/sci/ - Science & Math

Good one, OP. Good one.

>>2359542
rerolling

>>1910216
#include <iostream>
#include <cstring>

using namespace std;

int main() {
char s[100];
cin>>s;
int n = strlen(s);
for (int i = 0; i < n/2; i++) {
swap(s[i], s[n-1-i]);
}
cout<<s;
return 0;
}

>>1374581
already posted by cereal guy
>>1374562

>>1308933
[math tag is shit, use [eqn

also, your difficult problem is easy as fuck

<div class="math">\sum_{n=1}^{\infty} (T_n - T) = -\sum_{n=0}^{\infty} \frac{(-1)^n \cdot n}{2n+1} = -\lim_{x\to 1}\sum_{n=0}^{\infty} {\frac{(-1)^n \cdot n}{2n+1} x^{2n+1}} = -\lim_{x\to 1} \sum_{n=0}^{\infty} {\frac{1}{2}((-1)^n - \frac{(-1)^n}{2n+1}) x^{2n+1}} = -\frac{1}{2}\lim_{x\to 1} (\sum_{n=0}^{\infty} {(-1)^n x^{2n+1}} - \sum_{n=0}^{\infty} {\frac{(-1)^n}{2n+1} x^{2n+1}}) = -\frac{1}{4} + \frac{\pi}{8}</div>

>>1166912
Must have taken a lot of attempts

Cereal reporting
<div class="math">\int_{0}^{2\pi}\sin^{2n}{x} \, dx= \int_{0}^{2\pi} -\sin ^{2n-1}x \, d{\cos x} = 0 + \int_{0}^{2\pi} {(2n-1)}\sin ^{2n-2}x
\cos ^2x \, dx = \int_{0}^{2\pi} {(2n-1)}(\sin ^{2n-2}x - \sin^{2n}x) \, dx</div>

<div class="math">{2n}\int_{0}^{2\pi}\sin^{2n}{x} \, dx = (2n-1)\int_{0}^{2\pi}\sin^{2n-2}{x} \, dx</div>

<div class="math">\int_{0}^{2\pi}\sin^{2n}{x} \, dx = \frac{2n-1}{2n}\int_{0}^{2\pi}\sin^{2n-2}{x} \, dx</div>

<div class="math">\int_{0}^{2\pi}\sin^{2n}{x} \, dx = \frac{1\cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n} \cdot 2\pi</div>

the average value is <div class="math">\frac{1\cdot 3 \cdot 5 \cdot \ldots \cdot (99)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 100}</div>

Cereal reporting
<div class="math">\int_{0}^{2\pi}\sin^{2n}{x} \, dx= \int_{0}^{2\pi} -\sin ^{2n-1}x \, d{\cos x} = 0 + \int_{0}^{2\pi} {(2n-1)}\sin ^{2n-2}x
\cos ^2x \, dx = \int_{0}^{2\pi} {(2n-1)}(\sin ^{2n-2}x - \sin^{2n}x) \, dx</div>

<div class="math">{2n}\int_{0}^{2\pi}\sin^{2n}{x} \, dx = (2n-1)\int_{0}^{2\pi}\sin^{2n-2}{x} \, dx</div>

<div class="math">\int_{0}^{2\pi}\sin^{2n}{x} \, dx = \frac{2n-1}{2n}\int_{0}^{2\pi}\sin^{2n-2}{x} \, dx</div>

<div class="math">\int_{0}^{2\pi}\sin^{2n}{x} \, dx = \frac{1\cdot 2 \cdot 3 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n} \cdot 2\pi</div>

the average value is <div class="math">\frac{1\cdot 2 \cdot 3 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n}</div>