In that final line, the middle term -Nx^(N-1) from the evaluated function seems to have been lost. If you evaluate the final formula for small n you find that it does not match the expected results. For instance, when n=2, N=1 and the formula result is 3 when the expected result is 1. Also, further simplification can be done between the result from the evaluated function and the extra term N(n-2^N+1) The result I get is (n+1)N-2^(N+1)+2 or, regrouped a bit, (n+1)N-2(2^N-1)
My math work = watching your content + professor Penn's content + SyberMath's content. I'll stop the video at the beginning and try to figure out what tools you're going to use.
Instead of tricky calculus derivatives you can change k multiplier into summation over some v from 1 to k, then exchange summation order which will give you sum over v of geometric series, which will surprisingly end up being geometric series itself.
Hey Michael! I'm from Brazil and I follow your channel here in UA-cam. I don't lose any video you post here and I love your explanations about advanced Math.
Nice! Another way to solve this is to use the lattice-point counting method you covered in an earlier video, the discrete version of integration by parts.
Thank you for this beautiful solution 👍👍👍 , which is further Sn= N(n+1)+2(1-2^N). I enjoy all your Math-videos. Greetings and best wishes from Munich - Wolfgang 😊😊😊
Doesn't you need a 'proposition' for the closed form for the series first of all for doing an induction to prove on? I don't see how can we just see the formula out of nowhere and do the induction. (Or maybe I just suck)
@@ornithology11 But you still need to know the formula before induction. You may guess it or simply you know the formula itself first of all but nonetheless you still need the formula which not all mathematicians will remember. No offense tho.
@@aidancheung7264 No offense, you are right. I was familiar with this sum because it was one of my “personal” induction proof when I was a toddler and I was very fixated on perfect numbers. Being a childhood memory I was convinced it was a fairly well-known lemma, but it actually took me a while to find a reference online now that you asked me for it. What is right is right.
a fun thing is that instead of explicitly working in terms of x^k for x = 2 one can just treat constants as variables and blow people's minds by saying that, e.g., d(2^k)/d2 = k·2^(k-1) -- because our symbols mean what we say they mean, so the symbol "2" can be a variable that we set equal to the value 2 when convenient
@@coreyyanofsky But if for example the expression was x^(2k) for x = 2, your method would get ambiguous. You’d be differentiating x^(x k), which is incorrect, or you’d have to remember that the 2 in the exponent is not the variable wrt to which you’re differentiating.
@@clusterforkit is true that one shouldn't use the same symbol to refer to things that can vary independently if one wants correct partial derivatives, but that's a general principle
In primitive Pythagorean triples, a or b must be a multiple of 3. Why? Since all three are squared, by Fermat's little theorem, a ^ 2 + b ^ 2 = C ^ 2 ... Is. 1 +1 = 1 Mod 3 ... Absurd. Unless a or b are multiples of 3. 1+ 0 = 1 ... So a or b has to be multiples of 3 ... No problem.
i don't understand how you can move from a sum to a differential. explain how🙏🏻, i didn't understand a word in the video because I don't know English. but I know about mathematics
May i suggest one problem to you, I've been thinking about it for 2 days but i can't wrap my head around it. The problem says that i need to calculate: I=∫ sinx/(3sinx+4cosx) dx and J=∫ cosx/(3sinx+4cosx) dx.I calculated 3I+4J but i don't know what should be the next step in solving this problem.I would appreciate any kind of answer from anybody.
Since you have two unknowns and one equation, you'll need one more equation. The key is to put the integrand in the form ∫ f'(x)/f(x) dx. So calculating -4I+3J you can get a closed form for the integral.
Engraçado você vir parar aqui. Na qualidade de ensino atual do Brasil, eh matemática de ensino superior. Não era pra aparecer pra você. Deve ter bastante brasileiro vendo isso. Kkkk
You've lost middle term in (N-1)x^N - N x^(N-1) + 1, when wrote last row, so the answer is not correct The correct answer is simpler: S_n = N n + N + 2 - 2^(N+1) The sum i*2^i could be found without derivatives: S' = sum i*2^i = sum (i-1)*2^i + sum 2^i = 2 sum (i-1)*2^(i-1) + sum 2^i sum (i-1)*2^(i-1) = sum j*2^j (other range of summation index than S'), that can be expressed from S'
HOMEWORK : Let f: [0,∞) → R be a differentiable function with |f(x)| ≤ M and f(x)f′(x) ≥ cos(x) for x ∈ [0,∞), where M > 0. Find the limit of f(x) as x → ∞ OR prove this limit doesn’t exist. SOURCE : The 23rd Annual Vojtěch Jarník International Mathematical Competition
SOLUTION *f(x) does not have a limit* Consider a function F: [0,∞) → R given byF(x) :=f²(x) − 2sin(x). Then: - |F(x)| ≤ f²(x) + 2|sin(x)| ≤ M + 2 - F′(x) = 2f(x)f′(x) − 2cos(x) ≥ 0. Hence we infer that F is increasing and bounded. Let x_n equal to: - nπ if n = 2k-1 - nπ + π/2 if n = 2k Then (F(x_n)) is increasing and bounded and hence convergent. Assume that lim f(x), as x → ∞, exists. In turn, this implies that lim f²(x_n), as n → ∞, exists. Thus the sequence F(x_n)−f²(x_n) is convergent. But F(x_n) − f²(x_n) = −2sin(x_n). Consequently we get that the sequence (sin(x_n)) is convergent. This contradicts the fact that (sin(x_n)) is not convergent since sin(x_n) = 0 if n is 2k-1, and 1 if n is 2k.
A "nice" function (also called an elementary function): A function that is closed under arithmetic operations (+, -, ×, ÷), exponents, roots, logarithms, and trigonometric and hyperbolic functions (along with their inverses). Most of the time, it's composed of more than one function.
In that final line, the middle term -Nx^(N-1) from the evaluated function seems to have been lost. If you evaluate the final formula for small n you find that it does not match the expected results. For instance, when n=2, N=1 and the formula result is 3 when the expected result is 1.
Also, further simplification can be done between the result from the evaluated function and the extra term N(n-2^N+1)
The result I get is (n+1)N-2^(N+1)+2 or, regrouped a bit, (n+1)N-2(2^N-1)
You are right, I found exactly the same result.
Yes, correct!
I would like how know how much math do you read on a daily basis.
For me it’s very little 😆
Hi BPRP!!!
Yay, I knew Michael would eventually draw the attention of my other favorite Math youtuber ever since I found his channel in early 2020.
I find your dexterity more impressive 😆
My math work = watching your content + professor Penn's content + SyberMath's content. I'll stop the video at the beginning and try to figure out what tools you're going to use.
Happy Sunday from Nigeria 🇳🇬
Happy Sunday! 🙌❤️
Happy Sunday from Brazil
13:00 "And that's a good place to stop".
Oops you didn't stop lol
Instead of tricky calculus derivatives you can change k multiplier into summation over some v from 1 to k, then exchange summation order which will give you sum over v of geometric series, which will surprisingly end up being geometric series itself.
Corrected partial sum: -2^(N+1) + N(n+1) + 2
Hmmm yes the floor is made out of log.
Jack HandMa: Very very very clever!
Hey Michael! I'm from Brazil and I follow your channel here in UA-cam. I don't lose any video you post here and I love your explanations about advanced Math.
Nice! Another way to solve this is to use the lattice-point counting method you covered in an earlier video, the discrete version of integration by parts.
May be you could use induction for this: how much does Sn increase at each step (Sn -Sn-1)
13:01
Thanks for another video. I enjoy them and learn a lot from you.
Thank you for this beautiful solution 👍👍👍 , which is further Sn= N(n+1)+2(1-2^N). I enjoy all your Math-videos. Greetings and best wishes from Munich - Wolfgang 😊😊😊
Every time you mentioned the seventh sum in the first part I couldn't help but start singing Iron Maiden along with it!
Lol, had the exact same thing!
9:06 you can use induction for the last sum evaluation, no geometric series needed.
Doesn't you need a 'proposition' for the closed form for the series first of all for doing an induction to prove on?
I don't see how can we just see the formula out of nowhere and do the induction. (Or maybe I just suck)
@@aidancheung7264 exactly
@@aidancheung7264 that sum it's just the derivative of this sum math.stackexchange.com/a/784252 proved by induction.
@@ornithology11 But you still need to know the formula before induction.
You may guess it or simply you know the formula itself first of all but nonetheless you still need the formula which not all mathematicians will remember.
No offense tho.
@@aidancheung7264 No offense, you are right. I was familiar with this sum because it was one of my “personal” induction proof when I was a toddler and I was very fixated on perfect numbers.
Being a childhood memory I was convinced it was a fairly well-known lemma, but it actually took me a while to find a reference online now that you asked me for it. What is right is right.
Nice work professor Penn!
We can simplify a bit, as the answer can be expresad as 2(n-1)2^n + 2 + n^2 - n 2^n + n = (n-2)2^n + n^2 + (n-2)
Note difference between the N and n terms: n is the Sum's term number, while N=floor(log-base2(n)):
S(n) = 2( (N-1)2^N - N.2^(N-1) + 1) + N(n - 2^N +1)
= (2N-2)2^N - N.2^N + 2 + N.n - N.2^N +N
= (-2)2^N + 2 + N.n +N
= -2^(N+1) + 2 + N.n + N
= 2 + N(n+1) - 2^(N+1)
E.g. S(7) = 2 + 2(7+1) - 2^(2+1) = 2+16-8 = 10 i.e. S(7) = floor(log-base2(k)) for k=1 to 7 = 0+1+1+2+2+2+2
I think in the final answer we miss a -N * 2^N
I agree
En el paréntesis de la respuesta va
N-(N/2)-1=(N/2)-1 es decir la respuesta es
((N/2)-1)2^(N+1)+N(n-(2^N)+1)+2
ok so it is nice because it is of basis 2, how would you do with natural log for example?
I just noticed that he forgot to distribute the 2 to the 2nd term in the 1st set of parentheses.
The result after simplification is N + Nn - 2^(N+1) + 2
kinda similar to the sum of the floor of sqrt(k)
Michael's few = infinite favorite things.
Thank you, professor!
You missed a term in the last step that will also cancel another term
a fun thing is that instead of explicitly working in terms of x^k for x = 2 one can just treat constants as variables and blow people's minds by saying that, e.g., d(2^k)/d2 = k·2^(k-1) -- because our symbols mean what we say they mean, so the symbol "2" can be a variable that we set equal to the value 2 when convenient
Uhh that's a bit fishy, I wouldn't get into the habit of that
@@clusterfork no it's fine actually, there's nothing magic about the Arabic numerals that makes them unsuitable for symbolic manipulations
@@coreyyanofsky But if for example the expression was x^(2k) for x = 2, your method would get ambiguous. You’d be differentiating x^(x k), which is incorrect, or you’d have to remember that the 2 in the exponent is not the variable wrt to which you’re differentiating.
@@clusterforkit is true that one shouldn't use the same symbol to refer to things that can vary independently if one wants correct partial derivatives, but that's a general principle
1st one was AGP
WTH was that last bit? Are we going to get outtakes now? I would really like that.
In primitive Pythagorean triples, a or b must be a multiple of 3. Why?
Since all three are squared, by Fermat's little theorem, a ^ 2 + b ^ 2 = C ^ 2 ...
Is.
1 +1 = 1 Mod 3 ... Absurd. Unless a or b are multiples of 3.
1+ 0 = 1 ... So a or b has to be multiples of 3 ...
No problem.
10:01 Interchange of summation and derivative?
it’s a finite sum so it’s allowed without restriction
i don't understand how you can move from a sum to a differential. explain how🙏🏻, i didn't understand a word in the video because I don't know English. but I know about mathematics
There is a mistake on thumbnail: sum is not to infinity
I think it's not a mistake but clickbait
Sum of my favorite things
WHOOOOO!!! That was a dooozy!!!
May i suggest one problem to you, I've been thinking about it for 2 days but i can't wrap my head around it. The problem says that i need to calculate:
I=∫ sinx/(3sinx+4cosx) dx and J=∫ cosx/(3sinx+4cosx) dx.I calculated 3I+4J but i don't know what should be the next step in solving this problem.I would appreciate any kind of answer from anybody.
Since you have two unknowns and one equation, you'll need one more equation. The key is to put the integrand in the form ∫ f'(x)/f(x) dx. So calculating -4I+3J you can get a closed form for the integral.
@@elephantdinosaur2284 Thanks a lot
Actually, the final answer should be N(n+1)+2-2^(N+1).
I don’t understand the part why log2 3 is equal to 1
log_2(3) is between 1 and 2. floor( log_2(3) ) is equal to 1 because the floor function rounds the input down to the nearest integer.
The thumbnail should be corrected, it's an obviously divergent series
Nice syllogism (Σ log m)
ohh
The final answer should be
N*(n+1)-2^(N+1).
+2
Detesto matematica fiz varios concurso e nao passei
Engraçado você vir parar aqui. Na qualidade de ensino atual do Brasil, eh matemática de ensino superior. Não era pra aparecer pra você. Deve ter bastante brasileiro vendo isso. Kkkk
You've lost middle term in (N-1)x^N - N x^(N-1) + 1, when wrote last row, so the answer is not correct
The correct answer is simpler: S_n = N n + N + 2 - 2^(N+1)
The sum i*2^i could be found without derivatives:
S' = sum i*2^i = sum (i-1)*2^i + sum 2^i = 2 sum (i-1)*2^(i-1) + sum 2^i
sum (i-1)*2^(i-1) = sum j*2^j (other range of summation index than S'), that can be expressed from S'
HOMEWORK : Let f: [0,∞) → R be a differentiable function with |f(x)| ≤ M and f(x)f′(x) ≥ cos(x) for x ∈ [0,∞), where M > 0. Find the limit of f(x) as x → ∞ OR prove this limit doesn’t exist.
SOURCE : The 23rd Annual Vojtěch Jarník International Mathematical Competition
SOLUTION
*f(x) does not have a limit*
Consider a function F: [0,∞) → R given byF(x) :=f²(x) − 2sin(x). Then:
- |F(x)| ≤ f²(x) + 2|sin(x)| ≤ M + 2
- F′(x) = 2f(x)f′(x) − 2cos(x) ≥ 0.
Hence we infer that F is increasing and bounded. Let x_n equal to:
- nπ if n = 2k-1
- nπ + π/2 if n = 2k
Then (F(x_n)) is increasing and bounded and hence convergent. Assume that lim f(x), as x → ∞, exists. In turn, this implies that lim f²(x_n), as n → ∞, exists. Thus the sequence F(x_n)−f²(x_n) is convergent. But F(x_n) − f²(x_n) = −2sin(x_n).
Consequently we get that the sequence (sin(x_n)) is convergent. This contradicts the fact that (sin(x_n)) is not convergent since sin(x_n) = 0 if n is 2k-1, and 1 if n is 2k.
Hi , after a long time
@@pardeepgarg2640 Hello 👋
Are you eautistic? xD
A "nice" function (also called an elementary function):
A function that is closed under arithmetic operations (+, -, ×, ÷), exponents, roots, logarithms, and trigonometric and hyperbolic functions (along with their inverses).
Most of the time, it's composed of more than one function.
The mistake is at the calculation in the end after he plug in x=2
Question itself in nice form. The answer is in ugly form
You should have ended this video of some of your favourite things with a backflip.
I think it should be -N(n-2^(N+1)+1)
I’m in year 9 and I got a C in maths why am I here
The mistake is that the sum( k2^k) is (N-2)2^N +2
Sum of these series is clearly infinity, no matter how much you try to prove the otherwise....
The correct answer is S=(N-1)2^(N+1)+2+N(n-2^(N+1)+1).
NICE PROBLEM
How are your comments from 2 weeks ago?
@@Popcorn31416 this video was uploaded to his playlist the floor is lava 2 weeks ago
@@Popcorn31416 he is time travelling