Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with trigonometric-sums
Search options not deleted
user 126661
1
vote
1
answer
136
views
Proof of an inequality $s_m(n) \le f_m(n)$
For fixed $m = 0, 1, 2, ...$
$$f_m(k) = \prod_{j=1}^{m}(k+j).$$
Some examples of $f_m(k)$ are as following:
$$f_0(k) = 1, \quad f_1(k) = (k+1), \quad f_2(k) = (k+1)(k+2).$$
The $s_m(n)$ is defined as …
- 161
3
votes
2
answers
229
views
Inductive proof of $s(n)≤n+1$
I was able to conclude, numerically, the following:
$$s(n) = \sum_{j=0}^n\frac{(-4)^j}{(2j+1)!}\left(\sum_{k=j}^n\frac{(k+1)(2k+1)(k+j)!}{(k-j)!}\right)x^{2j+2}\le n+1$$
for $x\in[0,1]$. For example …
- 161
1
vote
Accepted
Proof of an inequality $s_m(n) \le f_m(n)$
For the people who are interested, inspired from this answer,
$$
\begin{align}
s_m(n)
&= \sin\left(\frac{t}{2}\right)\sum_{k=0}^nf_m(k)\sin(k+0.5)t\\
&=\frac{1}{2}\sum_{k=0}^nf_m(k)2\sin(k+0.5)t\sin\ …
- 161