| e2x+1 | ||
napisac rozwinięcie funkcji f(x)= | w szereg Maclaurina, a następnie obliczyć | |
| e3x |
| e2x+1 | ||
f(x)= | =e−x+e−3x= | |
| e3x |
| 1 | 1 | (−1)n | (−3)n | |||||
=∑ | (−x)n+∑ | (−3x)n=∑ | (x)n+∑ | (x)n= | ||||
| n! | n! | n! | n! |
| (−1)n | (−3)n | (−1)n+3n*(−1)n | ||||
∑( | + | )(x)n=∑( | +)(x)n= | |||
| n! | n! | n! |
| (−1)n*(3n+1) | ||
∑( | )(x)n | |
| n! |
| f(101)(0) | (−1)101*(1+3101) | ||
= | |||
| 101! | 101! |