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here's where I am stuck:

$$\lim_{n \to \infty} \sum_{k=0}^{n-1} \cos(kx/n)\frac{1}{n}$$

so...it looks like at this point I could convert to a Riemann integral, but to which one?

Maybe $$\int_0^{\infty} \cos(t) dt?$$

That wouldn't converge, so it wouldn't be good.

How could I proceed? It's easy to accept that $\large \frac{1}{n}$ maps to $dt$, when going from a Riemann sum to a Riemann integral, but I particularly have trouble figuring out how to convert the discrete variable to the continuous, integration variable, t.

Any ideas are welcome.

Thanks,

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  • $\begingroup$ The limit is $1/x \cdot \int_0^x \cos(t)dt $ $\endgroup$ – Joel Dec 29 '15 at 2:00
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    $\begingroup$ The limit corresponds to the integral $\int_0^1 \cos(tx)\, dt$. $\endgroup$ – levap Dec 29 '15 at 2:00
  • $\begingroup$ @levap I doubt it since for $x=0$ it's the harmonic series, which diverges. $\endgroup$ – Gregory Grant Dec 29 '15 at 2:02
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    $\begingroup$ @GregoryGrant For $x = 0$, the sum is identically one (note that the summation is over $k$, not $n$). $\endgroup$ – levap Dec 29 '15 at 2:08
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    $\begingroup$ Hi @Joel, my answer agrees with yours: $\large \frac{sin(x)}{x}$. Thanks so much for your comment. I feel a bit better now with converting the discrete variable into the continuous variable :-) $\endgroup$ – user301446 Dec 29 '15 at 2:32
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Hint: How would you write $$\int_0^1 \cos(xt) \,dt$$ as a Riemann sum?

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  • $\begingroup$ Hmmm @Snow, something like $\sum_1^{N} cos ( ?) \frac{1}{N}$ , so that the total interval length is 1. What do you think? I am confused about the input for cosine, though. Would it just be... $xn$? Seems like such an...arbitrary choice...and that I'm not following any hard rules to make that conversion. (the only rule I am aware of in such conversions is that mesh(p) must go to zero, which we clearly have from 1/n...) Thanks, $\endgroup$ – user301446 Dec 29 '15 at 2:08
  • $\begingroup$ @user301446, Almost. You have $\sum_{k=0}^{N-1} \cos(x t_k) \frac{1}{N}$. The first time most students encounter Riemann sums, they see it not in the generality of partitions and mesh, but with equally sized subintervals. Can you write an expression for $t_k$ if you want subintervals to be of equal length and $t_k$ is the left end point of the $k$-th subinterval? $\endgroup$ – Snow Dec 29 '15 at 2:12
  • $\begingroup$ Hi @Snow ooh...hmm...$\frac {k}{N}$? so that $\frac{k}{N}$ belongs in the k-th sub-interval where it is to be the input of cosine - to give the height of the rectangle...I think... $\endgroup$ – user301446 Dec 29 '15 at 2:20
  • $\begingroup$ ....so that...letting mesh(p) go to zero, such an input becomes the continuous variable, $t$. so, it's the mapping $\frac{k}{N} \to t$, so that in total the mapping is from $\frac{k}{N}x \to tx$ $\endgroup$ – user301446 Dec 29 '15 at 2:23
  • $\begingroup$ What do you think @Snow? :-) $\endgroup$ – user301446 Dec 29 '15 at 2:25
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Consider the function $f(t) = \cos(tx)$ on $[0,1]$. Let

$$ 0 = t_0 < t_1 = \frac{1}{n} < t_2 = \frac{2}{n} < \ldots < t_{n-1} = \frac{n-1}{n} < t_n = 1$$ be a partition of $[0,1]$ into $n$ sub-intervals of length $\frac{1}{n}$. In the interval $[t_i, t_{i+1}]$ choose the point $\xi_i = t_i$. Then the Riemann sum corresponding to the partition $\{t_i\}_{i=0}^{n-1}$ and the choice of points $\{\xi_i\}_{i=0}^{n-1}$ is

$$ \sum_{k=0}^{n-1} f(\xi_k) (t_{k+1} - t_k) = \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \frac{1}{n} = \sum_{k=0}^{n-1} \cos \left( \frac{k}{n} x \right) \frac{1}{n}. $$

Hence, the limit of the expression (as $n \to \infty$) converges to

$$ \int_{0}^1 \cos(tx) \, dt = \begin{cases} [\frac{\sin(tx)}{x}]^{t = 1}_{t=0} & x \neq 0, \\ 1 & x = 0. \end{cases} = \begin{cases} \frac{\sin(x)}{x} & x \neq 0, \\ 1 & x = 0. \end{cases} $$

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  • $\begingroup$ Thanks for your answer, @levap :-) $\endgroup$ – user301446 Dec 29 '15 at 2:34

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