Please Register or Log in to view the hidden image! Well, I tried a change of variables, substituting x = nt/2 so f(t) -> f(2x/n). Taking the limit of n we then have: \( \lim_{n \to \infty} \frac{f(0)}{\pi} \int\limits_{-\infty}^{\infty} \Bigl [ \frac {sin(x)}{n\,sin(x/2)} \Bigr ]^2\, dx \) Now I need to show that the integral equals \( \pi \). I can do that if I can show: \( \lim_{n \to \infty} \int\limits_{-\infty}^{\infty} \Bigl [ \frac {sin(x)}{n\,sin(x/2)} \Bigr ]\, dx = \sqrt {\pi } \) Since sin(x) is symmetric about 0, this is equivalent to showing: \( \lim_{n \to \infty} \int\limits_{0}^{\infty} \Bigl [ \frac {sin(x)}{n\,sin(x/2)} \Bigr ]\, dx = \frac { \sqrt \pi }{2} \) I can see that the limit of n and the upper limit of integration are the same, but not sure how that helps.
I don't follow how you removed f(t) from the integral and don't think you've done the substitution uniformly.
Well, I assume I can do: \( lim_{n \to \infty} f(2x/n) = f(0) \), and still take the limit of n in the rest of the integrand. Ah, I see that I have made a mistake in the substitution, it should be: \( \lim_{n \to \infty} \frac{f(0)}{\pi} \int\limits_{-\infty}^{\infty} \Bigl [ \frac {sin(x)}{n\,sin(x/n)} \Bigr ]^2\, dx \)
Substitute x = nt so f(t) -> f(x/n). I was a bit unsure about the factor of 2 disappearing in the limit. \( \lim_{n \to \infty} \frac{f(0)}{2\pi} \int\limits_{-\infty}^{\infty} \Bigl [ \frac {sin(x/2)}{n\,sin(x/2n)} \Bigr ]^2\, dx \) Trigonometric substitutions. \( sin^2(x/2) = \left[1 - cos(x)\right]/2 = 1 - cos^2(x/2) \) So I can write \( \frac{sin^2(x/2)}{n^2\,sin^2(x/2n)} = \frac{1 - cos(x)}{n^2\left[1 - cos(x/n)\right]} = \frac{1 - cos(x)}{n^2\left[1 - cos^2(x/2n)\right]}\)
I think your source has a typo that might be made clear in context. \(\frac{1}{2 \pi n} \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } = \frac{1}{2 \pi} \sum_{k=1-n}^{n-1} \frac{n - | k |}{n} e^{ikt} \\ \quad \quad \quad = \frac{1}{2 \pi} \sum_{k=1-n}^{n-1} \frac{n - | k |}{n} e^{ikt} \) Thus \(\lim_{n\to\infty} \frac{1}{2 \pi n} \int_{-\pi}^{\pi} \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt = \frac{1}{2 \pi} \sum_{k=1-n}^{n-1} \frac{n - | k |}{n} \int_{-\pi}^{\pi} e^{ikt} dt = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{0} dt = \frac{2 \pi}{2 \pi} = 1\) But \(0 \leq \frac{1}{2 \pi n} \int_{\epsilon}^{\pi} \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt \leq \frac{1}{2 \pi n} \int_{\epsilon}^{\pi} \frac{1 }{\sin^2 \frac{t}{2} } dt = \frac{ \cot \frac{\epsilon}{2} }{ \pi n } \) for \( 0 \lt \epsilon \lt \pi\). Since \( \lim_{n\to\infty} \frac{ \cot \frac{\epsilon}{2} }{ \pi n } = 0\) it follows that \(\lim_{n\to\infty} \frac{1}{2 \pi n} \int_{-\epsilon}^{\epsilon} \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt = 1\) no matter how small \(\epsilon\) was. Since \(f(t) \approx f(0) + t f'(0)\) in a small neighborhood of 0, it follows that \(\lim_{n\to\infty} \frac{1}{2 \pi n} \int_{-\pi}^{\pi} f(t) \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt \\ \quad \quad \quad = \lim_{n\to\infty} \frac{1}{2 \pi n} \int_{-\epsilon}^{\epsilon} \left( f(0) + t f'(0) \right) \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt \\ \quad \quad \quad = \lim_{n\to\infty} \frac{1}{2 \pi n} \left ( \int_{-\epsilon}^{\epsilon} f(0) \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt + \int_{-\epsilon}^{0} t f'(0) \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt + \int_{0}^{\epsilon} t f'(0) \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt \right) \\ \quad \quad \quad = \lim_{n\to\infty} f(0) \frac{1}{2 \pi n} \int_{-\epsilon}^{\epsilon} \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } dt \\ \quad \quad \quad = f(0) \)
I can appreciate why engineers have lookup tables.Please Register or Log in to view the hidden image! Thanks for all that, it could take me a while to follow your working.
I would like to know how the expansion is done, in the first line of equations: \(\frac{1}{2 \pi n} \frac{\sin^2 \frac{n t}{2} }{\sin^2 \frac{t}{2} } = \frac{1}{2 \pi} \sum_{k=1-n}^{n-1} \frac{n - | k |}{n} e^{ikt} \\ \quad \quad \quad = \frac{1}{2 \pi} \sum_{k=1-n}^{n-1} \frac{n - | k |}{n} e^{ikt} \)
\(e^{ix} = \cos x + i \sin x\) \(\sin x = \frac{e^{ix} - e^{-ix}}{2i} \) \(\sin \frac{n t}{2} = \frac{e^{\frac{i n t}{2}} - e^{\frac{-i n t}{2}}}{2i} \) \( \frac{\sin \frac{n t}{2}}{\sin \frac{t}{2}} \\ = \frac{e^{\frac{i n t}{2}} - e^{\frac{-i n t}{2}}}{e^{\frac{i t}{2}} - e^{\frac{-i t}{2}}} \\ = \sum_{k=0}^{n-1} e^{\frac{i ( 2k + 1 - n) t}{2}} \) The last equality is a topic for proof by induction, but even without that it will soon be obviously true. \(\frac{\sin^2 \frac{n t}{2}}{\sin^2 \frac{t}{2}} \\ = \left( \sum_{j=0}^{n-1} e^{\frac{i ( 2j + 1 - n) t}{2}} \right) \left( \sum_{k=0}^{n-1} e^{\frac{-i ( 2k + 1 - n) t}{2}} \right) \\ = \sum_{j=0}^{n-1} \sum_{k=0}^{n-1} e^{\frac{i (2j-2k) t}{2}} \\ = \sum_{j=0}^{n-1} \sum_{k=0}^{n-1} e^{i (j-k) t} \\ = \sum_{\ell = 1-n}^{n-1} \sum_{m=1}^{n - | \ell |} e^{i \ell t} \\ = \sum_{k=1-n}^{n-1} ( n - | k | ) e^{i k t} \) Just a bit of heavy lifting for fun. See how we got from \( \sum_{j=0}^{n-1} \sum_{k=0}^{n-1} f(j-k) \) to \(\sum_{\ell = 1-n}^{n-1} \sum_{m=1}^{n - | \ell |} f(\ell) \) ? \( \begin{matrix} f(0) & f(1) & f(2) & \dots \\ f(-1) & f(0) & f(1) & \dots \\ f(-2) & f(-1) & f(0) & \dots \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \)
I haven't been ignoring this thread, my excuse being that my partner's laptop is in for repairs and mine has been commandeered for higher purposes. But I am trying to prove by induction, rpenner's equality and see where the lesson is in the question--the sentence with ? at the end.
What rpenner's solution does is use a closed form, a sum over n-1 terms, which is convergent. I thought it was some kind of known expansion at first, related to maybe the Taylor expansion of the sine function. Anyway, replacing variables doesn't seem to lead anywhere useful, but I can at least save some typing by replacing t/2 with x. So we want to first show something like: \( \frac {sin\,nx}{sin\,x} = \frac {e^{nx} - e^{-nx}}{e^x -e^{-x}} = \Sigma(n,x) \) where \( \Sigma(n,x) \) is a closed form, a function of n and x. This is equivalent to showing: \( e^{nx} - e^{-nx} = \Sigma(n,x)(e^x -e^{-x}) \) Which means we want a sum where everything cancels except the two terms on the left. And rpenner has: \( \Sigma(n,x) = \sum_{k=0}^{n-1} e^{i (2k + 1 - n) x} \) I can work backwards and show that it is a closed form solution, then work out how the induction on n gives that solution. Except I have to step away from the laptop just now.
If I begin counting at 0, there are n terms. If there are finite number of terms, then it is always defined. "Convergent" applies to infinite sums which are defined. I think it would be simpler to replace \(e^{\frac{it}{2}}\) with x and then you have something like a polynomial in both positive and negative powers of x.
Yes. I don't know why I posted that, since I had already worked out there are n terms in the sum. Or more concisely, there is a zeroth term when n = 1 in your expression. Ok, but it's a delta distribution sequence (function), the limit of n in the integration is +infinity. Are you saying there needs to be a proof that the expression is a closed form? Ok, so you define: \( x = e^{it/2} \) So then \( e^{int/2} = x^n \), and \( e^{i(2k + 1 - n)t/2} = x^{(2k + 1 - n)} \). So now we write: \( \frac {x^n - x ^{-n}}{x - x ^{-1}} = \sum_{k=0}^{n-1} x^{(2k + 1 -n)} \Rightarrow x^n - x ^{-n} = \sum_{k=0}^{n-1} x^{(2k + 1 -n)}(x - x ^{-1}) = \sum_{k=0}^{n-1} \left[ x^{(2k + 2 - n)} - x^{(2k - n)}\right] \)
Then as k varies from n - 1 down to zero: \( \sum_{k=0}^{n-1} \left[ x^{(2k + 2 - n)} - x^{(2k - n)}\right] = x^n - x^{n-2} + x^{n-2} - x^{n-4} + x^{n-4}... - x^{n-j} + x^{n-j} - x^{n- (j + 2)} + ... \) When j = 2n - 2, the last term will be \( x^{n - 2n} \).
Inductively you have a kth pair of summands, for some k, in the series: \( ... + x^{n-j} - x^{n-j-2} + ... \,\,\) where j = 2n - 2k - 2. These have a (k+1)th and (k-1)th pair (of summands) either side, where I treat k as decreasing left to right, each pair differs by ±2 because of the 2k in the exponent. \( ... + x^{n-j+2} - x^{n-j-2+2} + x^{n-j} - x^{n-j-2} + x^{n-j-2} + x^{n-j-2-2} + ... \) The middle four terms cancel, leaving the "first" and "last" terms. Rewrite the original sum in terms of j: \( \sum_{j=0}^{2n-2}\, x^{n-j} - x^{n-j-2} \). So j now defines the first and last terms in the series, everything in between vanishes, and we're pretty much there. With the equality . . . And that's not really showing how to find a closed form, I've only written one out.
Ed: this should go here: we have j = 2n - 2k - 2; j = 0 when k = n - 1, k = 0 when j = 2n - 2. And: 0 ≤ j ≤ 2(n-1), 0 ≤ k ≤ n-1. \( \sum_{k=0}^{n-1} x^{(2k+1-n)} = \sum_{j=0}^{2n-2} x^{(n-j-1)} \) So I can rewrite rpenner's product of two sums: \(\Bigl( \frac {x^n - x^{-n}}{x - x^{-1}}\Bigr)^2 = \sum_{j=0}^{2n-2} x^{(n-j-1)} \sum_{k=0}^{2n-2} x^{-(n-k-1)} \) I'm not sure though, why the second sum has negative exponent(s). Wait, yes I am. It's because you need the complex conjugate.
I wanted to know a bit more about why you increase the exponent by 2k (resp. decrease by j an even positive number). Since n can be odd or even; is it connected to the Taylor expansion (McLaurin series) after all? The expansion of sin(t) is: \( \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} t^{(2k+1)} = t - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}\, +\, ... \). So \( sin(nt) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} (nt)^{(2k+1)} \). But we also have Euler's identity: \( 2i\,sin(t) = e^{it} - e^{-it} \) Which we have as: \( 2i\,sin(t/2) = x - x^{-1} |_{x=e^{it/2}} \), and \( 2i\,sin(nt/2) = x^n - x^{-n} \). But n is a finite number, so we expand from k = 0 to n-1 in the McLaurin series, and we have a function sin(nt/2)/sin(t/2) which we want to express as the ratio of two (finite) sums. Let's try setting n to 3, so we have: \( x^3 - x^{-3} = (x - x^{-1})(some\, series\, \Sigma). \) Assume the first term in \( \Sigma\) is \( x^2 (= x ^{n-1}) \), and assume the last term in \( \Sigma \) is \( x^{-2} \), so the remaining terms have to cancel: \( (x - x^{-1})(x^2 + ? + ... + ? + x^{-2}) = x^3 - x^{-3} \).
Since n is 3, there has to be three summands in \( \Sigma \). For \( \Sigma \) to be symmetric, the only possibility is the middle term is: \( x^0 (=x^{n-3}) \). If n is odd, the first and last terms of \( \Sigma \) will have even exponents and the middle term will be = 1. If n is even, the first and last terms of \( \Sigma \) will have odd exponents and the middle two terms will be \( x + x^{-1} \). Since we have: \( x^3 - x^{-3} = (x - x^{-1})(x^2 + x^0 + x^{-2}) \), then whether n is odd or even, exponents differ by a factor of ±2, in the ordered series. I should have said, above, that each value of n has a McLaurin series expansion. And I should have said that \( \Sigma =\Sigma(n,x) \) having n terms hasn't been rigorously shown. If you assume that there are more than n terms, and that all terms cancel under multiplication: \( (x - x^{-1})\Sigma \), except for two terms, then you can easily show the first assumption leads to a contradiction.
So I think, what I've done there is use an argument about multiplication of exponents to see what \( \Sigma(n,x) \) is like, and so derive a closed form for a function like (sin nt)/(sin t). I can write something along the lines of: \( (x^n - x^{-n})/(x - x^{-1}) = x^{(n-1)} + x^{(n-1)-2} + x^{(n-1)-4}+ ... + x^{-(n-1)} \). So I define j = 0,2,4, ... 2(n-1). Then I have the sum over n terms like: \( x^{(n-1)-j} \)
And with this latest bit of fiddling around, I see my earlier statement about "needing" the complex conjugate, which is how you square a complex number, doesn't apply since the exponents in any \( \Sigma \) are symmetric about zero for even n, and about \( x^0 \) for odd n. If you multiply all the exponents in some \( \Sigma \) by -1, you have the same series. For example, \( (x^2 + x^0 + x^{-2}) \) will be unchanged after multiplication of the exponents by -1. In which light \(\Bigl( \frac {x^n - x^{-n}}{x - x^{-1}}\Bigr)^2 = \sum_{j=0}^{2n-2} x^{(n-1-j)} \sum_{k=0}^{2n-2} x^{-(n-1-k)} = \sum_{j=0}^{2n-2}\sum_{k=0}^{2n-2} x^{(k-j)} \) The "heavy lifting" and so some kind of answer to the question will involve inspecting the table of j rows and k columns in the actual multiplication, but immediately we see there are n \( x^0 \)'s on the diagonal. We now want to replace j and k with a single variable so the sum is unchanged. It appears that Taylor expansions aren't the reason the exponents in a closed form series are all odd or all even and different by a factor of ±2, it's Euler's trig identities (as polynomials with both positive and negative exponents).
