Wow, this looks absolutely intriguing. I cannot yet arrive to this same conclusion, but I have some math that may be of interest. Please note that I figured this all out on my own (including the Euler-Maclaurin formula. I only just recently learned that it was already discovered). Anyways, this does not at all answer your question, but I think you will appreciate this math, especially if you understand the Riemann Zeta function.

Use the Euler-Maclaurin formula to show that:
sum(sin(n),n,0,x)=sum(sin(n),n,1,x)=integral(sin(x))+sin(x)/2+cos(x)/12--cos(x)/(30*4!)+cos(x)/(42*6!)--cos(x)(30*8!)+...
=-|B₀|cos(x)+|B₁|sin(x)+|B₂|cos(x)/2!+|B₄|cos(x)/4!+|B₆|cos(x)/6!+|B₈|cos(x)/8!+|B₁₀|cos(x)/10!...
(where B_n is the Bernoulli numbers)
Then we have that sum(sin(n),n,1,x)=B₁sin(x)+cos(x)(-B₀+|B₂|/2!+|B₄|/4!+|B₆|/6!+|B₈|/8!+...)=B₁sin(x)-cos(x)+cos(x)sum(|B_2n|/(2n)!,n,1,x)

To be mathematically tricky, I will make that -cos(x) go into the sum, even though all the other terms are positive. How? Note that 2^0=1, but 2^n, where n is an integer greater than 0 is even. Then, (-1)²ⁿ=-1 for n=0 and 1 for all the rest. So:
sum(sin(n),n,1,x)=sin(x)/2+cos(x)sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)

Note that this is a constant: sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)

Now we step completely away from this. How do we write sum(sin(n),n,1,x) in an alternative way? First, look at sum(aⁿ,n,0,x)=a⁰+a¹+a²+a³+...+a^x. If we add a^x+1, we will have the following:
a^x+1+sum(aⁿ,n,0,x)=sum(aⁿ,n,0,x+1)
Reindexing the right side a few times:
a^x+1+sum(aⁿ,n,0,x)=a⁰+sum(aⁿ,n,1,x+1)
a^x+1+sum(aⁿ,n,0,x)=1+sum(aⁿ⁺¹,n,0,x)
Now the sums have the same index, so we can combine them:
a^x+1-1=sum(aⁿ⁺¹,n,0,x)-sum(aⁿ,n,0,x)
a^x+1-1=sum(aⁿ⁺¹-aⁿ,n,0,x)
a^x+1-1=sum(aⁿ(a-1),n,0,x)
a^x+1-1=(a-1)sum(aⁿ,n,0,x)
(a^x+1-1)/(a-1)=sum(aⁿ,n,0,x)


Now how is this useful? Recall that e^bi=cos(b)+isin(b) (if you want, I can give a clever little proof of that using the Maclaurin series of e^x). Then, if we take only the imaginary part of that, we will have sin(b). This means, if we want sin(0)+sin(1)+sin(2)+...+sin(x), we can simply do this:
sum(sin(n),n,0,x)=Im(sum(eⁿⁱ,n,0,x))=Im((e^i(x+1)-1)/(eⁱ-1))

So reindexing, we get:
sum(sin(n),n,1,x)=Im(sum(eⁿⁱ,n,1,x))=Im((e^i(x+1)-1)/(eⁱ-1)-1)
sum(sin(n),n,1,x)=Im(sum(eⁿⁱ,n,1,x))=Im((e^i(x+1)-1)/(eⁱ-1))          (this is because "-1" is not imaginary, so it doesn't matter)

Now we look back
Im((e^i(x+1)-1)/(eⁱ-1))=sin(x)/2+cos(x)sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)
Im((e^i(x+1)-1)/(cos(x)(eⁱ-1)))=tan(x)/2+sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)
Im((e^i(x+1)-1)/(cos(x)(eⁱ-1)))=tan(x)/2+sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)


And all I can get from this is that the constant "sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)" changes values This is why I love the Bernoulli numbers when manipulating them with alternating functions.

Hi! I'm a bit of a math nerd, too I'm not very good with math, but I definitely enjoy it and get obsessed with it
Here are some more!

Necro update!
So, if you look at the image attached, that is a formula I was working on for the past few years. I made a rough estimate of how many pages of notes I put into this problem and it was over a hundred Well, guess what? In the time that I have worked on it, I was able to prove the formula worked and that it extended to any function that you can find derivatives and integrals for (and since you can use Fourier analysis to find a function for just about anything, you are good for just about anything). However, I was still having doubts that it was original. Well, last week I was looking up something only slightly related when I noticed a particular sequence:

1
1/2
1/6
-1/30
1/42
-1/30

At this point, I got excited. I know these numbers! Then the next term: 5/66. I was ecstatic! These were the coefficients to my function and they are known as the Bernoulli numbers. See, the issue is that when I was writing down the numbers, I computed the coefficients, but when you look them up on the internet and whatnot, most people simply refer to them as B_n. Not very helpful for me. On top of that, my function is a generalisation of the Euler-Macluarin formula which cannot be google searched very easily (unless you know the name) and most people don't know of it or have never worked with it! So what have I done with it in the past few months?

I have made some pretty cool identities including one dealing with the coefficients and "e" (so now I have gone back and rewritten e in terms of the Bernoulli numbers). I now have two representations for sum(sin(x),x,1,n) which can be compared and I can derive more relationships (one of them uses imaginary numbers, the other uses sine, cosine, and tangent. We know where this is going Another proof of Euler's identity). But more importantly to me, I can apply this to analysing the Riemann Zeta function which I did a bunch of research on over the summer (and I've even presented about it a few times).

I also use this quite a bit in my statistics class and it has been very useful. For the curious, any polynomial with positive integer powers will result in a finite closed form expression. Any polynomial using exponentials, trig functions, or non-integer powers will have an series with infinitely many terms. Luckily, it converges pretty rapidly to the desired sum.

If you want to have fun, look at "identity 0" below. If you replace a with e, all of those ln(a) terms get cancelled to 1

Also, you can use this to derive that the "first Bernoulli numbers" divided by the "second Bernoulli numbers" (the only difference is that B₁ is negative in the first, positive in the second) will equal e

Wow, this is awesome o.o

That's pretty cool! All we need is 5 more DJ_Os and we will be at 500000

Yeah, I live about 90 miles south-east of Buffalo and it looks rather normal outside and about the snow, I said the same thing, almost. I was hoping for a bunch of snow because then there would be something to do after the storm (like sledding and whatnot). With rain, all you have is mud and puddles :[

When we first saw Omnimaga, we actually first recognised our deity in your images:


Next, we saw the most beautiful formula that we pass down from generation to generation. It was a formula found by our ancestors and is pivotal in rising to adulthood:


With the music of the gods streaming from the most beautiful technology, we knew we had to preserve this segment of the population.