If you dont know dirichletss test isnt there another way to solve this?? Because I doubt most people know that test and i dont think you can deduce it or would when doing a sum like this..
@@leif1075 You can use Fouriet series for f(x)=x over [-pi; pi]. It is an odd function, so it's expansion consist entirely of sines, and you can use theorems for Fouriet series convergence to prove convergense
@3:35 Felt like you were about to through the Chalk, the way our teachers do longtime ago to make us focus better. 🙏 Amazing Demonstration that looks like an acrobat over a trapeze . Thanks
A cool proof that this converges could involve noting that sin(n)/n converges, and perhaps prove the value with Fourier series, and then use the AST. Cool sum!
actually I don't believe in that shit as if number of terms are even then sum is 0 if odd then 1 and you take the mean to get the avg sum but there is nothing like that
x_1=1, the others are 0 squaring both sides twice leaves us with (x_1)(x_2+...+x_n)=(x_2)(x_1+x_3+x_4+...+x_n) multiplying out the first terms and canceling them we get 0=(x_1)(x_3...x_n)-(x_2)(x_3+...+x_n) now we can factor out the second term and use the fact that x_1-x_2=0 to see that 0= x_3+...x_n since all terms are non-negative, they must equal 0 now consider the equation sqrt(x_2)+sqrt(x_1)=sqrt(x_3)+sqrt(x_1+2*(x_2)) and after some rearranging we see that x_1=0, if we can divide by x_2 now if x_1 is equal to 0, x_2, by the condition, would be negative, so thats a contradiction meaning x_2=0 and it follows that x_1=1
SOLUTION *x_1 =, x_2 = ... = x_n = 0* Let us square the equality √x_1 + √(x_2 + ... + x_n) = √x_2 + √(x_3 + ... + x_n + x_1), subtract the sum x_1 + ... + x_n from both sides, and square again. We obtain x_1(x_2 + ... + x_n) = x_2(x_3 + ... + x_n + x_1), hence (x_1 - x_2)(x_3 + ... + x_n) = 0. Since x_1 - x_2 = 1, we have x_3 + ... + x_n = 0. Since our equations contain square roots of x_3, ..., x_n, these numbers are nonnegative, and since their sum is 0, each of them is 0. Suppose x_2 is not zero 0, that is, x_2−x_3 is also nonzero. Considering the sums which contain √x_2 and √x_3 and arguing as above, we get x_1= 0. Then x_2=−1, but since there exists √x_2, we obtain a contradiction. Thus x_2 = 0, hence x_1 = 1, and then all conditions are satisfied.
at 14:03 the series diverges at z=-e ^i does that not matter here, because the point where z=-e^i is a null set on the line youre intergrating over? or is there a different reason?
Very nice ! The most unexpected thing in my opinion is to get such a simple result when the angular variable is simply a natural number in “sin(n)”, not even a multiple of pi like in Fourier Series...etc
To get the argument at the end, it is a bit easier to use the identity 1+exp(ia) = exp(ia/2)[exp(-ia/2)+exp(ia/2)] = 2 cos(ia/2) exp(ia/2), or to do a bit of geometry as you have a neat inscribed angle situation.
Just wondering. When you finish a subgoal you always knock on the board to mark the finished result. Is that sound when "knocking" added later when editing?^^
log(1 + z) = sum_{n=1}^{\infty} (-1)^n z^n/n does hold for |z| = 1 except for z = -1, by Abel's Theorem. (Technically "Abel's test" gives convergence, and "Abel's Theorem" gives the limit.) Here you use the branch of the log with imaginary part between -pi and pi. So your series is the imaginary part of log(1 + e^i) = log(r e^{it}), where e^{it} = e^{i/2} and r = (e^{i/2} + e^{-i/2}) = 2 cos 1/2. Since log(re^{it}) = ln r + it, the answer is just t = 1/2.
14:10 i dont understood why you can use the geometric series formula without Z being in between -1 and 1, or Z is in this intervale? If so how can i prove that?
Look at integral limits. Z is between 0 and -exp(i), and the modulus of the latter is 1. The geometric series converges when |z| < 1. It seems that |z| = 1 is outside the range, but that's one of the integral limits, so we'd never reach that value (to be honest, I'm not 100% sure why we can do it. Could someone explain?). Therefore we can use the geometric series formula.
... Int_{0}^{e^i} \sum z^n = lim_{r --> 1} Int_{0}^{re^i} \sum z^n = lim_{r --> 1} Int_{0}^{re^i} 1/(1-z) = ... Guess this says something to you. Haha. What I thought wierd was a complex number in the limits of integration. But it is something easy to understand in the end.
@@SciencePhysics I have a memory from college (20 years ago) that to be rigorous when going from the sum to the integral, that the integral is the limit of the sum when x tends to the upper bound from the left (something about Riemann Sums...) and therefore is never 1 and you can use the geometric series formula.
For the Durichlet's test, couldn't we just choose a_n = sin(n)/n and b_n = (-1)^(n+1)? The limit of a_n is really easy to find (for all natural numner n, -1
Yeah it would work if the fact that (a_n) is monotonic wasn't an hypothesis, but he actually forgot this property in the test statement. (in your case, sin(n) / n is not monotonic) That said, there are still easier ways to prove that the sum of sin(n) is bounded.
@@Jinbmt656 which easier ways? If you are talking about using some formula of sum of sines of an arithmetic progression, the formula comes exactly from the same reasoning he used.
Some commenters have pointed towards the Fourier series of x/2 or x, but neither of those are even periodic functions so not sure how that would work. What does provide a valid solution is a rectangular pulse train. For a train centered on the origin with frequency and amplitude 1 and duty cycle delta we have f(x) = delta + (2/pi) * sum_{n=1}^{infty} [sin(n*pi*delta)/n * cos(2*pi*n*x)]. Now set duty cycle delta=1/pi and evaluate at x=1/2; this reduces the sines in the sum to sin(n) and the cosines to cos(n*pi) = (-1)^n = -(-1)^(n+1). x=1/2 is a midpoint between two pulses where f(x)=0 and so f(1/2) = 0 = 1/pi - (2/pi) * sum_{n=1}^{\infty} [(-1)^(n+1)*sin(n)/n] => sum_{n=1}^{\infty} [(-1)^(n+1)*sin(n)/n] = (1/pi)*(pi/2) = 1/2.
I solved this in 2 minutes by realizing this was a Fourier expansion of an odd function evaluated at 1. With a little bit of work, you can show f(x) = x/2 = Sum[(-1)^(n+1)*Sin[nx]/n,{n,1,inf}] which of course means f(1) = 1/2.
Another way to see that arg(1+cos1 + i sin1) = 1/2 is just to draw a triangle in a complex plane. z = 1+cos1 + isin1 is the vertex of an isosceles triangle with vertices 0,1,z. It is isosceles since |1-0| = 1 and |z-1| = 1. The arg is the measure of any two of the equal angles in the triangle. By definition, we are given that the angle z-1 makes with the horizontal is 1 rad, and so the measure of the interior non-identical angle in the isosceles triangle is 180-1 = 179. This means that each identical angle has measure (180-179)/2 = 1/2 rad.
Nice proof! (Except you're mixing degrees with radians at the end, but that has no bearing on the final result.) An alternate view (of essentially the same proof) is this: consider the triangle formed by -1, 0 and w = cos 1 + i sin 1, three points on the unit circle. Angle (+1, 0, w) is 1 rad, therefore angle (+1, -1, w) must is 1/2 rad (a nice theorem of circular geometry: the angle from the center to an arc of a circle is twice the angle from any(*) point of that circle to the same arc), and angle (+1, -1, w) is also (by translation) angle (+2 or +1, 0, w+1) = arg(1 + cos1 + i sin1) [(*) edit: actually only valid for points outside of that arc, as in this case]
This the sum of the integral values of the f(x) = sinc(x(1+pi))*(-(1+pi)), the fourier transformation of f is -pi * 1(|v| < (1+pi)/(2pi)). According to Poisson Summation formula, the sum of the function in the integers is the same as the sum of the fourier transformation of the function on integers. Let the sum in the problem be A, we need to calculate the sinc function summed over the integers, which is 2A + f(0) = -pi. f(0) = -1 - pi, so A = 1/2.
Kind of amazing this worked: at 14:10, you replaced the sum with 1/(1-z), but the sum only converges for |z| < 1, and in this case, |z| = 1. Of course, as a formal power series, the convergence is irrelevant, and 1/(1-z) is the proper generating function, so my comment is probably just as asinine as it looked the first time I read it back...
I think Dirichlet's test works a little easier with a_n=sin(n)/n and b_n=(-1)^(n+1). a_n converges to 0 because |a_n|≤1/n, and abs(sum(b_n, n=1,...,N)) is bounded by 1.
@@f5673-t1h fair point. But there will always be problems.... and tricks required to solve them. And it will usually be beautiful when a solution is found.
You and @f 5673 have different aesthetic perspectives. Haha. After 12 years dealing with people saying "this is beautiful", "that is beautiful", I realised it became a addiction from people, some weird psychological phenomena related, in the majority of times, to bad judgement or to be part of the group.
1:35 I freaked out when you wrote (1/n) -> 0. Special limited edition converging harmonic series? Then I realized you were talking about the individual terms rather than the sum.
Hi Michael You’ve done a great work I’ve seen a paper ( Life on the edge) by Alfa Poorten which generalize this series even in terms of cosine The idea is to consider the infinite series whose nth term is (-exp(inx)) which is exp(inx) /( 1+ exp(inx)) then integrate both sides wrt x from x to minus infinity and then separate the real and imginary parts. Have a nice day!
Not sure how inscribed angle theorem applies, but if you draw 0, e^i, 1+e^i, and 1 on the complex plane, you get a parallelogram. Then, finding that the angle to 1+e^i is 1/2 can be done with simple geometry.
i notice that (-1)^n = cos(n*pi) and sin(n*pi) = 0. So (-1)^n sin(n) = cos(n*pi)sin(n) + sin(n*pi)cos(n) = sin(n(pi + 1)). and sigma(((-1)^(n+1)) * sin(n)/n) = - sigma(sin(n(pi + 1))/n) Using Fourier series , easy to compute that sigma(sin(nx)/n) = (pi - x)/2. Using x = pi + 1, we found the result = - (pi - (pi + 1))/2 = 1/2
I'm not a mathematician, so can someone explain to me why the equation at 4:40 becomes the series sum at 5:22 where each term is multiplied by (-1)^n+1
Interesting, how in a usual series 1/n^2 all partial sums are a rational number, but the whole sum is irrational; while in this sum all partial sums are most likely irrational due to sin(n) , but the limit is rational
Find the sum of an alternative series is not really relevant in the sense it can have different values, since when you're summing into infinity you lose commutativity.
Would anyone really ever think of all this? Especially if you don't know dirichlets test, isnt there another waybto do it more logical and intuitve using just sum and difference of sine fornulas maybe and factoring out common terms..
why wouldn't you make An = sin(n)/n and Bn = (-1)^(n+1), the partial sums of Bn alternate so it's bounded and An limits to zero because sin(n) is bounded but 1/n diminishes to zero.
Stupid question - do you need to prove convergence if you can find the sum? Also alternative "solution": Take Bronstein or another suitable book and flip open the table of Fourier series, see that this is related to the sawtooth waveform and use that to calculate the sum.
“Stupid question - do you need to prove convergence if you can find the sum” Yes. Consider, for example, the sum 1-1+1-1+1-1+1-1... which obviously doesn’t converge. However, we can “compute” it’s “sum” by saying: S=1-1+1-1+1-1+1... S-1=-1+1-1+1-1+1... S-1 = -1*(1-1+1-1+1-1+1...) S-1=-S 2S=1 S=1/2 We’ve computed a value, yet the series doesn’t converge.
On 1:25, I really liked your pick for a(n) and b(n) and the turn it gave to the video, but if the goal was simplicity, just pick: a(n) = sin(n)/n , b(n)=(-1)^(n+1) . That way we can use L'Hopital's Rule to prove a(n) converges to 0 and the partial series of b(n) is clearly less than or equal to 1.
he factors an additional -1 out of the term he says that n+1 becomes a n+2 and taking this additional minus sign out leaves the term unchanged but an additional - appears in front of the Im
Not sure what is our doubt. He changed the index of the sum only for the first sum and adjusted the (-1)^(n+1) factor. For the second sum there is adjustment necessary, because he is not changing it. So (-1)^(n+1) stays the same in the second sum, and he rewrites it as -(-1)^n.
Hello Michael, I would really love it if you attempted some problems from the INMO 2021(Indian National Mathematical Olympiad) Regards, Adarsha Chandra (Fan from India)
This must be the most complicated way to write 1/2 I've ever seen.
Can't wait to use it as a substitution.
spoilers... =(
@@Cpt.Zenobia sorry :/
though, what do you expect when you read the comments before watching?
Ya gotta love these broken English flixzone scammers in the replies
@@Cpt.Zenobia the comment section is for discussing the video, deal with it
No, but really, we want a "C joins the battle" playlist.
engineering students: sin(n) = n, sin(n)/n = n/n = 1
sum becomes
1-1+1-1+1-... = S
S-1+S = 0
S = 1/2 🙂
Fucking engineering then
(-1)^n+1 ruins this becuase the series is -1+1-1+1... which equals -0.5
@@deenseen it starts with n=1 so not really
It's amazing that this equals a half plus a half in both of these places!
In Dirichlet's test, you missed the additional hypothesis that the (a_n) sequence needs to be monotonic. Great video tho
Yeah, also wanted to write this one, as there is a weaker test that only states that (-1)^n*a_n (a_n>0) converges iff a_n is monotonic and tends to 0
If you dont know dirichletss test isnt there another way to solve this?? Because I doubt most people know that test and i dont think you can deduce it or would when doing a sum like this..
@@leif1075 You can use Fouriet series for f(x)=x over [-pi; pi]. It is an odd function, so it's expansion consist entirely of sines, and you can use theorems for Fouriet series convergence to prove convergense
one of the coolest solutions i’ve ever seen, very well done
At 4:53 there's a (-1)^(n+1) missing in front of sin(n)..
Thank you! I thought so, too, and came to the comments to see if I was right about that.
"My glass is half full" expands nicely using this series sum.
8:36 Audi famam illius
17:58 Good Place To Stop
No "ℂ joins the battle" marker?
@@xCorvus7x Yeah, forgot about that one. Edited, thanks
@@goodplacetostop2973 you're welcome
"I'm gonna be a little bit sketchy here and change the order of summation and integration" absolutely amazing lol
Why is that amazing?
@@leif1075 are you familiar with the word on an everyday context?
@@pedromooregaissler6378 yea just wondering what you found amazing about it?
@@leif1075 I liked the sentence, not a usual place to use the word sketchy
@3:35 Felt like you were about to through the Chalk, the way our teachers do longtime ago to make us focus better. 🙏
Amazing Demonstration that looks like an acrobat over a trapeze . Thanks
Absolutely astonishing !!
Can you make a video on that 1 = 1/2 + 1/2 theory???
A cool proof that this converges could involve noting that sin(n)/n converges, and perhaps prove the value with Fourier series, and then use the AST. Cool sum!
Reminds me of the “proof” of Grandi’s series: 1 - 1 + 1 - ... + 1 = S , S=1/2
actually I don't believe in that shit
as if number of terms are even then sum is 0
if odd then 1
and you take the mean to get the avg sum
but there is nothing like that
@@pranjalsrivastava3343 well infinity is not even and odd so it's neither EZ.
HOMEWORK : Solve the system of equations (n > 2)
• √x_1 + √(x_2 + .... + x_n) = √x_2 + √(x_3 + .... + x_n + x_1) = √x_n + √(x_1 + .... + x_(n-1))
• x_1 - x_2 = 1
SOURCE : 2008 Tournament of Towns
x_1=1, the others are 0
squaring both sides twice leaves us with (x_1)(x_2+...+x_n)=(x_2)(x_1+x_3+x_4+...+x_n)
multiplying out the first terms and canceling them we get 0=(x_1)(x_3...x_n)-(x_2)(x_3+...+x_n)
now we can factor out the second term and use the fact that x_1-x_2=0 to see that 0= x_3+...x_n
since all terms are non-negative, they must equal 0
now consider the equation sqrt(x_2)+sqrt(x_1)=sqrt(x_3)+sqrt(x_1+2*(x_2)) and after some rearranging we see that x_1=0, if we can divide by x_2
now if x_1 is equal to 0, x_2, by the condition, would be negative, so thats a contradiction meaning x_2=0 and it follows that x_1=1
SOLUTION
*x_1 =, x_2 = ... = x_n = 0*
Let us square the equality √x_1 + √(x_2 + ... + x_n) = √x_2 + √(x_3 + ... + x_n + x_1), subtract the sum x_1 + ... + x_n from both sides, and square again. We obtain x_1(x_2 + ... + x_n) = x_2(x_3 + ... + x_n + x_1), hence (x_1 - x_2)(x_3 + ... + x_n) = 0. Since x_1 - x_2 = 1, we have x_3 + ... + x_n = 0. Since our equations contain square roots of x_3, ..., x_n, these numbers are nonnegative, and since their sum is 0, each of them is 0.
Suppose x_2 is not zero 0, that is, x_2−x_3 is also nonzero. Considering the sums which contain √x_2 and √x_3 and arguing as above, we get x_1= 0. Then x_2=−1, but since there exists √x_2, we obtain a contradiction. Thus x_2 = 0, hence x_1 = 1, and then all conditions are satisfied.
at 14:03 the series diverges at z=-e ^i does that not matter here, because the point where z=-e^i is a null set on the line youre intergrating over? or is there a different reason?
8:31 suddenly, Brawl.
Love this complex themed production. Awesome.
This is verified with the Fourier series of x.
Wasn't it much easier to take a_n= sin(n)/n and b_n=(-1)^n?
(-1)^n is not bounded
@@phythematics2188 yes, it is.
Like some other guy said above, the test he is using demands the sequence a_n to be monotonic.
@@sarthaktiwari3357 nah a_n must be monotonic, b_n doesn't need to
(-1)^n is not monotonous.
Very nice ! The most unexpected thing in my opinion is to get such a simple result when the angular variable is simply a natural number in “sin(n)”, not even a multiple of pi like in Fourier Series...etc
This is a mind-blowing summation!
To get the argument at the end, it is a bit easier to use the identity 1+exp(ia) = exp(ia/2)[exp(-ia/2)+exp(ia/2)] = 2 cos(ia/2) exp(ia/2), or to do a bit of geometry as you have a neat inscribed angle situation.
Just wondering. When you finish a subgoal you always knock on the board to mark the finished result. Is that sound when "knocking" added later when editing?^^
Very intresting, finding this sum with complex numbers.this kind of sums are often found by fourier series
I think we must precise the path of the integral from 0 to -e^i wich is the segment between the 2 values..
log(1 + z) = sum_{n=1}^{\infty} (-1)^n z^n/n does hold for |z| = 1 except for z = -1, by Abel's Theorem. (Technically "Abel's test" gives convergence, and "Abel's Theorem" gives the limit.) Here you use the branch of the log with imaginary part between -pi and pi. So your series is the imaginary part of log(1 + e^i) = log(r e^{it}), where e^{it} = e^{i/2} and r = (e^{i/2} + e^{-i/2}) = 2 cos 1/2. Since log(re^{it}) = ln r + it, the answer is just t = 1/2.
Yeah sure , you’re are right Michael
This exactly what Alfa Poorten said in his paper entitled (Life On The Edge)
We found an unlisted video!
14:10 i dont understood why you can use the geometric series formula without Z being in between -1 and 1, or Z is in this intervale? If so how can i prove that?
Look at integral limits. Z is between 0 and -exp(i), and the modulus of the latter is 1. The geometric series converges when |z| < 1.
It seems that |z| = 1 is outside the range, but that's one of the integral limits, so we'd never reach that value (to be honest, I'm not 100% sure why we can do it. Could someone explain?). Therefore we can use the geometric series formula.
... Int_{0}^{e^i} \sum z^n
= lim_{r --> 1} Int_{0}^{re^i} \sum z^n
= lim_{r --> 1} Int_{0}^{re^i} 1/(1-z)
= ...
Guess this says something to you. Haha. What I thought wierd was a complex number in the limits of integration. But it is something easy to understand in the end.
@@SciencePhysics I have a memory from college (20 years ago) that to be rigorous when going from the sum to the integral, that the integral is the limit of the sum when x tends to the upper bound from the left (something about Riemann Sums...) and therefore is never 1 and you can use the geometric series formula.
Yeah, that's true :)
For the Durichlet's test, couldn't we just choose a_n = sin(n)/n and b_n = (-1)^(n+1)?
The limit of a_n is really easy to find (for all natural numner n, -1
Yeah it would work if the fact that (a_n) is monotonic wasn't an hypothesis, but he actually forgot this property in the test statement. (in your case, sin(n) / n is not monotonic) That said, there are still easier ways to prove that the sum of sin(n) is bounded.
@@Jinbmt656 which easier ways? If you are talking about using some formula of sum of sines of an arithmetic progression, the formula comes exactly from the same reasoning he used.
Some commenters have pointed towards the Fourier series of x/2 or x, but neither of those are even periodic functions so not sure how that would work. What does provide a valid solution is a rectangular pulse train. For a train centered on the origin with frequency and amplitude 1 and duty cycle delta we have f(x) = delta + (2/pi) * sum_{n=1}^{infty} [sin(n*pi*delta)/n * cos(2*pi*n*x)]. Now set duty cycle delta=1/pi and evaluate at x=1/2; this reduces the sines in the sum to sin(n) and the cosines to cos(n*pi) = (-1)^n = -(-1)^(n+1). x=1/2 is a midpoint between two pulses where f(x)=0 and so f(1/2) = 0 = 1/pi - (2/pi) * sum_{n=1}^{\infty} [(-1)^(n+1)*sin(n)/n] => sum_{n=1}^{\infty} [(-1)^(n+1)*sin(n)/n] = (1/pi)*(pi/2) = 1/2.
I solved this in 2 minutes by realizing this was a Fourier expansion of an odd function evaluated at 1. With a little bit of work, you can show f(x) = x/2 = Sum[(-1)^(n+1)*Sin[nx]/n,{n,1,inf}] which of course means f(1) = 1/2.
Another way to see that arg(1+cos1 + i sin1) = 1/2 is just to draw a triangle in a complex plane. z = 1+cos1 + isin1 is the vertex of an isosceles triangle with vertices 0,1,z. It is isosceles since |1-0| = 1 and |z-1| = 1. The arg is the measure of any two of the equal angles in the triangle. By definition, we are given that the angle z-1 makes with the horizontal is 1 rad, and so the measure of the interior non-identical angle in the isosceles triangle is 180-1 = 179. This means that each identical angle has measure (180-179)/2 = 1/2 rad.
Nice proof! (Except you're mixing degrees with radians at the end, but that has no bearing on the final result.) An alternate view (of essentially the same proof) is this: consider the triangle formed by -1, 0 and w = cos 1 + i sin 1, three points on the unit circle. Angle (+1, 0, w) is 1 rad, therefore angle (+1, -1, w) must is 1/2 rad (a nice theorem of circular geometry: the angle from the center to an arc of a circle is twice the angle from any(*) point of that circle to the same arc), and angle (+1, -1, w) is also (by translation) angle (+2 or +1, 0, w+1) = arg(1 + cos1 + i sin1) [(*) edit: actually only valid for points outside of that arc, as in this case]
I am Time Traveller
how the fuck?
@@MelonMediaMedia probably cause he replied 5 days ago but the video is uploaded 40 mins ago 🤣🤣
What a satisfying answer!
It's a great problem and great explication 👍👍👍
This the sum of the integral values of the f(x) = sinc(x(1+pi))*(-(1+pi)), the fourier transformation of f is -pi * 1(|v| < (1+pi)/(2pi)). According to Poisson Summation formula, the sum of the function in the integers is the same as the sum of the fourier transformation of the function on integers. Let the sum in the problem be A, we need to calculate the sinc function summed over the integers, which is 2A + f(0) = -pi. f(0) = -1 - pi, so A = 1/2.
Kind of amazing this worked: at 14:10, you replaced the sum with 1/(1-z), but the sum only converges for |z| < 1, and in this case, |z| = 1. Of course, as a formal power series, the convergence is irrelevant, and 1/(1-z) is the proper generating function, so my comment is probably just as asinine as it looked the first time I read it back...
Actually I'd like some explanation here as well. And I think I have some idea from other comments.
Nice series, beautiful
A very elegant solution!
I think Dirichlet's test works a little easier with a_n=sin(n)/n and b_n=(-1)^(n+1).
a_n converges to 0 because |a_n|≤1/n, and abs(sum(b_n, n=1,...,N)) is bounded by 1.
That.... was.... beautiful.
This is why we study maths folks...
Not really. Math is more beautiful than random calculus problems (that are meant to be solvable) that you throw a bunch of tricks at.
@@f5673-t1h fair point. But there will always be problems.... and tricks required to solve them. And it will usually be beautiful when a solution is found.
You and @f 5673 have different aesthetic perspectives. Haha. After 12 years dealing with people saying "this is beautiful", "that is beautiful", I realised it became a addiction from people, some weird psychological phenomena related, in the majority of times, to bad judgement or to be part of the group.
chill tf out yall
chill tf out yall
1:35 I freaked out when you wrote (1/n) -> 0. Special limited edition converging harmonic series? Then I realized you were talking about the individual terms rather than the sum.
Pls fourier series for (sin(x))*lncos(x/2)
We do love when C joins the battle, don't we?
Hi Michael
You’ve done a great work
I’ve seen a paper ( Life on the edge) by Alfa Poorten which generalize this series even in terms of cosine
The idea is to consider the infinite series whose nth term is (-exp(inx)) which is exp(inx) /( 1+ exp(inx)) then integrate both sides wrt x from x to minus infinity and then separate the real and imginary parts.
Have a nice day!
For arg(1+e^i), you can also use the inscribed angle theorem :)
What's that theorem?
Not sure how inscribed angle theorem applies, but if you draw 0, e^i, 1+e^i, and 1 on the complex plane, you get a parallelogram. Then, finding that the angle to 1+e^i is 1/2 can be done with simple geometry.
@@axemenace6637 Use the circle of radius 1 centered at 1
@@shlokgupta9353 en.m.wikipedia.org/wiki/Inscribed_angle
@@Baruch785 I don't see how that helps anything. Maxim Enis' observation that we get a paralleogram and a diagonal in it is the correct approach
Well the version of it that I like most, is showing that the sum of the absolute value : \sum |\sin(n)|
is about 2\(pi) \ln(n)
That's a cool freaking sum
It's incredible that wolfram alpha can solve such a difficult problems in under 10 seconds
i notice that (-1)^n = cos(n*pi) and sin(n*pi) = 0. So (-1)^n sin(n) = cos(n*pi)sin(n) + sin(n*pi)cos(n) = sin(n(pi + 1)). and sigma(((-1)^(n+1)) * sin(n)/n) = - sigma(sin(n(pi + 1))/n)
Using Fourier series , easy to compute that sigma(sin(nx)/n) = (pi - x)/2. Using x = pi + 1, we found the result = - (pi - (pi + 1))/2 = 1/2
I would try mellin transform for harmonic series.
I recognized this result as 1/2*Fourier series of x evaluated at 1. Which tells you the answer is 1/2 immediately
Interesting. If you adapt the fourier series proof for this problem, it might even simpler.
Wonderful!!! Nice job :)
How can you use the geometric series formular for z = -e^i where abs(z) = 1 ?
Using limits.
Fourier sin series of $\frac{x}{2}$, then plug $x=1$. Done.
I did not know that Complex numbers were in the Red Guard.
Now prove Dirichlet’s test.
I'm not a mathematician, so can someone explain to me why the equation at 4:40 becomes the series sum at 5:22 where each term is multiplied by (-1)^n+1
I think u got confused bc he forgot the (-1)^n+1 in front of sin(n)
@@mboisne thanks, I thought I was going nuts.
12:22
Very nice!
Pardon me if I'm wrong, but can't you simply use the alternating series test to prove convergence?
good job ......see Article 0.3 : What are Kepler Primes?
Wow .. I hope that's not on the test 😂🤣
Don’t we need 2*pi*n for complex logarithm?
Interesting, how in a usual series 1/n^2 all partial sums are a rational number, but the whole sum is irrational; while in this sum all partial sums are most likely irrational due to sin(n) , but the limit is rational
Most definitions of irrational numbers involve a sequence of rationals (e.g., continued fractions).
At 6:00 approx. You said the finite sum of sin(n) rather than (-1)^{n+1}sin(n)
But why the sum of the geometric series is written like that. I mean z could be sup to 1
I didn't understand the calculation at 5'
Why don't you take a_n=(sin(n))/n and b_n=(-1)^(n+1) to make use of Dirichlet test easier?
this was great!
Someone knows the criteria to inverse the Imaginery function and a serie ?
and inverse the integral with the serie?
Arg(z) is defined modulo 2pi. Why the answer is 1/2? Why it’s not 1/2+2pi for example?
meme avec FOURIER si on pose fonction egale x sur -pi a pi on trouve ce resultat et avec moin d operation
Can we use the triangle inequality for showing the boundedness of the finite series?
Find the sum of an alternative series is not really relevant in the sense it can have different values, since when you're summing into infinity you lose commutativity.
I've never encountered Dirichlet's test before. What branch of mathematics would this fall under?
Just real analysis. You can find the test on Wikipedia. The proof works by summation by parts.
Yes that's so obvious I already knew that but obviously you can find it without C
wow, i wouldn't have guessed the sum will be this nice since sin(n) are not nice numbers in any sense
a(n)=sin(n) /n
b(n) =(-1)^n
Lim a(n) =0
|Sum of b(n)| bounded
As stated above, for the dirichlet test there is a third requirement for a_n to be monotonic, which sin(n)/n is not
Wait what
Real Analysis
elegante demostrecion
0:39 TIL that it's pronounced deer-ish-LAY, not deer-ik-LET
Would anyone really ever think of all this? Especially if you don't know dirichlets test, isnt there another waybto do it more logical and intuitve using just sum and difference of sine fornulas maybe and factoring out common terms..
why wouldn't you make An = sin(n)/n and Bn = (-1)^(n+1), the partial sums of Bn alternate so it's bounded and An limits to zero because sin(n) is bounded but 1/n diminishes to zero.
Because there is a third condition to Dirichlet's test, and that is that {a(n)} should be monotonic. Which 1/n is, but (-1)^(n+1) is not.
Stupid question - do you need to prove convergence if you can find the sum?
Also alternative "solution": Take Bronstein or another suitable book and flip open the table of Fourier series, see that this is related to the sawtooth waveform and use that to calculate the sum.
“Stupid question - do you need to prove convergence if you can find the sum”
Yes. Consider, for example, the sum 1-1+1-1+1-1+1-1... which obviously doesn’t converge. However, we can “compute” it’s “sum” by saying:
S=1-1+1-1+1-1+1...
S-1=-1+1-1+1-1+1...
S-1 = -1*(1-1+1-1+1-1+1...)
S-1=-S
2S=1
S=1/2
We’ve computed a value, yet the series doesn’t converge.
On 1:25, I really liked your pick for a(n) and b(n) and the turn it gave to the video, but if the goal was simplicity, just pick:
a(n) = sin(n)/n , b(n)=(-1)^(n+1) .
That way we can use L'Hopital's Rule to prove a(n) converges to 0 and the partial series of b(n) is clearly less than or equal to 1.
i feel cheated. so much trouble for writing 1 over 2. but liked it.
lmao dirichlets test is so overkill for this problem
Was expecting zero!!!
wow nice sum
How dose (-1)^(n+1) remains (-1)^(n+1) when u change the bound of sum with n to n+1 ?
he factors an additional -1 out of the term
he says that n+1 becomes a n+2 and taking this additional minus sign out leaves the term unchanged
but an additional - appears in front of the Im
Not sure what is our doubt. He changed the index of the sum only for the first sum and adjusted the (-1)^(n+1) factor. For the second sum there is adjustment necessary, because he is not changing it. So (-1)^(n+1) stays the same in the second sum, and he rewrites it as -(-1)^n.
¡Orales!
this sum seems interesting
Good
👍👍
Wow
Hello Michael, I would really love it if you attempted some problems from the INMO 2021(Indian National Mathematical Olympiad)
Regards,
Adarsha Chandra (Fan from India)