 is insignificant) = 250(12/12 + 6/12 + 4/12 + 3/12) = 250(25/12) = 6250/12 Dividing 6250 by 12 gives us 520.8... ≈521 is our final answer

[@kennethgee2004]

well the notation gets a little weird but the question is asking for rounding to the closest whole number this i usually done with the brackets. If we wanted ceiling or floor then we use the upper or lower arms on the bracket. The original form is better as there is only one step in each summation. We can see that as n approaches 10000 that the number is going to very tiny, so will not effect the round very much. just calculate from there since you ended up using a calculator anyways.

[@gamefacierglitches]

I got the correct answer, but didn't see about that cancellation. Instead I used the harmonic series approximation, since the gamma-constant would cancel, leaving: 1000/4 *[ln(9998/10002) + 25/12] Without a calculator, you can say the ln function is approximately ln(1) = 0, so you would have the 25000/48 = 250.833; with a calculator the ln function is the -0.1

[@er63438]

Nice! Thought you were going to take the limit and work with that but you did something completely different. Great video, as usual.

[@Grecks75]

He did exactly that. 25/48 is the limit of the series, and he basically used that as his approximation.

[@dem_hauke]

i really liked this video it was a really interesting solution

[@tygrataps]

Solving for B wasn't right... how does that affect everything afterwards? Or does it?

[@doctorb9264]

Agree but he magically got the correct values.

[@Grecks75]

It was only a small typo in the equation, n-1 instead of n-2; the values for A and B were still correct. It didn't affect anything further down the stream.

[@dirklutz2818]

Pure Magic!

[@IOwnKazakhstan]

I have no idea what the solution is but the first pattern I recognise is that the denominator is 1x5, 2x6, 3x7, 4x8, 5x9 etc.

[@LovePullups]

It is actually the same of 1/[(n-2)(n+2)]

[@souverain1er]

Fun as always despite the small arithmetic mistake

[@rohangt1]

Hello sir, I love your videos. If you can make a video on the following problem, it would be great -: Why 1^0 = 1^1 => 0 = 1 is false based on equality of power for the same base?

[@narangfamily7668]

Fun stuff mate like always ❤

[@copter7013]

Hey, can you solve this problem from a middle school thai competition? i want to know the approach; Find the sum of the squares of all the roots of the equation: x^2 -8floor(x) + 9 = 0

[@user-ut5ic6ox7s]

They have Floor concept in middle school????

[@copter7013]

@@user-ut5ic6ox7s this was a hard middle school competition, and the questions contained conbinatorics, functions, modular arithmetic. i have some more questions that are harder, but ive never seen this one before

[@copter7013]

@@user-ut5ic6ox7s i participated this year and got 6 / 100 haha, which was like top 200 out of 500

[@sisarka468]

x1 = sqrt(39), x2 = - sqrt(39) Solution: 78

[@copter7013]

@@sisarka468 looking at the graph, there's 5 solutions, so no

[@Jason-ot6jv]

Easy python solution 😄 sum = 0 for n in range(3, 10000): sum += 1/(n**2 - 4) answer = round(1000*sum) print(answer) This prints 521

[@555amry]

Did a race between mr newtons and a Casio calculator, the calculator won by 4 mins 😔😔😔

[@HridikKanjilal4444-k5f]

Sir I have sent you a problem in your gmail which was quite hard for me but it's really irritating to see a problem left unsolved. So can you please give it a try?

[@annoyingbstard9407]

Is it Africa by Toto?

[@pvpexagon4721]

Coolest guy on KZbin!

[@makehimobsessedwithyou6412]

Why when n=2 B(n-1) is zero?

[@tygrataps]

Good question!

[@adulis73]

Use Riemann Sum!

[@nicolasb11]

Well done

[@Shreyam-q6b]

[@doctorb9264]

I think you plugged in the wrong values and still got the correct values for A and B and hence he correct answer. Anyway cool problem !

[@georgeh1352]

Very interesting

[@badfriendlygamer9334]

Easy sum for people who learned sequence and series in class 11 india

[@Blacksmith1959]

Unless I have lost the plot and misunderstood, at 4.07 when n = 2, shouldn't we get 1 = 4A + B???? Also, when n = -2 isn't it -3B?

[@davez8816]

You have to factor An + Bn + 2A - 2B = 1 ie (A+B)n +2(A-B)=1 As this should be true for all n, means A+B must be =0 ie B = -A The other term is 2(A-B) = 1 ie 2(A - (-A))=1 and 4A=1 then A=1/4 while B=-A=-1/4

[@SalmonForYourLuck]

Oh i know he was supposed to write (B-2) but instead he wrote (B-1) thats why it looks wrong. He is thinking with (B-2) so the solution is still correct

[@maxhagenauer24]

Not only that but it was supposed to be n - 2 and not n - 1 multiplying the b.

[@peskarr]

tried before waatch and failed :)

[@michaelkeffer504]

B(n - 2)

[@maxmanolov3077]

more aime problems

[@荻野憲一-p7o]

not equals to 0.1

[@Jono4174]

I used the wrong method, but I got the right answer. 1/5+1/12+1/21+ integral of (1/(x^2-4)) from 5.5 to 10000.5 = 0.5213 Ans = 521 😅

[@oldjoec3710]

IMO, not the wrong method. Great use of calculus to solve the same problem. All that you would need to make the answer rigorous is some estimated bound on the error between the sum and the integral.

[@janavdhingra2409]

method of difference

[@barackfrans6777]

Mind yr decision used yr vedio to solve a math problem

[@ferashamdan4252]

رائع جدا

[@gf-xy2of]

>> round(1000*sum(1./([3:10000].^2-4))) ans = 521

[@AbhayPratapSinghRajawat-ch4uz]

Maths of prime Newton is as good as maths of Newton....

[@sowndolphin5386]

you cant take n = plus or minus 2 in an equation like 1/(n-2)(n+2) = A / (n-2) + B/(n+2), doesnt it create undefined numbers? if n = 2 1/0(undefined) = A/0 (undefined) + B/4 if n = -2 1/0(undefined) = -A/4 + B/0 (undefined)

[@robertpearce8394]

That is why the problem started at n=3.

[@sowndolphin5386]

​​@@robertpearce8394 he found A and B by taking n= 2 3:54

[@ahmedashraf1343]

You always say: "Never stop learning because those who stop learning stop living" so I liked to tell you why don't you learn about Islam? That's actually one of the main concepts in Islam and one of the core concepts in the start of the religion was the first word in the religion before all the religion concepts were done after about 23 years, and the word was "read", so What about trying to learn and read about this religion?

[@ChinmayPlays]

521 answer

[@alphabeta2903]

We iitians and jee aspirant solving it in less than 1 min. ,u taking 14 minutes to explain this question hates off to u, that's difference in TRUE teacher and students, I cant explain it for so long

[@xinpingdonohoe3978]

I got it, through complete overkill. After partial fractions, I rewrote it as S=250(H9998-(H10002 -25/12)) =250(H9998-H10002+25/12) Hm≈ln(m)+γ S≈250(ln(9998)-ln(10002)+γ-γ+25/12) =250(ln(9998/10002)+25/12) =10000/40 ln(9998/10002) +6250/12 =1/40 ln((1-4/10002)¹⁰⁰⁰⁰) +3125/6 3126/6=1563/3=521 3125/6=520.8333... S=1/40 ln((1-4/10002)¹⁰⁰⁰⁰)+520.8333... ≈1/40 ln(e^(-4)) +520.8333... =1/40 (-4) +520.8333... =-0.1+520.8333... =520.7333... round(S)=521

[@alexicon2006]

Why did bro do all that 😭😭😭

[@xinpingdonohoe3978]

@@alexicon2006 normally the ideas just flash in my mind. That's where it naturally took itself, so I followed it on paper, and missed the obvious route.

[@alexicon2006]

@@xinpingdonohoe3978 Yeah that happens to me too sometimes. We get so locked in on a route because we dont wanna risk losing sight for the sake of optimizing.

[@satyapalsingh4429]

You are the best teacher .I like your teaching .❤❤❤