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Oct
16
accepted Show from every positive sequence decreasing to zero, we can build a sequence satisfying a concavity condition
Oct
16
revised Show from every positive sequence decreasing to zero, we can build a sequence satisfying a concavity condition
per robert comment
Oct
16
comment Show from every positive sequence decreasing to zero, we can build a sequence satisfying a concavity condition
@RobertIsrael I see your point, I relax this requirement in the question. However, as $x_n$ is a decreasing sequence, I think your example doesn't apply, but your point is still valid.
Oct
16
revised Show from every positive sequence decreasing to zero, we can build a sequence satisfying a concavity condition
edit per Robert comment
Oct
16
accepted Derivating $f(t)=\int_0^t x dx$ using measure theory
Oct
16
accepted Convergence of the Fourier series of $f(t)=(t-\pi)\chi_{\left(0,2\pi\right)}$
Oct
16
asked Show from every positive sequence decreasing to zero, we can build a sequence satisfying a concavity condition
Oct
15
revised Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
previous typo were not corrected in title
Oct
15
comment Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
@joriki a) You are right. b) I agree. c) I see your point, but I can't find my error. Writing $\hat f_n$ for the $n$-th Fourier coefficient of $f$, I have $\hat f_0=0$ and $\hat f_n=\frac{1}{2 \pi}\int t e^{-int}dt=\frac{1}{2 \pi}\frac{te^{-int}}{(-in)}+0$.
Oct
15
revised Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
typos in last per joriki comment
Oct
15
revised Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
fourier series vs fourier sum
Oct
15
comment Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
@joriki I tried to clarified the notation, and you were right for $0 \le k \le 2n$. Thank you. Sorry for the many typos, Fourier analysis is a jungle for which my machete is not sharp yet.
Oct
15
revised Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
edit per joriki comment
Oct
15
accepted Inequality for term of a positive sequence : show $\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$
Oct
15
comment Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
@did The title was inacurrate. Thank you.
Oct
15
revised Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$
error in title per did comment
Oct
15
comment Inequality for term of a positive sequence : show $\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$
@sasposcat Good eyes, thanks.
Oct
15
revised Inequality for term of a positive sequence : show $\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$
typo from saspocat comment
Oct
15
revised Inequality for term of a positive sequence : show $\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$
typo on sequence space
Oct
15
asked Inequality for term of a positive sequence : show $\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$