Hi! I'm the organizer for SMT 2022! I'm glad you enjoyed our tiebreaker problems, and thank you for reviewing them!! For those of you who are high school students we encourage you to participate in this year's iteration of the contest, we're very excited :)
For the second question, removing all of the high level notation you could say, how many numbers less than 40 are perfect squares but not perfect fourth powers.
@@blackpenredpen Thanks. Also to finish the interpretation, it comes down to finding how many perfect squares there are below 40 and how many prefect fourth powers below forty and subtracting the later from the former
@@cosmicvoidtree Alternatively, you can find/generate a list of squares of numbers that are not squares themselves. This works for n≤40 because we don't really have to deal with sixth powers and etc.
Luckily there are no perfect 8th powers less than 40 or what you have said would not hold. Eg. 256 is 4^4 but it would satisfy the condition (f(256) can = 16 as f(4) = 2) perfect 16th powers don't hold as f(65536) != 256 since that is f(16). Will hold for 32nd powers but not 64th and so on.
I am from India , i study in 12th standard. when i study integration, i have some questions having a problem then i search in TH-cam then i show your channels good working teachers. I really motivated your teaching skills , i love mathematics 🥰🥰🥰🥰🙏🙏🙏🙏🙏🙏🙏🙏
That was a fun one!! It's so cool to see problems that I initially have no idea how to approach, solved with such a simple, easy-to-understand technique! Thanks!
As a 6th grader, our math club took the questions from the Stanford-Math League Tournament. Even though our group was comprised of 6th, 7th and 8th graders, I ended up winning with a t-shirt which has “Stanford-Math League Tournament”. I’ve already learned logarithms and even some calculus.
I don’t know if I ever got one right but I remember seeing questions like the last one all the time in these extracurricular math competitions lol The idea of dividing by x to get a binomial where 2ab removes the variable… so simple yet so crazy to come up with if you’re not used to these types of approaches
For the last one, I just multiplied both sides by x^2 + 3x + 1, which cancels out all the third and first degrees, and gives x^4 -7x^2 + 1 = 0. Rinse and repeat twice more. Basically the same process without dividing by x first.
Very fun! I did the first the same! For the second, I said that f(a)^2 = a can only be true if a is a perfect square, so a = 1,4,9,16,25,36. For a = 1, f(1) = 1^2, so we can get rid of that. And then 16 is the square of a square, so f(2) = 4 but f(16) = 4 breaks the bijectivity of f. No other is a 4th power so f(a) ≠ a^2 for a = 4, 9, 25, 36. For the third, I used the quadratic formula! Since x^2 - 3x + 1 = 0, we have x = (1/2)(3 ± sqrt(5)). And since x^16 - kx^8 + 1 = 0 is a quadratic in x^8, we get that x^8 = (1/2)(k ± sqrt(k^2 - 4)). raising (1/2)(3 ± sqrt(5)) to the 8th power (using the binomial theorem) gives (1/2)(2207 ± 987sqrt(5)) so (1/2)(2207 ± 987sqrt(5)) = (1/2)(k ± sqrt(k^2 ‐ 4)), which tells us k = 2207.
For the 3rd question, you can also notice that the k = sum of the roots of the equation. Then, sum the two roots of the staring equation (found via quadratic formula), each to the power of 8. This will be equal to k.
Yes you can compute the sum of the 8th power of the roots using a similar method of squaring, adding/subtracting repeatedly. But, bprp's method is quite elegant in a way.
Hi Jay. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
In my highschool in Poland we did problems like the first one on daily basis when we had logarythms in 11th grade but instead of doing it by factoring it out we did t=log2(n) then solve delta and substitute t1 and t2 and then solve for log2(n)
I see that a lot of these, high school juniors could solve, If they had the knowledge, I tried the problem with all the solutions less than 40, I actually did it and it was more logic than anything in my opinion
In the first problem: If you simply substitute “x” for “log (base 2) n”, you get (3/x) - (x/2) = (-5/2) Solve the quadratic and plug back in the 2 solutions (x = -1 & 6), setting them each = to “log (base 2) n” Same answer in the end. Just a less complex approach. You can actually solve it just by looking at it. It’s pretty clear what X is equal to. Regardless, more than one way to skin a cat. Per usual with the maths. 🤘😝🤘
@@natebobdog23 “EXCEPT you let x = log_2 (n)”. THATS the difference. Same answer-yet the slightest of variety in approaching and yielding that same solution. Your comment just defeated itself. Congrats. 👏👏👏 You also conveniently neglected the end of my original statement. Conveniently. No wonder you’re struggling over there. But hey, you’re learning! Good for you! I’ll be right here if u ever need me. ✌️
Alternative Solution to #3: Assume a and b are the root of x² - 3x + 1 = 0. Then a and b must also be a root of x¹⁶ - kx⁸ + 1 = 0. So, a¹⁶ - ka⁸ + 1 = 0 b¹⁶ - kb⁸ + 1 = 0 --------------------------- - (a⁸ + b⁸)(a⁸ - b⁸) - k(a⁸ - b⁸) = 0 k = a⁸ + b⁸ = 2207
In the second problem, it's given that f is bijective. But I wonder if it makes sense to think about how you would prove such a function would actually be bijective.
Consider any sequence n, n², n⁴, n⁸..., where n itself is not a square. N is a disjoint union of such sequences, so it suffices to prove that f is a bijection on each sequence separately. On any such sequence the function is given by f(n^(2^k)) = n^(2^(k+1)) if k is even, n^(2^(k-1)) if k is odd. From this it is clear that f(f(x)) = x for each x, so f is its own inverse and it is bijective.
Fiat Lux! As you were doing the question I thought in my head to make a common denominator on the left side and see that lon_2{n}=1. I was stunned the answer didn't line up. But then I saw you didn't switch the sign on when you moved 5log_2{n} to the other side. Maybe that will get my answer of n=1 and there is probably another one too. PS I made it to Cal because of you and Professor Leonard. TYSM
I did the first one another way, probably more complicated: () does not represent the base First, we can write the eqn as 1/log2^3(n) + 1/logn(2^-2) = -5/2 As logn(2^-2) =-2logn(2),,, And log2^3(n) =1/3log2(n) We get 3/log2(n) - 1/2logn(2) = -5/2 Then As logn(2) =1/log2(n) 6/log2(n) =log2(n) -5 Which simplifies down to (log2(n)) ^2 -5log2(n) -6=0 Which yields the solutions n=2^6 And n= 1/2
The trick of dividing by x is very nice but you end up with something very similar by just doing x^2 = 3x -1 square both sides x^4 = 9x^2 -2.3x + 1 and notice that you have an equation for 3x above: 3x = x^2 + 1 giving you x^4 = 7x^2 + 1 Now do that 2 more times and you get the answer. It's less elegant but doesn't require as big a flash of insight, I think.
Squaring 47 in your head. There is a cool way of working out squares of numbers between 41 and 59. 47 is 50-3 so 47^2 is (25-3)(100)+3^2=2209. Why does this work? (50+n)^2=2500+2(50n)+n^2= 2500+100n+n^2=(25+n)(100)+n^2. So for example 56^2=3136.
x^2 - 3·x + 1 = 0 x - 3 + 1/x = 0 x + 1/x = 3. We know that x^16 - k·x^8 + 1 = 0, so we want x^8 - k + 1/x^8 = 0, hence x^8 + 1/x^8 = k. The idea is to compute x^8 + 1/x^8 in terms of x + 1/x solely. This is doable, by considering the binomial theorem. This is the key to the question.
in the second question is there a generalized way to solve if, for example, it asked less than 500 or a larger number? how would we approach it? we surely can't go about counting every such pair...!?
Hey i have a question for you. This is from an IITJEE prep book by RD Sharma. Lo and behold: y'•y'"=3y"² I solved it (somehow lol) but later thought, "well doesn't mean that y', sqrt3 y, and y"' are always in GP?" That didn't quite work out well though. I urge you to do this!
#questions For the third problem, could you do a u-sub with let u = x^8? Also, I don’t get why we can’t just do the binomial expansion (x^2-3x)^8 = 1^8
Because you have a ton of terms such as x^15, x^14 etc with a binomial expansion of your expression. Yeah you could set up a system so that all the unwanted coefficients equal zero but at best it will be too messy and time conseuming and at worst it may not even work (I’m not gonna give it a try).
hi i have a calculus question that i really hope you'll answer because it's annoying me so bad when you take the derivative of ln x or ln 2x or any natural log of nx it'll always be 1/x right? so why when we integrate 1/x do we just say the integral is ln x and not some ln ax because it can be any constant multiplied by that x please answer and ty
Let me ask you a simpler question, the derivative of x^2 is 2x, right? But so is the derivative of x^2+1, and x^2+2, so why then is the integral of 2x equal to x^2 and not x^2 +a, the answer? It is, that's we have the +C in indefinite Integrals, similarly as we have the integral of 1/x is lnx +C, where C is any real numbers, which is also equivalent to adding ln(a) for any positive a
3rd question lai can\t just ake the two roots of the first expression, alpha and beta. so you get alpha = (3 +sqrt5) / 2 and beta = (3-sqrt5)/2 use substitution y = x^4 so the sum of roots of y = -k or ((3+sqrt5) / 2 )^4 + ((3-sqrt5)/2)^4 = 2207 or k =-2207
Who is the target participant for the SMT? I can see some good high school (grade 11 or so) students being able to do 1 and 3, but I don't think 2 would be suitable - maybe if "bijection" was replaced with "1-1 and onto".
Hi Shreeji. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
let's compute the solutions of 1st equation : x1 = (3+sqrt(5))/2 and x2 = (3-sqrt(5))/2. Then k = x1^8 + x2^8 = 2207. (directly, or by Girard-Newton formula)
Hi Gill. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
I did get all three, but made it harder than needed for #1 (took common log instead of log base 2, then had to substitute k=log4 and factor a multivariable quadratic expression before back-substituting) and for #3 (actually solved for x and found the 8th power--you don't get more and more terms, but you do get larger and larger coefficients--then compared that to the quadratic equation result for the 16th-degree polynomial).
For the last one, x² - 3x + 1 = 0 => x¹⁶ - kx² + 1 = 0, I just said x = y⁸, which means (y⁸)² - 3(y⁸) + 1 = 0 y¹⁶ - 3y⁸ + 1 = 0 since the symbols arbitrary, we can just replace y with x x¹⁶ - 3x⁸ + 1 = 0 k = 3
The symbols are not arbitrary though, since we have specific values of x that solve the quadratic. You can check that the roots of the quadratic don't satisfy the new equation you wrote.
But for the second question, can we rule out that f(1) =/= 1^2? Since it can go either way, isn't it a "maybe"? I would have written 4, and 5 if you count N = 1. But maybe I'm just stupid.
Definition of Injection: when all elements in the domain of a function f(x) has a unique image in its codomain. Ex: f(x)= x^3. Every x in the set of all real numbers give unique values of f(x). But say a function like f(x)= x^2 for R->R is not an injection since x=1 or x=-1 both give the same values of f(x) which is 1. Surjection: when all elements in the codomain of a function f(x) have preimages. Ex: f(x)= x+2 is a surjective function since all values of f(x) are defined for some x.
I think it means that, for a certain function, every one input can only have one unique output. For example: for the function y=x^2, 3^2 is nine, but so is (-3)^2, so it's not a bijection. Visually, if a function is bijective, then it passes the horizonal line test (it's like vertical line test, but horizontal)
Hi! I'm the organizer for SMT 2022! I'm glad you enjoyed our tiebreaker problems, and thank you for reviewing them!!
For those of you who are high school students we encourage you to participate in this year's iteration of the contest, we're very excited :)
Wow, great to have you here! As I said in both of my SMT videos, you guys have some really nice problems. Keep up the great work!
@@blackpenredpen amogus
@@theuserings no
@@kepler4192 👽
Yes, we would like to.
For the second question, removing all of the high level notation you could say, how many numbers less than 40 are perfect squares but not perfect fourth powers.
Yes!!! That’s a very simple way to interpret it. The hard part was definitely all the fancy math language.
@@blackpenredpen Thanks. Also to finish the interpretation, it comes down to finding how many perfect squares there are below 40 and how many prefect fourth powers below forty and subtracting the later from the former
@@cosmicvoidtree Alternatively, you can find/generate a list of squares of numbers that are not squares themselves. This works for n≤40 because we don't really have to deal with sixth powers and etc.
Wonderful insight! Nice.
Luckily there are no perfect 8th powers less than 40 or what you have said would not hold. Eg. 256 is 4^4 but it would satisfy the condition (f(256) can = 16 as f(4) = 2) perfect 16th powers don't hold as f(65536) != 256 since that is f(16). Will hold for 32nd powers but not 64th and so on.
lol ever since i picked up dummit and foote the only things that come to mind whenever i see or hear "algebra" are rings, groups, fields, modules etc.
@@shamcallado8947 haha yea
I am from India , i study in 12th standard. when i study integration, i have some questions having a problem then i search in TH-cam then i show your channels good working teachers. I really motivated your teaching skills , i love mathematics 🥰🥰🥰🥰🙏🙏🙏🙏🙏🙏🙏🙏
That was a fun one!! It's so cool to see problems that I initially have no idea how to approach, solved with such a simple, easy-to-understand technique! Thanks!
My 6th graders would probably wonder why lumber is involved in these 'log' math problems.
As a 6th grader, our math club took the questions from the Stanford-Math League Tournament. Even though our group was comprised of 6th, 7th and 8th graders, I ended up winning with a t-shirt which has “Stanford-Math League Tournament”. I’ve already learned logarithms and even some calculus.
@@thegoldlightning 🏅
@@thegoldlightning wut
@@thegoldlightningwait 8th graders don't learn log and calc yet?
@@windowsxpmemesandstufflol a lot of people there are in the top advanced program at our school. Some do even higher outside.
I don’t know if I ever got one right but I remember seeing questions like the last one all the time in these extracurricular math competitions lol
The idea of dividing by x to get a binomial where 2ab removes the variable… so simple yet so crazy to come up with if you’re not used to these types of approaches
For the last one, I just multiplied both sides by x^2 + 3x + 1, which cancels out all the third and first degrees, and gives x^4 -7x^2 + 1 = 0. Rinse and repeat twice more. Basically the same process without dividing by x first.
Another SMT problem (rotating y=x^2) : th-cam.com/video/gYAQg7xn-Xo/w-d-xo.html
cant you just do the quardtaic formula for the last one and then plug in x?
@@yodaimpostor4781 if I'm assuming what I understood from that
I got (3+-√5) ÷ 2 for the first equation
I wouldn’t be able to solve these in a half hour but you just make it looks so easy and make so much sense. Ty!
I really enjoy your channel. I don't understand half of what you are saying, but it stretches my brain in the right direction.
Very fun!
I did the first the same!
For the second, I said that f(a)^2 = a can only be true if a is a perfect square, so a = 1,4,9,16,25,36.
For a = 1, f(1) = 1^2, so we can get rid of that. And then 16 is the square of a square, so f(2) = 4 but f(16) = 4 breaks the bijectivity of f. No other is a 4th power so f(a) ≠ a^2 for a = 4, 9, 25, 36.
For the third, I used the quadratic formula! Since x^2 - 3x + 1 = 0, we have x = (1/2)(3 ± sqrt(5)). And since x^16 - kx^8 + 1 = 0 is a quadratic in x^8, we get that x^8 = (1/2)(k ± sqrt(k^2 - 4)).
raising (1/2)(3 ± sqrt(5)) to the 8th power (using the binomial theorem) gives (1/2)(2207 ± 987sqrt(5)) so
(1/2)(2207 ± 987sqrt(5)) = (1/2)(k ± sqrt(k^2 ‐ 4)), which tells us k = 2207.
For the 3rd question, you can also notice that the k = sum of the roots of the equation. Then, sum the two roots of the staring equation (found via quadratic formula), each to the power of 8. This will be equal to k.
Yes you can compute the sum of the 8th power of the roots using a similar method of squaring, adding/subtracting repeatedly. But, bprp's method is quite elegant in a way.
this man explained the entire paper within 15 minutes time limit lol
The best almost 15 mins. spent today. Thank you for the video.
Hi Jay. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
I love the first question, it helps my understanding for logs tats coming for my mocks, thanks!
If i was in a math class and the 2nd question was in the test... I'd have passed out 😂
Edit: how you tackled the 3rd problem was satisfying. No lie.
That is the kinda problems I, as a math major, get asked as homework.
the second one was by far the easiest imo
@@bonjour7209 definitely.
That 3rd problem was amazing!!
I am glad that I was able to do the 3rd problem 😁
In my highschool in Poland we did problems like the first one on daily basis when we had logarythms in 11th grade but instead of doing it by factoring it out we did t=log2(n) then solve delta and substitute t1 and t2 and then solve for log2(n)
Nice approach to solve problems 👍
I see that a lot of these, high school juniors could solve, If they had the knowledge, I tried the problem with all the solutions less than 40, I actually did it and it was more logic than anything in my opinion
BPRP: *Points at 9*
Also BPRP: "Five"
9:24
😂
All three are really great problems!
More math competition thanks Prof)
Your channel is truly awesome! I just wish you diversify the kind of problems you solve to include more stuff outside calculus :) Rock on!
UC Berkeley ?? That’s awesome
I tried doing it in 15 minutes and messed everything up with haste xD. Thank you for the solving methods exposed in this video.
Satisfying explaination ..
I liked the 3rd question and the 1st question a lot.
I don't know why I'm smiling on entire video, especially on the last part. Please help me lol.
i love knowing every step used in the video but not being able to string them together to get a solution:))))))))))
In the first problem:
If you simply substitute “x” for “log (base 2) n”, you get (3/x) - (x/2) = (-5/2)
Solve the quadratic and plug back in the 2 solutions (x = -1 & 6), setting them each = to “log (base 2) n”
Same answer in the end. Just a less complex approach. You can actually solve it just by looking at it. It’s pretty clear what X is equal to. Regardless, more than one way to skin a cat. Per usual with the maths. 🤘😝🤘
this isn't a different way to do the problem. it's literally the same exact thing except you let x = log_2(n). it doesn't save any time
@@natebobdog23 “EXCEPT you let x = log_2 (n)”. THATS the difference. Same answer-yet the slightest of variety in approaching and yielding that same solution.
Your comment just defeated itself. Congrats. 👏👏👏
You also conveniently neglected the end of my original statement. Conveniently. No wonder you’re struggling over there.
But hey, you’re learning! Good for you! I’ll be right here if u ever need me. ✌️
I looked at the problem and thought u-sub by defining int(f(x)dx)=F(x). Pretty much the same thing you did but you get nice algebraic steps
Alternative Solution to #3:
Assume a and b are the root of x² - 3x + 1 = 0. Then a and b must also be a root of x¹⁶ - kx⁸ + 1 = 0.
So,
a¹⁶ - ka⁸ + 1 = 0
b¹⁶ - kb⁸ + 1 = 0
--------------------------- -
(a⁸ + b⁸)(a⁸ - b⁸) - k(a⁸ - b⁸) = 0
k = a⁸ + b⁸ = 2207
That’s beautiful!
How did you get 2207 in the last step? Do you use quadratic formula to find a and b?
@@topgearfan2596 a and b were by assumption the roots of the first quadratic, so you just have to solve that one
@@Prxwler Yeah I get it. I feel like calculating that is more tedious than the method presented in the video.
that last question was satisfying
In the second problem, it's given that f is bijective. But I wonder if it makes sense to think about how you would prove such a function would actually be bijective.
Sure. I imagine it would be a pretty basic proof by cases. "Suppose f(a) = b. Then either b = a^2 or b = sqrt(a)." Follow from there.
Consider any sequence n, n², n⁴, n⁸..., where n itself is not a square. N is a disjoint union of such sequences, so it suffices to prove that f is a bijection on each sequence separately. On any such sequence the function is given by f(n^(2^k)) = n^(2^(k+1)) if k is even, n^(2^(k-1)) if k is odd. From this it is clear that f(f(x)) = x for each x, so f is its own inverse and it is bijective.
Wah new look 🤘🤘
Takk!
Fiat Lux! As you were doing the question I thought in my head to make a common denominator on the left side and see that lon_2{n}=1. I was stunned the answer didn't line up. But then I saw you didn't switch the sign on when you moved 5log_2{n} to the other side. Maybe that will get my answer of n=1 and there is probably another one too. PS I made it to Cal because of you and Professor Leonard. TYSM
I did the first one another way, probably more complicated:
() does not represent the base
First, we can write the eqn as
1/log2^3(n) + 1/logn(2^-2) = -5/2
As logn(2^-2) =-2logn(2),,,
And log2^3(n) =1/3log2(n)
We get 3/log2(n) - 1/2logn(2) = -5/2
Then
As logn(2) =1/log2(n)
6/log2(n) =log2(n) -5
Which simplifies down to
(log2(n)) ^2 -5log2(n) -6=0
Which yields the solutions n=2^6
And n= 1/2
Solved this equation in a similar way to yours. Thought it'd be harder, but that's school grade maths
But notice that solution n=1/2 is not integer, so this equation has only one solution n=2^6=64
Not -1/2, but 1/2
The trick of dividing by x is very nice but you end up with something very similar by just doing
x^2 = 3x -1
square both sides
x^4 = 9x^2 -2.3x + 1
and notice that you have an equation for 3x above: 3x = x^2 + 1
giving you x^4 = 7x^2 + 1
Now do that 2 more times and you get the answer.
It's less elegant but doesn't require as big a flash of insight, I think.
Quick tip you can use the asterisk symbol instead of decimal point for multiplication like 2*4
Squaring 47 in your head. There is a cool way of working out squares of numbers between 41 and 59. 47 is 50-3 so 47^2 is (25-3)(100)+3^2=2209. Why does this work? (50+n)^2=2500+2(50n)+n^2= 2500+100n+n^2=(25+n)(100)+n^2. So for example 56^2=3136.
Beautiful!!
x^2 - 3·x + 1 = 0 x - 3 + 1/x = 0 x + 1/x = 3. We know that x^16 - k·x^8 + 1 = 0, so we want x^8 - k + 1/x^8 = 0, hence x^8 + 1/x^8 = k. The idea is to compute x^8 + 1/x^8 in terms of x + 1/x solely. This is doable, by considering the binomial theorem. This is the key to the question.
for the 3rd question, cant you find out the value of x from the quadratic and then substitute in the 2nd eq?
Make a video on how to get the value of d/dx[erf(x)]
in the second question is there a generalized way to solve if, for example, it asked less than 500 or a larger number? how would we approach it? we surely can't go about counting every such pair...!?
yeah just find numbers that have natural square roots but not natural fourth roots
so take the number of squares and subtract the number of tesseracts
Beautiful, that was quite fun. Well explained, and funny too
Love from india🇮🇳🇮🇳🇮🇳🇮🇳
Hey i have a question for you. This is from an IITJEE prep book by RD Sharma. Lo and behold:
y'•y'"=3y"²
I solved it (somehow lol) but later thought, "well doesn't mean that y', sqrt3 y, and y"' are always in GP?" That didn't quite work out well though. I urge you to do this!
Amazing
8:24 this should be a meme! (The pause)
You should try to differentiate the function f(x)=x!
Thank you. guide me about: Integral (e^-x)/x Thank you
I solved only first one log question and rest two no idea
But i. Amazed 😮😲 by seeing solution of 3 rd one
#questions For the third problem, could you do a u-sub with let u = x^8? Also, I don’t get why we can’t just do the binomial expansion (x^2-3x)^8 = 1^8
Because you have a ton of terms such as x^15, x^14 etc with a binomial expansion of your expression. Yeah you could set up a system so that all the unwanted coefficients equal zero but at best it will be too messy and time conseuming and at worst it may not even work (I’m not gonna give it a try).
Proud of you sir i also want to be an mathematician like you👍👍👍👍🇮🇳🇮🇳🇮🇳
First was nice bro.😍
I am very glad while you change blackpenredpen
Another excellent video! Thanks Professor!
Cool way to do the last one.
Thanks.
I knew there had to be a more elegant solution to 3 than just solving the quadratic and making the second expression an equation for k
best math sam a reader is convert is it
sam math best
Try solving ln(i)
hi i have a calculus question that i really hope you'll answer because it's annoying me so bad
when you take the derivative of ln x or ln 2x or any natural log of nx it'll always be 1/x right? so why when we integrate 1/x do we just say the integral is ln x and not some ln ax because it can be any constant multiplied by that x
please answer and ty
bc of the +C
^ This. I was going to say, "We already do!"
Let me ask you a simpler question, the derivative of x^2 is 2x, right? But so is the derivative of x^2+1, and x^2+2, so why then is the integral of 2x equal to x^2 and not x^2 +a, the answer? It is, that's we have the +C in indefinite Integrals, similarly as we have the integral of 1/x is lnx +C, where C is any real numbers, which is also equivalent to adding ln(a) for any positive a
In second question 1 can be taken in both a² or √a form ...so the question should have mentioned least number of numbers less than 40
The question is fine...
Tutorial on partial differentiation please🥺
Can you integrate (1/dx) ?
that's not an integral. dx isn't a term to be wielded arbitrarily
Very good
3rd question lai can\t just ake the two roots of the first expression, alpha and beta. so you get
alpha = (3 +sqrt5) / 2 and beta = (3-sqrt5)/2 use substitution y = x^4 so the sum of roots of y = -k or
((3+sqrt5) / 2 )^4 + ((3-sqrt5)/2)^4 = 2207 or k =-2207
Sir plz have some problem from inmo
Who is the target participant for the SMT? I can see some good high school (grade 11 or so) students being able to do 1 and 3, but I don't think 2 would be suitable - maybe if "bijection" was replaced with "1-1 and onto".
Oh I am lovin' it!
Hi Shreeji. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
2:47 Shouldn't it be 5(log2(n)) instead of -5(log2(n)) because you're shifting to the other side?
nah the -5log2(n) wasnt the one he was shifting, he shifted -(log2(n))² and 6
A very nice sol !
The first two problems are somewhat doable, but I would absolutely have no chance tackling the last problem.
For question 1, that was a 5? I thought that was an S, and figured they were asking about arithmetic progressions or something
let's compute the solutions of 1st equation : x1 = (3+sqrt(5))/2 and x2 = (3-sqrt(5))/2. Then k = x1^8 + x2^8 = 2207. (directly, or by Girard-Newton formula)
I didnt know Saitama took over this channel
Busy man. He's been doing videos for CalisthenicMovement for a long time now.
I like #3
Me too
Awesome!
can we use the quadratic formula on the 3rd question?
yes but what a nightmare that would be
you should get marks for simply understanding question 2
Is there an even quicker way to solve the 2nd one and if the 40 was replaced by 400 or something ........just wanted to ask ........pls reply
Sorry but 6*^ isn't it 36? I did not fully understand the step from logn2 to finding n. Thank you for your answer
If log_2(n) = 6, then n = 2^6, not n = 6^2.
6:17 why isn’t zero in there? 😢
4:48
To do 47^2 in your head quickly, try this: (50-3)^2 = 2500 - 300 + 9 =2209
You are amazing
Hi Gill. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
You look like a math master just came down from the mountain or something XD.
I did get all three, but made it harder than needed for #1 (took common log instead of log base 2, then had to substitute k=log4 and factor a multivariable quadratic expression before back-substituting) and for #3 (actually solved for x and found the 8th power--you don't get more and more terms, but you do get larger and larger coefficients--then compared that to the quadratic equation result for the 16th-degree polynomial).
Wow, outside my country everyone is on different level
i put my bets on this dojo.
For the last one, x² - 3x + 1 = 0 => x¹⁶ - kx² + 1 = 0, I just said x = y⁸, which means
(y⁸)² - 3(y⁸) + 1 = 0
y¹⁶ - 3y⁸ + 1 = 0
since the symbols arbitrary, we can just replace y with x
x¹⁶ - 3x⁸ + 1 = 0
k = 3
The symbols are not arbitrary though, since we have specific values of x that solve the quadratic. You can check that the roots of the quadratic don't satisfy the new equation you wrote.
Dang, I was lost as to what that second question had been asking.
that was fun
What was your major .?
But for the second question, can we rule out that f(1) =/= 1^2? Since it can go either way, isn't it a "maybe"? I would have written 4, and 5 if you count N = 1. But maybe I'm just stupid.
Huh? For a = 1, we obviously have f(a) = a². So this is not a number we have to count when looking for all numbers for which f(a) != a².
I'll start with a simple question, what's a bijection?
It is when a function is both Injective and surjective
Definition of Injection: when all elements in the domain of a function f(x) has a unique image in its codomain. Ex: f(x)= x^3. Every x in the set of all real numbers give unique values of f(x). But say a function like f(x)= x^2 for R->R is not an injection since x=1 or x=-1 both give the same values of f(x) which is 1.
Surjection: when all elements in the codomain of a function f(x) have preimages. Ex: f(x)= x+2 is a surjective function since all values of f(x) are defined for some x.
I think it means that, for a certain function, every one input can only have one unique output.
For example: for the function y=x^2, 3^2 is nine, but so is (-3)^2, so it's not a bijection.
Visually, if a function is bijective, then it passes the horizonal line test (it's like vertical line test, but horizontal)
Nice one
Sir. Kindly give me answer- which University is the best for doing msc in mathematics (abroad) and may i know from where you had passed out