cos(1) + ... + cos(n)
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- Опубликовано: 25 авг 2024
- In this video, I calculate the sum of cos(1) + ... + cos(n) as well as sin(1) + ... + sin(n) by doing a quick excursion into the complex world. This is one of those instances in math where, in order to solve your problem, you turn it into a hard problem! Enjoy!
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and some people say complex numbers aren't useful
literally no one important has ever said that.
Well if you define as "unimportant" people who would say that you are certainly right.
AndDiracisHisProphet
What would make someone say that complex numbers aren't useful?
Who says that? Obviously excluding certain high schoolers who think math past algebra II isn't important.
It depends on how you define "useful".
Nice fact: This is used in trigonometric interpolation. A rollercoaster in spain is construct by using this series.
The advantage of trig interpolation is the curve is "infinity-smooth".
I had a lecture of "interpolation" and I got a paper about trig. interpolation to work on.^^
This is great.
Would you consider doing videos on past Putnam questions? That would be awesome.
The Zelda pun was great, i love your videos, they have the perfect balance for fun and learning
I was able to figure out this by using vectors. I imagined each term as the projection of a vector of length one onto the x axis, where the value inside the cosine is the angle between the vector and the x axis. To sum up these vectors, I figured out the magnitude of the resultant vector and projected that onto the x axis. It's also easy to find the sum of sines using this method because all you need to do is project the magnitude into the y axis.
I like this because if you sum up 1+...+n, you get n(n+1)/2, and if you only look at what's inside the trig functions, you get 1+...+n = ((n+1)/2)×((n/2)/(1/2)) which simplifies to n(n+1)/2.
Beautiful!!!
Amazing!!! I love complex tricks!!
Amazing!!! I love it xD
Love how complex makes things easier!
I just want you to know you are a savior. tons of love from Iran.
This series is actually very useful for phased array antennae.
No birds were harmed in the making of this video. :--)
And when you solve for tan(1) + ... + tan(n), everything cancels nicely ^-^
Oh, I should try that out, great idea!
Why?
Wow that solution was very imaginative!
Awesome as usual! I immediately wanted to know the generalization cos(x)+...+cos(N*x) and of course you exceeded my expectations by answering it at the end!
he teaches with light smile.it makes my mind consolation
It is usually a tedious task to prove sine and cosine addition laws through geometric constructions, if you accept the existence of complex numbers things get a lot easier, you just need euler identity and the proof unfolds by itself.
Woww amazing how complexe world can solve a real problem !!!!!!!!!!!!!!!!!!!!!!!!!!!
Very useful beautiful and understandable for everyone.
Not only that, if u divide those sums u precisely get tan((n+1)/2) . How cool is that!!
Ah, the Dirichlet kernel. Fun fact: If you sum all the odd or even up to n of Chebyshev polynomials of the first kind T_n(x), you get the nth Chebyshev polynomials of the second kind U_n(x), which happens to be the polynomial expression of the Dirichlet kernel when composed with cos(x).
Conversely you, work backwards and get T_n (x) from U_n(x) by taking the difference of two U polynomials since all the cumulative sums cancel except for the last term. But another interesting thing is that U_n-1(x)*n is the derivative of T_n(x),
So taking a finite sum in one domain gives you a derivative in the other domain, and taking a finite difference gives you an antiderivative in the other domain. I always thought that inverse relationship was interesting, Sort of like fourier duality but with polynomials.
OMG, that was awesome!
I really love your videos, especially this one :O ❤
Awesome video + awesome channel!
OMG, hi Anand!!! 😄 Hahaha, I hope you’re the same Anand I’m thinking of 😂 Thanks for watching!!!
A very nice solution. Thanks a lot👍
It's been too long since I've watched... Thank you!
Nice! If you don't love complex numbers, you don't love math :)
This could be hard to made for tan(n×x) if some cases the tan(j×x) (0
This is amazing!
Why didn't you use the cos and sin sum of angles formulas?( in this case sum of n/2 and 1/2) They may could be much easier to use in terms of tan and cot. I actually don't know, this is a question.
i graphed the sum and it seemed to align suspiciosly well with sin(x)+1/2
Love this video!
Euler formular is.... just fantastic.
I love your energy. Great video!
Incredible!
I studied this in GSEB textbook of this formula but the proof was given by PMI but I needed real proof.
can you put an example up of how to apply chebyshev's inequality ?
Thanks
Hi Dr Peyam and thank you for taking the time out to illustrate, from the simple to the hard on some math areas we all have struggled with in our studies. I think it would true that a lot of UG's in math, physical sciences and engineering struggled with Green's function. It would be good if you could give them your imitable treatment from easy through to grad ?
There’s a video on that!
This is so cool!!!
By using brut force i got to:
[-1+cos(1)-cos(n+1)+cos(n)]/(2-2cos(1))
But your answer is much more beauty.
I got (-1 - cos(n+1)+cot(1/2)sin(n+1))/2 , which is the same, by just subtracting one from 1-e^i(n+1) / 1-e^i instead of factoring out one e^i
Less complex in complex....
Nice video!
that was totally enjoyable,thank you very much DR Peyam :)
Wowww....
great video thank you soo much ^^
I love your videos, they are really helpful, could you please do a video on complex numbers factorial, for example i!
I searched everywhere there is no content on that topic, please do a video on that.
So the analog of the factorial function is the Gamma function, but unfortunately there isn’t much to say about Gamma(i) (you can’t simplify it more), but stick around, there’ll be something about it in a couple of weeks!
Dr. Peyam's Show thank you so much, i sure will
Good stuff. Needs more lighting on the whiteboard. Always following your lecture.
Wow
Can we make more of it? Like other strange series which can be solved in an unusual way (not necessarily with complex numbers)
I’ll try to think of some!
Depends on n being degree or radian
What happens if we use the definition of dirichilet kernel directly?...maybe the answer will come in a different form and we won't get the bonus formula for sin
Nice approach and works fine. An example starting from zero, i.e. Sum=cos(0)+cos(x)...+cos(Nx) = 1 + cos((N+1)/2.x).sin(Nx/2)/sin(x/2). This series is particular interesting since it gives for x=pi, the alternating series +1-1+1-1+1... with result 1 for N is even and 0 for N is uneven. This is the Grandi's series truncated to N.
This is super cool! Thanks so much!!!
Your welcome.
'complex'ification - absolutely worth it!
Nice proof
I used trigonometric identities to proof this when i was at highschool but that was messy. This proof much more elegant
What this made me think of: Assuming we're mathematicians of assumed liberty (or, in short, assume n!=gamma function of equal output for integer N) what is generally cos(n!)? And cos(sin(cos(...)))?
As soon as I saw that, I knew the right way was Euler :)
Euler is always the right way 😉
its,Dr peyam.
I couldnt wrap my brain around this. If you pick some arbitrarily large number the left side of the equation cos(1)+..+cos(n) can approach infinity and the right side, i thought, had a maximum value of 1 and a minimum of -1 because the functions have a range of [-1,1].
After thinking about it for a moment the division by sin(1/2) makes the range of the right side [-1/2,1/2] and the left seems unbounded. Something seems off about this. and I'm not sure how to reconcile it.
Oh, the point is that the left isn’t necessarily unbounded! There are lots of positive and negative values which seem to cancel out
How come its reasonable to use the geometric series on e^i? I'm not sure how it works with phasers, but doesn't r have to be less than 1? I feel like yhe naturl extension to phasers would be |z|
What you’re saying is true for a geometric *series*, an infinite sum. Here we’ve evaluating a finite sum, so the formula is valid for every r (except for 1)
The modulus of your number only needs to satisfy that condition if you want the infinite series to converge, as has been pointed out. Here he simply used the formula for the finite sum.
I tried this before watching the video and got some nasty sums. Didn't realize you can factor out half the angle so just sin remains
if you multiply the arguments you get 1/2(N+1)*(N/2)/(1/2)=N(N+1)/2, which is the sum og integers up to N. Coincidence? I hope not.
Hey Dr. Peyam, can you solve the integral from -infinity to +infinity of (sqrt(x^4+1)-x^2)?
Just do a u sub?
(X⁴+1)^½-x²?
Better yet break them into separate integrals and solve
The first one is done with simple u sub
Now thats cool
I'm not good at infinite series but
Can someone explain why it isn't just cos (1)
I mean
If we have infinitely terms that go from one to infinity
We could stipulate that we basically have an infinite amount of circles with an extra cos(1) thrown in...?
Waaaooowww..very beautiful derivation.!! Is that method Dr.Peyam's Idea?
Nice.
For Real!
sir you are my inspiration
Wow im so impressed
Wow i got the heart as soon as i commented
Sooooooooooo coooooooooooollllllllllllllll !!!!!!!!!!!!!!!!!!!!
Great! Is it possible to get that formula by induction?
I think so! Although it’s harder to guess what the formula is, but once you have it you can prove it by induction!
There is no such thing. Induction is used to PROVE a formula, not to get it.
@@znhait
Why?
To proof a formula by induction u have to "guess" a formula. So first "get" your assumption and then u can proof it.
Cool 👌🏻👍🏻
むぅーーーん なるほど
これがスっとできちゃうには、sinやeに相当馴染まないといけないねえ！
Does this mean, that sum from 0 to infinity of cos(n) will converge since cos(infinity) is not convergent, but it won't diverge aswell?
Beautifully solved.Is there any other way of solving without using complex numbers?
Indeed there is! One commenter below gave a non-complex proof
So good:-D
thank you so much!
Hi
I have a great problem -- I love maths but I don't know if I'm good enough at it to study it. Would you have you any advices for me?
Can give other solution without complex number?
Can you do a video, showing that the series diverges when it goes to infinity and why
Nice video D beyam
Can u make a video doing the Laplace transform of ln(x)
Flammable Maths Sounds like a job for you :)
Incroyable!!!!! Merci!
De rien :)
I don't understand how it diverges, since it seems you can bound the sum thanks to the relationship you show ?
First of all, the denominator might very small, so the right hand side might not be uniformly bounded. Also divergent here means not convergent, so strictly speaking (-1)^n diverges.
I see, thank you very much ! I'm looking forward your next video :)
So amejing
Let me solve this in my mockery way 😜 we know that cos(nx)=(-1)^n then cos(1)=-1
cos(2)=1
cos(3)=-1
cos(4)=1 and so on
Since so consider the series as
-1+1-1+1.....upto n
So when the value n is even then the sum of the series is zero
And when the value of n is odd then the sum of the series is -1
Not true?
@@drpeyam but it may be a solution
This is just for natural numbers right?
Yeah, but I think you can adapt it to alpha, 2 alpha..., n alpha, where alpha is any nonzero real number
Love the thumbnail!
Since the defined domain for the geometric series is |z| < 1, is it really fine to use the geometric series for e^ix, which has a modulus of precisely 1?
I was thinking that.
its finite sum, for finite sum you can use whatever common ratio you want
@@Fokalopoka Oh yeah
HAHA! This is amazing! I never thought about this way!
I also thought about using power series but seeing the result i guess it will be nightmare to do it that way.
by the way, I once thought about a similar
let's define the sequence (a_n) such that:
a_0=c for arbitrary c,
a_n=sin(a_(n-1))
what is the sum from k=0 to n of a_k.
if you know how to solve it i would love to hear
(this series does not converges when n goes to infinity, the proof: math.stackexchange.com/questions/2482637/is-c-sin-c-sin-sin-c-sin-sin-sin-c-cdots-convergent )
That’s a great idea! I doubt that there’s a nice formula for the sum, but I like the divergent series idea! :D
They are complex and complicated numbers
I'm from bihar and also knew these trick
Are you when putting x multiplying N you don't have cos((Nx + 1)/2) instead of cos ((N + 1) x /2) ?
I think you’ll still have cos((n+1)/2 x), because it’ll be a geometric sum with e^(ix), try it out :)
Yeah, I'll take a look at it when I get some time. Good channel!
Omg, could you use this thing to calculate the integral of cos(x)? Or sin(x). Off course this wouldn't be usefull as a proof for the derivatives of sinx because it would be circular. Because Eulers identity uses the derivatives and integral of sinx and cosx already.
I think so!
Can you proof that the geometric series formula works when r = e^i? Like when r = 2, it doesn't work.
It also works with r = 2. It’s not a geometric series here, just a finite sum
Oops... Good work!
Where did the isin(x) terms go from e^i
We’re just comparing real parts, so the imaginary parts get ignored. It would give you an identity with sin maybe
@@drpeyam ohh so because there’s no imaginary part of cos(n) when n is R, then the isin(n) terms would just evaluate to 0 anyways?
Uau !
is it just me who thinks Dr payam uploads a video on a topic that's like the advanced one of what BPRP does?
Haha, it’s actually a pure coincidence! I’ve made this video based on a comment on my video from a couple of weeks ago!
It's BPRP but four times longer and with more terrible but endearing jokes
Jobb kéz érted? Csak próbáld ki ok! Tudom nem fog menni! Tanuld meg! Csak egy kérdés! Mit csinálsz ha medvével találkozol? Nos:? :
Did you kill two birds with one stone?
Something wrong here...
Cos[(n+1)/2]Sin[n/2] < 1
Cos[(n+1)/2]Sin[n/2]/Sin[1/2] < 1/Sin[1/2] < 2
So the RHS cannot ever exceed 2, but I see no reason why Cos(1) + Cos(2) + ... + Cos(n) would be bounded. Since Cos is periodic, wouldn't it in fact be divergent?
Infinity < 2 ?!?!
We’re not taking a series here, just a finite sum!
+Dr. Peyam's Show
Yes, but my point was the RHS is bounded, while the LHS doesn't seem to be. For example, if you choose some very large but finite 'n', then surely Cos(1) + Cos(2) + ... + Cos(n) will surpass 2 right? How can something less than 2 equal something more than 2?
Not necessarily, since cos(n) can become negative, so there might be a cancellation which makes the whole sum < 2
Ah yes, thank you you're awesome btw!