1+2+3+...+n = ½n(n+1) So what you have is 6n·2/n(n+1) = k k,n being nat. nos., and n > 0 to avoid division by 0. Continuing, 12/(n+1) = k So n+1 can = {2,3,4,6,12}; consequently, n = {1,2,3,5,11}, and ∑n = 22
9:16 is there some theorem about the sum of all the factors of numbers, given the number has certain properties, either that its factors are obscenely numerous or specifically minimal for the local group of numbers, or that it is a power of a certain number etc?
Solved in my head in around 1 min, one good thing about this question is that it is asking for the sum of all possible n but not listing them out. Which you won't argue if 0 counts while it will make the denominator 0 because it won't change the result.
I solved this also in my head maybe in two minutes, triangular numbers, complex fraction and factors of number 12. To easy. I'm not math freak. I just like and use damn thing.
I'm not even sure what the original question was supposed to be? Brilliant doesn't seem to know what "Let" means or how quantification works. I mean, you probably did choose the right interpretation, but the statement is bad. "What is the sum of all natural numbers n such that is an integer." They just threw "Let" at the front to make it sound more mathy, but it makes the statement literally incoherent, but with an interpretation that makes sense.
I don't get the connection between the little Gauß and the binomial coefficients. Does this always work when the sums "level down" reach +0,+0,+0,...? And why?
From Brilliant.org....this was fun! Let G(n) be the expected value of the geometric mean of real numbers independently and randomly chosen between 0 and 1. What is Lim G(n) as n goes to infinity?
Isn't it 1-1/e? My math is rusty af, but isn't this argument correct? - If you divide the range by n and you pick a random number you have 1-1/n probability of not picking it. The intersection of events not picking it gives you the probability p of not picking it. As n aproach infinity 1-p is the probability of picking which is 1-1/e. If you multiply the probability by the span you get the average value 1-1/e?
that is simply to denote that the sum goes to a certain number n... not the fact that n>3... only criteria is n is natural number which includes 1,2,3 also if you were to exclude it, you'll be left with 16 not 12
Jimmy Actually, the answer won't even be complex: it will simply be "false". No complex solutions. An inconsistent equation. Because by definition, the absolute value of any complex number must be a non-negative real number. (big-@ss explanation and side-question ahead:) tbh though, I actually stumbled at it when I read it, so it's not a bad question at all. But when I gave it a go, I realized the equation |x|=-1 contradicts the very definition of the absolute value function: *given a complex number x of the form a+bi, |x|=√(a²+b²).* Since a & b are both real numbers (yes, bi is the imaginary component, but b alone is a real number all byself), a² and b² will both be non-negative real numbers. So if you add them together and square root that sum, you will end up with another non-negative real number. That result is also the result that you get from solving fit the absolute value of a complex number: that very same non-negative real number. So for any complex number x, |x| can never be equal to -1, nor could it even be equal to +i or -i. It could be any non-negative real number, from zero onwards, but it can only be a non-negative real number. No exceptions, by definition. Definitely a good problem to think about though, thanks for asking. That said, I can't say the same applies to quaternions, octonions, etc., nor do I know if the absolute value function is even defined for elements of those sets of numbers tbh, but I hope someone else can tackle that one and let us know 👍
Can someone show me how to solve this ODE ? 4y'' + y = 2*sec(t/2) I tried 10x and I did not find the correct answer. If possible, step by step solution.
I wanted to say pretty much as Kenneth has said. A similar question and solution has been posted on Mathforum. The solution to y" + 4y = sec(x) can be found here (I don't think a link I post will be visible): mathforum (dot) org / library / drmath / view / 52117 (dot) html remove all spaces and replace (dot) with . to get the link and the solution.
I don't understand how, from the sum of all natural numbers, you end up with those binomial coefficient. I understand the equality, it is easy to verify, but i don't understand how you came to that conclusion.
Zero is a whole number, not a natural number. You can't have k=12, n=0. Since you are summing the ns, it doesn't impact your final answer to add an extraneous zero, but it is still incorrect.
This was not a terribly difficult series problem on brilliant but still an interesting one, reminded me of Max's work on your channel a couple months ago. brilliant.org/practice/sequences-and-series-level-1-challenges/?problem=a-calculus-problem-by-danish-ahmed-3
True, n=0 is not valid, but *only* because 0÷0 is indeterminate. In fact, a natural number can either be a positive integer or a non-negative integer, and 0 is one of the latter.
Traditionally in combinatorics that just evaluates to the empty sum. To make that a little more particular, sum_{i=a}^{b} f(i) means "add up all instances of f(i) over the set of all integers i where i>=a and i
1+2+3+...+n = ½n(n+1)
So what you have is
6n·2/n(n+1) = k
k,n being nat. nos., and n > 0 to avoid division by 0. Continuing,
12/(n+1) = k
So n+1 can = {2,3,4,6,12}; consequently, n = {1,2,3,5,11}, and ∑n = 22
N=0 is not allowed because it would result in division by 0, but I guess that didn’t influence the final solution.
I solved it using the formula for quadratic equations: kn^2 + (k-12)n + 0 = 0 giving n = (12-k)/k.
Ok, i got two things to say
1. i wish my math teacher was like you, you're awesome.
2. Can you make a proof for the product rule please?
thanks!
Product rule of derivative?
blackpenredpen Yup, that one, its the foundation for almost every derivative proof so i thought it will be nice to know why its like that... thanks!
Here is a proof, Patrick JMT
th-cam.com/video/kNhYfmC_UPU/w-d-xo.html
binnacle true-north
It here th-cam.com/video/sP4vqks16cs/w-d-xo.html
I love how you always come up with very clever ways to solve these problems !
The way to sum up 1 + 2 + 3 + ... n is to notice the rows to the left of the diagonal of a square made up of n+1 x n+1 dots.
That's literally the most basic and probably the first way someone will ever find it. This was just an alternative way. Duh
9:16 is there some theorem about the sum of all the factors of numbers, given the number has certain properties, either that its factors are obscenely numerous or specifically minimal for the local group of numbers, or that it is a power of a certain number etc?
Ryan Roberson geometric sum
Solved in my head in around 1 min, one good thing about this question is that it is asking for the sum of all possible n but not listing them out. Which you won't argue if 0 counts while it will make the denominator 0 because it won't change the result.
Also the website is so coool
Boming Wang thanks! I like that site too!!
I solved this also in my head maybe in two minutes, triangular numbers, complex fraction and factors of number 12. To easy. I'm not math freak. I just like and use damn thing.
denis milic The problem was really stupid
We can't take n=0 as n belongs to N, but ans is still same, nice ques n solution
please explain how the binomial coefficients came in
Combinatorics YEAH ! Could you do something about the Catalan number ?
Olivier L. Applin I will try when I have time later on.
I'm not even sure what the original question was supposed to be? Brilliant doesn't seem to know what "Let" means or how quantification works. I mean, you probably did choose the right interpretation, but the statement is bad.
"What is the sum of all natural numbers n such that is an integer." They just threw "Let" at the front to make it sound more mathy, but it makes the statement literally incoherent, but with an interpretation that makes sense.
I don't get the connection between the little Gauß and the binomial coefficients. Does this always work when the sums "level down" reach +0,+0,+0,...? And why?
Fematika proved it here: th-cam.com/video/DcP4kKioDww/w-d-xo.html
8:50 0 IS NOT A SOLUTION
From Brilliant.org....this was fun!
Let G(n) be the expected value of the geometric mean of real numbers independently and randomly chosen between 0 and 1.
What is Lim G(n) as n goes to infinity?
Yea I saw that too! The answer is 1/e, which is ironically the same as my first video on brilliant. : )
You should make a video
Zonnymaka
I will see
Isn't it 1-1/e?
My math is rusty af, but isn't this argument correct?
- If you divide the range by n and you pick a random number you have 1-1/n probability of not picking it. The intersection of events not picking it gives you the probability p of not picking it. As n aproach infinity 1-p is the probability of picking which is 1-1/e. If you multiply the probability by the span you get the average value 1-1/e?
Hi there!
great channel!
Cna you explain where you came up with that trick for finding formulas for sequemces using combinations??? Seems exciting
How you made the combination? You may make a new video to show how to make this combination.
Have you solve problems from Skanavi book? They are very cool :)
Could you please do a video on how you can use n choose 0 to find the formula of a sequence?
Here's another example: th-cam.com/video/S7vnkAu2lGI/w-d-xo.html ,
And fematika proved it: th-cam.com/video/DcP4kKioDww/w-d-xo.html
Can you do a video on the integral from -inf to inf of cos(x)/(1+x^2). Also love the videos, keep up the good work.
Vanish I will have to think hard on that
Love this vid!
WOW! I arrived at 12/(n+1) = k immediately, what got me was "the sum of all n" statement...
unknown360ful Yes, that seems a bit random.
Nice work,but how to change your formula into binomial expression??plz help me to describe
can you do this one? find area under '1-e^(-x^-2)'
Sum of all n... So all valid 12/k -1where k is whole as well as the result of 12/k, with said result also being greater than -1
why is the combination done like n chooss 0, like what is the reason?
Amitakshar Biswas Because it is more concise and easy to understand, and it also gives us permission to borrow from Pascal’s triangle.
make a video on integral of x!
If we're being strict. Wouldnt the answer be 12? Because the equation clearly implies that n>3 ( since it writes 1+2+3...n)
that is simply to denote that the sum goes to a certain number n... not the fact that n>3... only criteria is n is natural number which includes 1,2,3
also if you were to exclude it, you'll be left with 16 not 12
Can you get a complex answer for |x|=-1?
Jimmy
Actually, the answer won't even be complex: it will simply be "false". No complex solutions. An inconsistent equation. Because by definition, the absolute value of any complex number must be a non-negative real number.
(big-@ss explanation and side-question ahead:) tbh though, I actually stumbled at it when I read it, so it's not a bad question at all. But when I gave it a go, I realized the equation |x|=-1 contradicts the very definition of the absolute value function: *given a complex number x of the form a+bi, |x|=√(a²+b²).*
Since a & b are both real numbers (yes, bi is the imaginary component, but b alone is a real number all byself), a² and b² will both be non-negative real numbers. So if you add them together and square root that sum, you will end up with another non-negative real number.
That result is also the result that you get from solving fit the absolute value of a complex number: that very same non-negative real number.
So for any complex number x, |x| can never be equal to -1, nor could it even be equal to +i or -i. It could be any non-negative real number, from zero onwards, but it can only be a non-negative real number. No exceptions, by definition.
Definitely a good problem to think about though, thanks for asking. That said, I can't say the same applies to quaternions, octonions, etc., nor do I know if the absolute value function is even defined for elements of those sets of numbers tbh, but I hope someone else can tackle that one and let us know 👍
Can someone show me how to solve this ODE ? 4y'' + y = 2*sec(t/2) I tried 10x and I did not find the correct answer.
If possible, step by step solution.
fseeletronicos © Solve for general 4y" + y = 0, then use the variation of parameter method. Let y1 and y2 be the solution to the general homogeneous. Then yp = C1y1 + C2y2, where C1' = -(F/a) * y2 / W where W is the wronski value of y1 and y2.
I wanted to say pretty much as Kenneth has said.
A similar question and solution has been posted on Mathforum. The solution to y" + 4y = sec(x)
can be found here (I don't think a link I post will be visible):
mathforum (dot) org / library / drmath / view / 52117 (dot) html
remove all spaces and replace (dot) with . to get the link and the solution.
but the answer is 16 coz we need to have N>3 no?
Why would N have to be larger than 3? What?
@@avanananaOP thought that since it says "1 + 2 + 3...+ n" in the problem, n would have to be greater than 3. I think.
I'm going to challenge my teacher with this question...hehe
This is obnoxiously easy. I dont think it would be a challenge for them
please make some videos about integral contour
Sorry, I haven't done those for a long time. So I will have prep for a while before I can make the vids.
blackpenredpen no problem,this is just my request. take your time
Rizky Agung peyam has some on his channel, check it out.
I think the answer should be 16 since it is implied that n is bigger than 3 in 1+2+3+...+n :)
blackpenredpenbluepen
for (n 3),is it n(n-1)(n-2)/3 or n(n-1)(n-2)/3! ?
with 3!
Hi! I was just wondering how can you do an integal like this: integral of ln|tan(x/2)|dx?
anarchostalinprimitividiagonalist partial
Not sure if trolling, because it didn't change the answer, but the n=0 solution doesn't work.
I don't understand how, from the sum of all natural numbers, you end up with those binomial coefficient.
I understand the equality, it is easy to verify, but i don't understand how you came to that conclusion.
n not equal to 0, coz 6n/(n x (n+1))/2 = 0/0
Zero is a whole number, not a natural number. You can't have k=12, n=0. Since you are summing the ns, it doesn't impact your final answer to add an extraneous zero, but it is still incorrect.
In the last is not true when n=0 & k=12 . ...we get 0/0=12 .
nice one
solve iit jee advance questions
Aditya Vijj
Try to make videos for us
This was not a terribly difficult series problem on brilliant but still an interesting one, reminded me of Max's work on your channel a couple months ago. brilliant.org/practice/sequences-and-series-level-1-challenges/?problem=a-calculus-problem-by-danish-ahmed-3
Black pen and red pen are crying because of blue pen :'(
great!
0 is not a natural number so the last solution where n = 0 is not valid.
True, n=0 is not valid, but *only* because 0÷0 is indeterminate. In fact, a natural number can either be a positive integer or a non-negative integer, and 0 is one of the latter.
Good video but it was too easy for my taste. I solved it in my head in 2 minutes
That's great. I wanted to use this question to show people another way to get the formula for 1+2+3+...+n
Ahsoka Tano
2 minutes?
Wow now i'm really proud of myself because i solved it in under 30 seconds and i'm 13😉
I also did it
Cubi Cardi
That's great, too!
It took me 10 to 15 sec (credits go to RMO classes AND BlackpenRedpen)
i wish you were a god
n can not be 0, because we are talking about natural numbers and so k will not be defined although it won't change the answer, just sayin' :)
Also sum from i=1 to n=0 just doesn't make sense
Traditionally in combinatorics that just evaluates to the empty sum. To make that a little more particular, sum_{i=a}^{b} f(i) means "add up all instances of f(i) over the set of all integers i where i>=a and i
It’s 0/0 which is indeterminate, the number exists but it is a continuation
I think sum is 28
one of your recent
22
明明就
I solved it in my head.
Difficulty: 1/10
Hope you become the new genius who can solve all the problems the World has.
anybody from 2021
luv zero