Another way to prove this would be to multiply the sum by sin x and use the product-to-sum identities. That gives sin(x)sin(kx)=(cos(k-1)x-cos(k+1)x)/2. When we add all the terms up, it telescopes and we end up with 1-cos 2Nx. That makes the original sum (1-cos(2Nx))/(2sin x).

@drpeyam

2 жыл бұрын

Wow, I really like that too!!!

@brendanlawlor2214

2 жыл бұрын

great simplicity thanks friend 😃

@jofx4051

2 жыл бұрын

Cool man like that alternative way

You’re missing a 2 in the denominator at the end. Because sin^2(x)=(1-cos(2x))/2

@drpeyam

2 жыл бұрын

Ok

@giovanniautobello2677

2 жыл бұрын

Good. I found it. Anyway I prefer the factorization sin^2(x)=(1-cos(x))*(1+cos(x)). Just to calculate sin(x) and cos(x).

@Happy_Abe

2 жыл бұрын

@@giovanniautobello2677 that’s cool!

I love these kind of results! I wonder if it would have been interesting to pair up sin(x) with sin((2n-1)x), sin(3x) with sin((2n-3)x), and so on, and then use sum-to-product identities to get a new sum involving cosines. Then you could apply the same method to the sum involving cosines, and perhaps prove two formulas by induction. Probably not quite as clean as your approach.

In the end you forgot 2 in the denominator

@drpeyam

2 жыл бұрын

Ok

I had a trigonometry class going on and I told my teacher this theorem At first she was astounded then she asked the validation and when I gave it to her I believe I got full marks in my internals 😂

Impressive!! Dr. Solved real life problems with complex world.😃

you can do the same to obtain cos(x)+cos(3x)+...+cos(2(n-1)x) = sin(2nx)/(2sin(x))

It's also an interesting point to see that the cosecant curve runs tangent to all iterations of this sum. Well, according to what Desmos showed me.

I found it easier to do it using just e^ix and then take the imaginary part of that sum at the end. It also gives the equivalent identity for cosine if you take the real part

Generating function approach can also be applied here OGF for sum of sines is A(x,t) = 1/(1-t)*sin(x)t/(1-2cos(x)t+t^2) OGF for sum of even values of sin is B(x,t) =1/(1-t)sin(2x)t/(1-2cos(2x)t+t^2) coeff(A(x,t),t,2n) - coeff(B(x,t),t,n) is your sum

You can use that finite sum formula, along with the Riemann-Lebesgue lemma, to show that the integral of f(x)/sin(x) from x=a to x=b (if it exists) is equal to twice the sum (from n= 1 to infinity) of the integral of f(x)*sin((2n-1)x) from x=a to x=b.

@drpeyam

2 жыл бұрын

So cool!!!

Amazing 😇🤩

I once found a pretty interesting identity: [sin(0) + sin(1) + ... + sin(n)] / [cos(0) + cos(1) + ... + cos(n)] = tan(n/2)

@DrBarker

2 жыл бұрын

I love this identity - I think it's my most viewed video haha!

@whozz

2 жыл бұрын

@@DrBarker Wow, interesting! I found it by myself like 3 years ago when I was tinkering with summations hahaha

@whozz

2 жыл бұрын

@@DrBarker do you know where it comes from?

@DrBarker

2 жыл бұрын

@@whozz It's one that I found by myself too - I started off by looking at sin(1) + sin(2) + ... + sin(n) and cos(1) + cos(2) + ... + cos(n), then realised they are connected. The results for sin(x) + sin(2x) + ... + sin(nx) and cos(x) + cos(2x) + ... + cos(nx) are called "Lagrange's trigonometric identities", so I guess Lagrange may have discovered them? I'm not sure if [sin(0) + sin(x) + sin(2x) + ... + sin(nx)]/[cos(0) + cos(x) + cos(2x) + ... cos(nx)] actually appears anywhere in the literature. It might just be too obscure a result to have been covered.

@whozz

2 жыл бұрын

​@@DrBarker Oh, nice! Your proof is similar to mine, though I used different trig identities.

denote S_N(x)=sin(x) + sin(3x) + sin(5x) + ... + sin((2n-1)x), N=1,2,3... . it's easy to show that for the problematic point x=0, S_N(0)=0 [w/o L'Hôpital's Lu]

Alternate the signs and divide by the coefficient and you get a square wave

One could also remark that since the imaginary part of the sum is the sum of the imaginary part, then the sum of sines become the imaginary part of the sum of complex exponential. The method is slightly different. Unfortunately, I do not find the same result, so I probably did a mistake, maybe somebody can point it out to me. sum(sin((2k-1)x)) = sum(Im(e(^i(2k-1)x)) = Im(sum(e^(i(2k-1)x)) and then you have your sum of q^k problem. This sum simplifies to [exp(-ix)*(exp(2i(N+1)x) - 1)]/[exp(2ix)-1] which further reduces to exp(i(N-1)x) cosecant(x) sin((N+1)x) The tricky part is actually going back to the Imaginary part. Cosecant is purely real, sinus is purely real, you just end up with: sin((N-1)x) sin((N+1)x) / sin(x) = (sin²(Nx) + sin²(x)) / sin(x) Which is strictly different from your result unfortunately.

JEE students remember a standard formula for [sin(A)+ sin(A+D)+... sin(A+(n-1)D)]=? to save time in exam and they don't even know how it has been derived.

quanti sono gli addendi nella parentesi?quella parentesi che definisce la somma della progressione geometrica?

This is good♥️

Is this construction by Fourier series?

A clever trig identity!

Very very very very good 👍😁

Multiply and simultaneously divide the specified sum by 2*sinx. (1/(2*sinx))*[2 (sinx)^2 +2*sinx*sin3x +2*sinx*sin5x +2*sinx*sin7x +...+ +2*sinx*sin(2N-1)x ]= (1/(2*sinx))*(1-cos2x+cos2x-cos4x+cos4x+...+ -cos(2N-2)x +cos(2N-2)x -cos2Nx)= (1-cos2Nx)/(2*sinx).)) 2*sinA*sinB= cos((A-B)/2)-cos((A+B)/2).

how cool is it

In Swedish, 'Nix' is a word, which could be translated as 'nope'.

@drpeyam

2 жыл бұрын

I had no idea!!! Hahaha

I know it’s pedantic but should we include everywhere but the zeros of sin(x). I think in the limit it does make sense

@drpeyam

2 жыл бұрын

But if sin(x) = 0 then you’re just summing up bunch of zeros

Please can you solution the summation of sinnk 😢 ?

4:22 "This is pretty complex" . Pun not intended?) Although, it's, indeed, pretty and complex:)

how about the infinite situation? is there any convergence?

@drpeyam

2 жыл бұрын

Such an interesting question!! If you let n go to infinity in the result, you’ll see that it can’t converge. But there is something called cesaro summation where you replace the series by r^n sin(nx) with r < 1 and sum this and let r go to 1, then that thing might converge

@zyctc000

2 жыл бұрын

@@drpeyam ah we have a rearrangement of the terms at 2:12. Maybe we could not do that to get the results if it is not converting in a infinite setup right?

@drpeyam

2 жыл бұрын

It’s still fine, the partial sums can be rearranged

@zyctc000

2 жыл бұрын

@@drpeyam thx for the explanation!

Its SUMthing I wish I was taught.

@drpeyam

2 жыл бұрын

Hahaha

Is this a Fourier series

@mathadventuress

2 жыл бұрын

Do you have any Fourier series

Why only 668views, this is amazing

@drpeyam

2 жыл бұрын

I knoooow 😭😭

Well if we take cos(x)+cos(3x)+...+cos((2n-1)x), then also we get interesting result which is equal to sin(2nx)/2sin(x)

@drpeyam

2 жыл бұрын

Niiiice and if you divide it by the sum of sin you get an even nicer result

First!