Hello. For the past 3 years I have been preparing for Polish Mathematical Olympiad. Ever since I started working on it, it was my dream to participate in the final round. You were one of the first youtubers i stumpled upon when doing research and I’ve watched you ever since. Last week I participated in the 2nd round of PMO and it turned out very well. I’m still not sure if I will qualify but it’s likely, given how I performed compared to other people. That’s why I wanted to thank you for all the good work you’ve done by running this channel and wherever you are right now, I hope everything is good for you. Best wishes from your fan
I am currently viewing the 2nd round of PMO on AoPS . Can you please tell me the questions which you managed to solve during the exam ? I am really curious , because at a glance only 1 and 4 seem doable to me . Also what was the cutoff , do the officials release it ? Also congrats on this huge achievement , almost forgot to say that lol
@@srijanbhowmick9570 Thanks! To get to the final round you needed 19 points this year (each questions having possible scores: 0,2,5,6). I managed to do tasks 1, 2, 4, 5 (each for 6 points but 2 for 5 points) and scored 23/36. I struggled the most with the 2nd question, however I came up with something reasonable in the end. You need to take into account that we had in total 5 hours per day to solve the questions, so I had the time to try many attempts. Fun fact: Problem no. 4 was a killer one and for some reason many really talented people struggled with it the most.
Very nice, albeit an easier one to what I’m used to here. Love these videos to self teach myself these math competition problems, even if I may not use the tricks that often lol. Do you take suggestions for problems? I was recently trying a couple imo shortlist problems and found algebra problem 1 in the 2021 shortlist to be a bit weird. They do have solutions listed on the page, however it’s very simplified and for some reason I couldn’t wrap my head around it. Not in the way that I didn’t understand what they wanted to do, but going from step to step was quite a big jump with it. Hope you’re good yourself!
Depends on the domain of the function. For real x, the limit exists (and is trivially 0) as sin is bounded for the numerator and has a simple order 1 pole for the denominator (i.e. Squeeze plus L'Hopital). For complex x, the limit does not exist (for instance, it diverges if we approach 1 from the imaginary direction). Also, the pole at 1 is essential, because of the nasty function sin(1/(1-x)).
That's what I have been trying to figure out. My guess is that letsthinkcritically probably uses his apple pencil and his iPad to make the videos. The software could be an app but I am not exactly sure what it is called.
Hey man @letsthinkcritically i really needed your help.I've seen in your channel you have a lot of content on APMO.Could you suggest me which country's problems should I practice to do good in APMO?
Sin(1/(x-1)) is bounded ([-1,1]) so you could just use lohpital for (x-1)^2/sin(pix) and you get 2(x-1)/(picos(pix)) which is 0/-pi =0 so the limit is 0. You can put any thing you want in the sin(1/(x-1)) and still get 0 .
After the substitution t = x-1 we get the expression t² sin(1/t) / sin(π(t+1)) which can be rewritten as -(t/π) (πt/sin(π(t+1))) sin(1/t). Now, -|(t/π) (πt/sin(π(t+1)))| ≤ -(t/π) (πt/sin(π(t+1))) sin(1/t) ≤ +|(t/π) (πt/sin(π(t+1)))| where both the first and last expressions have limits 0*1 = 0. Therefore, by the squeeze theorem, also -(t/π) (πt/sin(π(t+1))) sin(1/t) has limit 0.
sin (pi x ) = sin (pi - pi x) Then we have lim x->1 ((x-1)²*sin(1/(x-1)))/sin(pi - pi x) The denominator becomes simply pi - pi x (approaching 1 this is equivalent to the original denominator). We can factor out a minus pi and get - pi (x-1), which can be simplyfied with the numerator to get: lim x->1 ((x-1)*sin(1/(x-1)))/-pi) In the numerator we have , by substituting, 0 times a sin function which oscillates between -1 and 1, so the product of those is still 0. 0 over a finite number like -pi is simply 0. So the limit is equal to 0.
4:23 ish I think you have to consider the fact that the lim as t approaches 0 u approaches + OR MINUS infinity. Don't think this matters because that limit would still approach 0 after evaluating the other case so I think the result is still 0 but I think you need this to be rigorous.
lim f(x) as x -> 0+ is the same as lim f(1/x) as (1/x) -> infinity. If you are taking lim f(x) as x -> 0, then you need to check both lim f(1/x) as x -> +infinity and also check lim f(1/x) as (1/x) -> -infinity and the values must match. In this case they are both 0 so the limit exists but in general that doesn't always happen
Maybe a simpler approach: The given expression becomes : (x-1) *sin (1/(x-1)) * (x-1)/sin(πx) but sin (1/(x-1)) is bounded and lim (x-1) =0 so lim (x-1) *sin (1/(x-1)) = 0 . On the other hand by d' Hospital : lim (x-1)/sin(πx) =0 .
pls solve challenges and thrills of pre-college mathematics quadratic equation question number 28....the question If a,b are positive integers and 1/x+ 1(x-a)+ 1/(x-b) =0 has real roots then prove that one of the roots lie between a/3 and 2/a3 and the other one between -b/3 and -2b/3. I will be very thankful if you do before sept 3 as i have IOQM exam....i googled this up but i was not satisfied. Thank you
This question is wrong. First off, after simplifying you get the quadratic 3x^2-2(a+b)x+ab=0. Now, the discriminant is indeed always greater than zero (by AM-GM inequality) giving us two real roots, and by Vieta's relations, the product of roots = ab/3>0 as a and b are positive reals. This means that either both roots are positive or both roots or negative. However, we are asked to prove that one of the roots lies between a/3 and 2/a3 and the other one lies between -b/3 and -2b/3, which would suggest that one is positive while the other is negative, a contradiction. Furthermore, you may consider an example, say, 1/x+1/(x-2)+1/(x-3)=0. On inspection, you will find that the roots do not fulfil said criteria. [I have qualified the IOQM 2023 conducted on Sept 3rd]
way too complicated with those backwards and forwards substitutions 1. *sin(πx) = sin(πx - π + π) = -sin(πx - π)=-sin(π(x-1))* ==> *(x-1)/sin(πx) = -(x-1)/sin(π(x-1) = -1/π * (π(x-1)/sin(π(x-1) --> -1/π* when x-->1 2. *(x-1)*sin(1/(x-1)) --> 0* (obviously, as sin is limited by [-1..1] and *(x-1)-->0* ) 3. So we have a product of 2 factors where the 1st factor *--> -1/π* and the 2nd factor *-->0* ==> The total limes is obviously 0 (and strictly speaking the deduction "lim(a/b) = lim(a)/lim(b)" at 4:09 is in general wrong and must be justified)
@@anshugupta793 well your question asks for integer solutions to 3*2^(x+1) =y(y+2) but any x which is an integer also x+1 is an integer so why not ask for integers s.t. 3*2^x = y(y+2) ?
@@richardfredlund8846 no that you're wrong because you literally converted the equation into something else (i.e you divided by 2 in lhs but not in rhs so, it very obvious that the solution to the equation which you have will not Nesesarily satisfy the one which I have)
You had one of the best math channels on youtube for learning. Please come back.
Hello. For the past 3 years I have been preparing for Polish Mathematical Olympiad. Ever since I started working on it, it was my dream to participate in the final round. You were one of the first youtubers i stumpled upon when doing research and I’ve watched you ever since. Last week I participated in the 2nd round of PMO and it turned out very well. I’m still not sure if I will qualify but it’s likely, given how I performed compared to other people. That’s why I wanted to thank you for all the good work you’ve done by running this channel and wherever you are right now, I hope everything is good for you.
Best wishes from your fan
I made it!
@@jakubwieliczko257 This is wonderful to hear. Great job!
@@kevinkuruvilla6435 Thank you 🥰
I am currently viewing the 2nd round of PMO on AoPS . Can you please tell me the questions which you managed to solve during the exam ? I am really curious , because at a glance only 1 and 4 seem doable to me . Also what was the cutoff , do the officials release it ? Also congrats on this huge achievement , almost forgot to say that lol
@@srijanbhowmick9570 Thanks! To get to the final round you needed 19 points this year (each questions having possible scores: 0,2,5,6). I managed to do tasks 1, 2, 4, 5 (each for 6 points but 2 for 5 points) and scored 23/36. I struggled the most with the 2nd question, however I came up with something reasonable in the end. You need to take into account that we had in total 5 hours per day to solve the questions, so I had the time to try many attempts. Fun fact: Problem no. 4 was a killer one and for some reason many really talented people struggled with it the most.
This is by far the greatest math channel on UA-cam. I really miss you
Where are you? Please, come back. You are one of the best math channel on UA-cam.
I'm still waiting to see you next time :)
Pls do comeback u are one the best math solving channels
Very nice, albeit an easier one to what I’m used to here. Love these videos to self teach myself these math competition problems, even if I may not use the tricks that often lol. Do you take suggestions for problems? I was recently trying a couple imo shortlist problems and found algebra problem 1 in the 2021 shortlist to be a bit weird. They do have solutions listed on the page, however it’s very simplified and for some reason I couldn’t wrap my head around it. Not in the way that I didn’t understand what they wanted to do, but going from step to step was quite a big jump with it. Hope you’re good yourself!
Depends on the domain of the function. For real x, the limit exists (and is trivially 0) as sin is bounded for the numerator and has a simple order 1 pole for the denominator (i.e. Squeeze plus L'Hopital).
For complex x, the limit does not exist (for instance, it diverges if we approach 1 from the imaginary direction). Also, the pole at 1 is essential, because of the nasty function sin(1/(1-x)).
Come back 🥺
Where you go man?
I do miss you ❤
Why did u stop uploading???! Your content is amazing and we all want you back if possible. Keep up the good work!!
There is way lot easy than this
But in general your videos are greats
What software do you use to create these videos? It's great!
That's what I have been trying to figure out. My guess is that letsthinkcritically probably uses his apple pencil and his iPad to make the videos. The software could be an app but I am not exactly sure what it is called.
Hey man @letsthinkcritically i really needed your help.I've seen in your channel you have a lot of content on APMO.Could you suggest me which country's problems should I practice to do good in APMO?
Sin(1/(x-1)) is bounded ([-1,1]) so you could just use lohpital for (x-1)^2/sin(pix) and you get 2(x-1)/(picos(pix)) which is 0/-pi =0 so the limit is 0.
You can put any thing you want in the sin(1/(x-1)) and still get 0 .
1 year and bro is missing 😢
Try solving this functional equation:
Find all f: R+ --> R+ such that
f(f(x)/y) = y•f(y)•f(fx))
for all x, y belonging to R+.
After the substitution t = x-1 we get the expression t² sin(1/t) / sin(π(t+1)) which can be rewritten as -(t/π) (πt/sin(π(t+1))) sin(1/t).
Now, -|(t/π) (πt/sin(π(t+1)))| ≤ -(t/π) (πt/sin(π(t+1))) sin(1/t) ≤ +|(t/π) (πt/sin(π(t+1)))| where both the first and last expressions have limits 0*1 = 0.
Therefore, by the squeeze theorem, also -(t/π) (πt/sin(π(t+1))) sin(1/t) has limit 0.
Can we just use bounded [-1,1] for sin (1/(x-1)) and remove it?
sin (pi x ) = sin (pi - pi x)
Then we have
lim x->1 ((x-1)²*sin(1/(x-1)))/sin(pi - pi x)
The denominator becomes simply
pi - pi x (approaching 1 this is equivalent to the original denominator). We can factor out a minus pi and get - pi (x-1), which can be simplyfied with the numerator to get:
lim x->1 ((x-1)*sin(1/(x-1)))/-pi)
In the numerator we have , by substituting, 0 times a sin function which oscillates between -1 and 1, so the product of those is still 0.
0 over a finite number like -pi is simply 0.
So the limit is equal to 0.
PLEASE COMEBACK, YOU HAVEN'T DONE COMBINATORICS AND GEOMETRY
We need you bro
4:23 ish I think you have to consider the fact that the lim as t approaches 0 u approaches + OR MINUS infinity. Don't think this matters because that limit would still approach 0 after evaluating the other case so I think the result is still 0 but I think you need this to be rigorous.
lim f(x) as x -> 0+ is the same as lim f(1/x) as (1/x) -> infinity. If you are taking lim f(x) as x -> 0, then you need to check both lim f(1/x) as x -> +infinity and also check lim f(1/x) as (1/x) -> -infinity and the values must match. In this case they are both 0 so the limit exists but in general that doesn't always happen
I was just gonna comment the same
Q. If (a+1/b) = tan59; (b+1/c) = tan60 & (c+1/a) = tan61. Then find
Sqrt[(abc)^2022 + 1/(abc)^2022 + 2211] =?
Solution :- ua-cam.com/video/dV3ILYbSW9I/v-deo.html
Beautiful! Next '? 😉😙👀🏆🙏🏆
What is Wolfram alpha? Is it same as L'hopitais method .
It is an on-line sight for soln
Maybe a simpler approach: The given expression becomes : (x-1) *sin (1/(x-1)) * (x-1)/sin(πx) but sin (1/(x-1)) is bounded and lim (x-1) =0 so lim (x-1) *sin (1/(x-1)) = 0 . On the other hand by d' Hospital : lim (x-1)/sin(πx) =0 .
Ακριβώς. Απλά το όριο στο τέλος είναι -1/π αντί για 0 αλλά το γινόμενο βγαίνει 0 πάλι
pls solve challenges and thrills of pre-college mathematics quadratic equation question number 28....the question
If a,b are positive integers and 1/x+ 1(x-a)+ 1/(x-b) =0 has real roots then prove that one of the roots lie between a/3 and 2/a3 and the other one between -b/3 and -2b/3.
I will be very thankful if you do before sept 3 as i have IOQM exam....i googled this up but i was not satisfied. Thank you
This question is wrong. First off, after simplifying you get the quadratic 3x^2-2(a+b)x+ab=0. Now, the discriminant is indeed always greater than zero (by AM-GM inequality) giving us two real roots, and by Vieta's relations, the product of roots = ab/3>0 as a and b are positive reals. This means that either both roots are positive or both roots or negative. However, we are asked to prove that one of the roots lies between a/3 and 2/a3 and the other one lies between -b/3 and -2b/3, which would suggest that one is positive while the other is negative, a contradiction. Furthermore, you may consider an example, say, 1/x+1/(x-2)+1/(x-3)=0. On inspection, you will find that the roots do not fulfil said criteria.
[I have qualified the IOQM 2023 conducted on Sept 3rd]
way too complicated with those backwards and forwards substitutions
1. *sin(πx) = sin(πx - π + π) = -sin(πx - π)=-sin(π(x-1))*
==> *(x-1)/sin(πx) = -(x-1)/sin(π(x-1) = -1/π * (π(x-1)/sin(π(x-1) --> -1/π* when x-->1
2. *(x-1)*sin(1/(x-1)) --> 0* (obviously, as sin is limited by [-1..1] and *(x-1)-->0* )
3. So we have a product of 2 factors where the 1st factor *--> -1/π* and the 2nd factor *-->0* ==> The total limes is obviously 0
(and strictly speaking the deduction "lim(a/b) = lim(a)/lim(b)" at 4:09 is in general wrong and must be justified)
neat!
I would like to share this good problem
x, y are integers 3*2^(x+1) =y(y+2)
Find all solutions
why x+1 ... surely x would be the same problem more simply stated?
@@richardfredlund8846 I didn't get it what did you wanted to say?
@@anshugupta793 well your question asks for integer solutions to 3*2^(x+1) =y(y+2) but any x which is an integer also x+1 is an integer so why not ask for integers s.t. 3*2^x = y(y+2) ?
i guess the two cases are where y==0 mod 3 or when y==1 mod 3
@@richardfredlund8846 no that you're wrong because you literally converted the equation into something else (i.e you divided by 2 in lhs but not in rhs so, it very obvious that the solution to the equation which you have will not Nesesarily satisfy the one which I have)
its lTC lmao you bruhed
hi edwin 💀💀
@@LusePancakeshi jeremy
hi walmart
hi walmart
hi walmart
I got 0.874 + 83/2 ???? anyone elses?
@@English_shahriar1 very thanks fr.iend for help🤗
I let x=1+h and limh-0 sinπx/πx=1