2x − 2−x | ||
y = | ||
2x + 2−x |
1 | 1 | |||
wtedy 2−x = | = | |||
2x | t |
| 1 | ||||||||||||
y = | /*(t+ | ) | |||||||||||
| t |
1 | 1 | |||
y * (t+ | ) = t − | /* t | ||
t | t |
−1−y | ||
t2 = | ||
y−1 |
−1−y | ||
t = √ | ||
y−1 |
−1−y | ||
2x = √ | ||
y−1 |
−1−y | ||
x = log2√ | ||
y−1 |
−1−x | ||
y = log2√ | ||
x−1 |
−1−x | ||
wlasciwie moznaby rozpatrzec jeszcze y = − log2√ | ||
x−1 |
2x − 2−x | ||
y = | = tanh(xln(2)) | |
2x + 2−x |
1 | 1 | 1 | 1 + x | |||||
y−1 = | artanh(x) = | * | ln( | ) | ||||
ln(2) | ln(2) | 2 | 1 − x |