2x+2 | ||
j(x) = | , dla: |x+1|<1; | |
x2+2x |
t | ||
j(t)=2 | =∑n=0∞−2t2n+1=∑n=0∞−2(x+1)2n+1 | |
t2−1 |
1 | ||
bo 1+x+x2+...= | dla |x|<1 | |
1−x |
x2 | x3 | x4 | ||||
mamy ln(x+1)=x− | + | − | +... | |||
2 | 3 | 4 |
t2n(−1)n+1 | ||
k(t)=ln(t2+2)=ln(2)+ln((t/√2)2+1)=ln2+∑n=1∞ | ||
2nn |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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